text:
Nitric acid decomposition
I have a problem with the decomposition reaction of Nitric acid($\ce{HNO3}$) by light.
My problem: is that reaction is a redox reaction?
If it is please tell me the oxidation and reduction (balanced) reactions seperately.

Why do you think it's photochemical? Nitric acid is sold in clear colorless bottles.

@Zhe no, nitric acid is subjected to decomposition due to light. See my answer below.

Interesting. Never seen the brown bottle before... Thanks! @NilayGhosh

Generally, nitric acid is a strong oxidizing agent which get reduce to nitrogen oxides on reacting with metals which act as reducing agent thus forming a redox reaction. But in this case, only nitric acid is involved which get reduced to nitrogen oxides on reaction with air/light. So, this may be called oxidation reaction where nitric acid gets oxidized. From wikipedia article of nitric acid:

Nitric acid is subject to thermal or light decomposition and for this reason it was often stored in brown glass bottles:
$$\ce{ 4 HN^{V}O3 → 2 H2O + 4 N^{IV}O2 + O2}$$
This reaction may give rise to some non-negligible variations in the
  vapor pressure above the liquid because the nitrogen oxides produced
  dissolve partly or completely in the acid.
The nitrogen dioxide ($\ce{NO2}$) remains dissolved in the nitric acid
  coloring it yellow or even red at higher temperatures. While the pure
  acid tends to give off white fumes when exposed to air, acid with
  dissolved nitrogen dioxide gives off reddish-brown vapors, leading to
  the common name "red fuming acid" or "fuming nitric acid" – the most
  concentrated form of nitric acid at Standard Temperature and
  Pressure (STP). Nitrogen oxides ($\ce{NO_x}$) are soluble in nitric
  acid.

There are some old papers(here and here) which discuss the mechanism of nitric acid decomposition. Give a quick read on the papers and you will understand the mechanism of the reaction.


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What makes wrinkle-free cotton, wrinkle-free?
Disclaimer:
No, I don't own any "wrinkle-free" articles of clothing. So I haven't been able to observe first-hand if "wrinkle-free" clothing is totally "wrinkle- free" (if it isn't, then perhaps "wrinkle-resistant" might be an appropriate term?).

Back to question:
At the molecular level, what is it that makes cotton "wrinkle-free"? What chemo-physical process(es) is cotton subjected to, here?
I ran a cursory Google-search, and this was the best article that I could find, albeit, it's sorely lacking in information on the actual processes involved in the production of "wrinkle-free" cotton.

Don't have time to read this, but I believe this is the patent for it: https://www.google.com/patents/US6123739

A short overview is provided on https://en.wikipedia.org/wiki/Permanent_press , too. And http://journals.sagepub.com/doi/abs/10.1177/004051756103100408 (DOI: 10.1177/004051756103100408) one of the first articles about it.

source: Stack Exchange

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Advanced Quantum Chemistry Reference for Reactivity Phenomena
I'm currently in organic synthesis seeking to develop quantitative "physical" (that of a physicist), rather than "chemical", intuition for areas where traditional VSEPR/ MO models cannot fully describe the reactivity of atoms. Even if these problems are ultimately computational, I'm looking for a first-principles basis of reactivity to aid my intuition.
Searching for answers has led me to questions like:
How exactly do orbitals hybridize? E.g., if I put OH- and H+ in a room together, will applying TDSE to both their independent wave functions eventually give me the desired linear combination of wave functions?
At short distances between two reactive partners, through what quantum mechanism do atoms' electrons influence their reactive partner's electronic structure. That is to say, undergraduate QM does not really discuss how wavefunctions exert EM forces and how EM forces in turn influence other wavefunctions.
And so on.
Using the search function, I've found several suggestions for quantum chemistry texts that discuss electronic structure, but have yet to find one that satisfactorily provides a step-by-step walkthrough of the various quantum phenomena involved in the kinetics of reactivity. I suspect these questions dabble into QFT areas, which I'm comfortable with.
Such a text, ideally, would discuss the mathematical origin of anti-bonding orbitals and other phenomena undergrad QM/ QChem courses do not generally cover.
Thanks!

Maybe Steven Bachrach's book "Computational Organic Chemistry" blog here would have this?

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Derive relationship between celsius and fahrenheit scale
My book derives relationship between Celsius and Fahrenheit scale as

F-32/212-32 = C-0/100
F-32/180 = C/100

but I don't understand why F-32 is divided by 180? What does 180 and 100 indicate in both respectively?

https://chemistry.stackexchange.com/questions/15488/determining-boiling-point-on-a-created-temperature-scale

Order of operations matter. You are supposed to do division before subtraction. Also, there are some parentheses missing, so the overall expression is wrong.

They don't indicate anything. That's just the way it is. There is no physical reason behind it.

I'm voting to close this question as off-topic because this is not a question dealing with chemistry in the slightest.

Water boils at $100\ ^{\circ}\mathrm{C}$ and $212\ ^{\circ}\mathrm{F}$, while it freezes at $0\ ^{\circ}\mathrm{C}$ and $32\ ^{\circ}\mathrm{F}$. The numerators in formula provides the correct shift based on freezing point. The denominators provide the correct scaling between units by noting that a $180\ ^{\circ}$ increase on the Fahrenheit scale is equal to a $100\ ^{\circ}$ increase on the Celsius scale.
Note that your formula is missing some parentheses for correctness:
$$\frac{F-32}{212-32} = \frac{C-0}{100}$$ should be written as
$$(F-32)/(212-32) = C/100$$

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Absorption HOMO-LUMO, canonical?
So I already asked yesterday about canonical and localized orbitals together with Koopman's theorem where we came to the conclusion that for for example ionization potentials localized orbitals cannot be used. So I assume a ionization would be from the HOMO which leaves me with the question, whenever we say an electron is excited from the HOMO into the LUMO of a system, are we talking about the canonical forms then? Because before I knew about those two I always imagined that there is a local position somewhere in the molecule like a double bond where one electron is exited into an antibonding MO creating a single bond or 1.5 bond and that you can really tell from where to where the transition is. But canonical orbitals do not really look like we imagine orbitals they are distributed over the whole molecule.

I am not allowed to correct this, because its less than 6 characters ... but the guys actual name is "Koopmans" with the "s" at the end.

Short answer is yes, HOMO and LUMO refer to canonical orbitals, since only those have at least some physical meaning.
Two more remarks about that:
Koopmans' Theorem does only work if you add or remove one electron.
If you change two or more electrons of your system (e.g. exciting one electron is first removing one, then adding one in another orbital) you get additional terms in energy difference between both states. Those terms are coming from the electron-electron interaction of the two changed orbitals/electrons.
In case of changing only one orbital, those additional terms do not appear since there is no electron-electron self-interaction.
Furthermore, the approximation of Koopmans' Theorem, that orbitals do not change (do not relax, remain stationary) on removing/adding electrons, gets worse the more electrons change.
Orbitals itself are only an approximation. Factorizing the total electronic wave function into orbitals is mathematically only possible for non-interaction electrons, or like in Hartree-Fock if you assume some mean-field for the electron-electron interaction.
The resulting canonical orbitals do not always match the simplified picture based on our chemical intuition.


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Decomposition of acetonide protecting group in chloroform
Today I purified my compound containing an acetonide protected diol. I took an NMR sample and it looked great, but because I have only little amount of the compound, I put the sample back into the flask and evaporated the $\ce{CDCl3}$.
In a while, the solution turned from yellow to brown, so I took another NMR and it looked a little bit more messy and the $\ce{CH3}$ signals from acetonide group had decreased intensity (integral from 6 to around 4.5).
Is it possible that the $\ce{CDCl3}$ is so acidic it causes acetonide removal? The compound survived silica chromatography. Have you ever experienced something similar?

How old was your deuterochlororm? Chloroform is known to decompose on standing over time to give HCl and phosgene. Google acidity of chloroform for more details.

This is why I always add a spatula end of powdered potassium carbonate to a new bottle of deuterochloroform.

Also note that phosgene mentioned by @Waylander is extremely toxic.

@PrittBalagopal, within reason. We're talking about minute quantities being produced by a bottle of chloroform under normal conditions, there is no actual risk of poisoning oneself with phosgene.

@NotEvans that may be so. However it's best to warn the OP anyway.

Deuterochloroform (and indeed 'normal' chloroform) are known to be mildly acidic if left standing for any length of time:

Chloroform: It reacts slowly with oxygen, or oxidising agents, when exposed to air and light, giving, mainly, phosgene, Cl2 and HCl.
Source: Purification of Laboratory Chemicals, 7 ed.

To compensate for this, its common to add molecular sieves and a small amount of potassium carbonate to a new bottle of deuterochloroform to keep it dry and to minimise any acid that may be present/may be generated as the bottle ages.
If you look up the entry for acetonides in Greens' Protecting Groups in Organic Chemistry, you'll see that acetonices can come off under quite mild conditions, though personally I've never had an acetonide come off in an NMR tube of chloroform (I assume you didn't leave it standing for months in the tube...)
In order to cleave the acetonide under acidic conditions, you generally also need a nucleophile (methanol is common, water would do the trick), which if present should be very obvious on the NMR, and might go some way to explaining how you lost the acetonide.  You should also consider other explanations, for instance you mentioned concentrating the chloroform- are you sure the rotary evaporator hadn't just been used to take off something wet and acidic?
Hopefully you can just put the acetonide back on without any issues, I suggest picking up a new bottle of chloroform-d and adding some sieves and potassium carbonate (you could try NMR-ing your old bottle to see if theres water/methanol etc in it, and then purify appropriately, but the stuff is sufficiently cheap that I wouldn't waste my time).


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Why sodium is the worst reducing agent in group 1?
Analysing the oxidation enthalpies for $\ce{Li}$, $\ce{Na}$ and $\ce{K}$, we have their respective values:

$\ce{Li} = \pu{162 kJ/mol}$
$\ce{Na} = \pu{202 kJ/mol}$
$\ce{K} = \pu{188kJ/mol}$

Sodium has the largest positive oxidation enthalpy, which means that its oxidation will be less likely to happen according to its Gibbs free energy. But, why does this happens to sodium?

