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How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine?

Compare the basicity of the following compounds:

Comparing compounds I and II
Since there is no conjugated system, there is no resonance at play and we only have to discuss hyperconjucation and inductive effect.
The conjugate acid of compound I has one α-hydrogen while the conjugate acid of  compound II has four α-hydrogens. So, according to hyperconjugation, the conjugate acid of compound II is more stable than the conjugate acid of compound I and hence the basicity of compound II is more than that of compound I.
Further, the conjugate acid of compound I has one α-carbon, two β-carbons, two γ-carbons and one δ-carbon. While the conjugate acid of compound II has two α-carbons, two β-carbons and one γ-carbon.
Alkyl groups are electron releasing groups and, since the conjugate acid of compound II has its carbon atoms closer to the amine than in the case of compound I, it is further stabilised, and hence compound II is more basic.
Comparing compounds II and III
The conjugate acid of compound III has an ether group attached to the amine while that of compound II has an alkyl group to the amine. Since ether is an electron withdrawing group and alkyl is an electron releasing group, the former will destabilise the conjugate acid compound III while the latter will stabilise the compound II, and hence compound II is more basic than compound III.
So, I know that compound II is the most basic, but I still don't know how to compare compounds I and III.
Can anyone help?

Hyperconjugation is not relevant for ammonium cations.

https://chemistry.stackexchange.com/questions/90661/basicity-of-nitrogen-heterocyclic-compounds

You are right that (I) < (II) and (II) > (III), but my explanation for (I) < (II) is different from yours: nitrogen atom in (II) has more alkyl group than (I), therefore the its conjugated acid is more stable.
As inductive effect decrease fast along the carbon chain, even though cyclohexyl group have more carbons, its effect is still weaker than two alkyl group (you can treat piperidine as ethylpropylamine, they're similar). Inductive effect is usually stronger than hyperconjugation.
In (III), both two branches have -C effect, as oxygen is an electronegative element, which destabilises the conjugated acid. In (I), the only alkyl group has +C effect, which stabilise the conjugated acid. Therefore, (I) is a stronger base than (III)
Hence, we have the sequence of basicity: (II) > (I) > (III).
We can look up pKb of each substance to check the result:
Cyclohexylamine (I): 3.36
Piperidine (II): 2.78
Morpholine (III): 5.64
So, my comparison is rational.

In this answer of mine, I outlined why hyperconjugation is helpful in stabilising carbocations but why it also does not contribute to the stabilisation of ammonium cations. Indeed, when comparing the inductive effects of neighbouring groups on ammonium cations, only the corresponding electronegativities should be considered.
Cyclohexylammonium has three hydrogens and one carbon connected to it; both morpholinium and piperidinium have two carbons and two hydrogens attached. Since carbon is more electronegative than hydrogen, nitrogen is polarised less negatively if it is attached to more carbon atoms. (The $\ce{N-H}$ bonds are polarised towards nitrogen more strongly than $\ce{N-C}$ bonds.) Therefore, cyclohexylammonium has the largest electron density on nitrogen, meaning the positive charge is best stabilised and therefore the cation is most stable; thus, it is the most basic of the three.
Between piperidinium and morpholinium, the only difference is the oxygen atom in the latter. This oxygen atom draws away electrons inductively across the entire σ bonded ring, meaning that nitrogen is slightly less negatively polarised in morpholine. Therefore, the morpholinium cation is less stable than piperidinium.

Hyperconjugation certainly isn't relevant for the octet-complete ammonium cations. However, cyclohexylamine (or its conjugate acid) has a pKa of 10.64; piperidine 11.22. (ref)

"Inductive" reasoning is generally bad for amines.

Here cyclohexamine (1) is more basic than piperidine (2) which is more basic than morpholine (3).
The $\mathrm{1^{st}}$ compound has more methyl groups than the $\mathrm{2^{nd}}$ compound. Alkyl groups are electron donating group and thus stabilizes the positive charge of nitrogen.
In the $\mathrm{3^{rd}}$ compound, oxygen is present which is an electron withdrawing group, and thus reduces the basicity of nitrogen as it attracts the electrons towards itself. Hence, the $\mathrm{1^{st}}$ is most basic and $\mathrm{3^{rd}}$ is least basic.

But piperidine is a secondary amine while cyclohexylamine is a primary amine. Inductive effect is distance dependent. Also cyclohexamine has only 1 alpha hydrogen while piperidine has 4. So hyperconjugation flavours piperidine

Hyperconjugation is not considered in case of ammonium cations.

@Srishti Inductive effect decrease fast in chains, therefore having more -CH2- groups makes barely any difference.  (II) compound has two inductive effects from two alkyl groups, which means it is more basic.


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If graphite's just a huge polymer of benzene why isn't it an aromatic compound (polymer?)
My teacher had once told me that graphite was just an insanely huge polymer of benzene (Those weren't his exact words but the gist). Today While we were studying aromaticity in organic compounds we steadily advanced from benzene to napthalene the annulenes and so on. What really stuck me was how as the number of  benzene rings increased the stuff tended towards graphite (atleast thats what the structure seemed to indicate). So the main question is why isn't graphite aromatic if its just a big collection of benzene molecules

Graphite/graphene is aromatic. The first image below is a common crystallographic representation of graphite and does not account for covalent bonding.  The second one shows the molecular structure of graphite as a kekule representation. Graphite is made of planar sheets of carbon atoms in 6-atoms rings with conjugated double bonds. This also satisfies the 4n+2 π electrons criteria needed within the aromatic ring for graphite to be aromatic. This aromaticity is why graphite is not only stable to high temperatures but also a more thermodynamically favorable allotrope than diamond under typical conditions.

Crystallogrphic Representation of Graphite.

Kekule Representation of Graphite.

+1 (possibly helpful http://ion.chem.usu.edu/~boldyrev/graphene.php) and so I've just asked How is the aromaticity in graphene different from the aromaticity in benzene?

+1. To OP: I copy a comment from the thread mentioned by uhoh. Indeed related arguments. Benzene is one cycle, no other cores have the same aromaticity. "I might be too naive but at the end it means that while in benzene there is a sextet delocalised in a ring, in graphene the delocalisation over the neighbours results in just two pi electrons delocalised but pinned onto a single ring. The other are described by extended orbitals and are those responsible for the in plane electrical conductivity. I am aware that worded this way might even sound trivial." Replace graphene with graphite.

@Alchimista "... the delocalisation over the neighbours results in just two pi electrons delocalised but pinned onto a single ring." 1- Does neighbours mean upper & below rings for that ring of graphene? 2- What do you mean by ... pinned onto a single ring?

@apurvium a) read on plane neighbors. Planes do interact, but yet they can "slide" on each other. The interaction, or bond, is less strong between plane (think also of the anisotropic conductivity). b) pinned in the sense that while as benzene has a delocalised sextet, only two remain for each ring of graphite. The third being the one spreading along the sheet as per a)

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Priority in naming organic compound
I'm confused if ethyl and methyl can get any of the number 4 and 5 then what will be the order?

Welcome to chem.SE. Your question isn't entirely clear right now, you can [edit] your post to add more details (diagrams are always good!)

You also might want to consider reading our guidelines on how to ask nomenclature questions.

Related: Preference of locants in organic nomenclature and Nearest branch numbering

The most important simplified criteria for the numbering of simple compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

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

Therefore, if no decision can be made about the locants for an ethyl and a methyl group using the previous rules, the lowest locant is given to the ethyl group according to Rule (g) because ‘ethyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).
For example, a correct systematic name is 4-ethyl-5-methyloctane rather than 5-ethyl-4-methyloctane since ‘ethyl’ is cited first in the name.

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Can solutions of polar covalent compounds conduct electricity?
I learned in class that solutions of polar covalent compounds are weakly conductive, while ionic solutions are strongly conductive.
But I'm getting different answers online. According to this lecture, "They do not conduct electricity in the liquid state, or when soluble in water, do not conduct electricity in aqueous solution."
I saw in lab, however, that a dextrose (D-glucose) solution weakly conducted electricity, and so did that of levulose (D-fructose).
So can solutions of polar covalent compounds conduct electricity? If so, how? Since it's covalent, it's not like anything can dissociate to form ions, right?

Could someone answer the question or leave a comment instead of just downvoting? Thanks. I'm just a high school chem student wanting to learn

Did you check the conductivity of the pure solvent?

@Loong In my data, I checked the conductivity of a tin "solution" (just a piece of tin metal in distilled water) and it was zero as expected.

@Loong Speaking of data, I recall that the pH was unusually low for all the solutions—5.0. I dismissed this as the pH strips being old and inaccurate, but it could be that CO2 dissolved in the water and made it weakly conductive.

The conductivity of highly purified water at a temperature of $T=25\ \mathrm{^\circ C}$ is $\kappa=5.50\times10^{-6}\ \mathrm{S\ m^{-1}}$. The conductivity of water that is saturated with atmospheric $\ce{CO2}$ is increased to $\kappa=1.10\times10^{-4}\ \mathrm{S\ m^{-1}}$. Pure water in contact with air can reach a pH of about 5.8 due to dissolved atmospheric $\ce{CO2}$.

The conductivity of  the glucose you described is likely due to trace amounts of ionization from the hydroxyl groups. This website gives a  pKa range for alcohols near 16-18, so they do ionize to an extent. Any molecule with hydrogens is prone to ionizations in aqueous solution to some extent, which could potentially cause conudctivity. This previous post is probably a good resource.

