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how electron can go towards nucleus and then go far apart from it as in p orbital?see discription
In p orbital shape is dumbbell shaped which means electron go in the shape of dumbbell means first towards nucleus then go away from it. How can be this possible because first electron is attracted towards nucleus now it is very close means attraction force will be more and how it can go away from nucleus?

Dont think of orbitals as an orbit of the electron around the nucleus. Thats a HUGE no no. The orbitals are solutions to the wave function for a specific value of energy that characterizes each shell. This purely means that an orbital captures the area where the probability density of finding an electron is high, not where the electron orbits. So the dummbell shape results from the probability density, not the orvit of the electron.

Consider the 'orbital' to be the electron - it is not a point particle when localized in an orbital. The whole Bohr model, and the subsequent use of 'orbital' as nomenclature clouds the minds of physics and chemistry students. Welcome to the world of quantum physics - it is not classical mechanics anymore.


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Why is it important that glucose’s third OH group points to the left?
My question pertains to the location of the OH group on the 3rd carbon in a glucose molecule. If atoms can rotate freely around single bonds, then why is this important? Why can't glucose be drawn with 'all H on the left'?

This is the concept called stereochemistry.
Each carbon with 4 bonds has an approximately tetrahedral geometry.  We assign a configuration to the situation by numbering the substituents in terms of priority. Hydrogen is lowest.  Oxygen is highest.  The top carbon is second, and the other carbon is third.  We then trace a route from 1 to 2 to 3.  This would be a clockwise circle, and we would call this configuration S.  If we switched sides between the H and the O we would follow this circle widdershins and call the configuration R.  So the substituents can't be exchanged because the configuration would change.
So what?
When there is only one of these tetrahedral centers (called stereocenters) in play for a situation, it does not matter at all which one you have.  But the moment a second stereocenter comes into the picture, chemical properties begin to change.  A fairly tractable analogy is feet and shoes.  When you aren't wearing shoes, your left foot and your right foot are about the same.  But when you try to put shoes on, putting the right shoe on the left foot is slightly more difficult than putting the right shoe on the right foot.  In this loose analogy, the 'reaction' of putting your shoes on is slowed based on how the stereocenters of foot and shoe interact.
Onto glucose ...  Because there are 4 stereocenters in glucose, it has a distinct shape from its isomers.  The other chemicals in your body also have their own shapes/configurations.  Glucose reacts slightly differently than other sugars in the body because of the way these shapes fit together.  Therefore you need to keep the configurations known and well marked.
Most people (about 94% of the rest of you) don't intuitively see stereochemistry concepts and need to make models.  However, pragmatically, if you can take this explanation, and remember that 'it just matters' to keep the notation consistent, you can probably get through most of biochemistry.
If you want to make models, at your next wine and cheese party get some toothpicks, and two pieces each of four different kinds of cheese.  Put four of the cheese chunks on the end of the toothpicks, and then stick the tooth picks in a sausage or something.  Then make an exact copy.  After that is done, exchange two of the cheese chunks on one model and investigate what you have.  There should be no way you can twist or turn or rotate one model to have its kinds of cheeses in the same orientation as the other.  Once you have done this, the following picture will make much more sense:

Alright, see if I get this straight: technically 'all H on the left' is a (structural) isomer, but because of these so called stereocenters (will read up on those, not mentioned in the book as its a biology book with a real short biochem intro) the position of the HO and H on $_3C$ does matter. Is that about right? The importance seems clear, the book explains this by (in)active entaniomer of Ibuprofen and the ability for morphine to attach to an endorphine receptor.

Same concept, you seem to understand it from the 10000-foot view.  Specifically, for ibuprofen, one stereocenter is in the drug, and the other is in the receptor the drug interacts with.  It just does not fit properly when the drug is the wrong entantiomer, and hence provides no pain relief

I guess, your actual problem (in understanding why the orientation of the $\ce{-OH}$ groups in glucose is relevant) is caused by the Fischer projection, which is used in your picture.
The Fischer projection may be misleading since the carbon chain is actually not a flat, straight line. The four bonding partners of each carbon atom are arranged spatially in the form of a tetrahedron:

In a Fischer projection, vertically drawn bonds are considered to lie below the projection plane and horizontal bonds to lie above that plane:

Therefore, the horizontal $\ce{-H}$ and $\ce{-OH}$ groups that are shown in the Fischer projection of glucose are also considered to lie above the projection plane, approximately like this:

When you now rotate the $\ce{-OH}$ group at the third carbon atom to the right side, the $\ce{-H}$ and $\ce{-OH}$ at the third carbon lie below the projection plane:

However, if you simply place the $\ce{-OH}$ on the right side in a Fischer projection, the horizontal $\ce{-H}$ and $\ce{-OH}$ groups still lie above the projection plane:

As you can see, swapping two groups attached to a chiral carbon atom in a Fischer projection results in a different molecule.

Lighthart gave the theoretical explanation of what enantiomers (‘right foot — left foot’) and diastereomers (‘right foot and right shoe — right foot and left shoe’) can be thought of. Here is the practical explanation for glucose.
You drew glucose in a chain structure. This is not wrong, but only $0.25~\%$ of all glucose molecules in any aquaeous solution are in this chain conformation. The vast majority form chair structures. There are two very important chair structures which between them make up $99~\%$ of the glucose molecules in solution, they are called α-D-glucopyransoe and β-D-glucopyranose. Their respective structures are shown in figure 1 below.

Figure 1: Three structures that glucose assumes in solution. Fltr they are α-D-glucopyransoe, the open-chain form and β-D-glucopyranose. Image taken from Wikipedia, where a full list of authors is available.
For nomenclature purposes, remember that the carbon which is either bonded to two oxygens (cyclic chair structure) or forms a double bond to a single oxygen (chain form) is carbon 1 ($\ce{C{1}}$), and that we number the carbons sequentially up to carbon 6 ($\ce{C{6}}$). Note how the hydroxy groups on carbons 2, 3 and 4 are all equatorial (in the plane of the ring) while the hydrogens bonded directly to the same carbons are axial (up and down).
Your question specifically asked what would happen if $\ce{C{3}}$’s hydroxy group would point to the right rather than the left. This would result in a different monosaccharide called allose. Its structure is shown in figure 2.

Figure 2: Three structures D-allose may assume in solution. fltr these are α-D-allopyranose, the open-chain form of D-allose and β-D-allopyranose. Furanose forms which are not displayed here make up $12~\%$ of the allose molecules in solution. Image self-drawn.
Note how the hydroxy group on $\ce{C{3}}$ is now axial (pointing down) and the corresponding hydrogen is equatorial (in the ring plane). What may look like a minor difference is in fact significant. Allose and glucose are different compounds with different properties as shown in table 1 below.
$$\textbf{Table 1:}\text{Experimental melting points of allose and glucose}\\
\begin{array}{cc}\hline
\text{Compound} & \vartheta_\mathrm{m} \left [\mathrm{^\circ C} \right ] \\ \hline
\text{D-allose} & 128 \\
\text{D-glucose} & 146 \\ \hline\end{array}$$
(Other properties would differ, too, but allose is a poorly studied monosaccharide so not many experimental properties have been determined.)

The wiki article on glucose states that more than 99% of the glucose molecules do exist in it's cyclic, pyranose, form.

If you have a look at the cyclic form, you'd see that all -OH groups are equatorial (not the anomeric, ie. position 1 of the ring, which can be  $\alpha$ or $\beta$).
Combined with the knowledge of cyclohexane conformation (which to large extent works also for pyran), it is clear that most stable form is that with oxygens being equatorial.
If you want to draw the non-cyclic form in Fischer projection, you just have to respect the facts stated above, leading to the figure you have presented.
Although I have not found any proof, my personal guess why glucose is the most common of all sugars is it's stability, compared to other isomers.

There is penalty for axial OH groups in cyclohexanes is small, (half of that for methyl), and if you have two of them you can get some intramolecular hydrogen bonding, so I don't think your guess is the full story.  But this is a nice addendum to above.

@Lighthart: Unfortunately, I spent weeks of my life calculating sugars. Hydrogen bond can be of course formed, but once you put in water molecules, the whole story gets more complicated and in principle it does not matter that much, if you can form intramolecular H-bonds or not.

Well doesn't the concept of dissolving sugar in water make me look silly then?  Heh.

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Dehalogenation of vicinal dihalides
Why, among different metals, only zinc is favourable for the dehalogenation of vicinal dihalides?

Magnesium + ether will also reduce a vicinal dihalide to an alkene.

I guess, mostly for safety and simplicity reasons. Magnesium tends to react vigorously with alkylhalogenides and aluminium forms oxide coating. Let's not even talk about such cases as sodium.

Zinc is used effectively in the dehalogention of vicinal dihalides because a carbon zinc complex is formed in the mechanism which is probably more stable. This reaction is also known as Simmon' Smith reaction.

The stability and usefulness of the Simmons-Smith reagent (zinc metal)
may be attributed in part to the higher covalency of the carbon-zinc
bond together with solvation and internal coordination of the zinc (which makes the zinc complex stable). (Source)
Reduction of vicinal dihalides by zinc metal proceeds by way of
oxidative insertion of Zn(0) into a C-X bond, producing an organozinc
species (stable) which undergoes dehalogenation:

(source)

The Simmons-Smith reaction is the dehalogenation of a geminal dihalide, and thus shouldn’t be the main point of interest here.

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Why there are nodes in orbital?
I know that because Schrödinger equations solution for probability of that region is 0, but I was looking for an intuitive explanation.
How can you intuitively explain the existence of nodes in an orbital?

There are nodes in orbital (like 2s orbital has one node i.e, probability of finding electron in that region is zero). So my question is why electron can not appear in node, just because we say electron can not be found in node. So is there any reason which sounds logical?

It's a result we obtain from Schrödinger equation. What's wrong in having any? We could help you visualize the nodes if you brought the correct diagram or picture of an atom.

Why electron can not go in nodal region? Why it is restricted to go there? Just because Schrödinger equations solution to this region is zero.

Yes, because of that. What's wrong with that?

Is there any logical answer which fits into mind that why orbital has nodal region

I have no way of knowing what would fit in your mind and what would not.

Nothing will fit into your mind if you don't open it. Some things in Chemistry just have to be absorbed. Just the way you absorb that atoms are there, without actually confirming their existence by seeing them physically.

That being said, if you'll solve the Schrödinger equation for a variety of situations where it can be solved (harmonic oscillator, etc), you'll start to see the pattern.

I don't think kids in high school have the tools to do that @Ivan.

Define "logical" answer? Why isn't " the wavefunction is zero at that point" not a logical answer?

