text:
What is the lattice structure of manganese?
A transition element is defined as the one which has incompletely filled d orbital in its ground state or in any one of its oxidation state. Zinc , cadmium and mercury are not typical transition elements because they have fill filled d orbital in their ground state as well as in their common oxidation state. These elements do not show none of the characteristic properties of transition metals like the lattice structure as shown below in picture. They probably do not have proper metal-metal bonding, a characteristic feature of transition metals. But, still they are studied along with the chemistry of transition metals. But what about manganese? It is a core and important element in transition series. Here, X is unknown lattice structure.

Wikipedia declares zinc to have a distorted hexagonal close packing structure. Mercury is, of course, a liquid at room temperature and does not exhibit crystal structures in liquid form. I couldn’t find anything about cadmium, but since it exists in pure form, it will have more or less defined structural phases.

@Jan, but, wikipedia says cadmium has hexagonal closed  structure. Also see this -  http://www.webelements.com/cadmium/crystal_structure.html

You know that Webelements =/= Wikipedia? ;) Yeah, I could have checked there. But my point is that the elements have a structure even if they have filled sub-shells.

Iron is also slightly off.  Within a limited temperature range it's FCC under ambient pressure.  Shakespeare knew it all along:  "There are more worlds, Horatio, than are dreamt of in [ordinary] philosophy."

Ah, good old Mn. This is one of the uglier ones in terms of phases.
$\alpha$-Mn, the room temperature phase, is CBCC (A12 family), fairly unusual. Near 1000K, Mn transitions to $\beta$-Mn, with a simple cubic (A13 family) crystal structure. $\gamma$-Mn is FCC, face-centered cubic (A1), and $\delta$-Mn is BCC, body-centered cubic (A2).
One excellent reference is A.T. Dinsdale, SGTE Data for Pure Elements, CALPHAD 15(4) 317-425 (1991). There is a readily available PDF of an updated version of that paper that can be found (Dinsdale). This paper gives the accepted Gibbs free energy functions for the elements in their various stable, and some unstable, forms.
Editing to add additional information on the $\alpha$- and $\beta$-Mn crystal phases. Both of these phases are pretty weird, and each is the first observed crystal with their structure (hence they are the 'prototype' in crystallography terms).
$\alpha$-Mn was described by A.J. Bradley and J. Thewlis in 'The Crystal Structure of $\alpha$-Manganese', Proc. Royal Society 115, 456-471 (1927). The unit cell is based on body-centered cubic, but contains 58 atoms representing 4 distinct positions. It can be thought of as a bcc Bravais lattice with a 29 atom basis. Mn is the only element that exhibits this crystal structure.
$\beta$-Mn was described by G.D. Preston in Phil. Mag. 5(33) 1207-1212 (1928). The unit cell is simple cubic, containing 20 atoms in 2 groups. Preston's unit cell drawing is significantly worse than that of Bradley and Thewlis, although representations of the unit cell can be easily found on the web these days. Again, Mn is the only element that exhibits this crystal structure.
The $\delta$-Mn and $\gamma$-Mn are normal, boring bcc and fcc crystals, so no need to point to references for them.
I would provide links to the papers if I could - my institute does not have direct access to them.
As yet further info on the crystal structure references.  The A12, A13, ... notation is the Strukturebericht notation used by crystallographers. Unfortunately, there are a number of other notations used as well, including Schonenflies and Hermann-Mauguin. The Strukturebericht is pretty basic, mostly a catalog of crystal types, while the others convey the actual symmetries of the crystal structure. Some of this confusion of notations seems to date from the early days of crystallography when it was done in many places and languages. There are seven crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups to be described to some level by the various notations. For this answer I stuck to the Strukturebericht because that is the notation used in the Dinsdale reference on Gibbs free energies of the elemental phases.

Could you provide a list of abbreviations? Since I studied chemistry in Germany, I am not familiar with CBCC, FCC and BCC. (And since I never truely studied solid-state chemistry, I don’t know what A13, A12 or the others are, either).