Consider for example this lab where polar covalent bonds are said to be strong, weak or non-electrolytes : ELECTROLYTES AND NONELECTROLYTES CHEM 151
The part of the molecule which dissociates has strongly ionic character but the molecule itself is described as being polar covalent. Not not all bonds in the molecular have the same level of covalent character, but most are covalent and it is polar.
There are also other ways for molecules to allow conduction, conjugated systems can do it by a very different process see for example the wiki on Conductive Polymers


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Why are snowflakes shaped as they are?
(It was the Winterbash icon that got me thinking on this.)
Snow flakes take on a variety of intricate shapes, but what processes are responsible for this?

(Got the image off of Wikipedia)
I mean, I wouldn't have raised this question if those water droplets, descending towards earth, assumed a... less fancy shape. Not only are these minuscule structures very intricate, they also display a high degree of symmetry (with most having multiple planes of symmetry).
Now, as I understand it, all processes proceed so as to maximize the "randomness" of its constituent particles. (Oversimplified version of the Second Law of Thermodynamics, yes, I know... just don't chew me out in the comments section...)
This Law can easily be observed in, and verified by, natural processes.
Now the formation of snow is a natural process, agreed? The way my brain sees it, is that water droplets ought to freeze into random, and by virtue of its "randomness", highly unsymmetrical shapes. But this is not the case here!
For some confounded reason, the shapes assumed by snowflakes are eerily symmetrical. So eerie, that I'm surprised it hasn't attracted the attention of conspiracy theorists yet, as in the case of crop circles. Now I'm not saying you don't find symmetry in nature, there're plenty of examples out there. But what I am trying to point out here, is that you don't typically see symmetry to such a high extent as in the case of snowflakes.
There are innumerable natural processes that give rise to weird shaped stuff, but I'm not really sure what makes snowflakes so different. However, I did notice a few points of difference between the formation of snowflakes and other natural processes:
1) As water droplets/snow descends, it's subjected to significant temperature changes as it passes through different layers of the atmosphere.
2) As water droplets/snow descends, it's subject to a significant increase in pressure as it passes through different layers of the atmosphere.
So, I reiterate my question.
Why are snowflakes shaped as they are?

Spin-off question:
I heard (not sure where) that each snowflake assumes a unique shape. How true is this?

Related: Why is it that every snowflake is unique?

@Loong Well, that answers the spin-off question. Thanks!

The discussion about why and how snowflakes get their shapes is quite involved; in general there are different gradients--temperature, vapour pressure, humidity etc. that a snow flake passes through as it descends and each of these parameters influence its dendritic structure. However, I can  assure you that there is no violation of the second law in this process, because maximum entropy doesn't suggest maximum asymmetry.

A good discription of how to grow snowflakes in this YouTube video.
watch?v=ao2Jfm35XeE

Now, as I understand it, all processes proceed so as to maximize the "randomness" of its constituent particles. (Oversimplified version of the Second Law of Thermodynamics, yes, I know... just don't chew me out in the comments section...)

Yes. True.

This Law can easily be observed in, and verified by, natural processes.

Sure. Still with you.

Now the formation of snow is a natural process, agreed? The way my brain sees it, is that water droplets ought to freeze into random, and by virtue of its "randomness", highly unsymmetrical shapes. But this is not the case here!

Nope. Let's consider a counter argument: chemical bond formation between atoms is a natural process, and natural processes seek to maximise entropy. Thus, all molecules should take on highly unsymmetrical shapes, shouldn't they?  The answer is obviously no. Same applies for snow flakes. First of all, "order" and "symmetry" aren't precluded by thermodynamics. Molecular and crystal structure emerges within the realms of laws of thermodynamics and quantum mechanics.
Like I said in the comments, the precise discussion of how a snow flake's shape merges will be quite involved. There will be a lot of variables that affect the growth of snow crystals: temperature, vapour pressure humidity, wind speed and direction etcetera.
The final shape of the snow flake would depend on the precise path it followed, and the conditions it experienced as it descended, and is impossible to predict/calculate.
Of course, the underlying six-fold symmetry comes from the arrangement of water molecules in the ice crystal lattice.
Also, there are no natural processes coordinating the growth of the arms to be perfectly symmetrical; their growth is governed by simple statistics.
The arms of a snow crystal all grow independently, and grow under randomly fluctuating conditions; however, the effects of these fluctuations are somewhat averaged over the entire snow flake as it tumbles through the sky, and you end up with a somewhat regular shape.
There is in fact, asymmetry in snow flakes. Just take a close, hard look at a photograph

Sure there is some symmetry: the snow flake has 6-major arms, and then each arm has sub-branches, and each sub-branch branches out further. The overall branching pattern is the same, because of random averaging--the snow flake randomly rotates and reorients as it falls, each arm feels (on average) the same conditions. But are each of the sub-branches or the sub-sub-branches identical in shape, number, size etcetera? No.


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Can the number of intermediates for a reaction be more than the total number of transition states?
I'm getting results from a transition state guess software, in which the number of intermediates are two and there's only one transition state for a reaction, is that possible?

In principle you could have a barrier-less process, but it sounds quite fishy. I guess it depends on your reaction. Can you say more?

I'm attaching an image of the reaction (Michael Addition reaction) and the energy profile diagram that I'm getting for this reaction, I'm asking here that does this energy profile diagram makes sense?

I've seen a couple of inorganic metal complexes reactions where the total no. of intermediates are more than the total no. of transition states, I'm trying to figure out if that's also possible for this particular reaction.

@Zhe like radical recombinations?

@EashaanGodbole That would be an example, but your reaction does not seem such a process. Bonds are breaking and forming at the same time, so I would assume there is an intermediate. You can try an IRC search.

What are the intermediate structures, in particular what's the difference between intermediate_2 and the product? What level of theory is being used? Where does the final number come from? This is a simple Michael addition. The first intermediate is just the pre-complex bringing the reactants together from infinity. Are these numbers BSSE corrected?

Let's assume your reactive complex (reactants interacting) are optimised, i.e. its gradient equals zero, and your potential energy surface is continuous.
Then any direction is either uphill or constant (otherwise the optimisation hypothesis would not apply).
There are three possibilities: (a) going from reactants to intermediate gives a transition state which has higher energy than both reactive complex and intermediate; (b) reactants and intermediate lie on a plateau where energy does not change; or (c) one hypothesis is wrong.
So yes, for barrierless processes.

Completely normal, especially for a bimolecular process forming and adduct like that. However, you shouldn't be applying transition state theory for this kind of process - variational methods (minimising the reaction rate) are more accurate.
These systems can be treated very efficiently using energy grained master equation with RRKM theory, for example software packages look at Mesmer or Paper.

This answer has nothing to do with the question since it doesn't answer the  if it's possible to go from reactants to product/intermediate without going through a higher energy state.


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Why Phenyl Carbocation is unstable?
Phenyl carbonation is said to be unstable because the + charge on it cannot be stabilized by Resonance or any other electronic effects, so my question is that if the resonance occurs in the direction shown in the below picture the + charge can be delocalized to some extent, so will it adapt that ?

see the description of the phenyl carbocation in this earlier answer

The carbocation is orthogonal to the pi system

Those are valid resonance structures that you've drawn.  They correspond to  the charge-separated resonance structures one can draw for benzene.  Since they involve charge separation, they contribute little to the description of the aromatic pi system.  Further, the pi system is orthogonal to the empty $\ce{sp^2}$ orbital, consequently such resonance structures provide little, if any, stabilization of the cationic center.

I disagree with your first statement that it cannot be stabilized by any electronic effect because electron donating groups could make it more stable than before

Your resonance structure doesn't do anything to stabilize the cation. You've created a singlet carbene at the top most carbon, so it has both the cation and the anion on the same carbon. These do not cancel out. As drawn, you have two hydrogens on the top carbon. You need to replace it with a single lone pair and an empty orbital.

Except for the fact that this is an inverse singlet carbene with the hybrid orbital being empty and the p orbital being populated … very interesting to think about the compound! =D

By the way, why would anybody downvote this (correct) answer?

@Jan Yeah, filled p orbital doesn't exist since it's really the resonance structure, but you're right that this is unusual.

I don't see it as a "singlet carbene at the top most carbon".  It is a series of charge-separated resonance structures of benzene which don't really contribute much to the overall description of the aromatic system.  Further, as I  pointed out in my earlier answer, the aromatic pi system and the unoccupied $\ce{sp^2}$ orbital are orthogonal. Consequently there is not much, if any, interaction between them.

@Zhe thanks for clearing my doubt. You are right , even if the bond breaks in the direction shown carbon is left with an unfilled octet.

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Why is atmospheric composition compared by volume?
The composition of the Earth's atmosphere by its constituent gases (nitrogen, oxygen, argon...), is often given by percent volume. (See this Wikipedia page, for example). Why are these compositions given by volume? I understand that a volume composition shows the mole fractions, which is the real point of interest. Volume compositions compare the volume taken up by different amounts of gases under the same hypothetical pressure and temperature. However, in any "container," each gas occupies the same volume as another; what differs is the partial pressure of each gas. Thus, why not base compositions off of partial pressures? It is physically correct to say that oxygen accounts for 21% of the pressure of the atmosphere, rather than saying that oxygen accounts for 21% of the volume of the atmosphere (which is thankfully not true). I understand that comparing pressure under equal volume and comparing volume and equal pressure both show the same thing, but why not use pressure if it is more physically correct?

You have to consider that lay-men are interested in these numbers too. Most people can easily grasp the idea of volume percent of each gas, but to present this data to them as partial pressures they would have to become informed of the concept of partial pressures. Given that the two methods give the same number, and the one is more complicated for the lay-man, it makes complete sense to use the simple measure.  This simplicity also biases the syndication of information to be represented in volume percent, not partial pressure.

The short answer is that (a) partial pressures have units of pressure, and (b) pressure changes with altitude and/or temperature.

The composition of the Earth's atmosphere by its constituent gases (nitrogen, oxygen, argon...), is often given by percent volume. (See this Wikipedia page, for example). Why are these compositions given by volume? I understand that a volume composition shows the mole fractions, which is the real point of interest.