I will repeat that question why electron can not appear in nodal region is there any reason like proton electron attraction or electron electron repulsion Ar any other factor which does not allow electron to be present there

I think you guys are not getting the point behind my question

@getafix they're perhaps looking for an intuitive explanation.

Thanx dead for your contribution

Well, in that case, quantum mechanics defies intuition (at least it defies mine).  Our intuition comes from observations in the macroscopic world, and will not serve us well in the microscopic world. However, you asked for factors..factors which affect the functional form of the solutions are the shape of the potential, and the boundary conditions of the problem

Its sound better...

@ahmad The point of all this discussion is, there is another property of matter, which is not intuitive at all. The "property" $F=ma$ is intuitive because we see it everywhere. Schrodinger's equation is another such "property", which is not at all intuitive. It is this property that gives birth to those nodes. You will have to study a little bit of quantum mechanics to understand more of it. Food for thought: "Properties" like the electrostatic force and so on also apply at that scale, along with this new one.

https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane

http://chemistry.stackexchange.com/questions/35672/how-do-electrons-cross-nodes-in-orbitals http://chemistry.stackexchange.com/questions/8655/structure-of-atom-and-nodes

@getafix Let's keep it friendly, please (re: your first comment).  Thanks!

It is simply because the wave nature of the functions.
Imagine a tight rope with fixed ends (like a string in a guitar). If you shake it it will vibrate. Both ends remains fixed. If you take a high frame rate video and see it in slow motion, at a fixed horizontal length (let say $x$ positions) the rope will oscillate up and down with some amplitude. It can be seen one or more intermediate $x$ positions that remains always with null amplitude (if it is in state close to a stationary one). There you have nodes in one dimensional string.
Visual example
You would have more complex patterns going to a two dimensional system, and even more in a three dimensional case. They all have nodes, like the 1D case. The 3D case can be represented mathematically by using the spherical harmonics. The angular part of the solutions for the hydrogen atom are those spherical harmonics. The only intuitive explaining that I found is their wave nature. Of course it is not as visual as in the 1D case which is easily found in the daily life. And clearly, what is more weird is the wave nature of matter.
I hope that this answer address your question.

Just to add to this, the 3D analogue of the "fixed ends" of the rope is the requirement that the wavefunction be square-integrable (aka normalisable), i.e. $\psi \to 0$ as $r \to \infty$. That would explain the presence of the radial nodes, angular nodes are a bit harder to justify imo.


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How small a peak could one identify by GCMS
I'm trying to characterize the impurity profile of an organic liquid sample that is already described as 99.7% pure (Boiling Point 230 C).
Would a GC-MS be a viable way to identify the smaller peaks? e.g. How small a peak could one expect to get a decent mass spec. pattern for? Say, a peak with 0.01%  of total peak area, would it be viable to identify it on a GC-MS?
I'm assuming the smaller peaks can be resolved satisfactorily. I don't need quantification of the peaks just identification. I can match against the NIST library of spectra but also have a clue based on the manufacturing process about what the likely impurities may be.

You can get GCMS peaks for very small components of the overall mixture.
However, most GCMS patterns are electron impact, and hence probabilistic, so if you don't have enough sample, the fingerprint won't be reliable.
In my experience 0.3% produces spectra which are 'good enough'... as long as you don't inadvertently overload your column.

Of course, the background noise is important for how small peaks you can confidently identify as a "signal", and for how small peaks you can actually confidently quantify. A bunch of factors come into play that affect the resulting limit of detection and limit of quantification.


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Natural Product class
I have a quick question: how can I distinguish between terpenoids compounds and polyketides?
For example artemisinin (see image below): it present a continuous carbon chain alternated with $\ce{O2}$, but it is a polycyclic structure as in terpenoid compound.
Which is the main criterion I should consider here?

It’s very hard to tell for some molecules a priori. For most, there are enough indicators.
Polyketides:

will typically have 1,3-functionalisation patterns, so hydroxy groups or keto groups are typically two carbons away from the next one. And most methyl groups also have a nonmethylated carbon in-between. Note that 1,3 can be expanded to 1,5; 1,7; 1,9; …
if cyclic, will often be only monocyclic or any second cycle will typically be something like a ketal (two oxygens bonded to a single carbon; formed by a hydroxy group attacking a ketone often five carbons away).
if cyclic are typically macrolactones, so large cycles (14, 20 or even more carbon atoms) and linked by an ester group.
can contain phenyl groups or other aromatic systems formed by elimination reactions. Often, these will be 3,5-bishydroxyphenyl groups.
typically contain an even number of carbon atoms in the chain (i.e. excluding any methyl groups mentioned above).

On the other hand, terpenoids:

very commonly have a carbon count that is a multiple of five (even often multiples of ten).
can very often be broken down into isoprene (2-methylbutane) units by paper chemistry.
typically have 1,5-methyl functionalisation (although this can change due to Wagner–Meerwein rearrangements in the biosynthesis).
can have 1,1-dimethylated carbons which polyketides cannot.
often form polycyclic structures composed of small all-carbon cycles (small being 5- to 7-membered rings).
rarely contain lactones or lactams (cyclic esters or amides).
are generally less oxygenated.
very rarely contain more than one or two double bonds, never contain phenyl rings to the best of my knowledge.

With that, you have some criteria at hand to attempt a classification. Artemisinin’s tricyclic structure with an all-carbon cycle looks a lot more like a terpenoid than a polyketide.

There are no ketone functionalities in artemisinin, so that kinda gives it away. The rest of the oxygens in the molecule also are not polyketide-like, here you have ethers and a peroxy bridge, you'd expect ketones and hydroxyls.
Otherwise the true distinction is of course in the biosynthesis, the classes use different building blocks (some isoprene derivative or  malonyl-CoA). That would require a bit more research.

So the presence of a ketone group can be considered as a criterion?

@chocoly It's a good indicator. Polyketides are normally built up of multiple acetyl subunits. While theoretically almost all of them could be reduced to ethylene moieties, in practice they're usually not. So you tend to end up with hydroxyls and ketones on a large number of carbons. Terpenes usually start off without oxygens in the structure, so typically they'll only have a few oxygens, which were added later.


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Solubility of glucose
On eating powdered glucose I felt a cold sensation in my mouth. Then I thought it would be a result of the reaction between saliva and glucose. Then I thought that this would imply that the reaction would be endothermic. I don't find myself to be pretty sure and need to know whether I am right or not. Why would the reaction be endothermic in spite of the fact that glucose is readily soluble in saliva?

Was this chemically pure glucose?

Why do you think dissolving something that is readily soluble can't be endothermic?

Dissolution can certainly be endothermic. Whether a reaction happens is controlled by $\Delta G$ not $\Delta H$

It was glucose-d and 99.4% pure.Do molecules use my mouth's heat tobreak their bonds and get salivated and as a result the cold sensation?

The dissolution of glucose in an aqueous solution is an endothermic process.  As stated in a comment, a reaction need not be exothermic to be spontaneous. Don't forget about the entropy term.
Simply put, the glucose dissolved endothermically in your saliva causing a heat transfer from your mouth to the glucose/saliva solution, resulting in a slight decrease in the temperature of your mouth, making it cooler than it was prior do dissolving the glucose.


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How to make by dilution a certain ion concentration
The question I'm trying to answer is this:

What volume of $0.25\ \mathrm{mol/l}$ calcium nitrate is required to make by dilution with water $500\ \mathrm{cm^3}$ of a solution with a nitrate ion concentration of $0.1\ \mathrm{mol/l}$?

My attempt:
Using $n=cv$, we find the the number of moles of nitrate ions to be $0.05$ moles. Now from the chemical formula of calcium nitrate, we know that $1$ mole of calcium ions gives two moles of nitrate ions. Therefore $0.05$ moles of nitrate ions gives us $0.025$ moles of calcium ions. Now, using $v=\frac{n}{c}$, where $n$ is the no. of moles of calcium ions, we obtain the volume of $0.1$ litres.
My attempt actually managed to give me the correct but I don't understand a part of it; I can't understand why when I'm trying to find the volume of calcium nitrate I have to use the number of moles of calcium ions and not the number of moles calcium ions added to the number of moles of the nitrate ions.
(Note: I asked this question before, deleted it, changed it and re-published it as of now.)

Hint - A molecule of $\ce{CaNO3}$ is 1:1 for anion and cation. So a 0.25 mol/l solution is 0.25 mol/l for both the cation and the anion.

@MaxW But isn't the formula $Ca(NO_3)_2$?

Geez, never delete and repost! Questions can be edited and reopened.

@Mithoron I thought that if I edited it it would just get lost...

@ReinhildVanRosenú - You're absolutely right. I had a major brain fart. A $\ce{CaNO3}$ solution is 1:2 for cation and anion. So a 0.25 mol/l solution of $\ce{CaNO3}$ is 0.25 mol/l for the cation $\ce{Ca^{+2}}$ and 0.50 mol/l for the anion $\ce{NO3^{-1}}$. Thanks for the correction.

Noting that the solution is 0.50 mol/l for $\ce{NO3^{-1}}$, you can use the equation: $$m_1 * V_1 = m_2 * V_2$$ or $$0.50 mol/l * 0.500 l = 0.1 mol/l * x l $$ therefore $$x = \dfrac{0.25}{0.1} = 2.5 l$$

When you use the number of moles of calcium ions, you are actually using the number of moles of calcium nitrate. You have done the 1:2 conversion already in your question. You found the amount of nitrate ions needed to be 0.05. But then instead of converting that amount to moles of calcium ions, you were actually converting it to a number of moles of calcium nitrate, since the ratio of moles calcium ($\ce{Ca}$) to moles calcium nitrate ($\ce{Ca(NO3)2}$) is 1:1.


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What happens to the Lithium halide salt when reacting Lithium metal with alkyl halides?
In organic chemistry, I learned a lot of useful reactions but unfortunately I didn't get to try many of them first hand (what's sad is I am a Chemistry major too).  I've searched the internet back and forth for the answer to this question which I am fairly sure I know the answer to, but I thought I would just double check.
When reacting lithium with an alkyl halide (the reaction requires 2 mol Li metal per 1 mol of the alkyl halide), one mol of lithium forms the organometallic compound while the other forms the lithium halide salt.  The reaction takes place in diethyl ether, so my question is does the halide salt fall out of solution or not?  I would assume so considering diethyl ethers low dielectric constant in combination with the ionic character of Lithium salts, but I thought I would ask in hopes someone on here has performed this reaction first hand.  Thanks!
Edit: So I read LiCl has minimal solubility in diethyl ether, but LiBr is moderately to highly soluble which makes absolutely no sense to me.  I guess perhaps the lone pairs on oxygen might play a role in this but then again why wouldn't this have the same affect on LiCl?

Never done it myself, but I'm fairly sure you are correct.

Lattice energy of LiBr is smaller than LiCl, so dissolution is more favourable.