Ah, the joys of crystallography notation... The A12 notation comes from the Strukturebericht  designation. Other systems used include the Schoenflies notation, Hermann-Mauguin, and various others. One gets used to dealing with them all at some level.  Now, FCC is face-centered cubic (Strukturebericth A1), BCC is body centered cubic (A2), HCP is hexagonal close packed (A3). CBCC is not common, and I tend to think of it as 'complex' BCC, but that is probably poor form on my part. I can edit to add more detail. Thanks for pointing that out.


text:
Compounds forming between silicon gallium and nitrogen
What are the possible compounds that can form between silicon, gallium and nitrogen?

I'm voting to close this question as off-topic because it is de facto a list-question.

The ASM Alloy Phase Diagram Database (ASM Database if your institution has access) shows GaN and Si$_{3}$N$_{4}$ as the only compounds in that system. Standard binary alloy references agree. No ternary compounds are listed. Given the difference in crystal structures between all end members and the two nitride phases, mutual solubility of any two phases seems unlikely (little to none exists in the binaries, even Ga in Si has a pretty low solubility).

Thank you. Your explanation seems to be reasonable, but I have seen papers discussing the formation of $SiGaN_{3}$ compound [1]. Can you please comment on that?
[1] T. Markurt et al. Phys. Rev. Lett. 110, 036103 (2013)

There calculation is for a surface reconstruction, not a bulk phase. The free surface changes the free energy, leading to, for example, dimers being formed on Si (100) surface. This is not a bulk thermodynamic phase, so will not show up on binary or ternary (or higher) diagrams.

As a side note: silicon tetrazide is known, aluminium azide is known, I guess gallium azide should form as well.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Multiple values for the magnetic quantum number and the spin projection quantum number?
The valence electron configuration of boron is $2\mathrm{p}^1$.
So, $n=2$, and $l=1$. $m_l$ ranges from $-l$ to $l$. Can I say $m_l$ (for the specified electron)
may be (-1 or 0 or +1) or should I just say (-1)?

As you know the $\mathrm{p}$ subshell has three orbitals ($\mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$).
Can I associate $\mathrm{p}_x$ with $m_l = -1$, $\mathrm{p}_y$ with $m_l = 0$ and $\mathrm{p}_z$ with $m_l = +1$?
(associate each magnetic quantum number with a particular orbital)
I tried to explain what I meant in the following illustration :

$m_s$ is either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Can I say $m_s$ (for the specified electron) equals ($+\frac{1}{2}$ or $-\frac{1}{2}$)? (both values)
or $+\frac{1}{2}$ because it's alone in the orbital (not paired)

Part of your question (regarding the assignment of $m_l$ to p orbitals) is answered here: http://chemistry.stackexchange.com/questions/33645

For the other part, at this level, I believe all you need to know is that the three p orbitals are degenerate, as well as the two spin states of the electrons. Therefore, regardless of which p orbital the electron is in, and regardless of whether the spin is +1/2 or -1/2, the energy is the same. As such, you would save yourself lots of time if you just wrote one quantum number instead of all the possible quantum numbers, since they all have the same energy. Of course, it's not that simple. There is something called spin-orbit coupling, which changes the energy levels slightly.

Please visit this page, this page and this one on how to format your posts better. Alternatively, visit this chatroom for further formatting guidance.

I am under no obligation to answer your question in the comments section. Nevertheless, if you want to point something out, you might wish to tell me exactly which part I did not answer.

@orthocresol Actually, you didn't answer my question which is:
What's the ml and the ms for the first electron in a p subshell?
In other words:
What's the ml and the ms for 2p1?
If it an exam question what's your answer will be?
Sorry, I'm new here I'm editing my comment many times.