You are right that the mole fractions are the real points of interest.  You may additionally be interested to know that many techniques that measure the percent volume of a certain gas actually measure its mole fraction.  Thus, reporting the results in percent volume is correct only if the gas is ideal.

Volume compositions compare the volume taken up by different amounts of gases under the same hypothetical pressure and temperature. However, in any "container," each gas occupies the same volume as another; what differs is the partial pressure of each gas. Thus, why not base compositions off of partial pressures?

Partial pressures have units of pressure.  So at 1 bar of pressure at 298 K, oxygen has a partial pressure of 0.21 bar and nitrogen has a partial pressure of 0.78 bar.  Let's say you had a closed rigid container of gas initially at this pressure and temperature.  Let's now say that it's a hot day so the temperature is 313 K.  Now the total pressure is $1\times\frac{313}{298}=1.05$  bar, and as a result the oxygen partial pressure is 0.22 bar and the nitrogen partial pressure is 0.82 bar.  But meanwhile the volume fraction (or equivalently the mole fraction) of each gas is unchanged.

It is physically correct to say that oxygen accounts for 21% of the pressure of the atmosphere, rather than saying that oxygen accounts for 21% of the volume of the atmosphere (which is thankfully not true).

I don't fully understand this sentence but do not believe it is correct.  Oxygen does account for 21% of the volume of the atmosphere.

I understand that comparing pressure under equal volume and comparing volume and equal pressure both show the same thing, but why not use pressure if it is more physically correct?

It is not any more physically correct than using volume fraction.  It additionally is complicated that partial pressures have units and will change with temperature or total pressure.

It will be more a long comment than an answer.

The only reason I found is that the pressure and the volume are two different types of physical properties. The pressure is an intensive property whereas the volume is extensive. See here for more information.
It means that if you take one liter of each two gas separately at atmospheric pressure and if we assume there are absolutely no reactions between each others then if you mix them you'll get two liters of a gas solution at atmospheric pressure. But then the partial pressure of both gas will be divided by two.
So instead of give the partial pressures for which you need the volume to calculate them, this is enough to know the relative amount of each gas in volume in the gas phase.

Quite simply, for the unit's use as a constant in measurments and expiriments. The measurement's frame of reference only needs to apply to volume, not pressure. But exceptions can always be made.


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Which is the currently accepted mechanism of a Wittig reaction?
The Wittig reaction is one of the most significant advances in synthetic organic chemistry in the 20th century and rightfully won its discoverer, Georg Wittig, the Nobel Prize in Chemistry.
A Wittig reaction is the addition of a phosphorus ylide (previously thought to be an ylene with a $\ce{C=P}$ bond) to an aldehyde or a ketone resulting in a $\ce{C=C}$ double bond.
Mechanistically, there is little debate that the final step is a cycloreversion of an oxaphosphetane liberating a phosphane oxide. However, as far as I know there is no agreement on the preceeding step(s) to form said oxaphosphetane; with two different competing mechanistic proposals:

Nucleophilic attack of the ylide onto the carbonyl carbon to give a charge-separated betain structure. This rotates around the newly-formed $\ce{C-C}$ bond until oxide and phosphonium are in proximity whereupon a bond is formed to close the oxaphosphetane (stepwise mechanism)
[2+2] cycloaddition of the ylide and the carbonyl to immediately form the oxaphosphetane in a concerted manner.

Wikipedia just presents these two mechanisms side by side hence it is not really helpful.
Which of the two mechanisms is, at present, considered the most likely reaction pathway? Which principal pieces of evidence point towards it and disfavour the competing mechanism?
Please answer including references to the corresponding journal articles.

tl;dr: I don't think there is any mechanism that is 100% correct, and, in cases like this especially, I think it would completely depend upon what set of carbonyl/ylid/base/solvent etc. was used. But, of course, we like being able to generalise, and to my knowledge theres a lot more evidence to support a concerted type mechanism.
General Background
The question is hopefully summarised in the scheme below (no geometry, that, in itself, is a pretty lengthy discussion). Reaction of a carbonyl (in this case a ketone) with a phosphorus ylid is able to give rise to two species, either an oxaphosphetane directly or a betaine (an internal salt for the benefit of Loong) which then goes on to form the oxaphosphetane which itself is able to eliminate to afford the olefin with generation of triphenylphosphine oxide as a thermodynamic driving force.
This answer will present the case for a cycloaddition mechanism, and evidence against the betaine pathway. Importantly, only Wittig reactions of unstabilised ylids, as with stabilised ylids the rules of the game change due to reversibility of intermediate formation (see a Horner–Wadsworth–Emmons).

Evidence against the initial formation of a betaine
Betaines have never been observed during the course of a 'normal' Wittig reaction, that is, spectroscopically, we are unable to see it (this may, of course, just be due to the fact that the formation of the oxaphosphetane is so rapid relative to the timescale of the methods used).
Many reports of betaines have been from accidental (or otherwise) opening of oxaphosphetane intermediates, and indeed this seems to have been what led Wittig down the betaine pathway in the initial reports (though in the first ever paper he suggested that oxaphosphetane was the sole intermediate, but subsequently couldn't find any evidence for it back in the days before NMR etc).
Wittig[2] and others[3] reported the isolation of crystalline betaine salts. Phosphonium bromides were treated with PhLi to give a ylid, to which a carbonyl was added; subsequent addition of hydrogen bromide afforded crystalline solids which were able to be elucidated, proving the formation of a betaine.
Whilst this was convincing at the time, we now know that the betaine isn't observed, but rather the oxaphosphetane is. Vedejs (whom we'll come back to), has gone on to show that the earlier findings could be explained by quenching of the oxaphosphetane rather than as a result of directly trapping the betaine[4], which is more consistent with the other data available to us.

Scheme 1: Generation of the crystalline betaine derivative from the oxaphosphetane.

There are more recent reports of 'stable' betaines for very specific substrates where subsequent collapse isn't possible, again generated from oxaphosphetane intermediates.
Stefan Berger's group at Leipzig reported a betaine stabilised by a bipyridyl group, allowing NMR data to be reported for the first time[1].

Scheme 2 Reversible formation of a stabilised betaine

In this case, the lithium holds a chelated structure together, preventing it from forming the alkene (and also to a certain extent preventing reformation of the oxaphosphetane on steric grounds). Interesting, with addition of 18-crown-6 to sequester the lithium, the reaction proceeded normally in the forward direction with alkene signals observed.
The fact Berger was able to do this, does of course not imply that they are real intermediates along the Wittig pathway, but it was a nice piece of mechanistic work that I thought deserved mention.
In addition to this mechanistic work, there are some empirical issues with invoking a betaine intermediate that cannot be easily explained. Namely, all Wittig reactions using non-stabilised ylids should be under kinetic control and irreversible (hence highly (Z)-selective due to the inability of the initial intermediates to reverse and hence equilibrate to the thermodynamic product), however this is frequently observed not to be the case, and in certain cases, high (E)-selectivity can be achieved.
Evidence for direct oxaphosphetane formation
Vedejs was the one of the first to propose direct (irreversible) cycloaddition of the ylid and carbonyl to give rise to the oxaphosphetane, quickly followed by cycloreversion to form the desired alkene and a phosphine oxide.

Scheme 3: cycloaddition/cycloreversion mechanism for Wittig

In his early reports, direct observation of the oxaphosphetane by P-NMR was conducted[5], the paper reasons that the high-field chemical shift observed for the phosphorus containing intermediate ruled out charged, 4-valent species such as betaines, but was instead more consistent with formation of the 5-valent neutral species such as an oxaphosphetane.

Scheme 4: Vedejs' observation of oxaphosphetane.

This NMR analysis isn't really too convincing by itself, however in the scheme above, compound 4 had previously been characterised by X-ray crystallography[6] and various NMR methods and as such gave some weight to the species observed by Vedejs being similar to the known compound. More recent work[7] has also observed directly the cis/trans oxaphosphetanes formed via competing cycloadditions.
Summary
Overall, direct observation of the oxaphosphetane but not the betaine does seem to suggest that mechanistically the cyclic intermediate is formed directly rather than via the betaine closing.
As stated previously, we can't 100% rule out the possibility that the betaine closure is just sufficiently rapid to appear invisible to our currently spectroscopic methods however every report of isolated/observed betaines that I've came across thus far has either intentionally been formed from an oxaphosphetane or can be explained by this without the authors recognising it.
One thing that is hugely missing from the story is computational work, whilst some studies have calculated the relative energies of the two intermediates, to my knowledge, the entire reaction coordinate hasn't been fully explored.
In conclusion, I think the cycloaddition mechanism is safest, and indeed it was the one I was taught as an undergraduate (and the one I know several other leading universities use), so if in doubt i'd go for that explanation, but on the firm understanding that more mechanistic work is needed.