These are all negative I'm afraid.

I don't think they are... Take Lithium, for example: Dh(sublimation) = 161 kj/mol. Dh(ionization) = 520 kJ/mol. Dh(hydration) = -519 kJ/mol.

Considering that Dh(oxidation) is: Dh = Dh(subli) + Dh(ioniz) + Dh(hydr), it will be positive...

Maybe, no idea why you didn't use standard electrode potentials.

Currently, i am making an analysing of the Dh, Dg and the standard electrode potentials to explain why the elements of the group 1 are the best reducing agents. This was easy to explain, but when i go through the elements of group 1, the Sodium is the problem, because he has a higher Dh(oxidation) and a lower Standard Electrode Potential. This is what i am really stuck, because those two factors will tell that Sodium is the worst Reducing agent on group 1

https://chemistry.stackexchange.com/questions/28568/why-is-lithium-the-most-reducing-alkali-metal-and-not-caesium

Here are the values http://imgur.com/a/6GIKx

Yes, this explains very well why Lithium is a stronger reducing agente. But, this still doesn't explain why Potassium, for example, is a better reducing agent than Soudium, you see? Considering that he has a larger size than Sodium.

Why are you looking at enthalpy?!

I am looking at it to make a analysis of the DGoxidation (Considering that DGox = Dhox - T*Ds). I was able to explain why group 1 have the best reducing agents, but now i have to see the case for sodium.

And even if i try to explain it using Standard Electrode Potential, it will still be lower for Sodium than for the other elements of group 1.

The type of analysis that i am making have to consider things like: What factor on the DH(oxidation) will be more relevant or primary to explain why the elements of group 1  are the best reducing agents? The conclusion that i brought up was: the Dh(ionization) and the Dh(sublimation) says that the elements of group 1 are the best reducing agents. But, the Dh(hydration) says they are the worst. So, Dh(ion) and Dh(subli) are the primary factors to explain this...

Reduction potentials are related to free energy not enthalpy. Looking at enthalpy is wrong unless you are going to undertake the impossible task of quantifying ion solvation...

Yes, true. I am just using the enthalpy because i have to make a analysis of it, really. I have just to consider that, if the enthalpy is more positive, it will contribute less to the Free energy, you know? I'm making this question based on the enthalpy just because i have to make analysis of the group 1 based on it, and relate this to the Free energy.

I think that this will make it more clear: When you analyse the Dh(oxidation) of Lithium you have: His Dh(ion) is the more positive between the elements of the group. His Dh(Subl) is the more positive between the elements of the group. So, what will make his Dh(oxidation) less positive? The DH(hydration), which is exothermic, right? So, when you sum those 3, the result is that Lithium have a less positive Dh(oxidation). This means that his Dh(hydration) was able to compensate the Dh(ion) and Dh(sublimation).

I don't understand why on sodium this compensation didnt happened so well. His Dh(hydration) is higher than for K, Rb andCs. But, still it wasn't enough to make his Dh(oxi) less positive than for K, Rb and Cs. I don't see why this happened to him, you see?

The general trend in the reactivity of the group 1 elements is that it increases down the group. This can be attributed to the decrease in 1st ionisation energy as the atomic radius increases.
Reducing power can be measured by the standard electrode potential E0. This refers to the half - cell:
$$M^+(aq)+e⇌M(s)$$
If you look at the values for group 1 we see a different pattern:
Electrode Potential (V) Li. -3.02 Na. -2.71 K. -2.92 Rb. -2.99 Cs. -3.02
Lithium has an anomalously large and negative electrode potential. After Li the values become more large and negative, indicating, as you rightly say, that sodium has the lowest reducing power.
The reason for the unusual value for lithium is that E0 values are measured in solution whereas ionisation energies are measured in the gaseous phase.
To turn a metal atom in the solid state into an aqueous ion we can think of a 3 step process:
The metal is atomized. $\ce{Li(s)→Li(g)}$
The metal is ionised. $\ce{Li(g)→Li+(g)+e}$
The ion is hydrated. $\ce{Li+(g) +(aq)→Li+(aq)}$
Steps 1 and 2 are both endothermic, i.e. they require energy. Step 3 is exothermic as it is a bond forming process. It is referred to as the enthalpy of hydration.
The small size of the lithium ion gives it a large charge density and so water molecules are strongly attracted giving a large, negative enthalpy of hydration.
This is the main factor responsible for lithium's large, negative electrode potential.
In aqueous conditions it is as reducing as cesium though much less reactive in anhydrous conditions.
This anomaly accounts for sodium having the lowest reducing power, but only in aqueous conditions.


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BCS class of nifedipine
So I had an exam today in which one of the questions was in what class (I-IV) of the Biopharmaceutical Classification System (BCS) the drug nifedipine falls in. Class I drugs are considered highly soluble and highly permeable, class II badly soluble and highly permeable, class III highly solube and badly permeable and class IV badly soluble and badly permeable.

Solubility
The aqueous solubility of a drug substance is considered as high according to the HHSFDA
  BCS criteria when:

The ratio of the highest orally administered dose (in mg) to the solubility (mg/ml)
  is less than 250 ml
This criterion is met over the pH range 1-7.5 at 37°C. [1]

Permeability
According to HHS-FDA a drug is considered a highly permeable, when more than 90%
  of the orally administered dose is absorbed in the small intestine.[1]

Obviously judging from literature it falls under BCS class II and you'd probably assume so if you look at the solubility (practically insoluble in water <1g/10000ml) and the permeability, which is good because it is absorbed well (>90%). [2] However the way the question was asked was interesting because of the information given. It explicitly said that a dose of 20 mg was the maximum dose and no exact solubility was given except that the solubility was "practically insoluble", meaning a solubility of < 1 g/ 10000 ml. Now whether the solubility is 1g /10001 ml or 1g/ 100002 ml, nobody knows... The extent of absorption given was 90% and thus high permeability was assumed. However when you calculate how much volume of water is needed to let this fully dissolve, you'd come with the answer 0,02 * 10000 = 200 ml water. This is smaller than 250 ml and thus one would be inclined to then assume the solubility is high according to the specifications stated above, which would lead you to think the drug falls under BCS class I. Of course in reality you'd have to take into account the exact solubility which probably isn't exactly 1 g/10000 ml and the exact volume of the gastrointestinal tract which probably isnt exactly 250 ml.
Now what am I supposed to believe in this situation? In this case the dose plays an important role in determining the BCS class and thus answering the question. Would you in this case have taken the rules for what they are and follow them like I did, taking into account dose and assuming the lowest solubility or would you have based it purely on the "practical insolubility in water" of the drug and conclude low solubility and thus BCS class II?
Maybe I'm thinking too hard about this, but I'm curious if anyone has something to say on this.
I personally think it would've been fair if the exact solubility (perhaps even a fictive solubility) of nifedipine and exact volume of GIT were given in this situation to make this question more clear.
Sources:

http://www.who.int/medicines/services/expertcommittees/pharmprep/QAS04_109Rev1_Waive_invivo_bioequiv.pdf
http://www.dissolutiontech.com/DTresour/200905Articles/DT200905_A01.pdf

The tl;dr answer is that, as the question is worded (and without an actual copy of the question or the data provided to you), its not possible to definitively assign a BCS class.

Nifedipine is a class II drug:

class II is low solubility and high permeability. Formulation typically is used to enhance solubility of compounds in this class.

The data in your question is a little confusing, but you state that the compound is 'practically insoluble' which means that it must be either class II or class IV (practically insoluble definitely suggests low solubility).
With the information provided (i.e. nothing about permeability), you cannot distinguish between II and IV. Permeability, whilst often correlated to solubility, is not directly linked, such that you cannot attempt (as I think you have tried) to calculate one based on the other (how permeable something is depends upon structure, and whether it can get through the membrane).

Oh yeah, I think I forgot to mention, but it was considered highly permeable since the extent of absorption was 90%.

You should [edit] that comment into your question for completeness

Yeah you're right I will in a second. Thanks for your thoughts by the way.


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Reducing strength of (i-Bu)2AlH vs Lithium aluminum tri(t-butoxy)hydride
Hi Im trying to figure out which of the following is the stronger reducing agent.
(i-Bu)2AlH vs Lithium aluminum tri(t-butoxy)hydride
From my understanding, the Lithium aluminum tri(t-butoxy)hydride should be the stronger reducing reagent since it has a negative charge on the aluminum which would help push electron density onto the hydrogen. But on the lecture slides it says that Lithium aluminum tri(t-butoxy)hydride is a milder reducing agent. Can someone provide reasoning why that is the case. Thanks


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Which of the following species has an enthalpy change of formation = 0 kJ/mol?
Below was a question in the 2012 Australian Chemistry Olympiads:

The enthalpy change of formation ($\Delta_\mathrm{f}H^\circ$) for a species at
  $\pu{298 K}$ is defined as the enthalpy change that accompanies the
  formation of one of the following of one mole of a substance from its
  constituent elements in their standard states. Which of the following
  species has $\Delta_\mathrm{f}H^\circ = \pu{0 kJ/mol}$?
a. $\ce{H2O(l)}$  b. $\ce{Na(g)}$  c. $\ce{Na(s)}$  d.
  $\ce{CO2(g)}$  e. $\ce{O3(g)}$

Original
Solutions stated the answer is b, but isn't the standard state of $\ce{Na}$ solid?

Clearly a typo. The correct answer is (c) just like you think it is.

The correct answer is (C) due to the fact that it is an element to its standard state

The correct answer is indeed c. $\ce{Na(s)}$
Since for the heat of formation to be zero, the molecule should be the must abundant chemical formula of the element.
$\ce{Na}$ is a metal and is predominantly found in solid state (I dont think it exists as gas anywhere in the atmosphere of earth).
$\Delta_\mathrm{f}H^\circ$ is a quantity which is taken relative to the heat of formation of the most abundant naturally occuring form of an element.