@orthocresol The lattice energy for LiCl is only ~50 kj/mol greater than the lattice energy for LiBr, yet the solubility of LiBr is far greater than LiCl (at least the few sources I could find portrayed it this way).  Is this a typical occurrence (indicating the realtionship between solubility and lattice energy is non-linear)?

If you just think about K = exp(-ΔG/RT) you'll see that 50 kJ/mol can potentially lead to a huge change in K already, so I wouldn't be surprised.


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Hydronium Structure
I was doing an acid-catalyzed mechanism and I ran into hydronium ($\ce{H3O}$) and I wasn't able to figure out why its structure is the way it is. If you look at the structure, you see the hydrogens, and an oxygen; however, the oxygen has two electrons remaining outside of it. This is confusing to me as If you add up all the electrons (3 from Hydrogens, 2 from the remaining oxygen), we have a total of 5 valence electrons, even though we should have a total of six. I would assume that each hydrogen would take up one electron from the oxygen each, thus three electrons would remain on the oxygen.
Could someone please explain why this is? Thanks.
Here is a little sketch I made to show what I mean:

Why should there be 6 valence electrons? There aren't 6 electrons around oxygen in, say, water. Have you heard of something called the octet rule? And do you know how to draw a dot cross diagram (Lewis structure) of $\ce{H3O+}$? A lousy mspaint mockup would suffice.

+orthocresol I have done a bit of a review of the Lewis dot diagrams, from a long time ago; thus, I would assume I may have some mess ups! I made a little diagram so that you can see what I mean

Okay. First things first: in $\ce{H3O+}$, the oxygen has a positive charge, so it should really have one electron less then what you drew. Next, if you count ALL the electrons in your diagram (minus the one that shouldn't be there), there are 8 in total: 2 from the oxygen lone pair and 6 from the O-H bonds (3 from oxygen, 3 from hydrogen). All in accordance with the octet rule.

There is no $\ce{H30}$, like you originally wrote. I assume you wanted to write $\ce{H3O}$, which still does not exist but at least gives us a hint as to where your problem is. The correct notation and formula would be $\ce{H3O+}$. Now what could the $+$ sign mean? ;)

+Jan Ah, I understand now! I was given the acid-catalyzed question verbally , and they didn't tell me that charge. Thank you so much!

Thanks to help from orthocresol & Jan, I was able to figure out my mistake. In the question, I was not given the charge nor the proper notation, and I was very confused. So anyways, the correct notation would have been $\ce{H3O+}$. Thus, there would actually only be two electrons remaining on the oxygen, and not three as I had thought.
Also, they had given me the octet rule (all electrons must add up to 8) which will definitely help me not run into this mistake again!
Thanks.

2 Hydrogen atoms would share their valence electrons with oxygen and forms covalent bond. The third Hydogen atom forms a coordination bond that is, it is actually a proton and both the shared electrons are given by oxygen. The net charge is +1.


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Is there a difference between energy levels and electron shells?
Is there any difference between energy levels and electron shells, or do they mean the same thing? Does the number of energy levels remain constant when progressing across a period?

Well, they are different, since the 2s and 2p orbitals have different energies but belong to the same "shell".

Shell is just a bunch of levels.

Introduction
The absolute energy levels of all electrons in an atom are different, and will shift subtly when electrons are added or taken away. However, it is true that, generally speaking, sets of electrons in an atom will tend to have energy levels very closely clustered together, with big differences in energy to other sets of electrons in the atoms. We say that these electrons are all in the same energy level. On the other hand, election shells have to do with the mathematics of the electrons 'orbitals', and are closely related, but are not the same. If you want to avoid 'hair-splitting' detail at this point, the difference is this -
Energy levels
Each electron in an atom belongs to an energy level in which all the electrons in that level have nearly exactly the same energy, but there are big energy differences between the levels. If you were to strip away all the electrons from an atom and then return them one by one, you would find that you could only put 2 in the lowest level and then 8 in the next. The capacity for the first five levels are 2, 8, 8, 18, 18 in that order. This corresponds to the number of atoms in each of the first five periods of the periodic table and helps explain the properties of elements in the periodic table. This makes 'energy levels' a very useful idea. So energy levels go from 1 to 5 for the first 5 periods of the periodic table, and in simple terms, full energy levels tend to be 'desired' by atoms.
Electron shells
Electron shells are slightly different, but closely related. They are numbered, like the energy levels. The first shell, 'is' the first energy level. Likewise, the second shell, is the second energy level. But after this, it becomes more complicated, and this is where the confusion can come in.
With the exception of the first shell, each shell is made up of sub-shells. These sub-shells make up each energy level, but sub-shells can belong to different numbered energy level from shell 3 onwards.
To add some detail. There are 3 kinds of shell, 3 ways of satisfying the quantum maths for bound electrons, in the first 5 periods. There are labelled, 's' shells, 'p' shells and 'd' shells. In the mathematics for these, there are 'shell numbers'. For shell number '1', the mathematics only allows for 's'-shells. For shell number 2, 's'-shells and 'p'-shells are allowed, and for shell number 3, s, p and d-shells are allowed. S-shells can take 2 electrons, p-shells 6 electrons and d-shells 10. So shell level 1 has 2 electrons, level 2, 8 (2+6=8) electrons, and level 3, 18 (2+6+10=18) electrons, giving the pattern 2, 8, 18. This is clearly different from the energy levels, and to know why, we have to know more about the sub-shells.
The sub-shells are written down with the shell number first, the shell type second, and the number of electrons actually in that shell, third, (usually as a superscript). So Hydrogen's electrons would be 1s1, and Helium's 1s2. For the second energy level, Li would be 1s2 2s1, and further across the period, Carbon, 1s2 2s2 2p2, having six electrons. At the end of the period, Neon would have two full energy levels, written as 1s2 2s2 2p6. (Often lower full shells are ignored, so Neon would be written 2s2 2p6.) So far, electron 'shell numbers' coincide with energy levels. However, the third energy level has only 3s and 3p shells, not 3d. This is because the 3d shell has an energy level between 4s and 4p. So the 3d sub-shell belongs to energy level 4. Similarly, the 4d sub-shell has an energy level between the 5s and 5p sub-shells. These extra 10 electrons in the energy levels 4 and 5 produce the transition metal block in the periodic table.
Conclusion
Shells and energy levels come from the properties of atomic electrons, but one comes from the mathematics of electon orbitals, and the other from the energy that the electrons take in these sub-shells. This means that referring to energy levels as ‘shells’ will produce confusion, when an understanding of ‘sub-shells’ is needed for a better understanding of the periodic table chemistry.

I am sorry but this is really not correct. Energy levels are not what you described at all. Those are shells. s, p, d, and f are subshells not shells. Shells refer to collections of subshells with the same principal quantum number.

Clear expressions, thanks. Want to add, that the confusion about 3s after 4s,... comes from the postulated number of electrons for each shell with 2*n^2. This equation is simply not helpful and doesn't matches the Periodic table of elements. It would be better to explain the electron number of the first three shells by the equilibration of the electrons magnetic dipole moment.

So is the term "shells" even used anymore or is it more correct to use the terminology 'energy levels' , 'sublevels' , and 'orbitals'?

They are different. Take iron as an example. If we using the term shell, the outermost shell should have 16 electrons, according to the "2 8 18" sequence. As for the energy level, the electron configuration of $\ce{Fe}$ should be $\mathrm{(1s)^{2}(2s)^{2}(2p)^{6}(3s)^{2}(3p)^{6}(4s)^{2}(3d)^{6}}$, as the $\mathrm{4s}$ sublevel (subshell) is the furthest from the nucleus, the outermost energy level should have 2 electrons. Shells are more likely to used from a mathematical view since it don't consider the actual energy of each subshell.

Iron has 4 electron shells, so the valence shell only has 2 electrons


text:
Why are the electronegativities of transition metals so similar to each other?
I was wondering why the electronegativity among transition metals are so similar. I can't find anything on the internet, just vague answers about the d orbitals that don't seem to provide a sufficient justification.

Working on it, but a very general answer is that since there is an xs sublevel and a (x-1)d sublevel that are very close to each other in energy, there are only small differences in the relative stability of different electron configurations of transition metals.  Rare earth metals gain much more stability by losing s electrons, but transition metals don't because they still have a partially filled d, but they are not as unstable in the first place because the xs and (x-1d) electrons have a variety of arrangements that are more stable than for a base metal that only has 1 or 2 s electrons.

For example, Chromium: $4s^1$ $3d^5$ is more stable than would be $4s^2$ $3d^4$.  By being able to use both sublevels there are more stable configurations "in between".  Interesting that the electronegativity tends to rise and then fall across the period, because you start to run out of different ways to configure the electrons.  By the way, this gave me a question.  Why are do the rare earth metals have lower electonegativity than the d-block.  It seems to be due to a lack of overlap in energy level between xs and (x-2)f sublevel, so fewer "in between" configurations.


text:
How does SOFC work?
The Solid Oxide Fuel Cell is a particular case in which an oxide is used as a conductor. Once i've heard that oxides usually aren't good conductors but it turned out that a lot of oxides are conductors. What i am asking is :

How can i know if a oxide is a conductor or not?

If there isn't any conduction band since it's not a metal, how do the          electrons move and how does the SOFC work?

Thank you.

Not all conductors work by conduction of electrons - ionic conductors are a thing. I think the most famous one is alpha phase of AgI. For oxides, off the top of my head I can think of YSZ and beta-alumina. Basically there are holes in the crystal structure which let oxide (or other) ions move. I'd say though that it's difficult to predict whether it's a conductor without knowing the system.


text:
where can I find experimental compressibility factor values?
I am interested in experimental compressibility factors for various gases, such as methane and carbon monoxide. Not only for standard pressures and room temperature, but in a range from 1 to a few 1000 bars and various temperatures. Where can I get them? I've done a quick google search, but they seem to be well hidden.
EDIT: Perhaps a better question is where can I find second, third, fourth, fifth... virial coefficient for these 2 gases? I am interested at 2 temperatures: 200 K and 287 K. I found some so far, but not at 287 K.

Googled "compressibility factor of methane", got this as first hit http://encyclopedia.airliquide.com/Encyclopedia.asp seems to have data you need for CH4 and CO as well as a ton of other gases. Seems like not every gas has compressibility factor info but at least for CH4 and CO it's there

Perhaps I am being dumb, but it only states the Z at 1 atm which is of little use to me, because I  am interested at Z from 1-1000 bar.

Oh, I see, that's different then.

Virial coefficients and compressiblity factors differ when calculated by different gas equations.

Why don't you just use the law of corresponding states and modifications thereof, based on reduced temperature, reduced pressure, and ascentric factors.  This is in practically every thermo book.

based on my research so far, law of corresponding states can easily produce 5% error. It is a nice law, but I don't think it is completely exact.

The NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) is a program that can calculate the thermodynamic and transport properties of some important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available as NIST Standard Reference Database 23 here.
For example, the following selected results have been obtained with REFPROP Version 9.0. The number of digits is set to the default value (5 digits), which are not necessarily significant.
Methane
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99369
2nd Virial Coef.: −0.0065173 m³/kg
3rd Virial Coef.: 0.000015219 (m³/kg)²
Methane
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99802
2nd Virial Coef.: −0.0029520 m³/kg
3rd Virial Coef.: 0.000010232 (m³/kg)²
Carbon monoxide
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99739
2nd Virial Coef.: −0.0015501 m³/kg
3rd Virial Coef.: 0.0000035524 (m³/kg)²
Carbon monoxide
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99952
2nd Virial Coef.: −0.00040865 m³/kg
3rd Virial Coef.: 0.0000027781 (m³/kg)²
REFPROP can also be used to plot the results, for example for carbon monoxide:

This is great. Is there an option to also get the 4th and 5th virial coefficient? And if so, do you mind posting it?

@sanjihan I can only find the second and third coefficient in REFPROP. I guess, in general, the value for the fourth coefficient is rarely available.

@sanjihan You might also be interested in the book Virial Coefficients of Pure Gases and Mixtures. (I did not check the content.)


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Hydrolysis equilibrium of urotropin
In a test I was asked why the hydrolysis equilibrium of Urotropine (Hexamethylenetetramine) in neutral solution lies far on the side of the reactants. Well I had no idea and I'd be very glad if somebody could give me an answer on that question.

Small amount of acid is required for hydrolysis.
As a result ammonium salt and formaldehyde are produced.
Update:
It can be explained with enthalpy of reaction Wiki Enthalpy of reaction.
To calculate it, enthalpy of each  reactant required.
From wiki: ΔH(C6H12N4)=-99.2kJ/mol, ΔH(H2O)=-285.8kJ/mol, ΔH(NH3)=-45.94kJ/mol, ΔH(CH2O)=-115.9kJ/mol;
ΔH(reaction)=6*ΔH(CH2O)+4*ΔH(NH3)-ΔH(C6H12N4)-6*ΔH(H2O)=918kJ/mol;
As ΔH(reaction) is large, positive number, a lot of heat required for this reaction, as a result, equilibrium is far on the side of the reactants.

Yeah, but were is the acid? In your equation, you use water as reactant. While water can be seen as a very weak acid, it does have a pH of 7 which is neutral. And the question was, why the equilibrium in neutral solutions (in Water) lies on the left side. So I don't see the question being answered here :/

Probably the "in neutral solution" mean that urotropin should react with water. Acid binds molecules of ammonia, so hydrolysis become possible.

Yes, it does, but that's not an explanation why the equilibrium is far on the side ot the reactants. If it would react with the water, the equilibrium would displace.

It can be explained with enthalpy of reaction Wiki Enthalpy of reaction. To calculate it, enthalpy of each  reactant required. From wiki: ΔH(C6H12N4)=-99.2kJ/mol, ΔH(H2O)=-285.8kJ/mol, ΔH(NH3)=-45.94kJ/mol, ΔH(CH2O)=-115.9kJ/mol; ΔH(reaction)=6ΔH(CH2O)+4ΔH(NH3)-ΔH(C6H12N4)-6*ΔH(H2O)=918kJ/mol; As ΔH(reaction) is large, positive number, so a lot of heat required for this reaction, as a result, equilibrium is far on the side of the reactants.

Tha'ts a pretty good argument, but a reaction happens voluntary if ΔG < 0. As we all know ΔG = ΔH - TΔS and ΔS should be pretty big in this case (7 molecules react to form 10 molecules). But you may be right because ΔH is really big and up to a significant temperature, ΔS will not be bigger, so the reaction shouldn't happen voluntary.


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Why do these reactions occur?

Identify the products $[A]$ to $[D]$:
$\ce{CH3COONa}+\ce{NaNH2} + \triangle \rightarrow [A]$
$[A] +\ce{CO_2} \rightarrow [B]$
$[B] + \ce{H2O} \rightarrow [C]$
$[C] +\ce{P2O5 } \rightarrow [D]$
$[A],[B],[C],[D]$ are the main product in each step.

According to my book's answer
$[A]=\ce{NaCH2COONa}$
$[B]=\ce{NaOOC-COONa}$
$[C]= \ce{(CH2COOH)}_2$
$[D]= \ce{C_3O_2}$
I can intuitively understand that product $[A]$ is correct.However I never learnt the next three reactions.Can someone explain/give a source from which I can learn why the next three reactions occur ?
Thanks.

You wrote B and C wrong.

That implies that the author of the study material wrote it wrong. BTW according to you what should be the answer? @Mithoron

B, C, and D should all have three carbon atoms. Your B has two, and your C has four.

@aventurin Bad reasoning.The equations may not be balanced.

Abstraction of $\ce{H+}$ by strong base $\ce{NH2^-}$
$$\ce{CH3-COO- + NH2^- -> ^-CH2-COO- + NH3}$$
Carboxylation to form malonate
$$\ce{^-CH2-COO- + CO2 -> ^-OOC-CH2-COO-}$$
Free malonic acid
$$\ce{^-OOC-CH2-COO- + 2 H+ -> HOOC-CH2-COOH}$$
Dehydration of malonic acid with $\ce{P4O10}$ yields carbon suboxide (unbalanced)
$$\ce{HOOC-CH2-COOH + P4O10-> O=C=C=C=O + H3PO4}$$


text:
Nitric acid plus Hydroiodic acid
Why $\ce{I2}$ is formed when $\ce{HI}$ and $\ce{HNO3}$ are reacted?
I know that $\ce{HI}$ is more acidic than $\ce{HNO3}$ so nitric acid will accept protons from $\ce{HI}$, so $\ce{I-}$ (iodide ion ) should be formed and nitric acid on accepting proton would form $\ce{H2NO3+}$ but that is not formed why?

http://chemistry.stackexchange.com/questions/600/why-is-nitric-acid-such-a-strong-oxidizing-agent

This is not about acidic strength and protons at all. This is a redox reaction.

@IvanNeretin  in an weak acid strong acid reaction products are formed such that weak acid gets proton and strong acid looses proton but in redox reaction what is rule of thumb if reduction potentials are not known?

Nobody remembers the exact potentials, but you are expected to know that $\ce{HNO3}$ is a pretty strong oxidant, while $\ce I$ is an element with relatively low electronegativity (for a non-metal, that is). That should suffice.

@IvanNeretin I2 formed will react with HNO3 to form hio3 and as seen in image in question this Hio3 will again produce I2 so where does overall reaction leads us to?

That depends on how much $\ce{HI}$ do you have.

@IvanNeretin can you please explain why image says iodination is reversible?

Because it is reversible; what's unclear about that? The C-I bond is not particularly strong and snaps easily.

@IvanNeretin could you please answer this: chemistry.stackexchange.com/questions/57714/selectivity-of-catalysts  and thanks for above.

Let us continue this discussion in chat.

Nitrate is a strong oxidant which oxidizes the iodide to iodine.
$$\ce{NO_3^- + 2I^- + 2H^+\rightarrow NO_2^- + I_2 \uparrow + H2O}$$
Note that the oxidation number of the nitrogen atom in $\ce{NO3-}$ is $+V$ and in $\ce{NO2-}$ it's $+III$, so over all, we have a reduction equation of:
$$\ce{NO_3^- + 2e^- + 2H^+\rightarrow NO_2^- + H2O}$$
On the oxidation side we want to form $\ce{I_2}$ out of $\ce{I^-}$, so the oxidation equation is:
$$\ce{2I^- \rightarrow I_2 + 2e^-}$$

Actually, I don't believe it will stop at I2. Also, pay attention to the formatting and use \ce{}.

In parts it will react further, but test it and you'll see, that a violet vapour ($\ce{I2}$) comes out of the beaker, so you definitely form iodine.

No, iodine is oxidized only by concentrated (fuming) nitric acid.

Well, I have made a different experience during my internship. When adding an iodide solution to an unknown, nitrate-containing substance violet vapors came out of my beaker, thus iodide was definitely oxidized to iodine. And that unknown solution was definitely not fuming $\ce{HNO3}$.

@Chwebo I mean that elemental iodine is difficult to oxidize with diluted nitric acid. Iodide ions are on the other hand easily oxidized.

@vapid Oh okay, then I misunderstood you. I've just briefly looked at my table works and yes to oxidize iodine further you would need at least fuming $\ce{HNO3}$. For the questioner therefore only elemental iodine remains.

Agreed. Diluted $\ce{HNO3}$ will give $\ce{I2}$, concentrated acid will push further.


text:
How to know the ionization power of an unknown metal? - difficult
Imagine having the 4 following beakers:

where P, Q, R and S are metallic plates. In beaker 1 and beaker 2 nothing happens. However in beaker 3 and 4 the metalic pieces become thinner.
How would you orden the ions $P^{p+}, Q^{q+}, s^{s+}, R^{r+}$ the following ions by oxidizing power?
I think it is one of the following possibilities:

$P^{p+} < Q^{q+} < s^{s+} < R^{r+}$

I think that P and Q clearly have the highest oxidizing power as in beaker 3 and 4 the metallic plates become thinner. But I have no idea whether R has a higher oxidizing power than S. Or is it the other way around?
Could someone help me and explain how I can order these ions by oxidizing power?

Hint: pay attention to the solution in the beaker

@getafix it's not that I don't want to think about it or so. But I have already posted this twice. Have never received an answer untill now. I have to understand this for tomorrow. Would you mind explaining please?