References
 Generally: Modern Carbonyl Olefination, Carey Advanced Organic A, and Comprehensive Organic Synthesis I.
[1] Neumann, R. A.; Berger, S. Observation of a Betaine Lithium Salt Adduct During the Course of a Wittig Reaction. Eur. J. Org. Chem. 1998, 6, 1085. Subramanyam had reported similar chemistry earlier, but the NMR data was incomplete.
[2] Wittig, G.; Haag, A. Über Phosphin-alkylene als olefinbildende Reagenzien, VIII. Allenderivate aus Ketenen. Chem. Ber. 1963, 96 , 1535. DOI: 10.1002/cber.19630960609. A very old paper in German.
[3] Schlosser, M.; Christmann, K. F. Olefinierungen mit Phosphor-Yliden, I. Mechanismus und Stereochemie der Wittig-Reaktion. Liebigs Ann. Chem. 1967, 708, 1. DOI: 10.1002/jlac.19677080102.
[4] Vedejs, E.; Meier, G. P.; Snoble, K. A. J. Low-temperature characterization of the intermediates in the Wittig reaction. J. Am. Chem. Soc. 1981, 103, 2823. DOI: 10.1021/ja00400a055.
[5] Vedejs, E.; Snoble, K. A. J. Direct observation of oxaphosphetanes from typical Wittig reactions. J. Am. Chem. Soc. 1973, 95, 5778. DOI: 10.1021/ja00798a066.
[6] Mazhar-Ul-Haque; Caughlan, C. N.; Ramirez, F.; Pilot, J. F.; Smith, C. P. Crystal and molecular structure of a four-membered cyclic oxyphosphorane with pentavalent phosphorus, PO2(C6H5)2(CF3)4C3H2. J. Am. Chem. Soc. 1971, 93, 5229. DOI: 10.1021/ja00749a044.
[7] Maryanoff, B. E.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108, 7664. DOI: 10.1021/ja00284a034.

The Wittig reaction has been known for decades, but research into its mechanism has continued to throw up surprises even until now. The literature is unfortunately very difficult to read, as numerous hypotheses have been proposed, each having subtle but important differences.
Historically, betaines were postulated as the main intermediates in the Wittig reaction, and supposedly underwent fast cyclisation and decomposition to give the products. This was likely mainly because betaines were detected under certain Wittig conditions, e.g. the presence of lithium salts led to the formation of lithium betaines. (Generally, our mechanistic understanding of lithium Wittig reactions is poorer, and for this reason the scope of this answer will be restricted to "salt-free" Wittig reactions.)
However, Vedejs' positive observations of oxaphosphetanes via 31P NMR1,2 (described in NotEvans's answer) is generally credited with changing chemists' viewpoints. Since then, the formation of an oxaphosphetane, and its subsequent cycloreversion to form the products, was not really the main point of contention. The issue was whether the oxaphosphetane is directly formed from the products, or whether it involves the intermediacy of the betaine:

Without going into too much detail about transition states and stereoselectivity issues, the general consensus is that for such reactions, oxaphosphetane formation is concerted and does not involve the intermediacy of a betaine. This applies to unstabilised, semi-stabilised, and stabilised phosphonium ylids alike.3

The observation of oxaphosphetanes alone cannot rule out the intermediacy of a betaine. It is possible to make betaines via the deprotonation of β-hydroxyphosphonium salts (the conjugate acid of the betaine), and it turns out that only oxaphosphetanes or products (alkene + Ph3PO) were detectable;4,5 so the cyclisation of the betaine is too fast for NMR investigation. As such, Vedejs' lack of observation of the betaine does not necessarily mean that it is not an intermediate (although it certainly rules out the possibility that it is a key intermediate, as was held for many decades).
Actual objections to the intermediacy of betaines were based upon a number of facts, including but not limited to:

Kinetic studies found that with stabilised ylids, there was a rate decrease in more polar solvents (DMF/DMSO/acetone vs benzene), indicating a non-polar transition state.6
No betaine or ionic intermediates have ever been detected under salt-free conditions.7,8
Early computational studies indicated that betaines were not an energy minimum9 (however, Vedejs does note that calculations in the gas phase would naturally disfavour zwitterionic species10).

One way to definitively prove the non-existence of betaines along the reaction coordinate would be to separately prepare the betaine by deprotonation of a β-hydroxyphosphonium salt and to investigate whether the product distribution was the same as that of the corresponding Wittig reaction.
However, one cannot use the alkene geometry as the evidence. The betaine mechanism was postulated to proceed via an open transition state (similar to a Mukaiyama aldol with BF3), which preferentially led to the (Z)-alkene. The formation of the (E)-alkene was said to arise from thermodynamically-driven interconversion of the erythro and threo betaine diastereomers. In order to compare the stereoselectivities, one would therefore have to synthesise a precise mixture of the two diastereomers.
A conclusive piece of evidence finally came from Vedejs' investigation of stereochemistry – not that of the product, but that of the intermediate oxaphosphetane.10 The idea behind the experiment was that if the substituents on phosphorus were not all the same, then there would be diastereomeric oxaphosphetanes 4 through 6 formed (all compounds are racemic). The beauty is that, since the phosphorus ylid 1 has a symmetrically substituted carbon, there is only one possible betaine that can be formed (3):

With ordinary substituted aryl groups, this experiment was impossible because the oxaphosphetanes 4 through 6 underwent rapid pseudorotation and interconversion. However, using the dibenzophosphole ylid 7 (shown below) made pseudorotation sufficiently slow on the laboratory timescale ($t_{1/2} = \pu{3 h}$ at -78 °C), and also eliminated the possible rotamer 6 which suffers from excessive ring strain. The requisite betaine 11 was actually formed by quenching the oxaphosphetane with acid (forming the β-hydroxyphosphonium salt 10, which can then be deprotonated again):

The diastereomers 8 and 9 were characterised by NMR studies. As might be expected, the ratios were vastly different. When starting from the ylid and carbonyl, the ratio of 8 to 9 was $6.5 : 1$, but when the β-hydroxyphosphonium salt 10 was deprotonated to form the betaine 11, the ratio was $1 : 4.2$. These results entirely disprove the intermediacy of a betaine. In Vedejs' words:

[...] the conventional betaine mechanism can play at most a minor role in the Wittig reaction of 7. [...] No evidence remains to support the conventional betaine pathway that is featured in virtually all textbooks.

A more recent DFT study by Harvey and Aggarwal – who, by the way, also put forth the currently accepted origin of (E)-selectivity with stabilised ylids11 – took into account solvation effects, which were expected to be significant in influencing the mechanism of a polar reaction. The work encompassed unstabilised, semi-stabilised, and stabilised ylids.12 They found that the mechanism was best described as (I simply quote as I can't summarise it any better):

a concerted, asynchronous, double nucleophilic addition occurring via a four- center transition state in which the C–C bond formation is well advanced, whereas the P–O interaction is mainly electrostatic, the P–O bond formation occurring after the TS. Although it is also possible in some cases to find a transition state leading to a betaine, such barriers lie higher in energy than the corresponding four-center transition state and are unlikely to play an important role in the mechanism.

Additional evidence that fully supports the concerted oxaphosphetane formation involves the increased (Z)-selectivity observed when aldehydes with β-heteroatoms undergo Wittig reactions.13 This is observed, again, for all three types of ylids (unstabilised, semi-stabilised, and stabilised), and is best rationalised by a secondary orbital interaction in the puckered transition state:

The authors did not investigate the possibility of rationalising this using a betaine mechanism. However, as far as I can personally tell, in the putative transition state leading to the betaine, there is no obvious way of explaining this. This does not prove, but is best explained by, and hence strongly points towards, the concerted and irreversible formation of an oxaphosphetane as the key step of the mechanism for all ylid types.

Beyond the question of concerted vs stepwise, there are also many details of the mechanism that have been elucidated over the years, most notably the reversibility of the process and the reasons behind the different stereoselectivities of stabilised and unstabilised ylids. Excellent summaries are provided in refs 12 and 13; Comprehensive Organic Synthesis II, chapter 1–15 also has an overview of recent research.

Notes and references

Vedejs, E.; Snoble, K. A. J. Direct observation of oxaphosphetanes from typical Wittig reactions. J. Am. Chem. Soc. 1973, 95 (17), 5778–5780. DOI: 10.1021/ja00798a066.
Vedejs, E.; Meier, G. P.; Snoble, K. A. J. Low-temperature characterization of the intermediates in the Wittig reaction. J. Am. Chem. Soc. 1981, 103 (10), 2823–2831. DOI: 10.1021/ja00400a055.
The Horner–Wadsworth–Emmons olefination uses phosphonate ylids, and does not fall under the scope of this answer.
Vedejs, E.; Snoble, K. A. J.; Fuchs, P. L. Phosphorus betaines derived from cycloheptene and cyclooctene oxides. Inversion of cyclooctene. J. Org. Chem. 1973, 38 (6), 1178–1183. DOI: 10.1021/jo00946a024.
Maryanoff, B.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108 (24), 7664–7678. DOI: 10.1021/ja00284a034.
Frøyen, P. Kinetic Study of the Wittig Reaction. Acta. Chem. Scand. 1972, 26, 2163–2168. DOI: 10.3891/acta.chem.scand.26-2163.
Vedejs, E.; Marth, C. F. Mechanism of the Wittig reaction: the role of substituents at phosphorus. J. Am. Chem. Soc. 1988, 110 (12), 3948–3958. DOI: 10.1021/ja00220a037.
Maryanoff, B.; Reitz, A. B. The Wittig Olefination Reaction and Modifications Involving Phosphoryl-Stabilized Carbanions. Stereochemistry, Mechanism, and Selected Synthetic Aspects. Chem. Rev. 1989, 89 (4), 863–927. DOI: 10.1021/cr00094a007.
Volatron, F.; Eisenstein, O. Wittig versus Corey-Chaykovsky Reaction. Theoretical study of the reactivity of phosphonium methylide and sulfonium methylide with formaldehyde. J. Am. Chem. Soc. 1987, 109 (1), 1–14. DOI: 10.1021/ja00235a001.
Vedejs, E.; Marth, C. F. Mechanism of Wittig reaction: evidence against betaine intermediates. J. Am. Chem. Soc. 1990, 112 (10), 3905–3909. DOI: 10.1021/ja00166a026.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. On the Origin of High E Selectivity in the Wittig Reaction of Stabilized Ylides:  Importance of Dipole−Dipole Interactions. J. Am. Chem. Soc. 2005, 127 (39), 13468–13469. DOI: 10.1021/ja0539589.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.
Byrne, P. A.; Gilheany, D. G. Unequivocal Experimental Evidence for a Unified Lithium Salt-Free Wittig Reaction Mechanism for All Phosphonium Ylide Types: Reactions with β-Heteroatom-Substituted Aldehydes Are Consistently Selective for cis-Oxaphosphetane-Derived Products. J. Am. Chem. Soc. 2012, 134 (22), 9225–9239. DOI: 10.1021/ja300943z.