I dare you to name one place on Earth where sodium exits naturogenically in a solid state.

Linear, the issue you describe is not the fault of sodium, but of the abundance of water. Sodium metal is perfectly happy when it isn't getting electrons ripped from its valence shell to form hydrogen gas. ; )

@Pete I think sodium is an electron pusher. Poor water didn't even want that electron!

@Pete I don't think you can blame one or the other in a reaction; that's the definition of the pot calling the kettle black :}. In any case, what I meant is that the predominant state in nature does not have direct bearing on what is taken as the relative zero state. IUPAC defines it generally as what is thermodynamically most stable given a standard state with some well-defined pressure, and additionally a temperature. Sometimes, kinetically stable (metastable) is used but has to be made clear. In short, need not be most abundant form in nature.

Of course it must be a typo. Correct answer is "c".
The $\Delta_\mathrm{f}H^\circ$ of $\ce{Na(s)}$ is $\pu{0 J}$ and the  $\Delta_\mathrm{f}H^\circ$ of  $\ce{Na(g)}$ is $\pu{107.8 kJ}$.


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Tetrahedrane Properties and Analogs
I took a look into some tetrahedrane ($\ce{C4H4}$) analogs, platonic hydrocarbon, basically very strained tetrahedral geometry at every carbon.  I was wondering what will be the predicted delta H of formation for the tetra nitro analog tetranitrotetrahedrane?  What will be the predicted detonation velocity too?  Could there be modified analogs that are even more energetic?

Have you heard of hydrazine? It's the most relaxed compound with a $\ce{N-N}$ bond. Even for that, Wikipedia says: Hydrazine is highly toxic and dangerously unstable unless handled in solution. Considering that fact, $\ce{N-N}$ bond is too weak to form tetrahedrane molecules.

If you are still interested, here is that article about cubane: http://pubs.acs.org/doi/abs/10.1021/jp511756v.  And the article about nitrocubanes: http://onlinelibrary.wiley.com/doi/10.1002/1521-4087(200203)27:1%3C1::AID-PREP1%3E3.0.CO;2-6/epdf

@pritt I know it's old comment, but OP's talking about tetranitro, not tetraaza compound.  Tetrahedral N4 is crazy, but tetranitro... might be possible (and very powerful).

Highly nitrated cage compounds are a subject of some research interest (at least for chemists who think the five fingers is a surplus) as Derek Lowe illustrates in his Things i Won't Work With column. Octanitrocubane is known.

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How to explain molecular geometry without the help of VSEPR, valence bond, or hybridization theories?
I was taught, at the high school level, how to rationalise molecular geometries with the help of VSEPR, valence bond, and hybridization theories.
However, I have recently also come to know that these theories are outdated and have many limitations. So, how can I rationalise these geometries without the help of these theories? Can I use molecular orbital theory for that, and if so, how?

see https://chemistry.stackexchange.com/questions/33879/what-is-actually-the-difference-between-valence-bond-theory-and-molecular-orbita

Short Answer: No, you can't. Reason being that so far this is the only way devised, which works most of the time.
Long Answer:
The Molecular Orbital Theory is, in practice, a complement to the Valence Bond Theory, Hybridization and VSEPR.
There are successes as well as shortcomings of both the Molecular Orbital Theory, and the other three.
In general, the VSEPR theory is used to predict shapes of molecules and ions, not the Molecular Orbital Theory. It can be extended to a large number of Covalent compounds, but it has a few limitations as far as ionic compounds and multi-centered bonds are concerned. The MOT tells us that when atoms combine, they form molecules orbitals, which are lower in energy than the atomic orbitals. It is generalizable, and can be extended to even semiconductors and superconductors. It is very useful when it comes to finding the actual energies associated with every single orbital.
The Valence Bond Theory, along with Hybridization and VSEPR, is the one that is generally used to find the shapes of molecules. The reason is simple: The Molecular Orbital Theory tells us the arrangements of delocalized electrons in Molecular Orbitals, whereas, the other theories tell us the possible positions of the combined atoms after considering that the electrons are localized and repel each other.
However, there is the case of multi-centered $\pi$-bonds, also known as banana bonds, and the 4c-3e bonds, which can be explained only by the Molecular Orbital Theory. A famous example would be Di-Borane, or $\ce{B2H6}$.
Thus, both the Valence Bond approach as well as the Molecular Orbital approach are often required for an accurate prediction of the shape of a molecules.
I quote an article from Chemistry LibreTexts:

Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and anti-bonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. Despite various shortcomings, complete disregard of one theory and not the other would hinder our ability to describe the bonding in molecules.

Thus, although the VB, VSEPR, and Hybridization theories are outdated and have many limitations, they, along with the Molecular Orbital Theory provide us with a toolbox for almost any molecule possible.
(If there is any place where I have erred, please correct me)

This is somewhat true, but maybe a bit too simplistic an answer. MOT offers many ways of predicting structure of molecules (or perhaps more often, rationalising structures of molecules once they have been determined experimentally). In terms of being able to explain structures, it doesn't matter whether the model used is localised or delocalised.

@orthocresol True that.... I was trying to provide a simplistic answer for the person asking, as he sounded like a high school student. In fact, I am not aware about the undermining details of the MOT

//the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and anti-bonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. //   How? How can MOT tell us that for example, methane is tetrahedral? How can MOT give us a more complete 'picture' of the structure of a molecule?

Fair enough. For methane you may wish to refer to Albright, Burdett, Whangbo "Orbital Interactions in Chemistry" 2nd ed. p 193 onwards. The correlation diagram indicates that a tetrahedral $T_\mathrm{d}$ geometry is favoured over a square planar $D_\mathrm{4h}$ one (for example). Essentially this is linked to better overlap between C and H orbitals in $T_\mathrm{d}$ geometry. More sophisticated calculations will allow one to find the geometry with the minimum energy - this is routinely done with computers nowadays. (Not always with MOT, though.)

@orthocresol I request you to kindly add that and more information to a more complete answer, and post it. I really want to know more. I am willing to delete my answer in favor of a more complete one.

@Abhigyan "whereas, the other theories tell us the possible positions of the combined atoms after considering that the electrons are localized and repel each other." I don't understand why u use 'localized' here.

In other theories, the underlying principle of electron sharing is that shared electrons are in the form of 'clouds', or there are regions of electron density. This hints at the idea that electrons don't move around the entire atom, at least as per these theories. On the other hand, the Molecular Orbital Theory is based on the concept that electrons move around the entire molecule. Thus, it is considered delocalized.


text:
Why is the tetraamminecopper(II) ion square planar?
The tetraamminecopper(II) ion is square planar in shape. But why? How can I rationalise this?
Can I rationalise this without the concept of hybridization? And even if I take hybridization into account, which hybridization takes place here and why?

Possible duplicate of Why is the hybridization of [Cu(NH3)4]2+ sp2d instead of sp3?

https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes?rq=1

It is not duplicate. I specifically asked about tetraamminecopper(II) ion. The question you referred doesn't discuss about tetraamminecopper(II) ion.

Well it's not really square planar, it's just a streched octahedron. For copper(II) you end up with a $d^9$-system, a $d^8$ system might be able to do a real square plane but that one electron in a $d^9$ would cost too much energy if it was in a true square plane. The effect is called Jahn-Teller effect. You can simply imagine the $z$-axis for example of the octahedron to be elongated. If you did this till it breaks, you end up with the square base of an octahedron. And this is also what happens in a real square planar complex. All $d$-orbitals with a $z$-component are lowered in energy so they become occupied by electrons and, due to repulsion with the ligand's electrons, ligands will not come from the $z$-axis anymore. On the other hand if you lower the $z$-part in energy due to lever principle you will have to increase the energy of other orbitals as well.
You can see this quite beautifully demonstrated here:

Let's imagine the right case with a square plane and a strong ligand so you start filling orbitals with two electrons before continuing to fill the next higher level or orbitals. Then you could add 8 electrons in total and leave the $d_{x^2 - y^2}$ empty. In total this gives you more energy than in the case of an octahedron. If you look to the left at the octahedron, you can see that for 8 electrons you would have to fill the $e_g$ as well and in total you'd end up with less energy which is why a plane geometry is favored by low spin $d^8$ systems.
Now copper(II) with its $d^9$ is in between that, a plane would be favored of course but it would require a lot of energy to put that extra electron into the $d_{x^2 - y^2}$. So you go in between, a really stretched octahedron, this is the second picture. Here the $d_{x^2 - y^2}$ is not as high as in the other case but the other orbitals you fill are already lower than in a symmetric octahedron so you end up with more energy already. And this is why copper(II) tends to do that. On the other hand this also changes the $d-d$ gap and explains that blue color of copper(II) compounds.
The other two ligands which are on the streched positions can be water, it could be a water and a sulfate, like in $\ce{CuSO4 * 5 H2O}$, but there is still something usually.

Which hybridization takes place here? sp3 or sp2d or dsp2 and why?

@Hisab Jahn-Teller Effect

@Hisab Hybridisation should not be used to describe bonding in metal ion complexes due to its inadequacies. Refer to https://chemistry.stackexchange.com/q/76726/44877 for more details.


text:
How do I eliminate the disagreeable odor of soy milk?
I've was making tofu from soy-milk, but noticed this disagreeable odor produced during the process.
I conducted a string of internet searches in an attempt to find a method to eliminate this odor.
I learnt that the enzyme lipoxygenase is primarily responsible for this odor, and is activated upon grinding soy beans in water. Additionally, it appears that lipoxygenase is inactivated in the presence of  D-glucose and the enzyme glucose oxidase.
Obtaining D-glucose is fairly simple, but I'm at odds as to where I can find glucose oxidase.
Further investigation suggests that commonly available honey contains glucose oxidase.
Armed with this knowledge, I want to know:
Can I add a measured quantity of dilute honey, along with some D-glucose, to the soy milk and inactivate the lipoxygenase enzyme this way?