I guess you made an attempt to answer your question, so it complies with the homework policy on Chemistry.SE. I hope I'm not doing your homework for you

I am guessing you know when a metal bar is becoming thinner, the metal is being oxidised, i.e $$ \ce{M_{(s)}} \rightarrow \ce{M^{n+}_{(aq)}}$$
$P^{p+}, Q^{q+}, s^{s+}, R^{r+}$
Let's start with beaker 3 and beaker 4
In beaker 3, you have $\ce{S_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and you observer the metal bar becomes thinner. Which means, $\ce{S_{(s)}}$ is being oxidised by $\ce{R^{r+}_{(aq)}}$, or in other words, the reduction potential of $\ce{R^{r+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{S^{s+}_{(aq)}}$
$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}}$$
In beaker 4, you have $\ce{Q_{(s)}}$ in $\ce{S^{s+}_{(aq)}}$, and again the same reasoning applies.
$\ce{Q_{(s)}}$ is being oxidised by $\ce{S^{s+}_{(aq)}}$, or in other words, the reduction potential of $\ce{S^{s+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{Q^{q+}_{(aq)}}$
$$\ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$
Thus, so far we have:
$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$
Let's look at beaker 1 and 2
In beaker 1 you have $\ce{P_{(s)}}$ in $\ce{Q^{q+}_{(aq)}}$, and oxidation doesn't take place. Which means the reduction potential of of $\ce{Q^{q+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$
$$\ce{P^{p+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$
And similarly, in beaker 2 you have $\ce{P_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and nothing happens. Which means that the Which means the reduction potential of of $\ce{R^{r+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$
$$\ce{P^{p+}_{(aq)}} > \ce{R^{r+}_{(aq)}}$$
So the final order becomes (in terms of reduction potential, or oxidising ability):
$$ \ce{P^{p+}_{(aq)}} >\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

Thank you so much! I have been thinking about it quite some time! Just one clarification. If the reduction potential of the solids in beaker 1 is higher than the liquid. Why doesn't the metal piece become thicker/bigger?

https://en.wikipedia.org/wiki/Electroplating
You need to supply electrons for that to happen.


text:
Effect of temperature in the free radical polymerization of acrylic acid
I am working with the model of a free radical polymerization process in a tubular reactor. In the simulations, I observe that by keeping the feed flow of the initiator (sodium persulfate) and the monomer (acrylic acid) constant and just by increasing the jacket temperature, the concentration of the monomer (for example of the product) increases, even though the reaction is exothermic.
I have used the method of moments to model the chain length distribution of polymers. Change of viscosity has not been modeled.
My question is: Is the increase in the monmoer concentration by the increase of temperature, an expected behavior in polymerization? If so, do you know some literature that I can learn more about this?

Can you please make clear what kind of simulation this is?

Actually, I am working on the design of a model-based controller for this. The controller works properly and there is no problem in that part. But in the simulation of the process, I observe the above mentioned phenomena which seems a bit wierd for me.

You are not explaining what your simulation takes into account and how it works. The way it is, this question should get closed. There is no way to answer it.

So you need those kind of details...
The model consists of 8 pdes, which describe the evolution of following states: Initiator and Monomer concentrations, zeroth, first and second moment of active polymers, first and second moment of inactive polymers and temperature.
The reation rate coefficients are modeled by Arhenius approach.
Viscosity is not taken into account. No fouling is considered.
If you need to know some other details, I can provide.

So it is a simple chemical reaction kinetics model. Please put that in the question. And what do you mean by "conversion reduces"? The rate? The final absolute conversion after some end condition you impose is reached? By making a secret out of essential parts of your problem, you waste everybodys time!

Yes, this is expected.
With higher temperature, there is a larger concentration of radicals in the polymerization mixture, and initially the polymerization goes faster than at lower temperatures.  This leads to a faster increase in viscosity, which slows the ability of molecules to diffuse in solution.  This means it is more difficult for unreacted monomer to find growing radicals, and for unreacted initiator to find monomer.  Radicals are highly reactive so when they can't react the have a habit of engaging in termination reactions which are nonproductive regarding conversion.
This is normally called autoacceleration, and is fairly well described in Odian's 'Principles of Polymerization'.  Generally, you do not need to heat radical polymerizations beyond the bare minimum to decompose your thermal initiator.

Thanks a lot for the response and clear explanations.
The thing which I still do not understand is that, in the simulations, the concentration of the monomer, under the conditions which I have pointed in the question, increases. If by increasing the temperature, the concentration of the monomer would remain constant, it would be more understandable for me. Can you imagine that this behavior is correct?
Thanks in advance,

You are going much too fast. At higher temperature, the initiator decays much more rapidly, increasing the radical concentration and thereby also the rate of recombination. The radical chain reaction itself is not so very much influenced by the temperature (being exothermic, it actually slows down at some point). The result is leftover, unreacted monomer, and very few radicals left to work on them. What you describe is the "Trommsdorff-Norrish effect", btw. I'll doubt that this "simulation" takes viscosity into account. Does it?

Thanks @Karl for the response. Actually, we do not consider viscosity into account in the simulations. But as a fact, we know that viscosity increases as the polymerization progresses.


text:
Hill's system alphabetizes elements symbol or name?
Take monosodium glutamate.
Its chemical formula is $$\ce{C5H8NO4Na or   C5H8NNaO4}$$
First way nitrogen, oxygen, sodium, with is in order of element's name.
Second way, N, Na, O, in order of element's symbol.
Have seen it both ways, hoping someone could clarify. Thanks.

Organic Chemists have a bias for C, H, N and O, so by habit I tend to write those ones first.  I don't actually know the formal rule.

Go straight to the original source, ON A SYSTEM OF INDEXING CHEMICAL LITERATURE; ADOPTED BY THE CLASSIFICATION DIVISION OF THE U. S. PATENT OFFICE, and you can see that Hill uses the element symbol to determine alphabetical order.

In your example $\ce{C5H8NNaO4}$ is correct.

The current IUPAC Red Book (2005) also refers to the original Hill publication. However, IUPAC recommends the order ‘B before C’ for carbaboranes as an exception to the Hill order.


text:
Why do ligands approach along the axes?
According to the Crystal Field Theory (CFT), it is given that ligands (their lone pairs) approach the central metal along the axes. When they approach, the degeneracy of the $\mathrm{d}$ orbitals is lost. The explanation:
The $\mathrm{d}_{x^2-y^2}$ and the $\mathrm{d}_{z^2}$ orbitals(the $\mathrm{e_g}$ orbitals) are oriented along the axes while others are oriented in between the axes. In octahedral geometry having complexes, when the ligands approach, they approach along the axes and thus the $\mathrm{e_g}$ orbitals which are along the axes experience more repulsion. Hence, these orbitals increase in energy more than the $\mathrm{t_{2g}}$ orbitals. Thus, the orbitals split in energy due to the approach of ligands.
My question is  why do the  ligands (their lone pairs) have to approach along the axes. They can approach from anywhere.

You basically said it.

They can approach from anywhere.

Well, I can define the axes to be anywhere.
An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. Hence, the $\mathrm{d}_{z^2}$ orbital does not have the ‘orientation’ you know from pictures — it only gets that once in some way axes are defined. You could now go ahead and define that one of your octahedral ligand approaches the central metal (placed at $(0,0,0)$) from the coordinates $(1,1,1)$. Get out a large sheet of paper and a new pencil and do the maths, calculating what the final orbital distribution will be. The maths will be a lot more complicated, but the result will be the same. Hence, chemists decide to not do complicated maths and just have octahedral ligands approach along the coordinate axes.
Note that in tetrahedral complexes, the ligands would indeed approach from $(1,1,1)$ and the symmetry-equivalent positions of a tetrahedron.

I get it. But until the ligands approach, don't the $d$ orbitals exist in specific orientations?

No, they don't.

@IvanNeretin Could you please explain why not? Are the dx2-y2 orbitals not uniquely identifiable from, say, the dxy orbitals, until a ligands approach?

@OneEyedMushroom This is already not the case for p orbitals. In fact, what we often think of as $\mathrm p_x$ and $\mathrm p_y$ would be a pair of complex-number orbitals $\mathrm p_{-1}$ and $\mathrm p_{+1}$ in isolation. Maybe there is a chance of identifying $\mathrm{p_0}$ (also known as $\mathrm p_z$) because of its quantum state but this would be outside of my knowledge of maths and quantum mechanics.

More fundamentally, the choice of p-orbital basis is arbitrary, because all three p-orbitals are eigenvalues of the Hamiltonian with the same energy. Consequently, any linear combination of p-orbitals, $c_x\mathrm{p}_x + c_y\mathrm{p}_y + c_z\mathrm{p}_z$, is also a perfectly valid 'orbital' with $n = 1$ and $\ell = 1$. The p-orbitals $\mathrm{p_x}$ (i.e. the choice of coefficients $(c_x, c_y, c_z) = (1, 0, 0)$), etc. are entirely arbitrary choices, and the fact that we typically choose them to be along interatomic axes is purely out of convenience.


text:
Why does diamagnetic current induce a downfield while the paramagnetic current induces an upfield shift?
I know that aromatic rings exhibit diamagnetic ring currents which causes the protons outside the ring to go downfield in H-NMR. Antiaromatic compounds exhibit paramagnetic ring currents which have the opposite effect. How does the difference in two electrons between aromatic and antiaromatic systems cause such a dramatic shift in the properties of the ring current? Why does diamagnetic current induce a downfield while the paramagnetic current induces an upfield shift?
I've found this question: Ring currents of anti-aromatic and aromatic systems but the answer doesn't address how the different currents give rise to different effects in H-NMR.

The second answer http://chemistry.stackexchange.com/a/57413/17952 to the linked question addresses exactly that.

What is unrealistic about it?  This is what molecular orbital theory predicts and the resulting diamagnetic/paramagnetic ring currents match the prediction.  If you study other magnetic systems, you always find the situation that diamagnetic substances have all electrons paired up (indeed, superconductors are perfectly diamagnetic, magn. susc. = -1, due to the cooper pairs) while all paramagnetic substances have unpaired electrons.

Because of Jan-Teller effect the electrons are paired in cyclobutadiene.

Good point!  (I clearly answered before thinking clearly about the question).  I would write off any discussion of anti-aromatic ring currents as fictitious/theoretical since you can't create a stable anti-aromatic system.  It's expected to behave a certain way, but kind of a pointless discussion since it can't be observed.  Although the ring-current model is a simple/approximate model, it does work quite well: many beautiful examples of very negative chemical shifts for protons dangling above a benzene ring or on the inside of annulenes, etc. vs. those around the perimeter of the ring.


text:
Is methanol really more acidic than water?
The question Why is methanol more acidic than water? deals with the reasoning of why methanol is more acidic than water. However, as mentioned in the comments of that question, the acidity constant of water is $14.0$, as confirmed by two sources$^{[1][2]}$, with one of them offering a very convincing explanation. The Wikipedia page$^{[3]}$ for methanol quotes its acidity constant to be $15.5$.
So it would seem, water is more acidic than methanol. But an answer to the question mentioned above gives a good reasoning for why methanol is more acidic than water (it doesn't deal with the pKa of water, though). Also, the answer mentions that water would be much worse an acid in DMSO. Another question implies that methanol and other alcohols are more acidic than water.
On the other hand, it is also true that we regard alkoxide ions as being more strongly basic than the hydroxide ion (thus we use them primarily for elimination reactions).
This leaves us in a fix. So what is the ultimate truth?

    [1]: What is pKa of water? - Chemistry ChemLibreTexts
    [2]: Water (Data Page)
    [3]: Methanol

It hardly makes sense to compare the pKa of methanol in water with that of water in water. I've heard this question before, and to me it is like splitting hair.