"As such, you would save yourself lots of time if you just wrote one quantum number instead of all the possible quantum numbers, since they all have the same energy." TL;DR: Just use any number you like.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why cannot the Schrödinger equation be solved exactly for systems in which more than two particles interact?
Is it so because it is impossible to solve mathematically? If so, why? Or is it possible to solve the equation mathematically, but no physical interpretation of the solutions is possible? Or is it because of practical issues, i.e. exact solutions would require more computer power than we have available today?
Several books state that the Schrödinger equation can only be solved exactly for systems of maximum two interacting particles, but fail to explain why. For example, Andrew R. Leach mentions the impossibility, but fails to elaborate on why that is so, in his "Molecular Modelling: Principles and Applications" (2nd ed., page 34).
If the answer could be explained in terms that a new master student in theoretical chemistry would understand, that would be great!

Because nobody has yet found a solution to the general three-body problem? Many have searched, none have succeeded (unless you can reduce the problem, such as fixing the nuclei positions in the H$^{+}_{2}$ ion.

Because there are very many functions, but relatively few analytic functions. Most interesting things (three-body, Navier-Stokes, etc.) can't be solved analytically, so what? That's the nature of our world.

related: chemistry.stackexchange.com/q/4438/4945

Is it so because it is impossible to solve mathematically?

Yes, in both classical and quantum mechanics there is no known general closed form analytical solution for the three-body problem. To be more precise, for the case of classical mechanics, an analytical solution in the form of a convergent power series exist, but is useless in practice due to very slow convergence.

If the answer could be explained in terms that a new master student in
  theoretical chemistry would understand, that would be great!

I doubt that is possible. For the case of classical mechanics, Poincaré had a simple argument: one has more degrees of freedom than integrals of motion, and thus, could not solve the equation of motion through the usual integration method. But, as I mentioned, that's not the whole story, and secondly, I'm not sure how to deal with quantum mechanics anyway. So, as Ivan Neretin mentioned in his comment, things are the way they are: for all practical purposes the three-body problems could not be solved analytically.

Why cannot the Schrödinger equation be solved exactly for systems in
  which more than two particles interact?

And besides, you have to be careful the wording here. I mean, you can easily have a two-particle system for which there is no known analytical solution. This is because the procedure to obtain analytical solution for a two-particle problem relies on the center-of-mass separation trick which works if the potential energy depends only on the relative positions of the particles. More formally, it works only if interactions in the two-particle system are invariant to the translation of the system as a whole. This is the case of two isolated particles that interact just with each other, for instance, an electron and a proton in an isolated hydrogen atom. For this system you can indeed use the center-of-mass separation, reduce the two-body problem to a set of two one-body problems, and get the well-known analytical solutions.
But what if you put that hydrogen atom in a inhomogeneous electric field? Center-of-mass separation won't work than, since potential energy terms due to interaction of both the electron and the proton with the external field depend on their actual position with respect to the field, not just on their relative positions. The same thing happens when considering two electrons in a field of a fixed proton, i.e. just the electronic subsystem of a helium atom. The electronic two-body problem is also unsolvable analytically in that case due to the presence of external (for the two-electron system in question) inhomogeneous electric field created by the proton.

+1 for mentioning Poincarè and integrals of the motion etc, really good answer!


text:
The difference between one pot synthetic method and one step reaction
I often read in the papers of organic synthesis that the reactions can be carried out in one-pot method, and the others can be realized in one-step. I want to know the difference between one pot and one step in organic synthesis.

"One step" refers to a process that produces the desired product by running a single reaction--especially in a situation where there are alternative pathways to the same product involving two or more reactions in a sequence.
"One pot" refers to a process involving a series of distinct reactions carried out sequentially and in a single reaction vessel to produce the desired product without isolating and purifying the intermediates after each step.


text:
Why are phosphate-phosphate bonds weaker than phosphate-water bonds?
According to the Wikipedia entry on ATP,

ATP is an unstable molecule in unbuffered water, in which it hydrolyses to ADP and phosphate. This is because the strength of the bonds between the phosphate groups in ATP is less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water

Why are the bonds between a water and a phosphate stronger than the bonds between two phosphates?
I'm studying first year chemistry, so an explanation that appealed to basic principles would be best.
(I've thought about this for a bit, and looked at the molecular diagrams in hope of seeing the answer. I'd include my thoughts on the matter in this question, however those thoughts amount to a half dozen inconclusive lines of speculation, so I opted to leave them out. Saying that to say, I have put effort into trying to figure out the answer, but I haven't gotten anywhere.)