text:
Suitable first-year experiment idea - stability of coordination complexes in TLC?
I'm a first-year in university and we have to carry out a "measurement" project next term. I'm fascinated by coordination chemistry (though I haven't taken the course yet), and while our equipment list is pretty limited it does include a thin-layer chromatography setup (and a colorimeter & a pH probe, which I would love to use if there was a way).
I was thinking of using TLC to look at the stability of different complexes (either different ligands or central metals), similar to the experiments in these papers:

Stability of mixed-ligand $\ce{zinc(II)}$ carboxylate complexes in thin-layer chromatography
Thin-layer chromatographic determination of the stability of complexes of substituted styrenes on silver nitrate-impregnated silica

However, I'm struggling to come up with examples of complexes that can be made easily (without a fume hood) and inexpensively. I'm not very knowledgable about this topic so I apologize if this is a bad question, I was just wondering if this seemed like a feasible yet worthwhile project given constraints in equipment/money and my knowledge. In previous years, people have mostly done biology or physics related projects so I'm a little doubtful if this idea—in particular, synthesizing a variety of complexes—is realistic. I'd appreciate any advice, specific or general! Thanks!

Ferrocene and acetyl ferrocene are the archetypal undergraduate examples.

@NotWoodward Thank you! I think we have those in our labs for the 2nd year student. I'm a little worried that it wouldn't be considered "novel", is there anyway to modify the project slightly?

You can replace iron with various other metals? There are many complexes you could buy/make really. I'm not sure the whole idea is novel anyway, nor would novel really be expected for an undergraduate.

@NotWoodward Thanks so much! Haha it definitely isn't a novel idea by any means, but "novel" was in the project description so I guess I'll try.


text:
Can the equilibrium constant be equal to 0?
Can $K_\mathrm{eq}$ (not $Q$) be equal to $0$?
My book says:

$K_\mathrm{eq}$ is always positive because the concentration at equilibrium, small or large, are always positive.

But it says $Q$ can be $0$, when the concentration of the products is $0$. Why can't $K_\mathrm{eq}$ be $0$? In other words, why can't a reaction exist such that the equilibrium condition doesn't produce any products?

because it wouldn't be a reaction.
I guess your question is OK if we are asking if there are sets of reactants and products that contain the same set of atoms but in which the reactants are so much more stable that we don't get any products.  I think that the answer would be that if a set of reactants could become a set of products, given a large enough sample at least some will be in the form of products.

Ah, good point.

It may be "possible" that since energy is quantized, you might have a situation where there is no product.  Energy in a system is not uniform so even unlikely rearrangements can occur with some frequency, but maybe since energy is also not continuously distributed, but is quantized at all levels so you could have a situation with low enough energy that you don't get any product, but heck if the universe can result from "nothing" then I don't think so.


text:
Why is Nitric Acid more reactive and caustic than Sulfuric or Hydrochloric?
Nitric acid reacts much more vigorously with living tissue and is a obviously more immediately hazardous on contact than sulfuric or hydrochloric.
What is the reason for this?  Is it due to the oxidizing potential of the nitrate anion?
Does the greater reactivity of nitric acid also hold for reactions of acids with metals? Perhaps the factors related to the acid anions that determine dissociation are not the same ones that determine their reactivity?

As you state, it is the oxidizing potential of the NO3- radical. Neutral nitrates also react with organic compounds. For example AgNO3 was called "lunar caustic", though the pH+ of an aqueous is about 6 or 7 (depending on dissolved CO2).

Nitric Acid is indeed what most often is referred to as a "strong acid" (at ambient temperatures), and its $\mathrm{pK_a} \approx -1$, i.e nitric acid in diluted solution is almost always fully dissociated except in extremely acidic solutions.
First of all, how does it compare to other "strong acids", for example sulfuric acid
$\ce{HNO3 + 2 H2SO4 ⇌ NO2+ + H3O+ + 2HSO4−}$
Clearly, in this reaction sulfuric acid acts as the "acid", while nitric acid acts as the "base", hence sulfuric acid is indeed a stronger acid.
Reactions of dilute nitric acid with some metals are typical of an acid, namely:
$\ce{Mg + 2 HNO3 → Mg(NO3)2 + H2\uparrow}$
Other metals that react in this fashion are manganese and zinc.
You do correctly note that a special feature of nitric acid's reactivity is its strength as an oxidising agent, and this is also evident in its reactions with metals, for example. However, do note, that the exact details depend on the concentration of the acid and the nature of the metal
For example, nitric acid can oxidise non-active metals such as copper and silver, however, the products depend on temperature and the acid concentration.
For instance, copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:
$\ce{3 Cu + 8 HNO3 → 3 Cu^2+ + 2 NO + 4 H2O + 6 NO3−}$
The nitric oxide produced may react with atmospheric oxygen to give nitrogen dioxide.
However, with more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:
$\ce{Cu + 4 H+ + 2 NO3− → Cu^2+ + 2 NO2 + 2 H2O}$
Metals like chromium, iron and aluminium readily dissolve in dilute nitric acid, but the concentrated acid forms a metal-oxide layer that protects the bulk of the metal from further oxidation; thus, concentrated nitric acid serves to passivate these metals towards further reactions. Common metals that are passivated by concentrated nitric acid are iron, cobalt, chromium, nickel, and aluminium.

I'm upvoting.  As far as the answer, what is nitric acid doing to organic material then?  Is it oxidizing them to more oxidized organic compounds or is it actually causing combustion.
And outside of the answer, is Silver Nitrate similarly formed with dilute Nitric acid.  Also, wondering as I know that Potassium Chlorate can react similarly to Potassium Nitrate as a rapid oxidizer, how does Perchloric acid compare to Nitric as an oxidizer.

@JosephHirsch Yes, it would act as an oxidant when it reacts with organic matter. Yes, silver nitrate can be formed by reacting metallic silver with nitric acid. However, I don't know how perchloric acid and nitric acid compare as oxidants, but I believe you can find relevant data online.


text:
Why are there 14 electrons in the third shell of iron atom?
The third shell must contain at least 18 electrons. But in iron there are 14 electrons in third and 2 electrons in 4th. Why are there 14 electrons in the third shell of iron atom?

The figure shows 12 in the 3rd and 2 in the 4th.  The term shell is probably inappropriate as there was a model of Bohr which used the term "shell" to correspond to "periods" in the periodic table.  As for the reason?  There is nothing that requires an energy level to fill up before another one starts to fill.  Its all about stability and energy.  It depends on many factors.  Transition metals will almost always have 1 or 2 electrons in the s sublevel above the d sublevel that corresponds to its place in the d block

The third shell must contain at least 18 electrons.

This statement is wrong. In fact, the third shell is only able to contain up to 18 electrons. This is a mathematical result of quantum chemistry that is, in principle, derived in every introductory quantum chemistry course. The reason being that there are three different subshells in the third shell (called s, p and d orbitals) which can contain 2, 6 and 10 electrons, respectively. After 18 electrons have been added to the third shell, there is no combination of quantum numbers to allow for a 19th electron within that same shell.

On to the question of why there are two electrons in the 4s orbitals (the fourth shell) although the third shell is not full yet. This is because the subshells in every atom with more than one electron have different energy levels. In general, the energy levels follow what is known as the aufbau principle: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, … However, this is more of a general observation. It works very well for general chemistry classes, but there are quite a number of edgy cases that these same classes will have to introduce special rules for.
A great general reference to check out would be this question. But I would also like to point you to this chemguide.co.uk guide, which explains the entire complication in more details. The gist is that the energy levels of 4s and 3d are so similar in an innocent case such as argon that depending on how many protons and additional electrons are added a different orbital may be the one lower in energy and leading to the more stable electronic configuration.
If, when coming from argon you add exactly one proton and exactly one electron (generating a potassium atom), the energy levels will shift relatively to each other in such a way that the 4s orbital is lower; thus, potassium’s electronic configuration is $\ce{[Ar] 4s^1}$. If you add another proton without an electron (generating $\ce{Ca+}$), the picture remains the same. Calcium(I) has the same electronic configuration. If you add a third proton to generate scandium(II) ($\ce{Sc^2+}$), the 3d orbitals will suddenly move below the 4s orbitals in energy so its configuration is $\ce{[Ar] 3d^1}$. (This is because the 3d orbitals are less diffuse or ‘closer to the nucleus’ and therfore get stabilised more.)
If you thought you could now easily add a second electron into another d orbital when generating scandium(I) ($\ce{Sc+}$), think again. Due to electron-electron repulsion (or more generally, the Hund rule), this additional electron is placed in the 4s orbital leading to $\ce{[Ar] 3d^1 4s^1}$. And the final electron to generate a neutral scandium atom would again be added into the 4s orbital to give $\ce{[Ar] 3d^1 4s^2}$.
(In reality, you do not simply ‘add electrons’ and all the electrons of an atom are indistinguishable. What you can do, however, is calculate or measure the electronic structure of all these atoms and ions to determine which levels are populated by how many electrons. And that will give you the result discussed above.)

Tl;dr: It’s complicated, but an easy rule of thumb would be to fill the next shell’s s electrons before filling up a given shell’s d electrons. Thus, for iron the 4s level is filled first, before an additional six d electrons are added. Read above as to why it is not that simple.