So the super uber ultimate answer is: it depends

@Karl But the question I have linked to, has compared the acidities of alcohols in aqueous solutions with water itself! And the same answer has indeed used pKa values to compare their acidities. My concern is they might be using the wrong pKa for water.

@Mithoron I already realised that when I read that question I have mentioned. But right now, I just want to compare acidities of their aqueous solutions!

I'm sorry, but this discussion is nonsense, imho. You can formally run the calculation that gives the pKa of something dissolved in water also for pure water, but the result has no sensible meaning at all.

@Karl So how would you compare the acidities of water and methanol? And then do you consider the concept of pKa not applicable for water or what? I do not understand your reasoning...I am guessing that you want to calculate the pKa values of water and methanol using equilibrium constants, in some other solvent like DMSO. But that doesn't explain why measuring pKa of water directly, with the pKa of methanol in water.

The concept of pKa is NOT applicable to the solvent, it is about the solute! If your actual question is "can a methoxy anion exist in water", then say so.

$\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$

If you want a clear-cut answer, get rid of all those complicated intermolecular interactions and study the compounds' acidity in the gas phase. The hydroxide anion releases more energy when protonated, compared to a methoxide anion. Thus, excluding all but the most fundamental factors, methanol is more acidic than water.

@NicolauSakerNeto  Why not instead compare the pure substances in the liquid phase, since that's the more common application?  I.e., why not compare the compare the $pK_a$ for pure liquid MeOH with that for pure liquid water?  I.e., compare the $pK_a$ for $\ce{CH3OH_{(l)} <=> CH3O^-{(l)} + H^+{(l)}}$ to that for $\ce{H2O_{(l)} <=> OH^-{(l)} + H^+{(l)}}$.  If the former is measurable, the complications from the interactions aren't an issue—they're subsumed into the measurement. It's easy enough to find the $pK_a$ for MeOH in water.  But I've not been able to find a $pK_a$ for pure MeOH.

@theorist There is some confusion regarding whether the equilibria you mention are represented by the acid dissociation or autoionization constants, which has been discussed several times across multiple Chem.SE posts, though it seems they are, in fact, one and the same.

(cont.) You could certainly compare them (the autoionization constant of MeOH at 25 °C is $10^{-16.6}$ ref), though there is room for ambiguity - what if MeOH is actually a stronger acid than water, but a much weaker base? That would also lead to a smaller autoionization constant. It would also be interesting to compare the $pK_a$ of water in methanol and vice-versa.

First, water’s $\mathrm{p}K_\mathrm{a}$ in water is $14$ as explained here. This means that $\ce{H2O}$ is slightly dissociated in liquid form, such that $[\ce{H+}] = [\ce{HO-}] = 10^{-7}$.
Second, I would say that caring too much about which compound is the best acid is a bit like arguing by definition. Acidity is a fuzzy concept used as a shorthand for deeper meaning. This means there is little point arguing about the shorthand when we have the actual properties it synthesizes.
In this case (sou rce):
$$
\begin{array} {lrrl}
\hline
\text{Solvent} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{MeOH}) & \mathrm{Interpretation} \\
\hline
\ce{H2O} & 14.0 & 15.5 & \text{Water is more dissociated than methanol} \\
\text{DMSO} & 31.4 & 29.0 & \text{Methanol is more dissociated than water}\\
\hline
\end{array}
$$
So you can argue (as in the answer you linked) that methanol is more acidic than water in the abstract sense, because $\ce{DMSO}$ doesn’t stabilize the anions by hydrogen bonding and so is closer to the “ideal” case (whatever that may be).
You already have the $\mathrm{p}K_\mathrm{a}$ values, any practical question you might have about the concentrations of theses solutes in $\ce{H2O}$ and $\text{DMSO}$ is already answerable.

Thank you for your answer! Just one more thing, can I then assume the methoxide ion to be a stronger base than hydroxide ion, (in DMSO)? I was intending to use this answer for that purpose also, but I realise that maybe that would be too simplistic!

No, it's the other way around. Methanol being more dissociated means that the methoxide is a weaker base than the hydroxide (in DMSO). In water the trend is reversed because of better hydrogen bonding with the smaller hydroxide ion.

In water, methanol is more acidic than water
The apparent contradiction comes from using a different standard state for water (activity of pure water defined as 1) and methanol (activity of approximately 1 mol/L defined as 1). So it is inappropriate to directly compare equilibrium constants or $\mathrm{p}K_\mathrm{a}$ values if in one system, a species is the solvent and in another one it is not.
Instead, consider the concentrations at a pH of 14 (e.g. roughly 1 mol/L $\ce{NaOH}$). At this pH, the ratio of water to hydroxide is 55:1, or roughly 2% dissociated. We can calculate the ratio of methanol to methoxide using the Henderson-Hasselbalch relationship (or straight from the equilibrium constant expression). It comes out to about 30:1, or roughly 3% dissociated. So methanol is the stronger acid because at a given pH, the ratio of deprotonated species to protonated species is higher. You could do the same at any other pH with the same conclusion that methanol is the stronger acid with water as the solvent.
Using $\mathrm{p}K_\mathrm{a}$ values for comparison
If you try to use the Henderson-Hasselbalch equation for water, you have to plug in "1" for the activity of water because that is what was used in the definition of the $\mathrm{p}K_\mathrm{a}$. If you do that, you find that the ratio of activities is 1:1, corresponding to 1 mol/L hydroxide and almost pure water, i.e. 55 mol/L. This is all internally consistent, but to compare it to methanol without comparing apples to oranges, it is appropriate to compare the degree of dissociation as done above.
In SCH's answer, the claim is made that in aqueous solution, "water is more dissociated than methanol". If you look at the degree of dissociation (mol fraction of undissociated vs dissociated) as I did above, this statement is incorrect. What is correct is that, using the more logical value of $\mathrm{p}K_\mathrm{a} = 14$ for water, the $\mathrm{p}K_\mathrm{a}$ of water is lower than that of methanol.
Solvent-dependence

[OP] ...we regard alkoxide ions as being more strongly basic than the hydroxide ion...

This statement was made in the context of organic chemistry. In this context, you will have solvents other than water. So you could compare methanol's and water's acidity in e.g. acetonitrile, DMSO or methanol or in solvent mixtures. This makes things complicated (see @Mithoron in the comments: "It depends"), and you would have to limit your question to a specific situation. The blanket statement that alkoxide ions are (always) more strongly basic than hydroxide ions is incorrect.

It seems the OP was curious about the acidity of methanol in water, which you addressed.  But if I myself were to compare the acidity of methanol and water, I think I'd want to compare the $pK_a$ for pure methanol with that for pure water.  It's easy enough to find the $pK_a$ for methanol in water—for a dilute soution, it's 15.5.  But I've not been able to find a $pK_a$ for pure methanol, i.e., for $\ce{CH3OH_{(l)} <=> CH3O^-{(l)} + H^+{(l)}}$

@theorist Not methanol but ethanol, and just the autoprotolysis constant:     DOI: 10.1016/S1658-3655(12)60002-8


text:
Does a messier room actually have higher entropy?
When my chemistry teacher was explaining entropy, she used the analogy of a clean vs. messy room. What I'm wondering is whether a messy room really does have higher entropy. Does this still work on a larger scale? If not, can I trust the analogy?

Rooms don't have entropy. Other than that, the analogy is good.

Calculating the entropy of a small number of objects is not very meaningful, neither is calculating the entropy of an assortment of macroscopic objects.You cannot predict anything sensible from the result.

Fun fact: Every letter in a common sentence contains roundabout one bit of information. Is the rest noise (=entropy), or is it structure, predetermined by the information?

A room can be a complicated system, (at least mine can be) so I shall assume a simplified version: an enclosed space bound by 4 walls, a ceiling and a floor and the only piece of furniture inside the room is a desk, and  two books on it -- book 1 and book 2.
In the this scenario, I would define a clean room as one with both the books neatly stacked on the desk, and in a messy room you can find one of the books on the floor.
Now, if I were to try enumerate the number of ways of organising the books in a clean and a messy room, I would find that there is exactly one way in which I can achieve a "clean room configuration", and that is by having both books on the desk. On the other hand if you wish to achieve a "messy room configuration", you can do so in two ways-- either have book 1 on the floor and book 2 on the desk, or have book 2 on the floor and book 1 on the desk.
The no. of ways in which you can achieve a configuration is called its weight and is defined as:
$$W = \frac{N!}{N_0!N_1!N_2!....}$$
In the first case, $W=1$ (clean) and in the second $W=2$ (messy)
The boltzmann formula for entropy is: $ S = k_b\ln W$
Plugging in $W$, we get $$S_\text{clean} = 0$$ and $$S_\text{messy} = k_b\ln2$$
Thus, the messier room has a higher entropy. This somewhat "silly" argument can be reproduced for a room with more furniture, and more items in it. Feel free to toss in some clothes, papers, chairs, book cases, toys, shoes, pets...go nuts.
So, I believe the analogy holds good. But again, what I did here is a silly exercise to illustrate a point, and it isn't meaningful as such.
If you wish to learn more, watch this lecture by Leonard Susskind

This is a tricky question. The answer is no.
From Wikipedia

In thermodynamics, entropy (usual symbol S) is a measure of the number of microscopic configurations Ω that correspond to a thermodynamic system in a state specified by certain macroscopic variables.

It can not be defined the messy and clean states thermodynamically. (Which would be the reasonable thermodynamic variables?) If it would be possible, after further assumptions, the problem is reduced to the @getafix answer. Well, it is also needed that all the book remains stacked in the desk (all of them with the same propertied but distinguishable).
It is more important to note that the non technical concept  of messy is not related to entropy.
EDIT:
I have decided to make it explicit (although at first I though it was) as the content of the answer was qualified as arbitrary and it was down-voted.
The OP said:

What I'm wondering is whether a messy room really does have higher entropy

The other questions turns to be meaning less if we can not assert that the messy room has higher entropy.

It is clear from the context ("When my chemistry teacher was explaining entropy...") that we are talking about entropy in thermodynamics.
It is also clear from the context that we are talking at an introductory level.
In order to compare the entropy of two systems, it is required that the states of both systems can be defined in terms of state variables.
The system state variables can not be established for messy nor clean rooms.
As the system states can not be specified, then it is not true that a messy room has higher entropy.

The usage of analogies is not just matter of taste. At first, there is a goal when an analogy is used while teaching a concept $X$, clearly it should be to help the students to learn $X$. In order to describe $X$, a well known $Y$ is used. $X$ and $Y$ must share a common set of properties. Aided by $Y$, some features of $X$ are inferred by inductive reasoning.
It is not evident which are the common features that links entropy with the level of cleanliness of a room, so a valid analogy can not be built. Of course there could be established some analogy when a lot of considerations are taking into account. But, if the student have reached a conceptual level to understand "why" they are made, then, the analogy will be of little use. If they did not, take for sure that they won't understand the concept (because they won't know when they can trust the analogy).