One more query...
I found somewhere that glucose and glucose oxidase in water, makes hydrogen peroxide and this H2O2 inhibit the lipoxygenase enzyme... So my query is.... What, if add hydrogen peroxide directly, in a measured quantity, in water and grind the soybeans with this solution.

A few things first:
Enzymes are proteins (usually) that promote or accelerate a biochemical reaction. The enzyme lipoxygenase in particular, promotes the dioxygenation of some lipids called Polyunsaturated Fatty Acids (PUFAs); for this, the enzyme requires molecular oxygen (which is present in the air, and gets dissolved in the soy milk) . You can think of this as, "it makes the soy milk go bad". The oxidation of the PUFAs in the soy milk doesn't need lipoxygenase...but having lipoxygenase around makes things faster ;  if the oxidation of over 90% of the PUFAs in a bowl of soy milk takes (say) 10 days without the lipoxygenase...then it would take (say) a day with lipoxygenase present . As to what specific substance(s) is responsible for the smell, I don't really know yet.
The enzyme glucose oxidase promotes the breakdown of D-glucose. Once again, it too requires molecular oxygen.
Now the basis of that patent you cited (I haven't actually read the thing) is apparently along the lines of:

If you dump in glucose and (a small amount of) glucose oxidase into the soy milk, the reaction that follows (oxidation of glucose) should hopefully decrease the molecular oxygen content on the soy milk. This should decrease the activity (not complete inactivation though) of lipoxygenase and hence should reduce the intensity of the odor (not complete removal of the odor)

Now you suggest using dilute honey in place of pure glucose oxidase (since the former's easy to get). But that would barely work.
The composition of honey depends on multiple factors; type of bee, region, season, etc. But from looking up the composition of various kinds of honey (courtesy: Google), apparently glucose oxidase constitutes a fraction of honey. So diluting honey here would make things worse (lower concentration of glucose oxidase). Obviously, adding concentrated honey is also out of question...since that would alter the flavor of the tofu altogether.
As @airhuff mentioned in the comments below: Use of honey is still an option...but a poor option, nonetheless. Why you ask? Because there are no simple means for one to go about quantifying the glucose oxidase content in some random honey-sample. The process(es) are time-consuming, pricey and hence, (probably) impractical for your needs. Additionally, even if it contains the "right" amount of glucose oxidase, I fear that even dilute honey would still alter the taste of the tofu (there's a reason why honey is the popular epitome for "sweetness").

What can you do now?
1) You can buy glucose oxidase online at Sigma-Aldrich. But if your tofu-making session is a one-time thing (i.e- You don't run a tofu-making factory) then this is a bit impractical.
2) Minimize contact with air! The primary reason the soy milk reeks is because the lipoxygenase is breaking down PUFAs using molecular oxygen. So if you want to limit the enzyme's activity, you must keep contact between the soy milk and air to a minimum.
Some ways you can do this:
a) As soon as your done grinding the soy beans with water, transfer it to an air-tight container (fill it all the way to the brim) until you need it.
b) When you're transferring the soy milk from one vessel to another, be sure to avoid air-bubbles. This answer by @Buttonwood may prove useful in that regard.

On a side note:
I'm genuinely pleased to see that you've gone to great lengths to find a solution to your problem before you approached us at Chemistry.SE. For someone who probably doesn't have much of a background in Chemistry (In the original post, you mentioned you were a "commerce student"), this post came as a pleasant surprise. I myself am only student, and I'm relatively new to Chemistry.SE (been here for under a year)... and you'd be surprised to know how so very few users (from a "non-Chemistry" background such as yours) proactively attempt to resolve their problems before resorting to Chem.SE; so kudos to you! We look forward to your future contributions on Chem.SE. Cheers!

I might not write off the use of honey. One statement from the link in the OP reads "During ripening of honey, glucose oxidase is inactivated but it regains its activity on dilution of honey". And it seems from the patent that pretty low levels of glucose oxidase were used (sorry, I'm not a biochem guy and had a tough time quantifying from "units" of the enzyme). Still, you are probably right that the dilution of honey would be too much, it just might be a possibility though. Also, I wonder if heating would significantly reduce O2 content w/o destroying product? Anyway, nice +1 answer :)

What is your opinion for: 1) blanching soybean in water with sodium bicarbonate... 2) Other Heat treatment like deep fat-fry, steam heating 3) using vacuum deodoriser.............. These are another methods for removing/reducing unpleasant odor (I am searching evryday more and more and more for solving my) :) Good Night :) It is 12:24 am in india :)

@air Whup! Thanks for the pointer, I've "fine-tuned" the post in light of your comment. As for heating the stuff to drive out most of the oxygen... I honestly don't know if it would do more harm than good(I've never made tofu before...wouldn't touch soy milk with a meter-long stick anyways). It's an idea, but not one I'm keen on :P

@ManojVerma , note that your original question was "Can I add a measured quantity of dilute honey, along with some D-glucose, to the soy milk and inactivate the lipoxygenase enzyme this way?" I think paracetamol answered this very nicely. Your comment above constitutes a new question altogether. I would suggest accepting this answer as you seem to like it, then posting the above comment as a new question. Your call though.

@Manoj As with your main post...it'd be a great help if you could cite sources for these methods. I've never made tofu before, but: Option (1) doesn't look like it'll help; option (2) might probably help [but I don't know if it'll affect the taste], and option (3) isn't a lasting solution, neither is is very practical (to do at home). ;)


text:
Stereoelectronics of the Houk model
In Clayden's Organic Chemistry it gives the Houk model as a way for predicting selectivity for alkene reactions where an existing stereocentre is present (diastereoselectivity).
Unlike the description of the Felkin-Ahn model, there is no real description of how the Houk model works, and I can't find this information in March Advanced organic chemistry either.
My feeling is that there must be some stabilising interaction between the pi system of alkene with a sigma orbital on the adjacent carbon, however this doesn't explain why one would be favoured over another.

Regarding the question - which parts of the explanation in Clayden is insufficient? In pp 866-868 (2nd ed) there are already two examples given.

I don't think there really is an 'orbital' explanation for this (in the same way as the Felkin-Ahn model), the Houk model really just concerns itself with the lowest energy conformation based on strain/steric (A1,2 vs A1,3) arguments, with this lowest energy conformation being the most reactive. As @orthocresol says, the explanation in Clayden is pretty much the whole of the Houk model

I don't see why this is the slowest energy conformation  he gives conflicting examples where the reaction doesn't take place away from the bulky groip

1. Origins of the Houk model
The initial idea behind the Houk model was infact provided by Kishi, who proposed that alkenes with a chiral centre on the adjacent carbon adopted a 'reactive conformation' in which the small group eclipsed the alkene in order to minimise unfavourable steric interactions.[1]

Fig1: Kishi's "reactant conformer" model, taken from ref [1]
Houk took the 'reactive conformation' from the Kishi studies and modelled it computationally, showing that the lowest energy ground state conformation for alkenes with substituents at the allylic position was indeed the one in which the small group eclipsed the alkene, as proposed.[*]
2. Allylic strain
The Houk model works largely on steric grounds, balancing A1,2  strain and A1,3 strain, as can be seen from the figure below:

Fig2: The reactive conformation, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
The 'reactive conformation' is the middle structure (now days often known as the Houk conformation), in which A1,2  strain and A1,3 strain is minimal. In the left hand structure A1,2  strain is present, in the form of a H-H clash (doesn't look too significant, but raises the energy of the conformation significantly enough, similar to in the conformations of ethane). In the right hand structure A1,3  strain is present, this A1,3 strain is far more significant than A1,2, making this conformation sufficiently high in energy as to be barely populated at room temperature.
Based on the arguments above, it's worth pointing out that the Houk model works best for cis alkenes, with trans alkenes often giving poorer selectivity:[2]

Fig3: cis and trans alkenes reacting via the Houk model, taken from Organic Synthesis- Strategy & Control, Wyatt and Warren (Wiley)
This result is consistent with the fact that in a trans alkene, A1,3 strain is already minimised by virtue of the olefin geometry, meaning that two possible 'low energy' conformers exist which have neither A1,3 strain nor significant A1,2 strain.
3. Reactions not following the Houk model
Not all reactions for which the Houk model could be used give the expected 'Houk product'. The most common of these (and indeed the one in Clayden that you mention in the question as not following the Houk model) is when the starting material is an allylic alcohol.

Fig4: Reactions not following the Houk model due to directing effects, taken from Organic Chemistry, Clayden and Warren (Oxford University Press)
In the example above, the hydroxyl group is acting as a directing group, guiding epoxidation with the mCPBA to the same face (the steric argument, which would always predict epoxidation away from a bulky group, does not apply here).

Notes and references
[1]: Houk, K. N.; Rondan, N. G.; Wu, Y.-D.; Metz, J. T.; Paddon-Row, M. N. Theoretical studies of diasteroselective hydroborations. Tetrahedron 1984, 40 (12), 2257–2274. DOI: 10.1016/0040-4020(84)80009-5.
[2]: Vedejs, E.; McClure, C. K. Hyperconjugative effects of allylic substituents are not important in osmylations. J. Am. Chem. Soc. 1986, 108 (5), 1094–1096. DOI: 10.1021/ja00265a048.
[*]: The Houk group has been instrumental in providing a bridge between theoretical predictions and experimental results, providing weight to proposed models, and giving confidence in being able to look at a reaction and predict what product it may be.


text:
Conventional heating compared to microwave heating of chemical reactions
In order to try to increase yield of my cross-metathesis, it was suggested to carry out the reaction in the microwave (a lab microwave, not a kitchen microwave).
The rationale for this was that microwaves accelerate the reaction, but thinking on how a microwave actually works, it's basically just heating in the same way as I could achieve with a PEG bath.
When I searched the internet, there are lots of reviews and even books published about microwaves in organic synthesis, but I can't find a coherent argument for what microwaves are doing that makes reactions 'magically work'.
The radiation in theory simply causes the molecules to move creating heat, as far as I know, it isn't like photochemistry where something mechanistically different can occur.