Your doubt is - Total number of electrons in 3rd shell  (i.e. electrons from 3s + electrons from 3p + electrons from 3d) is not yet 18 but still how we filled up the 4th shell?
You have a general misconception than atomic orbitals always follow the 2,8,18... electron filling rule. According to this 2,8,18.. electron filling rule, 2 electrons should get filled up in 1st shell, then 8 electrons in second shell, then 18 electrons in 3rd shell........
Remember this rule mentioned above is completely elementary! Just remember 2,8,18... electron filling rule is actually a waste. Simply I would recommend forget about that 2,8,18... rule. It is very elementary rule... simply forget it. In 2,8,18... electron filling rule, "lower energy level orbitals gets filled up first"- this statement is not considered in that elementary 2,8,18..rule. But this statement which that elementary rule doesn't consider is very important statement to be followed while filling up of electrons. So remember 2,8,18... electron filling rule is an elementary rule for beginners who  just start to read about atoms and orbitals! Forget about that rule.
Its not necessary that 3rd shell has to have 18 electrons for 4th shell filling up to start.
The electron filling depends upon the energy levels of the orbitals. Lower energy orbitals get filled up first.
Iron's configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6. You can note from the configuration, 4s gets filled up before 3d . The reason behind this is lower energy orbitals are filled up first. Even though 4s belongs to 4th shell, still its energy is less than 3d which can be reasoned out by (n+l) rule. 4s energy is less than 3d, hence the two electrons which you expected to be going to 3d goes to 4s before 3d filling starts. Hence you end up having 14 electrons in 3rd shell and 2 electron in 4s. Remember 4s got filled up before 3d and the reason is 4s energy is less than that of 3d.

While the 2-8-18 shell model is elementary, the Aufbau filling is also a flawed concept.  I won't vote down because Aufbau is consistent with 4s2 3d6, but the 4s electrons present in Iron are actually higher energy than the 3d electrons and will ionize first and enter last.  It is still generally true that the 3d will "get full" after the 4s as you move across the periodic table, but it doesn't fill last in a given transition metal.  Iron is now typically shown as having a configuration of 12s 2s2 2p6 3s2 3p6 3d6 4s2  http://www.tabulka.cz/english/elements/configuration.asp?id=26 and others

So "filling first" is ambiguous.  In Iron, 4s is full while 3d is not.  4s tends to be occupied and or also full while 3d is not full as you move across the transition metals.  The electrons that are IN the 4s orbitals of a given transition metal are higher than the electrons in the 3d orbital.  http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
Ultimately the reason is that the 4s and 3d sublevels are close to each other in energy and depending on the nuclear charge and the exact number of electrons, different orbitals become relatively more or less favorable for the next e-.

So do you mean to say in Iron 3d fills first and then 4s? Can you please elaborate where you think I went wrong.

Yes IN IRON (and almost all transition metals) the 3d gets its electrons first and then the 4s if you are literally looking at the energy of the electrons, or stripping electrons and putting them back.  Modern electron configurations now show Iron as 3d6 4s2.
This is not the same as saying that 3d FILLS first because 3d is not FULL for Iron, but the 6 3d electrons in Iron are lower energy, ionize later, refill preferrentially to the 2 4s electrons in Iron.
The misconception comes from thinking that since Mg has 4s2 and no 3ds, that the 4s are lower energy for transition metals.

You answer the question but include additional information that is wrong, in part because the Aufbau is an oversimlification and in part because Aufbau is WRONG in terms of placing energy levels in a given element.  The best way to correct it is to say that the 3d does not have to have all of its orbitals full for some electrons to go into the 4s.  The ones that are in the 3d still are lower energy than the 4s present in transition metals.  Some sources will still list the 4s below the 3d in Iron.  It depends somewhat on what the order means to you-bookkeeping electrons or ordering energies.

Transition group is such a fascinating dilemma lol

Electronic configuration of $\ce{Fe}$ : $\mathrm{1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^6 \ 4s^2}$
Add the electrons of 3rd shell $\mathrm{3s^2 + 3p^6 + 3d^6} = 14$  electrons
And 4th shell $\mathrm{4s^2}$ contains $2$ electrons.

According to the Bohr model of atoms, the maximum number of atoms which can  be accommodated in the outermost shell is 8. That's why in the atomic shell diagram the 2 electrons are in the outermost shell.


text:
Derivation of relation between acid and base equilibrium constants
While deriving the relation between $K_a$ and $K_b$ in ionic equilibrium topic, say I take an example of
$$
\ce{NH4+ + H2O -> NH3 + H3O+} \tag{1}\label{1}
$$
The $K_a$ of this reaction is:
$$
\frac{\ce{[NH3][H3O+]}}{\ce{[NH4+][H2O]}}
$$
Now $\ce{NH3}$ is a conjugate base of $\ce{NH4+}$. Its $K_b$ can be written as:
$$
\ce{NH3 + H2O -> NH4+ + OH-} \tag{2}\label{2}
$$
$K_b$ of the reaction:
$$
\frac{\ce{[NH4+][OH-]}}{\ce{[NH3][H2O]}}
$$
In my textbooks while deriving relation between $K_a$ and $K_b$, it added the two equations. In the addition of two equations we get:
$$
\ce{NH4+ + 2H2O + NH3 -> NH3 + NH4+ + H3O+ + OH-} \tag{1 + 2}\label{1p2}
$$
My textbook cancelled out both the $\ce{[NH4+]}$ and $\ce{[NH3]}$ concentration terms from the LHS and RHS. $\ce{[NH4+]}$ on the LHS is from equation $\eqref{1}$ and $\ce{[NH4+]}$ on the RHS is from equation $\eqref{2}$.
My question is, how can we cancel them out? How can you say the $\ce{[NH4+]}$ formed from hydrolysis of $\ce{NH3}$ in reaction $\eqref{2}$ is equal to that of $\ce{[NH4+]}$ initially taken in reaction $\eqref{1}$? Even if you initially take same concentration of $\ce{[NH3]}$ for reaction $\eqref{2}$ as formed from reaction $\eqref{1}$, that concentration of $\ce{NH3}$ on hydrolysis can never give same concentration of $\ce{[NH4+]}$ as initially taken in reaction $\eqref{1}$. Then, how can you cancel these different $\ce{[NH4+]}$ concentrations when you are adding reaction $\eqref{1}$ and $\eqref{2}$?

$K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.
When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations

$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$
as well as
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$
must simultaneously hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium constants, after all.
So, in any given single solution with one well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain
$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$

If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.


text:
Would molybdenyl chloride or tungstenyl chloride work as an alternative to chromyl chloride test?
(This is a follow up of this question)
Chromyl chloride test is by far the best test to determine the presence of chloride ions in a salt because no analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides. This test is considered selective for $\ce{Cl−}$.
But can molybdenyl chloride work similar to that of chromyl chloride?.  I couldn't find enough information in the internet about the feasibility of the compounds as a replacement to chromyl chloride. What about tungstenyl chloride?
In this link, I found a preparation process of molybdenyl chloride but that doesn't follow the same reaction used to make chromyl chloride. It uses common salt, sulfuric acid and molydenum trioxide. This made me curious. Common salt contains chloride that ultimately reacts with other reagent to form molydenum chloride. If I use another chloride containing compound and react with the other reactants, will molydenyl chloride form?. I am aware of thermodynamic or kinetic aspect of the reaction for its feasibility but still will the reaction count as a form of 'molybdenyl chloride test'?

You should have at least find https://en.wikipedia.org/wiki/Tungsten_dichloride_dioxide before asking.

Maybe because there are various tungsten oxychlorides.

In the chromyl chloride test, chromyl chloride is not a reagent that can be added to the sample to detect any chloride present. It is made in situ through a reaction of chloride containing sample, potassium dichromate and sulfuric acid.
$$\ce{NaCl + K2Cr2O7 + 3H2SO4 -> 2Na2SO4 + K2SO4 + 2CrO2Cl2 ^ (red fumes) + 3H2O}$$
Now in order to make molybdenum chloride work like chromyl chloride, you have to actually make molybdenyl chloride. For this, you need potassium dimolybdate. But it is quite unstable and is prone to decomposition. From here:

Potassium Dimolybdate, $\ce{K2Mo2O7}$, is formed by fusion of suitable quantities of $\ce{K2CO3}$ and $\ce{MoO3}$, but is unstable; the aqueous extract of the fused mass yields large crystals which immediately begin to decompose.

Making tungstyl chloride is not feasible because in order to make it, you need potassium ditungstate which is costly, rarely used reagent and can be hardly found in chemical vendor or supplies. So, this method is out of scope and reach.

Potassium ditungstate or paratungstate has a unique formula that does not follow the general formula of potassium dichromate or potassium molybdate. Its formula is given as $\ce{5K2O.12WO3.11H2O}$ or $\ce{3K2O.7WO3.6H2O}$


text:
Is palladium really that poisonous?
In Iron man 2, the protagonist Tony Stark was anxious about his blood toxicity due to palladium core that was keeping him alive was slowly poisoning his blood. It is written here that:

The palladium core in the arc reactor that keeps Stark alive and
  powers the armor is slowly poisoning him, and he cannot find a
  substitute.

The palladium core

Palladium poisoning Tony Stark

But in the wikipedia article of Palladium, it is clearly written that:

Palladium is a metal with low toxicity. It is poorly absorbed by human
  body when digested.[...] High doses of palladium could be poisonous;
  tests on rodents suggest it may be carcinogenic, though no clear
  evidence indicates the element harms humans.

So, Palladium has no clear evidence that it is harmful to humans, why it was chosen for the film?
To quote user paracresol's question from a very, very related question:

Is there anything realistic[...], or is it just another movie gaffe?