Does this still work on a larger scale? If not, can I trust the analogy?

I think I was no clear when I said "no". I really should have said "sometimes, entropy has nothing to do with cleanliness".

It's your wilful decision to drive the analogy to the point where it breaks. I don't like it very much either. But saying it is just wrong is, well, arbitrary.

@Karl , I expanded my answer in order to address the points of our comments. Thanks for sharing your thoughts. It help to improve the contents and clarity of the answers.

"whether a messy room really does have higher entropy".

One messy room left untouched does not have a higher entropy than a clean one.  Both represent one way of arranging the things in the room (a microstate), so the entropy is zero for both.
If you have a messy person living in the room then, yes, over time the messy room will have a high entropy.  Things will move around but never be in the same place (many microstates), so you will have many different arrangements of things that always will be characterised as "messy" (macrostate).
A collection of several messy rooms, e.g. identically furnished but unmade rooms in a hotel, has a higher entropy than the same collection of rooms after house keeping has been to them all and arranged the things in exactly the same way.

+1 You touched an interesting point where an analogy can be drawn, if the internal structure of macroscopic objects is not considered. I would like to read "ergodic" scattered somewhere in point 3, to turn the statement a little more accurate.


text:
How long would it take for liquid oxygen to evaporate at room temperature?
I watched a video on youtube in making liquid oxygen. How long would it take for liquid oxygen or liquid of another gas (e.g. nitrogen or hydrogen) to evaporate and why would it not be instant? Could one touch liquid oxygen without getting hypothermia? In making oxygen a solid or even metallic oxygen with enough pressure and low temperature, if one was to remove the pressure and leave the solid oxygen on a counter, would it evaporate instantly?
More generically, if one was to apply enough pressure to a gas to make it a solid at room temperature and then remove the pressure, would it stay in the solid form?

Here is a bit of fun with solid argon in ambient conditions.

Your post is too broad - you ask many different questions. At least cutting out this (bad) hypothermia question might help. This signifies deeper problem - it looks you don't get kinetics of phase transitions and specific examples muddle it.

As already pointed out, thermodynamics tells us about the feasibility of processes whereas, kinetics tell us about their speeds. Also, the phase transition solid to vapour is called sublimation, and can happen below the triple point of the substance.

Instantly? Just like if you go outdoors when the temperature is below zero (centigrade) you freeze instantly? Are you from a location where temperatures never drop below freezing and simultaneously unaware that people live in places where the temperature is below freezing? Otherwise, I can't understand your confusion. Given perfect insulation (which doesn't exist) a material can theoretically maintain its temperature indefinitely. With real world insulation colder materials will slowly warm up, but the rate its temperature increase depends on the quality of insulation. A thermos of a hot liquid can still burn skin many hours after it has been filled. All substance have some ability to insulate, so a liquid or gas can insulate itself (that is can insulate material further away from the heat source). In beginning physical science, you should have learned that a material's temperature is an indication of its energy content. You should also have learned that energy can be neither created nor destroyed; which means that for a material to warm up, energy to do that has to come from somewhere. In terms of heat flow, there are three types of flow: Conduction, convection and radiation. A cup of lox (liquid oxygen) on a table, say, will mostly be warmed by conduction. (Unless you place it under intense lighting or near some other energy source.) As the lox heats up, the table cools down, meaning it will take longer and longer for heat to move into the lox from further away-heat conductivity is slow compared to the speed of light, speed of sound, or even a speeding bullet. (Depending on the shape of the cup, the air above the lox might also supply heat via turbulent flow) One good example of heat flow differences is when you walk barefoot over carpet, it feels warmer that walking over ceramic, wood, or metal even though they are at the same temperature! (or if the surface is hotter than your skin, the carpet will feel cooler than the other surfaces) The heat conductivity (not really related to electrical conductivity!) is faster in metal, and slower in carpet (metal>ceramic>wood) (plastics vary because their composition can be quite different, but organics are generally slower than metals and ceramics (including rock)). As far as what happens to a solid under pressure when the pressure is relieved, there is no single answer. It depends on the strength of the material. If the material is weaker than the forces of expansion, then it will puff up a little and this will probably turn a solid piece into a pile of powder. If the material is strong enough, it won't change its shape much until it begins to warm up. I don't know what the tensile strength is of solid oxygen. Note that compression heats a material and expansion COOLS a material (in general), so releasing the external pressure on an object will tend to cool it. As a material warms, its dimensions change. Perhaps you've seen the failure of a glass container when it is subject to too much temperature change too quickly? It cracks. Stress may cause the material to, drum roll...crack. Finally, for a gas (or liquid) which is above the temperature and pressure at which it exists (at equilibrium) as a gas (or liquid) but has been compressed to a solid, then releasing the pressure will allow mechanical stresses to be relieved and at the same time will cool the solid. The forces involved in containing a material as a solid when it is at a temperature at which is would normally be a gas are quite high (usually). How fast these forces are released depends not only on the material but on the amount of material. Usually the  compression required means the sample is really small, fractions of a gram. With such small samples, the phase change will be very rapid (since conduction depends on the surface area, and small samples have high surface area per unit mass. LNG (liquified natural gas) is a counter example. In many parts of the world, tanks of LNG can be seen almost anywhere. A (small) opening will allow the liquid to slowly gassify. If a large opening were provided, I don't know how fast a tank's content would evaporate; the evaporation front would be exerting a compressive force on the underlying liquid, and along with the cooling due to expansion, the material would PROBABLY not evaporate "instantly". Given that a material moving by a surface will tend to create static electricity, and given the explosive nature of alkane/air mixtures, that's NOT an experiment I'd want to try. Finally, note that a material which doesn't decompose as it is heated eventually reaches a temperature, called its critical temperature, above which the liquid state doesn't exist - a solid above this temp will sublime as it changes phase. Throwing lox into the air vastly increases its surface area, and allows the small droplets to rapidly warm (and turn into gas).

Excellent explanation! Regarding the walking outside analogy, I guess I assumed that if the temperature was cold enough say -273 C (close to absolute zero), that phase transition would be instant (in this case, I would instantly freeze) and applied that thinking to if temperature was hot enough for a solid that it would instantly sublime.

@jth_92 Yes, it still wouldn't be instant. The basic problem is still there - you need to move that heat. Think about what happens in space - even though the temperature is very close to absolute zero (about 3 K in the shade), it'd probably take days for something like a human to freeze - and only if you were dead. A living human actually produces more heat than it would radiate - you'd cool a bit at first (through evaporation) and then you'd be cooked alive, despite the frigid ambient temperature :) There's a reason we use phase change for cooling - it can move lots of heat very quickly.

@jth_92 And note that if you took a normal Earth room, perfectly isolated it and cooled it to near absolute zero, it would also be essentially a vacuum - all the air would condense into either a liquid or all the way to solid. If you avoided contact with the room itself (tricky when not in free fall :)), you wouldn't freeze at all before you died.


text:
What type of property is molar mass?
This maybe a really simple question, but I've been trying to find the answer and it's getting kind of hard.
Is molar mass a physical or chemical property?
My teacher says it's a chemical one, but I don't think it matches the definition for chemical properties (there's a chemical reaction or a change in the composition of the substance).

Molar mass is the mass of the chemical substance per unit mole. Or in other words, mass of one mole of a substance expressed in grams/mole is molar mass. It's the mass of 6.023x 10^23 unit particles(atoms/compounds/ions) in a chemical substance.
As it's at molar level, I don't think there's any direct way of measuring it. So, it may account as a chemical property.

Molar mass is simply the atomic mass expressed in grams/mol numerically,and atomic mass is actually a comparison of the mass of an element with c-12 isotope,we are not measuring it here but are comparing it with another element and to be a physical quantity a parameter needs to be measurable.Now imagine that atomic mass is a comparison and if you take a mole of the mass of a substance we are taking a collection of a lot of atomic masses which is not a physical quantity.therefore molar mass should also be a chemical quantity.i guess you should also refer to your teacher who would explain it to you in detail but i hope this should help you.:>


text:
Allred-Rochow vs Pauling Electronegativity scale
While observing the trends in electronegativity of the elements, I found a discrepancy between Allred-Rochow and Pauling scales. In Pauling's scale, chlorine is the third most electronegative element ($\ce{F > O > Cl > N}$). In Allred-Rochow's, it is the fourth ($\ce{F > O > N > Cl}$).
While of course the numeric values of the two scales are different, I thought that the rank of the elements would be the same, being electronegativity related to the properties of the atoms (for instance, the partial charge of molecules such as nitrogen trichloride relies on the electronegativity).
So, I was wondering: which scale is more accurate? And, if one is wrong, why are both scales still in use?

Both scales define electronegativity differently, perhaps that's why? Just venturing a guess. I wouldn't say either is wrong.

Yes, they are different things. There's no inherent "electronegativity" just artificial parameters, which people overestimate.