Ouch it hurts how wrongly Wikipedia is describing this >__<*

ATP is an unstable molecule in unbuffered water, in which it hydrolyses to ADP and phosphate. This is because the strength of the bonds between the phosphate groups in ATP is less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water acid anhydrides are generally prone to hydrolysis into two acid molecules.

Ignore that description Wikipedia just gave you, it is just blatantly bad from every perspective.
Phosphorous(V) is generally a strong electrophile and will be readily attacked nucleophilicly by water molecules — think of an $\mathrm{S_N2}$ mechanism. On the other side, $\ce{ROPO3^2-}$ is a good leaving group, too.
While the exact details may or may not be like I presented here, all acid anhydrides can easily hydrolysed similarly, be it disulphonic acid ($\ce{H2S2O7}$), organic acid anhydrides ($\ce{R-CO-O-CO-R}$) etc. Technically, the $\ce{P-O-P}$ bond in ATP/ADP is no different from the bonds in $\ce{P4O10}$ which we all know to be very readily hydrolysed.

My class has gotten to organic chemistry yet, so $S_{N}2$ bonds aren't comfortable terrain for me. However, I watched the relevant Khan Academy videos after reading your answer. It seems to me that in the phosphate groups in ATP, the phosphorous atom is overloaded with electrons. Consequently, it's nucleophilic. The hydrogens in water are partially positive, so they're electrophilic. As such, bonding with the hydrogen atoms is a more appealing option (so to speak) for the phosphorous atom. Is that, very roughly, the idea?

@Hal No. The phosphorous is extremely deprieved of electrons because it has four very electronegative oxygens clinging onto it, making it very positive. On the other hand, the oxygen in water is, of course, rather negative. So the oxygen's lone pair interacts with phosphorous.

Many thanks. I really wanted to understand this.


text:
How to interpret the activation energy calculated via the Arrhenius equation?
In solution phase, the activation energy of a reaction is usually determined through the study of kinetics and the Arrhenius equation. However, I am confused by the notation, and have trouble interpreting the Ea term, almost universally called "activation energy". Is it free energy? Or is it enthalpy?
This is important, because I want to compare some experimental and theoretical values and in the output of quantum chemistry software the differences between energy/enthalpy/free energy are quite clear, and I want to be sure that I compare the right ones.
So my question is:
The Ea calculated via the Arrhenius eq. is the enthalpy change going from the reactants to the transition state, or is it the free energy change?


text:
Calculating enthalpy change
My friends and I were talking casually about electrolysis, fuel cells, and whatnots when the following question arose:
Is $\ce{H2 (g) -> 2H+ (aq) + 2e-}$ endothermic or exothermic?
I have the following questions (by the way I am a physicist):

Is the reaction sensible? (I think it probably is, but no one is sure.)
I said the reaction was endothermic, with the following reasoning. According to this table, the standard enthalpy change of formation $\Delta H_\text{f}º$ of the reactant is zero, so if $\Delta H_\text{f}º$ of the products are positive then the reaction is endothermic.
Electrons are fundamental particles so their $\Delta H_\text{f}º$ is zero. As for $\ce{H+ (aq)}$, its $\Delta H_\text{f}º$ is the atomization enthalpy of hydrogen (218 kJ/mol [ref]) + ionization enthalpy (first ionisation energy: 1312 kJ/mol [ref]) + proton hydration enthalpy (-1150 kJ/mol [ref]). Therefore,
$2 \Delta H_\text{f}º[\ce{H+ (aq)}]$ = 2 (218 + 1312 - 1150) kJ/mol = 760 kJ/mol
which is a positive number, so the reaction is indeed endothermic. Am I right?

The equation would be correct assuming that you using the "standard hydrogen electrode" so that the oxidation reaction is at the interface and solvation of the H2(g) is not required.