Apart from reducing temperature gradient in the reaction mixture when using microwave there are also non-thermal effects.

Your question about the 'magic' microwave effect is one that I think most chemists who've used a microwave have asked. Anecdotally, some reactions do often just work better when carried out in a microwave. The whole debate is quite nicely summarised in this quote from a (fascinating...) book on the use of microwaves in chemical synthesis:

Since the early days of microwave synthesis, the observed rate accelerations and sometimes altered product distributions compared to oil-bath experiments have led to speculation on the existence of so-called “specific” or “non-thermal” microwave effects. Historically, such effects were claimed when the outcome of a synthesis performed under microwave conditions was different from that of the convention- ally heated counterpart at the same apparent temperature. Reviewing the present literature, it appears that today most scientists agree that in the majority of cases the reason for the observed rate enhancements is a purely thermal/kinetic effect, i.e., a consequence of the high reaction temperatures that can rapidly be attained when irradiating polar materials in a microwave field
Source: Microwaves in Organic and Medicinal Chemistry

Microwave radiation:

Source: Microwaves in Organic and Medicinal Chemistry
From the table above, its clear that microwave radiation is not strong enough to cause any actual chemical change itself (other than via the temperature it creates), for microwave radiation used by common household and lab microwaves (2450 MHz, which is important as it avoids interfering with communication equipment), the energy is sufficiently low that it couldn't in theory even over-come a hydrogen bond.
Thermal/pressure effects:
There are several reasons why the use of a microwave may be beneficial compared to a standard reflux setup in a PEG bath (or any other conventional heating method):

the microwave is able to quickly reach a given temperature, with only a short induction period. The result of this is that the reaction reaches the desired temperature quickly, rather than slowly warming from room temperature. This means that the reaction gets going immediately, which can be especially beneficial in cases where the catalyst/reagents being used aren't terribly stable (your metathesis is one such example where the catalyst can quickly degrade in solution).
the microwave heats the solution directly and uniformly, compared to heating conveniently where the glass is being heated and in turn the glass heats the solution.

Source: Biotage. Left image = microwave heating, right image = oil bath, both taken after 1 min of heating

the vials used in microwave reactions are sealed (they're closed systems), which means in addition to temperature, they're able to carry out reactions under increased pressure, allowing the reaction to be heated above the boiling point of the solvent being used.

Source: Biotage
Overall, the easiest way to show all of this is by looking at the read-out from a microwave reaction, showing how quickly the desired temperature can be reached and how this remains constant, along with the pressure, throughout the entire reaction run:

Non-thermal effects:
Non-thermal effects, or 'specific microwave effects' are defined as accelerations of chemical transformations in a microwave field that cannot be achieved or duplicated by conventional heating, but essentially are still thermal effects. These have been considered by many groups and indeed a wikipedia page exists discussing them, however they're difficult to show/prove experimentally.
For solution phase organic synthesis, its generally accepted that there is no specific microwave effect. Only in solid phase systems (ionic lattices for example) is there anything to suggest something unique going on that cannot be achieved using standard heating methods alone.


text:
Can the idea of entropy be extended to orbitals?
The forward reaction for: $\ce{HF<->H^+ +F^-}$ is entropically favourable; but energetically unfavorable: as there is too much electron density for $\ce{F^-}$ ion to cope up. That's a reason why $\ce{HF}$ is a weaker than $\ce{HI, HCl}$ etc.
But, can there be another explanation for weak acidity of $\ce{HF}$?
How about we introduce the idea of entropy to orbitals. Let me explain, in $\ce{HF}$ the elctron is in a molecular orbital of greater region than in $\ce{F^-}$. So, the valence electron pair is in an entropically more favorable state in $\ce{HF}$ than in $\ce{F^-}$: where it is contained in little space. That's why, $\ce{HF}$ is more favorable than   separated $\ce{H^+}$ and $\ce{F^+}$.
Maybe, for other hydracids of same kind eg. $\ce{HI, HCl}$ this effect isn't so important. That's why they are stronger acids.
Share your thoughts.

https://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

No.
The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally likely to be populated, then the number of possible microstates for the system, $W$, is large and hence $S = k \ln W$ is large.
An electron in an orbital isn't whizzing about, and an orbital most certainly isn't a container of variable size for the electron to move around in. An orbital is simply one (electronic) state of an electron.
Therefore, it doesn't make sense to talk about the translational states of an electron that is moving about in an orbital. It also doesn't make sense to talk about the translational entropy of an electron as being larger in a larger orbital.


text:
Why does viscosity decrease when divalent cations are dissolved into aqueous solutions of sodium carboxymethylcellulose?
Sodium Carboxymethylcellulose (CMC/NaCMC/Aqualon) is a common product used in many instances for its unique properties, especially its ability to make solutions viscous. It is well known that the charge of metal ions directly affects CMC's viscosity.
According to the pdf: AQUALON Sodium Carboxymethylcellulose (ResearchGate), it states that:

Compatibility of CMC with inorganic salt solutions depends largely on
  the ability of the added cation to form a soluble salt of
  carboxymethylcellulose.

Then it proceeds with:

As a general rule, monovalent cations from soluble salts of
  carboxymethylcellulose, divalent cations are borderline, and trivalent
  cations form insoluble salts.

It then continues on to discuss the effect of this on the viscosity of aqueous solutions with these cations mixed with CMC, but doesn't explain why.
It says that divalent cations (which are "borderline" yet still soluble into CMC solutions) decrease the viscosity of CMC solutions, yet monovalent cations (which fully dissolve) do not decease the solution's viscosity (or not to the same extent). Why is this?

I think it has to do with coiling in solution. When the polymer chains are tightly coiled, the viscosity is lower, because the tight coils don't interact with other polymer chains.


text:
Why is Cl2 only formed in acidic media?
If aqueous $\ce{NaCl}$ and $\ce{NaClO}$ are added together no reaction occurs, but if you add dilute acid to the solution $\ce{Cl2}$ gas is formed. Why does this reaction only happen in acidic media? I know that there's something to do with the reduction potentials but I couldn't formulate a full explanation for this effect.

Basically this is due to two reasons.
The disproportionation of chlorine in water looks like this:
$$
\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}
$$
So you need the $\ce{H+}$ to shift the equilibrium to the halogen site and the halogen produced is a gas and will leave the reaction.
The other explanation is that, as $\ce{HOCl}$ is an oxidizer and a $\ce{Cl+}$ transfer-agent, both of these actions happen in an $\ce{S_{N}2}$-type fashion. For the $\ce{Cl+}$ transfer it is called associative activated substitution of a nucleophile, here $\ce{Cl-}$. So it's pretty much like an $\ce{S_{N}2}$ reaction, where
$\ce{H-Cl}$ and $\ce{Cl-OH}$ react.


text:
Small Propellanes Stability
Why is [1.1.1] propellane more stable than other small members in the propellane family even though it seems to be under the greatest strain?  What makes a small propellanes stable, or in general, any propellane stable?

Interesting question. Strangely, Wikipedia contradicts itself here: On the propellanes page it says: Surprisingly, the most strained member [1.1.1] is far more stable than the other small ring members ([2.1.1], [2.2.1], [2.2.2], [3.2.1], [3.1.1], and [4.1.1]).,  but on the page for [2.2.2] propellane it says This ([2.2.2] propellane) compound is unstable (although not as much as [1.1.1]propellane). The bond angles on the shared carbons are considerably strained: three of them are close to 90°, the other three to 120°. The strain energy is estimated to be 93 kcal/mol (390 kJ/mol)..

Then what is the general consensus?

I think Carey A. F. Advanced Organic Chemistry: Part A: Structure and Mechanisms, 2000, p. 164 pretty much answers your question.

@andselisk Can you write an answer for this. The link is not working on my phone.

@Mockingbird Done, sorry for the inconvenience, that was a link to Google Books which can be wacky sometimes to get access to.

Fulfilling the request from @Mockingbird, there is a quote from (1, p. 90)

Surprisingly, [1.1.1]propellane is somewhat more stable to thermal
  decomposition than the next larger propellane, [2.1.1]propellane,
  indicating a reversal in the trend of increased reactivity with
  increased strain. To understand this observation, it is important to
  recognized that the energy of both the reactant and intermediate
  influence the rate of unimolecular reactions that lead to
  decomposition. In the case of propellanes, homolytic rupture of the
  central bond is expected to be the initial step in decomposition. This
  bond rupture is very endothermic for [1.1.1]propellane. Because
  relatively less strain is released in the case of [1.1.1]propellane
  than in the [2.1.1]- and [2.2.1]-homologs, [1.1.1]propellane is
  kinetically most stable.(2)

Another manifestation of the relatively small release of strain
  associated with breaking the central bond comes from MP4/6-31G∗
  calculations on the energy of the reverse ring closure.(3)

The thermal decomposition of [1.1.1]propellane has been studied both
  experimentally and by computation(4)

The book chapter further discusses reactivity of [1.1.1]propellane, if OP is really interested in this area, I would suggest to grab the book in the library or focus on the papers (2-4) the book references to.
(1) Carey, F. A.; Sundberg, R. J. Advanced organic chemistry. Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
(2) Wiberg, K. B. Angew. Chem. Int. Ed. Engl. 1986, 25 (4), 312–322. DOI: 10.1002/anie.198603121
(3) Adcock, W.; Binmore, G. T.; Krstic, A. R.; Walton, J. C.; Wilkie, J. J. Am. Chem. Soc. 1995, 117 (10), 2758–2766. DOI: 10.1021/ja00115a011
(4) Jarosch, O.; Walsh, R.; Szeimies, G. J. Am. Chem. Soc. 2000, 122 (35), 8490–8494. DOI: 10.1021/ja994043v


text:
Purification of the Hünig Base
My group uses a lot of Hünig base which needs to be dry and pure. The common procedure we use is to first distill from ninhydrin then from calcium chloride.
Apparently the problem is that Hünig base decomposes to give primary amines (which are removed by the ninhydrin). What I don't quite understand is why we need the second distillation (why can't we just distill from ninhydrin in the presence of a mild drying agent) but also why for triethyl amine we can just distill off calcium hydride (in told this drying agent causes Hünig base to decompose but I can't see how this would work given that the calcium hudride isn't a strong enough base to take a proton from the groups on the amine).