It is hard to answer why it was chosen in the film/comics, but if you focus on the part if pd is toxic, it os a fine question

@NilayGhosh Maybe it catalyses unwanted reactions?

Most likely not a movie gaffe, rather on purpose.  Entertainment producers often use fake toxins or reactions because publicizing real ones could pose liability risks if some people get the wrong idea.

Intoxication can be acute (short time of exposure, effects appear fast), or chronic (longer exposures, the onset toxic effects will take days, months or years). For Iron Man it seems clear that this would be a chronic intoxication.
I found this safety guideline from the EMA, which states:

Mice given $\ce{PdCl2}$ in drinking water (5 mg palladium/l) for a lifetime
  showed suppression of body weight and an increase in amyloidosis of
  several inner organs. In another study of questionable value
  (chloropalladosamine was given enterally to rats for about 6 months) a
  dose of 0.08 mg/kg body weight (corresponding to 0.041 mg $\ce{Pd^2+}$/kg) was
  determined as the no-observed-effect-level (NOEL), above which changes
  in body weight, hemoglobin content, blood serum parameters as well as
  functional and morphological changes have been reported. There is
  insufficient histological information available from this study.

Extrapolating human toxicity from mice data is a long shot, but since this is all we have it will have to do. Also, bear with me for pretty inaccurate math from here on (it's fiction anyway).
It's been a while since I watched the movie, but if I recall correctly each palladium piece was roughly 7cm x 4cm x 1cm. That's a volume of 28cm$^3$. The density of palladium is 12.023g/cm$^3$, therefore this piece weighs a bit under 350g.
I'm going to assume that he burns through a palladium piece every two days. Good times.
The NOEL (see above) is 0.08mg/kg/day. For an 85kg human, that's 6.8mg/day,  roughly 0.02% of the mass in the reactor. I recall that the palladium was heavily corroded when he extracted it, so a loss of 0.5-1% of palladium seems pretty reasonable. Being conservative, the reactor would release about 0.9g a day. If more than one tenth of a thousandth of this palladium is absorbed by Tony's bloodstream (which doesn't sound like much), then he will be in trouble.
To sum up, yes, I do believe it possible to suffer from palladium poisoning if you have a chunk of that thing dissolving inside your chest day and night.
Answering the other half of your question, Tony Stark used palladium for his first arc reactor in the first Iron Man. I don't think it was planned, they just went with it.

Mmm I don't think that the core was made of PdCl2... Palladium it self is not poisoning

Agreed, but I assumed that it was being oxidized so the behavior of whatever  ionic compound is being released might be similar.

Palladium as metal is a bio-compatible material used in many medical devices nowadays (cf. Platinum Metals Rev., 2012, 56, (3), 213–217) problems, however, may arise from compounds containing palladium e.g. if you ingest palladium because it can be easily bio-methylated but in general, this doesn't kill you but cause hypersensitivity to the metal.
If we take a look at references the poisoning was due to the fact that:

... palladium was continuously eroded by the neutron
bombardment and released deadly poisons into Tony's bloodstream.

If somehow you use neutron bombardment over palladium you will get  $\ce{^{103}Pd}$ that is a radioactive isotope of the palladium with a half-life of 15 days that would be definitely very harmful to your health hence is more probable that the risks come from the radioactivity of the metal.
So summarizing performing neutron bombardment inside your body (not the best way to produce energy, because actually is an energy-consuming process) definitely will kill you someday but this not because of Palladium corrosion but because of its radioactivity.


text:
Give some examples of additional work in thermodynamics
dU=dq + dW(expansion) + dW(additional), give some examples of additional work...

and gravitational, and magnetic work

Firstly, dq= du + dw. In thermodynamics, we assume all thermodynamic processes to be quasi-static (i.e. very very slow process). Assuming a process to be very very slow, you won't have any extra addition work done i.e. whatever heat you supply to the system is used up for increasing the internal energy of the gas and for work done by the gas against the external surrounding.
Now lets say if we didn't assumed thermodynamic process to be quasi-static. In this situation, if you apply heat to the system, it would increase the internal energy of the gas, and some part of the applied heat will be used up for the work done by gas against surrounding , and now as the process is not quasi-static, when the gas starts doing work by expansion of the piston, as the process is no longer slow, the piston will acquire some velocity and hence it would acquire a kinetic energy. And hence you will now have another work done. This additional work done is work done for the movement of piston. So if the process is not quasi-static, whenever you apply heat, that heat is used up in increasing the internal energy of the gas, used for work which is done by the gas against the surrounding and also an additional work which is work done to move the piston.
If you assumed the process to be quasi-static, then velocity of the piston during the expansion is approximated as zero and hence no kinetic energy of the piston and hence work done for the movement of piston is neglected. So whatever heat you applied will be utilised for increasing the internal energy of the gas and for the work done by the gas against the surrounding (and no extra additional work done)


text:
How to determine whether a organic compound is diamagnetic or not?
I have faced several questions regarding the diamagnetic behaviour of certain organic compounds. I know that one has to check the electronic configuration of the atoms in the compound and see if there is an unpaired electron or not? Of all aromatic compounds (benzene, cyclopropene cation, cyclopentene anion) that I have come across one thing was common they were all diamagnetic. I have the following question in my mind

Are all aromatic compounds necessarily diamagnetic and if it is true what may be the proper reasoning for that?
Moreover, the compounds which are stabilised by resonance (hyper conjugation,+M effect) are necessarily diamagnetic. I have this doubt because by the electronic configuration tertiary butane cation should be paramagnetic ($2p^1$ configuration) but is still diamagnetic.

The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it
 is $1s^{2}2s^{2}2p^{1}$.So it must be paramagnetic. This is where my confusion is.

Nonsense, t-butyl cation has no electrons on p-orbital.

Mithoron is right. The central carbon atom is $\mathrm{sp^3}$ hybridized.

@DHMO It isn't sp3, simply empty p orbital...

Sorry, I meant $\mathrm{sp^2}$ hybridized...

As only a slight oversimplication, if you have an even number of electrons, the system is diamagnetic for simple organic molecules. There are enough exceptions: diradicals, diradicals created from filling degenerate antibonding orbitals, excited states. But as a rule for most applications, this statement is true.

If all the electrons are spin paired then the molecule will be in a singlet state$^*$ and will be diamagnetic, if they are not so paired then it will be  will be paramagnetic. The rules are the same whether or not the molecule is 'organic'.
Most 'normal' organic compounds, e.g. methane, benzene, etc. have paired electrons in each of the orbitals from the lowest one to the highest occupied one (HOMO), and thus are diamagnetic. The orbitals including the LUMO and up are unoccupied.
Radicals such as $\ce{CH3\cdot}$ have an unpaired electron (doublet state) and so are paramagnetic.
If a biradical is formed e.g. carbenes (see IUPAC Gold Book 'biradical' for an example structure) then it will depend upon whether the spins are paired in a singlet state (diamagnetic) or parallel as in a triplet state which will be paramagnetic. Often singlet and triplets are close in energy with the triplet being lowest. Similarly, dioxygen has two unpaired electrons and forms a triplet ground state so is paramagnetic. Nitric oxide  has one unpaired electron and is paramagnetic.
If a molecule is put into an electronically excited state by absorbing a photon then both singlet and triplet states can be produced, however, theses are short lived and decay back to the ground state.
$^*$(singlet /doublet/triplet refer to spin multiplicity, spin quantum number  $\pm 1/2$ and total spin e.g. $S=s_1+s_2$ gives multiplicity g  as $g=2S+1$)
example of biradical from IUPAC Gold Book.

Thanks for the answer,but What about aromatic organic compounds?Are they necessarily diamagnetic?

What about tertiary butane cations.according to theory it must be paramagnetic but actually it is diamagnetic.

No, but almost all are. Have a look at the structure in the IUPAC Gold Book biradical page

@aniline. If experiment gives diamagnetic then thats what it is. It does not mean basic theory is wrong as this is supported by lots of data, but just that this particular calculation gives wrong answer possibly because singlet and triplet energy are close to one another and within error of calculation.

@aniline Go back to the drawing board with the tert-butyl cation. Remember that it is not a tert-butyl radical.

I see your point of confusion!

The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it is $\mathrm{1s^2\,2s^2\,2p^1}$.

Well yes but no. Remember that you don’t have a lone atom in space but a carbon bound to three other atoms. These three σ bonds must originate from three orbitals pointing towards these three atoms. If we assume the electronic configuration to start off as (emphasis important) $\mathrm{2s^2\,2p^1}$, we cannot form three bonds, we only have one ‘ready’ orbital.
Therefore, this atom hybridises — or so our simplified view of the world. We mix the $\mathrm{2s}$ and two $\mathrm{2p}$ orbitals to give three $\mathrm{sp^2}$ hybrid orbitals. These now nicely align in $120^\circ$ angles and in one plane and are ready to bond to three other carbons.
There is a fourth p orbital on carbon. It remains there as a p orbital in the final structure. The three electrons we had are delivered into the three $\mathrm{sp^2}$ orbitals; thus the final $\mathrm{p}_z$ orbital is empty; completely electron-free.
Each of the electrons in the $\mathrm{sp^2}$ orbitals is paired up as a result of forming the σ bond. Therefore, there are no unpaired electrons in the tert-butyl cation and the species is diamagnetic.
The electronic configuration could be taken as $(\sigma_{\ce{C{1}-H}})^6\,(\sigma_{\ce{C{2}-H}})^6\,(\sigma_{\ce{C{3}-H}})^6\,(\sigma_{\ce{C{1}-C+}})^2\,(\sigma_{\ce{C{2}-C+}})^2\,(\sigma_{\ce{C{3}-C+}})^2\,\mathrm{2p^0}$. But nobody would ever do that.