@MaxW You mean that on a standard hydrogen electrode $\ce{H (g)}$ is ionised into $\ce{H+ (aq) + e-}$ right away?

Yes. The point is that the reaction is taking place at the surface (boundary) not "inside" the solution. Think of the reverse reaction. You're electrolizing H+ ions to get H2(g). The reaction takes place at the surface of the electrode not in the bulk of the solution.

@MaxW OK. Thanks. And then by removing the proton hydration enthalpy from question 2 I'll get the right number for enthalpy change now?

You don't need to remove the hydration enthalpy. I think your analysis is perfectly fine as it is. You don't even have to consider the reaction in the context of an electrochemical cell. I could invent a fantasy planet 4000 light-years away from Earth, where aliens carry out this reaction on a daily process via miniscule tweezers that pluck out the electron from the nucleus, and the enthalpy change of the reaction (at a given $T$ and $p$) would still be the same.

@Taiki - Sorry, I just don't remember how to work the rest of this off the top of my head. It's been 40 years since I took P-Chem. This is just sort of funny reaction.

@orthocresol - Re remark "You don't even have to consider the reaction in the context of an electrochemical cell. "  I surely don't understand this. This is definitely a half-cell reaction. Obviously  there has to be oxidation too. You can't just generate H+ i0ns to get a positively charged solution.

@MaxW yes, and that is part of the reason why $\Delta H$ for the reaction above is conventionally specified as 0 (because whatever difference there is will be cancelled out by the oxidation half-equation). But if you are just interested in the reaction as written (as OP is), I doubt it has to be physically carried out. Basically I am saying you can do all these calculations on paper and that half-reaction would still be endothermic regardless of what the oxidation half-equation is.

@orthocresol - Well yes using the "standard hydrogen electrode" as the stake makes sense for relative electrochemical measurements since the hydrogen half-cell would be reproducible. When you add reduction and oxidation half-cells whatever bias there is cancels. But the half-reaction itself must have some absolute value. I doubt that the "real" value could be zero.

The point being that if you tried to use a copper electrode as the standard you could probably reproduce the solution to the required degree of accuracy. But the electrode itself would probably be very hard. How pure is "pure" copper? Having done a lot of XRF work no metal is "pure." With XRF you can easily find impurities in anything. You can even tag a lot of things to the geographic source by the distribution of trace impurities.

Occurs to me that in order to have some sort of "absolute" value then you need to know H2(g) pressure and concentration of H+ ions and Temperature. As I remember standard hydrogen cell uses 0.1N acid and  H2(g) at STP.

@orthocresol The discussion between you two is interesting. So what I'm doing is calculating the 'absolute' enthalpy change of the reaction occurring at the standard hydrogen electrode? And conventionally such change is defined to be zero?

Actually, I thought about it and we are not really disagreeing about anything. :) It is just that in order to define a "standard" reaction enthalpy, we need to define the pressure of the H2 gas and the concentration of the H+ ions, which is of course taken to be the standard state - 1 bar for H2 and 1 M for H+. Max was talking about the reaction in the context of an electrochemical cell, but my only point was that as long as you specify the initial and final states sufficiently clearly, the enthalpy change doesn't depend on how the reaction is carried out

Also, you're absolutely right about the "absolute" enthalpy change. Conventionally it's defined to be zero because the formation of the other counterion will "cancel out" the absolute enthalpy change of your reaction.

@orthocresol Now it's all clear. Thank you very much!


text:
Why does silver sulfate not precipitate in this double displacement reaction?
Aqueous silver nitrate is mixed with aqueous sodium sulfate.  We were asked to perform this reaction in lab in my chemistry class, and we did not observe a precipitant being formed. However, the equation indicates otherwise:
$$\ce{2AgNO3 (aq) + Na2SO4 (aq) -> Ag2SO4 (s) + 2NaNO3 (aq)}$$
This shows the formation of $\ce{Ag2SO4}$, silver sulfate. I found in the Wikipedia article for silver sulfate a reference to "ruby red illumination" and thus am led to believe that, like some silver-based substances used in photographic film, the precipitant might have been immediately been affected by light in the room and thus decomposed into something soluble in water. I have not, however, been able to find references to this anywhere else; I have found here a reference to it being classified as "sparingly soluble," but I haven't the slightest idea what this means.