Neither scale is 'more accurate'. As The_Vinz said, each scale may be applied in different areas, and there is a moderately good correlation between the two scales once conversion factors are introduced.
Firstly, let's use the equations for the two scales to explain why the halogen electronegativity orders are different:
Pauling electronegativity is essentially based on the idea that an atom's electronegativity can be defined by the degree to which it polarizes bonds. Pauling's theorem states that the difference between the experimental bond dissociation enthalpy (BDE) of two atoms and the geometric mean of the BDE of each of those atoms' covalent homonuclear bonds is proportional to the two atoms' difference in electronegativity:
$$\Delta=|\chi_{\ce{A}}-\chi_{\ce{B}}|\propto|BDE_{\ce{A-B}}-\sqrt{D_{\ce{A-A}}\cdot D_{\ce{B-B}}}|$$
Essentially, this is a measure of the ionic character of a bond. The geometric mean term in the above equation represents a hypothetical nonpolar heteronuclear bond, so the difference between that BDE and the experimental BDE of the polar heteronuclear bond describes the degree to which the bond is polarized- its ionic character. A scaling factor is introduced and everything is measured with respect to fluorine's Pauling electronegativity value, which is arbitrarily set to 4.0:
$$|\chi_{\ce{A}}-\chi_{\ce{B}}|=0.102 \sqrt {\Delta}$$
So for chlorine (compared to fluorine): (1)
$$\Delta=|250.548-\sqrt{242.58016 \cdot 156.996}$$
$$|\chi_{\ce{F}}-\chi_{\ce{Cl}}|=0.102 \sqrt {55.396}=0.7596$$
$$\chi_{\ce{Cl}} \approx 3.2404>3.04=\chi_{\ce{N}}$$
(mileage may vary with these calculations)
Allred-Rochow (AR) electronegativity, on the other hand, is defined directly by the force resisting removal of an atom's valence electron. This force is equivalent to:
$$F=\frac {(Z^*e)(e)}{4\pi r^2\epsilon_o}$$
where $Z^*$ is the effective nuclear charge, either measured experimentally or calculated using Slater's rules. Introducing some more scaling factors, which make things line up with Pauling electronegativity values, the equation for Allred-Rochow electronegativity is:
$$\chi_\text{AR}=0.359\left(\frac{Z^*}{r^2}\right)+0.744$$
where r is the atom's covalent radius, in Angstroms. $\chi_{\text{AR},\mathrm{Cl}}=2.83 < \chi_{\text{AR},\mathrm{Cl}}=3.07$[2]
The truth is probably somewhere in between. The two scales are often used interchangeably, or a combination of the two values is used. Originally, the advantages had to do with the experimental basis. Bond dissociation energies may be easier to measure than exact effective nuclear charge or ionic radii. Based on their definitions, it seems to me that AR electronegativity would be more useful for looking at physical properties of an atom in a molecule, like ionization. Pauling electronegativity is really a measure of a bond's polarity.
In the case of day to day estimates about polar bonds in organic chemistry, I think Pauling electronegativity is more accurate. Ultimately, though, these are all quantum mechanical phenomena and exact values must be calculated using quantum methods that implicitly "understand" the real phenomenon that is electronegativity.
In the case of nitrogen and chlorine, though, Pauling electronegativity is closer to the "real" situation. For example, B3LYP/6-31G** calculations shown that nitrogen trichloride has a net dipole of -0.584 Debye, pointing towards the trichloride base of the trigonal pyramid[3], so the $\ce{N-Cl}$ bond is indeed polarized in the $\ce{Cl}$ direction as Pauling electronegativity seems to imply.

References:

Properties of Atoms, Radicals, and Bonds: Table 4.11 (https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf)
Gong-Du Zhou, Fundamentals of Structural Chemistry, World Scientific, Singapore, Malaysia, p. 85
Computational Chemistry Comparison and Benchmark DataBase; CCCBDB: "Calculated electric dipole moments for NCl3 (nitrogen trichloride) at B3LYP/6-31G**" (cccbdb.nist.gov)

Pauling's electronegativity scale is determined by considering the strengths of A-A, A-B, B-B bonds.
Allred-Rochow's electronegativity is determined by considering the charge experienced by an electron on the "surface" of the atom.
While the correlation between the two scales is not perfect, it is quite good, since the scales are calculated with different methods:

The correlation between Allred–Rochow electronegativities (x-axis, in Å−2) and Pauling electronegativities (y-axis).
  By Physchim62 GFDL, CC-BY-SA-3.0, from Wikimedia Commons

Also the partial charge is a model, useful in order to make rapid evaluations on real things. Sometimes, when molecules with almost covalent bonds (such as nitrogen trichloride) are modeled with different electronegativity scales, the partial charge may be different, but the actual electron cloud may not agree with the model.
The real electron density can be then calculated with quantomechanical methods, which are much more accurate and yet complicated than the simple model of electronegativity.

still the answer given confuses me, when I look the value for IVA with catenation. Pb is less in cattnation A-A bond strength but is the second by value given in pauling scale.May you explain this. Thank you.

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


text:
Core cutoff radius and energy cutoff radius in pseudo potential DFT
From what I understand the core cutoff radius is the distance where the pseudo wavefunction and the full electron wavefunction start to resemble each other. In other words, the pseudo wavefunction and full electron wavefunction have the same energy and electron densities for distances larger than the core cutoff radius. The energy cutoff radius is the energy that determines the size of the plane wave basis set.
My question is how they are related. Most computational software control the energy cutoff radius, and the energy cutoff radius is lower for ultrasoft pseudopotentials. My question is why is this so? I understand conceptually that ultrasoft pseudopotentials are smoother. Hence the Fourier series is less complicated. Hence the basis set does not need to be large. However, I have trouble putting this vague (if correct) concept into concrete reasoning. What are relevant equations that directly relate the energy cutoff and core cutoff?

I assume by "energy cutoff radius" you mean plane wave cutoff? And by "core cutoff radius" you mean "core radius"?

The reason pseudopotentials are use is because (at low pressures), core electrons do not participate in chemical bonding or reactions, and more over have high kinetic energies. If you expand and electronic wave function in a plane wave basis, high energy wave functions require high frequency basis functions, therefore in order to capture the wave function of a core electron accurately, you would need a high plane wave cutoff, which is computationally expensive.
To avoid this, the core electrons are replaced by a pseudopotential, which is by construction smooth. The core radius is to some extent a measure of the 'hardness' of the pseudopotential. A soft pseudopotential will generally have a larger core radius, and will be smooth. A hard pseudopotential will have a lower core radius, and in order to capture the rapidly varying (more jagged) potential near the core, will be less smooth, and therefore require higher frequencies in its Fourier transform to recover the correct electronic wave functions. Harder pseudopotential are necessary, for example, at high pressures, when the nuclei are squeezed together --- if the core parts of the pseudopotential overlap, then the calculation will fail catastrophically.

For simplicity consider an atom. If the atom has m energy levels (wavefunction), you can convert this system into n energy levels problem where
m > n by replacing the innermost (core) electrons with an effective potential.
For e.g. Ca has 20 electrons
1s   2e
2s   2e
2p   6e
3s   2e
3p   6e
4s   2e
In pseudopotential you can replace the 1s by an effective potential. So, in this two approach AE (all electron) and PSP (pseudopotential) both describe the same atom where the AE will be used as reference. Hence the energy (eigenvalue) of the AE wavefunction and pseudo wavefunction are same.
There are two types of PSP generally used the Norm Conserving and ultrasoft.
In both the approaches, a cut off radius (Rcut) value is set beyond which it is imposed that the AE and PSP wavefunction exactly fall on top of each other.
Below this cut off radius you can have the pseudo wavefunction in any possible shape you want. However the shape of the wavefunction below Rcut
is optimized such that it is smooth (i.e. you need small basis set to represent the wavefunction). If the norm of the charge within this Rcut coming from pseudo wavefunction is constrained to be equal to charge of AE wvafeunction then it is norm conserving PSP. In case of ultrasoft this constraint is not enforced. The loss in charge due to removing this constraint called augmentation charges is complemented through additional complictated calculations. Consider 3d orbital. It is localised. Hence, in case of norm conserving psp you will need a large basis set and hence higher cutoff compared to ultrasoft where the wavefunction can be made much smoother.
Generally in the DFT codes the cutoff radius is set/fixed in the psp as they can significantly effect the results. The energy cut off entirely depends on what system studied (atoms), what kind of potential used. If the Rcut is increased, one has more freedom to play with the shape of wavefunction and hence make smoth (small basis set -> cut off). In other words it can be correlated to the kinetic energy of the wavefunction. If the KE of the pseudo wavefunction is reduced you need small Ecut.

Thank you for your effort, but I am not sure whether it addresses the question.

Can you un-abbreviate AE and PSP at least once? An outside reader such as myself might not know what they stand for.


text:
Number of electrons in pseudo wavefunction and all electron wavefunction in pseudopotentials
In pseudo potential theory most pseudo wavefunctions are specified to be the valence pseudo wavefunction. These pseudo wavefunctions are often compared with an all-electron wavefunction. My question is whether the all-electron wavefunction is the wavefunction for a valence electron as well. Also is the all-electron wavefunction a one electron wavefunction? Is the valence pseudo wavefunction a one electron wavefunction? Part of my confusion stems from the fact that some journals state that the charge of the pseudo wavefunction and all electron wavefunction have to be the same.

In pseudopotential approach the core electrons are replaced by the pseudopotential. In case of all electron calculation we obtain wavefunction for both core and valence electrons. In case of pseudopotential only the valence electron remains. So when we compare the all- electron and pseudo wavefunction we compare only the valence ones. All the wavefunctions are one electron wavefunction. The charge is basically the norm of the square of the
wavefunction. The condition of charge equivalence between the pseudo and all electron wavefunction is imposed to get correct scattering properties. This condition is known as norm conserving pseudopotential.


text:
Explain!"An orbital of smaller size means there is more chance of finding an electron near the nucleus"
how can orbital of smaller size increase the probability of finding the electron near the nucleus
Please explain in simple language.

What do you mean by the nucleus of an orbital?

i mean the central part of an orbital,we were told in our coaching classes that it is known as the nucleus of an orbital

I'm afraid that as far as I know there is no such thing as the nucleus of an orbital.

oh thank you,i guess the teachers were speaking about it for explanation purpose,but i still do not understand how can smaller size  mean an increased probability of finding electron near the nucleus.

As user getafix pointed out in the comments, the "nucleus of an orbital" (as your post suggests) isn't really a thing. By "nucleus", I think it'd be safe to assume your teacher is referring to the nucleus of the atom/ion. Well, to me, I don't see anything blatantly incorrect, since from what I've learnt (as a high-school student) all orbitals in an atom/ion are centered around the nucleus of the said atom/ion.
Of course there are variations in shape [spherical(s-orbitals), dumbbells (p-orbitals), Double dumbbells (d-orbitals), etc] and size depending on which orbital, sub-shell and shell/energy-level you're considering. As you would already know, an orbital is a 3-D electron density "map" (of sorts), and what it pretty much does is demarcate the most probable locations an electron can be found in at a given instant of time. Chances of finding electrons outside of the orbitals is pretty slim.
Have a look at this diagram on atomic orbitals:

 Source. I found this via a Google Images search. I don't endorse that site though

It should be pretty obvious, that the smaller the size of the orbital, the better your chances are of finding the electron (present in that particular orbital) closer to the nucleus.

Perhaps an analogy might help?
Suppose I have a little plastic box with a light bulb centered inside of it, say a fruit fly also managed to end up in it as well. (Note the picture sans the fly)

Think of the bulb as your nucleus, the fly as the electron, and the box (which restricts the movement of the fly to within the volume of the box) as an orbital. Now naturally the fly will follow a random flight path (obviously within the constraints of the box), however its movement is largely centered around the bulb (take a walk in a nearby park in the evening, and that observation shouldn't surprise you)) as it is drawn to the light (compare it to the nucleus-electron attraction, albeit much weaker).
So at any given instant of time, you have better chances of locating the fly closer to the bulb than anywhere else in the box.
Now as I progressively keep on increasing the size of the box, the fly gets more room to fly out further, and although (for any given size of the box) the best probability of finding the fly is closer to the bulb than anywhere else within the volume of the box.
Even though the probability steadily decreases as the box's size increases, I reiterate, it is still the highest probability (relative to any other position in the same box).
Hope this helped :-)