Ref: Purification of Laboratory Chemicals, 7 ed.
In the first case, your labs 'common procedure' isn't the 'common procedure' that most people tend to follow. Generally, it should be carried out as follows

Distill the DIPEA from ninhydrin at atmospheric pressure. The ninhydrin reacts with any primary/secondary amines, removing these impurities from the DIPEA. This ninhydrin step liberates water, as it works by condensing the primary/secondary amines.  Since the boiling point of DIPEA isn't that distinct from that of water, people tend not to try fractionally distilling the two away from one another. The DIPEA that comes across in this first distillation is free of primary/secondary amines, but contaminated with up to ~1% water (which is far too much to get away with using in moisture sensitive chemistry).

Ref: Chemistry LibreText

Re-distil the 'wet' DIPEA from KOH, again at atmospheric pressure, which removes water and provides clean and dry base.

The reason why calcium hydride is often not used is because it is thought to promote the degradation of DIPEA by cleavage of the isopropyl groups (they can eliminate to give stable products. Certainly, distillation from calcium hydride is usually fine, but the DIPEA shouldn't be stored over it for prolonged periods.

This doesn't answer why I can't just mix the ninhydrin and KOH in one flask and do just one distillation not two distillation as described, or even just distill on to a lot of KOH and allow that to do the drying in tbe receiving flask


text:
Epoxide ring opens, or not?

Does the epoxide ring open in this case? I guessed it does, but it really doesn't. The answer given (among the four options) is another epoxide, with some sort of substitution. Could someone please explain the mechanism for this reaction?
Also, what exactly is the condition for ring opening of epoxides? Thanks for the help!

Epoxide gets opened at the least hindered side by the ethoxide (acting as a nucleophile rather than as a base in this case).
The resulting alkoxide intermediate can itself act as a nucleophile. In the middle intermediate could either kick out ethoxide (which would give back the starting material), or kick out the chloride (to give a different epoxide). Since the chloride is a better leaving group, it is this one that probably leaves (theres an element to which the whole thing is reversible away, and if it does kick out ethoxide, the ethoxide can just attack again to re-form the middle alkoxide intermediate)

The product on the right hand side is probably what the mixture would converge upon, given that chloride is  a better leaving group than ethoxide.

Is there a name for this? I could've sworn this was called the Payne rearrangement, but Wikipedia tells me it's something different.

@orthocresol- it's only a Payne rearrangement (I think...) where its a 2,3-epoxyalcohol and the epoxide is 'walked' to the end of a chain before being intercepted by a nucleophile (or at least thats the common textbook variant, using a bulky thiol nucleophile to trap)

It's technically not rearrangement, just standard epoxy making reaction.

@Mithoron You guys fon't seem sure. What's the real answer?

"2,3-epoxyalcohol and the epoxide is 'walked' to the end of a chain before being intercepted by a nuclueophile " Nucleophile or electrophile?

@Mockingbird, I think the discussion is about what should be called a Payne rearrangement. If you look it up in Clayden it might be more clear

@NotEvans Did you see my 2nd comment?


text:
Why is nitrosobenzene electrophilic on O in organocatalytic oxygenations?
In 2003 Zhong1 and MacMillan2 reported the use of nitrosobenzene, PhNO, in the organocatalytic asymmetric α-oxygenation of carbonyl compounds:

which is all nice, except for the slightly counterintuitive fact that PhNO is acting as an electrophile on oxygen rather than on nitrogen.
Logically PhNO should be electrophilic on nitrogen (less electronegative than oxygen), and indeed, a 1994 review on the reactivity of PhNO lists numerous examples where this is the case.3 On top of that, if the enamine were to ad to nitrogen instead of oxygen, the phenyl group could be placed equatorial in the transition state:

Is there an explanation for the regioselectivity here?

References

Zhong, G. A Facile and Rapid Route to Highly Enantiopure 1,2-Diols by Novel Catalytic Asymmetric α-Aminoxylation of Aldehydes. Angew. Chem. Int. Ed. 2003, 42 (35), 4247–4250. DOI: 10.1002/anie.200352097.
Brown, S. P.; Brochu, M. P.; Sinz, C. J.; MacMillan, D. W. C.The Direct and Enantioselective Organocatalytic α-Oxidation of Aldehydes. J. Am. Chem. Soc. 2003, 125 (36), 10808–10809. DOI: 10.1021/ja037096s.
Zuman, P.; Shah, B. Addition, Reduction, and Oxidation Reactions of Nitrosobenzene. Chem. Rev. 1994, 94 (6), 1621–1641. DOI: 10.1021/cr00030a007.

Using proline as the catalyst provides enantioselectivity to the process, but also allows N vs O regioselectivity to be controlled (though not always perfectly).
Houk (and subsequently others) applied computational methods to the reaction in an attempt to understand why the nitrosobenzene was acting as an electrophile at oxygen rather than nitrogen, as would traditionally be expected (and as is often the case when no proline is used). The computational studies showed a clear preference for the transition state in which the basic nitrogen of nitrosobenzene was able to interact with the acidic proton of the proline:

Proline catalysis occurs via a proline enamine attack on the oxygen of the nitrosobenzene monomer with a simultaneous proton transfer from the carboxylic acid. Computational evidence shows that the observed heteroatom selectivity materializes through the difference in basicities of the oxygen and the nitrogen atoms.
Source: Cheong, P. H.-Y.; Houk, K. N. J. Am. Chem. Soc. 2004, 126 (43), 13912–13913. DOI: 10.1021/ja0464746.

This study further indicated that the transition state for attack at N (which you drew) is disfavoured by 2.5 kcal/mol:

Ref: Enantioselective Organocatalysis, 2007, Dalco (Wiley)

So in a nutshell, nitrogen is more basic and therefore preferentially accepts th proton?


text:
Reasoning for baking soda treating poison ivy rashes
A quick Google search will show that baking soda (sodium bicarbonate) mixed with water in a 3 to 1 ratio will make a paste to be applied on poison ivy rashes. This is commonly advertised as a home remedy to poison ivy rashes.
Treatment for poison ivy ranges from mechanically removing the rash (scratching, which can make it worse), drawing the urushiol oil out, drying the rash, and simply waiting it out as your immune system takes care of it. Which method does this sodium bicarbonate paste use?
Basically, why should someone cover a poison ivy rash with a sodium bicarbonate paste other than it allegedly being an affective home remedy? Is there an scientific/chemical explaination as to why it might work?

@Close-voters Please avoid hasty close-voting. Just because the post has something to do with medicine doesn't mean it should be closed as a "Personal Medical Question". Geez guys, this (topical application of baking soda solution) is a general and common mode of treatment, hence, this is a perfectly valid question and should remain open.

@paracetamol ""why should I cover my poison ivy afflicted arm" does feel personal

@Mith Meh... I guess that's still okay. I mean, putting the "paste" over the afflicted piece of skin is a commonly followed (apparently) method of first-aid. Wanting to know why it should be followed seams reasonable to me :P

It still does look like question for dermatologist.


text:
Reforging tungsten/tungsten carbide temperatures
I was wondering about the reforging of tungsten and the temperature needed.
As I understand it, tungsten melts at $\pu{3442 ^\circ C}$ and tungsten carbide at $\pu{2870 ^\circ C}$. However, once a jeweler makes the rings they will never take them back to alter them.
Does the temperature needed to melt a pure tungsten ring change after the initial forging (with all heat treating and cooling included) or is there no change?
If there is a change:

What temperature is needed for tungsten?
What temperature is needed for tungsten carbide?
What atomic/chemical changes took place?

https://chemistry.stackexchange.com/questions/16640/how-do-you-melt-metals-with-super-high-melting-points

Avoid asking multiple questions at once please.


text:
Finding number of orbitals in nickel whose |m| = 1
I am new to quantum mechanics, and I need to find out total number of orbitals in nickel , which have $ \vert m \vert =  1 $ and at least one electron is present.
m = magnetic quantum number
I need to understand whole thing, so I need full explanation.

Write all possibilites and count them.

I'm voting to close this question as off-topic because itcan be considered homework with no input from OP

@Mithoron How can be considered a homework? What's the meaning of no input from OP?

Conventionally, the location of an electron is written in the following way,
$n,~ l,~m_l,~m_s $
$n$ represents the principal quantum number,
$l$ represents the azimuthal quantum number,
$m_l$ represents the magnetic quantum number
$m_s$ represents the spin quantum number.
The absolute value of the magnetic quantum number must be less than or equal to that of the azimuthal quantum number. This means that the azimuthal quantum number must be greater than or equal to 1 (according to your question). Because the azimuthal quantum number must be less than that of the principal quantum number, the principal quantum number must be greater than or equal to 2.
Remember that the azimuthal quantum number corresponds to the subshell.
0 - $s$ subshell,
1 - $p$ subshell,
2 - $d$ subshell,
3 - $f$ subshell,
The ground state electron configuration of nickel is
$1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^8
$
Now, time to find electrons in orbitals that fit our requirements.  First, $n ≥ 2$, this gives us:
$2s~
2p
~3s
~3p~
4s
~3d
$
Second, $l ≥1$, this gives us:
$2p
~3p~
3d
$
Because we want the magnetic quantum number to be only 1 or -1, there are two possibilities for orbitals each that fulfill the requirements.
This gives us 2 + 2 + 2 or
6 possibilities.


text:
Does a chiral centre have to have 3 or 4 different groups attached?
I am seeing conflicting answers on the net
e.g.

A carbon atom is chiral if it has four different items bonded to it at the same position. (Wiki)

compared to

Stereocenter (chiral center): An atom with three or more different attachments (Chem.Ucla)

Why does one say 3 or more, and one say 4?
Also, there is a double bonded carbon atom here that is said to be a stereocentre, which I cannot understand.  Isn't the double bond counted as two of the "same" attachments (so not even satisfying the 3 different groups criteria). How is that still a chiral centre?