Also I remember from my ion lottery lab course that the precipitation of silver halides is not clouded by sulphate present … i.e. one can detect the absence of halides in the presence of sulphate with silver nitrate.

You can check solubilities Wikipedia article you linked...

As far as "sparing soluble" I"ll add that "soluble" and "insoluble" are really absolutes that don't really exist. We'd say that NaCl is soluble but only so much will dissolve. We'd say that SiO2 is insoluble but water flowing over it on a geological time scale will dissolve it. So as Einstein said - It's all relative.

Actually, "complete solubility" does exist; it's called miscibility.  E.G. water and ethanol, or silver and gold, are miscible in any proportion.

@DrMoishe-Pippik - With the definition of solvent and solute there is the notion that the solute dissolves in the solvent and that I can evaporate the solution to get the solute back. So which is the solute, the ethanol or the water??

The solvent must be the solute and solute must be the solvent -- all is one.  A koan?

Since the lab is past due now, I'll give what I think is the answer.
$\ce{Ag2SO4}$ is somewhat soluble in water. It is most likely that you simply didn't get enough to cause precipitation.
The other possibility, which I don't think applies here, is that you have a supersaturated solution. There are some precipitates for which crystals are just stubborn to nucleate.

I think it really does not matter if the lab is done or not. I think the OP showed reasonable effort and subsequently asked for help to understand the concept of solubility better.

The solubility product constant for silver sulfate is: $6.0\times10^{-5}$.
The concentrations of the reagents used (i.e. silver nitrate and sodium sulfate) has to be high enough: $[\ce{Ag+}]^2[\ce{SO4^2-}] > 6.0\times10^{-5}$.
A better answer can be found here.
However, here, the $K_\mathrm{sp}$ for silver sulfate is said to be $1.2\times10^{-5}$.

@Coco Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have reformatted your answer and converted it to a comment, as it does not really answer the question on its own.

Because in the lab, for this experiment, we used the concentrations of AgNo3 0.1M solution or Na2so4 0.1 M solution is not enough strong to make the reaction happen. Therefore, the precipitates aren't formed.

It could have been the temperature and conc. At 100C 1.33g of silver sulphate will dissolve in 100mL of water. Maybe if your cooled the solution you would see the precipitate. I just dissolved metallic silver in a solution of H2SO4 and H202. I got a white precipitate that I would assume is silver sulphate.


text:
Forces on piston in gas expansion
I'm slightly confused by a common situation in thermochemistry.
Consider a container with gas inside and a movable, massless piston. The pressure of the gas inside is 6 atm and the external pressure is 5 atm. Thus the gas will expand. However since the piston is massless, the two forces on the piston must be equal by Newton's 2nd Law. But the pressure from the gas and the pressure from the outside atmosphere are clearly unequal, so what's the problem here?

http://chemistry.stackexchange.com/questions/38345/why-do-we-use-external-pressure-to-find-force-exerted-by-gas-on-a-piston-in-a-cl

Also your reasoning makes no sense... So you should probably elaborate

Which part doesn't make sense?

What @orthocresol is saying is not correct.  If the piston mass gets very small, its acceleration obviously does not become infinite.  This is because the force per unit area exerted by the gas on the piston will approach the pressure exerted externally.  But the gas pressure in the cylinder is not uniform, and viscous stresses are also present.  For more details, see my answer in this thread:http://chemistry.stackexchange.com/questions/38345/why-do-we-use-external-pressure-to-find-force-exerted-by-gas-on-a-piston-in-a-cl/38405#38405

@ChesterMiller I'll just go ahead and delete it then

Your point - that forces should be equal. I have no idea what's your problem.