You can add an electron into the $\mathrm{p}_z$ orbital to get a paramagnetic compound. Since you added an electron from the tert-butyl cation to arrive there, you no longer have a cation. Instead, you have created the tert-butyl radical. The electronic configuration from above would end in $\mathrm{2p^1}$.
Whenever you see the word radical, you can strongly assume that you have unpaired electrons. If you do not see it, you very likely have a diamagnetic organic molecule.
Note that NMR spectroscopy is intensely popular in organic chemistry. Because it bases on magnetic fields (and resonance therewith), it doesn’t really work with paramagnetic compounds. But more than $99.999~\%$ of known organic molecules have an NMR spectrum published.

Or a better answer that shows up in the next 3 days ;)  His (and yours) will be tough to beat though, very thorough.

@airhuff That too, of course; I am very open for that option ;)


text:
Enthalpy of solution
If NaCl dissolves in water, it is said that it happens because the hydration energy is greater than the lattice energy. If so, the enthalpy of solution attained is about +3.88kJ/mol, implying that it is endothermic.
From what I have understood, the hydration enthalpy is exothermic and thus its effect should be noticeable. That is, when we add up the lattice and hydration, the negative sign should have come in our final answer.
Why does this not happen? Where have I gone wrong?

You don't add up. You destroy the lattice, so the lattice energy should be subtracted. If you think of it, the answer (and even sign thereof) is not all that obvious.

Spontaneity of a process (dissolution in this case) is determined by Gibbs free energy, not enthalpy. So your first statement is categorically wrong. There are very many salts with endothermic dissolution enthalpies that still dissolve. I don't know off-hand if NaCl is one of them, but the point still stands. See: Why does a substance with an endothermic heat of solution dissolve?

For any salt dissolving in water, its not compulsory that enthalpy change needs to be negative. Yes, if it is negative then its better dissolving for sure, But it doesn't mean the enthalpy change has to be compulsorily negative for salt to dissolve in water.
Enthalpy change when we dissolve Nacl in water is positive. This is because breaking of ionic bonds in the crystal lattice of Nacl requires more energy than that of provided by hydration of Nacl. Yes, hydration energy for Nacl is less than the energy required to break the Nacl apart. Hence the extra energy which can add up with the hydration energy to break the bonds of Nacl apart comes from the heat of the water. So the extra amount of energy which hydration could not provide comes from the heat of the water. Hence water looses heat, becomes slightly cold and also hydration energy (exothermic process) is less than the required energy to break the bonds, hence the net enthalpy change is positive.
So till now we learnt that for Nacl hydration enthalpy is not greater than enthalpy required to break the bonds and hence the extra energy required to satisfy the bond breaking comes form the heat of the water, and as hydration enthalpy is less than required energy to break the bond of Nacl, the net enthalpy change is positive, hence endothermic process.
Now the question arises how comes then Nacl dissolve in water?
As already said its not compulsory that enthalpy change needs to be negative for a salt to dissolve in water. Remember my friend, dissolving depends upon ΔG. ΔG is the most important factor which decides weather the reaction will happen or not. If ΔG is negative and ΔH is positive, salt will for sure dissolve because dissolving of salt (or any reaction) depends on ΔG. So if ΔG is negative for the dissolving of salt then salt would dissolve. Infact for Nacl ΔG is negative. Lets see how.
ΔG= ΔH −TΔS.
In dissolving Nacl in water ΔS is very much positive because is dissolving Nacl, Na and Cl gets apart, increasing the disorderedness in the system, hence ΔS is highly positive. So practically TΔS term is higher than ΔH term and hence we end up having ΔG negative.
Hence as ΔG is negative, Nacl dissolve. Remember dissolving of Nacl depends on ΔG. Negative ΔG makes Nacl dissolve in water.

If the enthalpy of dissosiation was influenced by the heat from the water, then why doesn't it add up to the effect of the hydration enthalpy?

I do feel that your answer is correct, although when we add up the lattice and hydration energy, the hydration energy is still lesser than the lattice energy. But then if the heat was being used, then the overall energy should surpass the magnitude of the lattice energy, right? In that way, the answer would have been negative!   However, I do feel that the equation in my textbook that says the enthalpy of solution equals enthalpy of hydration and lattice is incomplete. I mean, it should have mentioned the heat from the water right?

sorry for my delayed reply! Ya in textbooks they don't go through detailed information. However what I mentioned is what exactly happens. Rather hydration enthalpy plays a very important role. For example in reduction potential of Co, Ni,Mn,Cr are all +ve. For reduction we require ionisation. Ionisation enthalpy of these transitive elements is very high. So think from where do they get the energy? Its from hydration energy. Infact hydration enthalpy is much higher than the ionisation enthalpies (1st and 2nd) and hence making electrode potential positive. So hydration enthalpy is imp factor.


text:
If the half-life of an isotope exceeds the age of the Universe, then how is it measured?
According to this Wikipedia article, the half-life of Bismuth-209 is 19 billion billion years, which exceeds the age of the Universe by factor on the order of a billion. How is the half-life of an element measured and experimentally verified if the half-life exceeds the age of the Universe?

You don't need to wait for an entire half of it to decay. Here's the original publication on Bi-209 (unfortunately behind a paywall).

Welcome to Chemistry StackExchange. Looking at the article the detector is a 'luminescent bolometer' which seems to be the same/similar as described at this site http://www.cresst.de/cresst.php.  You may need to ask the question on the physics site to get more detailed information. The result must involve some serious statistical estimation as they observed only 128 events over 5 days.

Could you measure the speed of a car in mph or km/h in less than an hour?  This is slightly more tricky as it is not linear but extrapolation is the same. If you found that the car had travelled 1km in a minute then you would probably happy to deduce that its speed was 60 km/h.

Well, half-life describes the time after which half of the substance has decayed.
This is all probability and statistics. If you look at a single atom you cannot make any prediction when it will decay. It might be within the next 3 seconds or it might be here even after hundreds of billion years. The probability that it decays during the half-life is 50%, so if we look at a very large number of such atoms we can use statistics to predict how much substance will actually decay in a given time.
Now bismuth-209 has $1.9×10^{19}$ years half-life, So if we have a lot of Bismuth-209 after that time we will only have half of it. But that doesn't mean we don't have any decay right now. 209 grams of Bismuth-209 are 1 mole so $6.022×10^{23}$ particles, a lot of those. Now the decay goes down exponentially but if we just calculate the average decay rate during the first half-life if we start off with 1 mole and we keep in mind that the rate will be much faster in the beginning we see that we have an average of around 16000 decays every year, which means around 44 a day. Again, in the beginning it will be more, approaching the time of half-life it will be less.
Now how do we know the half-life? It's simple in theory, some french researcher put 31 grams bismuth-209 in a box with very sensitive equipment and measured the alpha particles which are send out during decay. This isn't a trivial task to do, but it's possible. They counted 128 particles over 5 days and then did the math to calculate the half-life.
The original paper can be found here.

So if I am understanding correctly, it is kinda like measuring its 1/1000-life instead of its 1/2-life.  Is that correct?

Yes, I think you are thinking correctly, but 1/1000-life would mean there's only 1/1000 left. So it's more the 99.9%-life (or 999/1000-life)

It is not possible to measure the decay profile vs time as this is too long so the average number of disintegrations /time interval can be counted instead. From a consideration of the distribution of radioactive events the mean number of events in time $t$ is $M=N_0(1-e^{-kt})$ where $k$ is the decay constant and $N_0$ the initial number of atoms. For small values of $kt$, which is the case here as the decay lifetime is vast then, $M = N_0kt$. The rate and hence half-life (0.693/k) can be obtained if $N_0$ and $t$ are measured, however, there is uncertainty in this value.
In the case of measuring particles the binomial and poisson distributions have the property that the standard deviation in the number of events measure is $\sigma=\sqrt{M}$. Thus from a single measurement the std dev is obtained. This is not generally true, e.g. reading a thermometer or measuring a voltage.  If $1000$ counts are recorded in $10$ seconds  the standard deviation is $\sigma= \sqrt{1000}\approx 32$ and the mean count rate $R=(1000\pm 32)/10 = 100\pm3.2$ /sec. This can then be used to assess the uncertainty in the half-life using the propagation of error method.
As the count rate is inversely proportional to time,  with longer counting times the std dev is reduced but only as $1/\sqrt{t}$

But for 128Te with T1/2=2.2.10^24 years..... 1 kernel in 1 mol decays in average every 5.3 years. I cannot imagine how they measured it. Unless they measured decays products of billions years old samples.

@Poutnik Yes, that is a critical fault for such half lives. I have looked about some more and it seems that it is possible to estimate the decay rate constant from a knowledge of the shape of the tunnelling potential (Gamow-Gurney-Condon model). The experimental and calculated decay are within a factor of about 5 for half lives $\approx 10^{10} $ years, but I don't know if this extends to such long half lives as you mention.


text:
What active mass actually is?
I'm so much confused about what active mass actually is, my text book states that, active mass is a dimensionless quantity, but for our convenience and purposes, we take it with dimensions of molarity, partial pressure, etc. But I searched, and I got that, historically, what Guldberg and Waage stated about active mass in the law of mass action, is " Parts of body present in 1cc of the total volume " (Rough translation), now it contradicts that, it's dimensionless. On Google I found, that active mass is actually a ratio of mass which participates in reaction and total mass and it goes true with the fact written in my text book, that active mass of pure liquid and pure solids is unity. So, what is actually the definition of active mass ? What is actually about the dimensions ? And why we take active mass of pure solids and liquids as unity ?

I have never heard this term used. Based on your context, you might be referring to several things. One of them is activity, which is unitless representation of the effective concentration. In $\Delta G^{\varnothing} = -RT \ln K$, the equilibrium constant must be unitless, so we need unitless concentrations.