It looks like the difference is that the UCLA definition doesn't include a hydrogen atom as an attachment, so 3 non-hydrogen attachments qualifies if the fourth "attachment" is a hydrogen.

How about the middle picture in the Ucla Link. There is a double bonded carbon, and it calls this a stereocentre/chiral centre. So there is no 4th attachment right? Also, isn't the double bond counted as two of the "same" attachments (so not even satisfying the 3 different groups criteria). How is that still a chiral centre? Thank you!

Yea, I missed that, so my comment doesn't make sense. Their (UCLA's) explanation just doesn't quite read right and I don't get it...sorry.

That's okay. Thank you anyway :) I thought I had chirality down pat until I saw this ... which is going to lead to my next question of why cis/trans isomers fall under the category of diastereoisomers, but I will save that for a separate question after this one has been answered.

That image doesn't make sense. In an alkene with E/Z isomerism the unit that leads to stereoisomerism is the alkene, not one particular carbon of the alkene. None of the alkene carbons are chiral centres. Clearly it seems that UCLA have their own idea of what a stereocentre is, but it doesn't make sense to me for the stated reasons above.

Regarding the diastereomer thing see https://chemistry.stackexchange.com/a/68709/16683 the definition of diastereomer is simply a stereoisomer that is not an enantiomer, which E/Z isomers fall into.

https://chemistry.stackexchange.com/questions/50106/can-heteroatoms-with-lone-pairs-be-chiral-centres - if you don't count lone pair as substituent... This is the thing with 3 or 4 and talking double bond  here is nonsense.

Very good question. I can confidently say I would not have learned the nuance between the two terms without this question.

@K-Feldspar I think specifying the difference between a chiral center and a stereocenter is important here. Stereocenter include both center for optical as well as geometrical isomerism ... So in ur picture the red carbon atoms are stereocenter as they can show G.I. but the blue ones can't..

The answer is that all chiral centers are stereocenters, but not all stereocenters are chiral centers thus the words are not interchangeable.
A chiral center requires four or more  different molecular groups attached to a single center atom (asymetrical). In the case of pnictogens (nitrogen, phosphorus, etc...) or chalcogens (sulfer, selenium, etc...) an electron pair can count as one of the molecular groups.
An alkene is a stereocenter that requires 3 or 4 different molecular groups such that no two common groups are on the same side of the bond (must be able to determine an E/Z configuration). In the case of the blue stereocenter, the methyl groups are on the same side and thus switching their position would create an identical compound meaning the alkene is not a stereocenter.  For the red center switching the position of the bromine and methyl groups would create the Z-isomer from the E-isomer thus is a stereocenter, but not a chiral center.  If however, the hydrogen of the blue stereogroup were to switch positions with a methyl group then the blue alkene would be a stereocenter with three distinct molecular groups.


text:
How to calculate average volume of a molecule?
The molecular weight of glucose-1-phosphate is $\pu{260 g/mol}$ and its density is $\pu{1.5 g/mL}$. What is the average volume occupied by 1 molecule of this compound?

Volume occupied by a single molecule can be roughly estimated as follows:
$$V_x = \frac{V_\mathrm m}{N_\mathrm A} = \frac{M}{\rho N_\mathrm A} = \frac{\pu{260 g mol^-1}}{(\pu{1.5 g ml^-1})(\pu{6.02E23 mol^-1})} \approx \pu{2.9E-22 ml},$$
where $V_\mathrm m$ is the molar volume, $M$ is the molar mass, $\rho$ is the density, and $N_\mathrm A$ is the Avogadro constant.
Note that "volume occupied by a molecule" and "volume of a molecule" are two separate entities.


text:
D-Threose: reduction and optical activity
I've bumped into this exercise:

Judging by the mechanism of reaction between D-threose and $\ce{NaBH4}$, do you think the final product will be optically active?

To me looks like a reduction, since $\ce{NaBH4}$ is a lightly reductive agent, but then why the answer is "No, the product is not optically active"?
Is it because of the presence of anomers or there are other reasons?

I have edited your question to include the ce for chemical equations. For example, typing $$\ce{2KClO3 ->[MnO2] 2KCl + 3O2}$$ will be rendered as: $$\ce{2KClO3 ->[MnO2] 2KCl + 3O2}$$

When you reduce the aldehyde (from the open chain threose, since the cycle cannot be reduced as it has only acetal-like functionality) you introduce a symmetry element (both ends of the molecule become $\ce{CH2OH}$).
The presence of a symmetry element means the reduced molecule is not chiral, but meso.


text:
How can scientists measure the electron affinity and the ionization energy of an element?
I am pretty curious about the method that the scientists use to measure the electron affinity and the ionization energy of an element. If someone knows about it, please tell me.

Did you consider consulting a reference like a textbook, or  https://en.wikipedia.org/wiki/Electron_affinity or https://www.britannica.com/science/ionization-energy?


text:
Gibbs free energy and electrical potential
The relation between the Gibbs free energy and the electrical potential is:
$$\Delta G=-zFE$$
$z$ is the valence of the electrochemical reaction and $F$ is the Faraday constant.
If at equilibrium the $\Delta G$ is zero this means that also $E$ is zero but this is not true because $E$ at equilibrium is the equilibrium potential of the electrochemical reaction considered.
This makes some confusion in my mind and for sure I'm not properly consider something.
Can someone help to clarify my mind?

Look in your textbook for the distinction between $\Delta G$ and $\Delta G^o$.

That $\Delta G0$ is evaluated at standard conditions. I don't get your answer

The equilibrium potential $E_\text{eq}$ is always zero -- at equilibrium we have no net flux and hence no potential difference.

In case of a galvanic cell it is not true that when the two electrodes are not connected the two half cells are at equilibrium? If I connect the electrodes to a voltmeter I measure a potential difference.

Your $\Delta G$ is really $\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated change in the electroststic energy. That is, it equals $zF$ (the amount of charge moving against the electric potential for 1 mole of reaction) times $E$. Therefore, at equilibrium, the total change of the Gibbs free energy satisfies
\begin{equation}
\begin{split}
\Delta G_{\mathrm{total}} &= \Delta G_{\mathrm{reaction}} + \Delta G_{\mathrm{charging}}\\
&= 0
\end{split}
\end{equation}
just as it should.
In fact, when deriving the expression for the cell potential $E$, the logic goes backwards. We know that the electrodes are charged until equilibrium is reached, in which case the sum of $\Delta G_{\mathrm{reaction}}$ and $\Delta G_{\mathrm{charging}} = zFE$ should vanish. Thus, the cell potential is given by
\begin{equation}
E = -\frac{\Delta G_{\mathrm{reaction}}}{zF}.
\end{equation}


text:
Why does the conductivity of a LiOH solution decrease when boric acid is added?
If I start with pure water and add $\ce{LiOH}$ the conductivity of solution increases. However, if I then dissolve boric acid in the same solution the conductivity falls. I would have thought adding more cations or anions to a system would always increase conductivity, but in this case it does not. Why is this?

TL;DR: The addition of boric acid effectively switches the conducting anionic solute from hydroxide to borate, which has a lesser contribution to the conductivity due to its larger size and corresponding lower diffusivity. Most solute systems will not behave this way.

In sufficiently dilute solution, the contribution $\kappa_i$ of each dissolved ion to the overall conductivity is proportional to the concentration, diffusivity, and squared charge of that ion:
$$
\kappa_i \propto z_i^2D_iC_i
$$
And, the total conductivity is just the sum of these contributions:
$$
\kappa = \sum_i{\kappa_i}
$$
In this case, you're starting with $\ce{LiOH}$, which dissociates  more or less completely into aqueous $\ce{Li^+}$ and $\ce{OH^-}$, and the conductivity rises as you add more of it. Once you start adding boric acid, though, you're no longer just dissociating ions into the solution.
Boric acid is a funny creature.  Even though it looks like it might be a triprotic Arrhenius acid if we were to write it as $\ce{H3BO3}$, in reality it's a monoprotic Brønsted acid that either exchanges a hydroxide ion with the water solvent:
$$
\ce{B(OH)3 + 2H2O <=> B(OH)4^- + H3O^+}
$$
or, at sufficiently high $\mathrm{pH}$ $(\mathrm pK_\mathrm a = 9.2)$, it just takes up a hydroxide ion from solution:
$$
\ce{B(OH)3 + OH^- <=> B(OH)4^-}
$$
So, when added to a solution of $\ce{LiOH}$, boric acid just substitutes dissolved $\ce{OH^-}$ for dissolved $\ce{B(OH)4^-}$, with no net addition of anions to the solution!  Thus, the $C_i$ contribution to $\kappa$ doesn't really change much. As well, both $\ce{OH-}$ and $\ce{B(OH)4^-}$ are monovalent anions, so there's no effect on any of the $z_i^2$ terms either.
The overall decrease in $\kappa$ comes from the change in diffusivity that results from the swap of hydroxide for borate.  The diffusivity of an anion is roughly proportional to its ionic volume—bigger ions experience more "drag" from the surrounding solvent molecules, so they diffuse (move around) more slowly. I couldn't find an ionic radius for borate on a quick search, but just from their compositions, it seems sensible to assume that the borate anion is considerably larger than the hydroxide anion, and thus its diffusivity should be correspondingly less.


text:
Purification of isoborneol
How can I convert isoborneol powder to isoborneol flakes?
I have made the isoborneol powder from reduction of camphor using sodium borohydride.

Recrystallisation ?

iso-Borneol is a fairly common reagent, and methods for its preparation and purification are readily accessible (even without journal access). For information, the entry in Purification of Laboratory Chemicals suggests recrystallising from ethanol or petroleum ether, melting point data is also provided as a way of assaying the resulting material  (though NMR would be better).

dl-Isoborneol [124-76-5] M 154.3, m 212° (sealed tube). Crystallise isoborneol from EtOH or petroleum ether (b 60-80°). It sublimes in a vacuum. The 4-nitrobenzoyl derivative has m 153°. [Yager and Morgan J Am Chem Soc 57 2081 1935, Beilstein 6 II 80, 6 III 299, 6 IV 281.]
Taken from Purification of Laboratory Chemicals, 7 ed. Armagero and Perrin

Its worth noting that a powder and flakes are essentially just two amorphous forms, and there isn't really any benefit of having one over the other (if anything, the powder is more convenient to handle).


text:
Quantifying reactivity in substrates and electrophiles
What is "quantitative treatment of reactivity in substrates and electrophiles"?
I read about it in Jerry March's Advanced Organic Chemistry textbook, but I'm finding it difficult to comprehend.

One large body of work aiming to quantify electrophilicity was accumulated by the Mayr group. Numerous reagents were tested against typical substrates, recording the kinetic rate constants allowed to attribute electrophilicity / nucleophilicity along the equation of
$$ \log k (\pu{20 ^\circ{}C}) = s_N (N + E) $$
with
$E$ = electrophilicity parameter
$N$ = nucleophilicity parameter
$s_N$ = nucleophile-specific sensitivity parameter
Beside the selection of publication mentioned on the group page, work was invested to create a map, and a public database with currently "1075 nucleophiles and 276 electrophiles".

So, testing various possible electrophiles on a given aromatic system, and quantifying the rates of reaction for each reaction is what you meant ?

@VishnuHaveri Yes, I did mean as one way to quantify electrophilicity.  An other, according to https://en.wikipedia.org/wiki/Electrophile, is the electrophilicity index $\omega$, as fraction of the square of electronegativity ($\chi^2$), and chemical hardness ($\eta{}$). This second approach leaves kinetics, and solvent dependency  on nucleopilicity or electrophilicty of "the Mayr approach" out.


text:
What are the organisms that are used for toxicology studies in water?
I have heard that zebra fish is used for toxicology studies . Is there any other organism that is actively used for toxicological studies?https://www.google.co.in/search?q=zebrafish+toxicity+studies&oq=zebra+fish+tox&aqs=chrome.2.69i57j0l5.11109j0j7&sourceid=chrome&ie=UTF-8

Not sure if on topic here... feels rather Biology to me.

Definitely more of a biology question. Also, https://en.m.wikipedia.org/wiki/Aquatic_toxicology

@orthocresol.I am sorry if it's off topic . But there are many projects going on in the branch of environmental engineering to determine toxicity of effluents(domestic and industrial)with these organisms. So i was curious to know about other organisms that are involved in this study like zebra fish

I'm voting to close this question as off-topic because the selection of  appropriate organisms for "toxicology-studies" is by and large, a biologist's problem. Burn this question with fire! On a more serious note, this ought to be migrated to the Bio.SE.

Well, sometimes daphnia (Daphnia Magna also known as water fleas) are used in toxicology bioassays a an indicator genus. I have  personally experimented with Daphnia before, and it is easy to see their internal organs through their transparent body. For example, this study uses Daphnia to test the effect of nanoparticles on Daphnia and this article shows the effects of heavy metals on Daphnia. For more information, visit Wikipedia.


text:
Using baking soda paste (alkaline) to "neutralize" honeybee venom (acidic). Is this scientifically sound?
I recall reading (courtesy: my 9th grade Chemistry text-book, chapter Acids, Bases and Salts), something along the lines of:

Bee venom is acidic. This is why the area of skin (stung by a bee) is treated (first-aid) by the topical application of a sodium bicarbonate paste. This method works by neutralization of the acidic venom.

The problems with this theory? I see three:
Firstly,
A honeybee's stinger can grow to be as long as 3mm. If bee venom were acidic, then how would the baking-soda paste neutralize it (entirely) by simply applying it on the surface?
Secondly,
Wikipedia provides the composition of honeybee venom;

Melittin
Apamin
Adolapin
Phospholipase A2
Hyaluronidase
Histamine
Dopamine and Noradrenaline
Tertiapin

As far as I know...none of these are actual acids, instead, they're pretty much all proteins. Which appears to seriously contradict the popular "Bee venom is acidic" notion.
Thirdly,
Some sources (online) suggest that application of vinegar (which is acidic) also relieves the pain. Additionally, even meat tenderizers (proteases), toothpaste and wet aspirin tablets seem to do the trick.

So I'll break up my question into two bits (to hopefully, make it easier to answer):
1) Is honeybee venom acidic? If so, what substance present in it is responsible for the "acidic" nature of the venom?
2) Why does applying baking-soda (paste) relieve the pain? Does it somehow denature/"neutralize" the venom?

Honeybee venom is acidic. It contains formic acid, also known as $\ce{HCOOH}$. However,
Rubbing baking soda probably isn't really reducing the pain because

The formic acid isn't causing the pain
The venom is injected under tissue
The ratio of venom to baking soda is way out of proportion

For more information, visit this website.

@Javascripter One interesting point in your linked webpage "applying hot water to coagulate the venom".. Ha... Ha..

Thanks for taking the time to post an answer! A tip on formatting though: For emphasis, you can italicize words or put them in bold; typing it out in UPPERCASE however, as you did with "IS", can sometimes yield other (more dubious) meanings ;)

@Mocking Yeah...which is pretty much why I'm not (completely) buying what that site says :D

As para shown it doesn't contain HCOOH, it's only "formic" not "beeic" acid.


text:
How does calcium carbide artificially ripen fruits?
Lately, use of chemicals to artificially ripen fruits is in full swing. Firstly, how do they do it? Next, Are there any ways to distinguish between those that are naturally ripened and those that are artificially ripened?

If it's cheap/not pricey and available off-season, then it's likely to be artificially ripened (<----common sense) ;P

Note that there is significant list of fruits, including apples and most berries that do not ripen at all after being picked

@paracetamol Then does famous soft drinks(like maaza )use artificially ripened fruits?

Chitra, I don't think they would tell me that  ;)

But can anybody make a guess.

Technically, there is no difference between naturally ripened fruit and artificially ripened fruit... they're both ripened fruit. I suppose you were looking for some sort of simple chemical test to distinguish between them; but like I said, there's no point in it since in both cases you end up with ripe fruit.
Fruits normally emit ethene (ethylene) which is essential to the ripening process. Which is why fruits picked before they're ripe are sometimes wrapped in a bag that seals in whatever little ethene they emit, thereby accelerating the ripening process.
As you would know, calcium carbide does not ripen fruit per se. Rather, it is the reaction of calcium carbide with water/moisture to release ethyne (acetylene) that's important here. Though I do not know of their mode(s) of action, apparently ethyne is almost a effective as ethylene in ripening fruits (I still need to find a reliable reference for this; though it is supposed to be a fairly well known fact).
If you let a fruit ripen naturally, or if you use ethyne instead... there's really no chemical difference at the end (apart from the usual arsenic contaminants that's associated with commercial-grade calcium-carbide).
The actual mechanism by which ethene (ethylene) and ethyne (acetylene) initiate, and thereafter, accelerate fruit-ripening is not really understood. Pectinases play a major role in the ripening process (they break down a structural component of plant cell walls called "pectin", which is why ripe fruit tend to be soft), however the scientific community has not yet arrived at a consensus as to how ethene and ethyne are linked to the activation/production of these enzymes.

However, there are a couple of subtle (non-chemical) difference between naturally and artificially ripened fruit:
Fruit that has ripened prior to plucking tend to be "fuller" than artificially ripened ones. This is because the fruit still gets an influx of nutrients while it's still attached to the tree...in other words; in the time it takes to ripen, the fruit grows as well. Artificially ripened fruit however, tend to be smaller, as they pretty much stop growing as they ripen (off the tree). ["Small", "big", "fuller" are crudely defined terms...you need some "experience" with fruits if you want to put this tip to good use].
Additionally, if you find the fruit being sold off-season (i.e- At a time of the year when it doesn't usually grow/ripen),  for a normal or somewhat cheaper price...then in all likelihood you have an artificially ripened fruit in your hand.

I reiterate:
It doesn't matter how it was ripened. It's ripe, that's all there is to it (from a consumer's PoV). Don't forget to wash and peel it before taking a bite :-)

"there's really no chemical difference at the end (apart from the usual arsenic contamina[tion]...)" That seems like... quite a significant difference.

@David But there's no significant change in the fruit itself. The arsenic contaminants stick to the surface...hence the "wash and peel" tip at the end of the post ;)

@DavidRicherby how does arsenic contamination occur here? since we aren't using the element anywhere?

@Chit Commercial grade calcium-carbide usually contains arsenic impurities : https://www.researchgate.net/publication/268226372_Quantification_of_arsenic_and_phosphorus_in_calcium_carbide_treated_mangoes

@ChitraLekha Calcium carbide is made by heating coke and lime in an electric arc furnace. Presumably the raw materials contain traces of arsenic -- nothing is ever pure. (I'd just drink the coke and lime but...)

Ethylene (C2H4) triggers enzymes including kinase (neutralizes acid), amylase (converts starch to sugar), hydrolase (degrades chlorophyll) and others to start ripening fruit. Calcium carbide, when in contact with moisture (such as on the surface of the fruit) produces acetylene (C2H2) which is quite similar to ethylene in starting ripening. For more information, visit Wikipedia.

One of the most simple tests include putting the mangoes in a bucket of water . The naturally ripened mangoes will sink while those that are artificially ripened would float. (because the density of artificially ripened would be lower than those that are naturally ripened.
And the color of the mangoes also indicate their ripening process . In some places there would be patches of green.
And I have also experienced burning taste while eating
And after we cut these mangoes certain places would be very ripe and other areas would still be hard(somewhat).
For more information click link https://www.youtube.com/watch?v=cwk-rtHPdes&t=42s

Is this specifically for mangoes? (Your question didn't specify "mango".)

yes this is for mangoes. And the color test may apply to bananas (where there would be PATCHES of green)