See also my answer which addresses this very question in:  http://chemistry.stackexchange.com/questions/39650/difference-in-work-done-by-a-gas-in-reversible-and-irreversible-isothermal-expan/39665#39665 and also see the following link, which examines the effects of viscous stresses in greater detail:  https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

Thanks @ChesterMiller I think your answer on the other question makes sense. I'll read your more detailed article when I have time.


text:
Why do hydroxyl ions migrate towards the anode in a Bacon Cell?
Why do hydroxyl ions in a Bacon cell migrate towards the anode? Is it because of the relative concentration of the different ions in the set-up? According to the set-up of the bacon cell, the anode is negatively charged, thus would actually repel the incoming negative hydroxyl ions, but does not seem to do so. Please let me know if any further details are required.
I read on a wikipedia link explaining cathodes that this kind of migration (i.e negative ions towards a negative electrode) is in fact typical of discharging cells, chemical energy being the driving force behind this sort of "uphill" motion. Could someone please elaborate on this?

Well, they are anions and this is anode...

https://en.wikipedia.org/wiki/Alkaline_fuel_cell aka Bacon Cell

@Mithoron-Yes,but the anode is negatively charged here hence the confusion.

@Martin-An alkaline fuel cell is also known as a Bacon Cell.I have added more details and an image which i hope would help readers understand the reason for my asking the question.

@MaxW-Thanks for the link.I did go through it but it still did not help understanding the reason for the migration of negative hydroxyl ions towards the negatively charged anode.I have added more details to the question.Would appreciate any sort of help here.


text:
What type of test can I do to show that water is one of the products in a neutralisation reaction?
I've completed an acid base titration experiment of $\ce{HCl}$ and $\ce{NaOH}$. This forms salt + water
$$\ce{HCl + NaOH -> NaCl + H2O}$$
Indicator was used and we evaporated the water to get the salt.
I've been asked to design a test to show that water was the other product of this reaction. I'm not sure what to do here. I'm thinking maybe a distillation type test but not sure where to go with this question …
I'm a middle school student and would like an answer that is of a middle to high school level if possible. Could you provide some examples of tests I could do, and explain how they work?
EDIT -
I really appreciate the answers guys/gals. I'm not going to pretend like they did not go over my head though!
I think the issue might be the wording of the question in the textbook. What I've written above is exactly how it was asked in the textbook but I get the feeling that the expected answer is much simpler considering the level of chemistry I'm dealing with. Even though I don't expect you to read all of this I've gone ahead and scanned the book page anyway. Below is the experiment I completed, and question 4 in the discussion section which is highlighted is what I am trying to answer.

I don't see how to do it as you described. By definition an "acid base titration" is done in water. You get just a tiny amount of extra water with the reaction. The only way that I could think to do it is to blow dry HCl gas over solid NaOH which is heated and trap the resulting H2O in a cold trap. Identification of H2O could be done via freezing point, melting point, and density (or instrumental techniques such as mass spec, IR, NMR). // The gist is that HCl would be blowing over NaOH in a "tube".

The point is that you have a restriction "I've been asked to design a test to show that water was the other product of this reaction." In other words, the reaction that is happening in the solution. So doing the reaction with gaseous HCl leads to another problem. The "reactions" are not occurring under the same conditions. How do you show the reactions are the same?

It appears you have created two accounts. Have a look at this link to obtain information what you can do.

As was written in comments, it is nigh impossible to detect the formation of more water in an aquaeous solution. Unless you coloured the oxygen atoms in $\ce{NaOH}$ pink or purple you would have no clue which water molecule is new and which isn’t.
Say you had a $1~\mathrm{M}$ solution of both acid and base and used $25~\mathrm{ml}$ each; you would create $25~\mathrm{mmol}~\ce{H2O}$ in the reaction or approximately $0.45~\mathrm{g}$. That’s just about $1~\%$ of the total water added, so you may be able to detect a slight increase in weight if you distilled away the water.
Do not try to get your hopes high, though.

I mostly agree, but there is a nasty wrinkle. How do you figure out the weight of the water in the HCl solution? For the NaOH  solution I could at least start with distilled water and solid NaOH.

@MaxW I assume a perfect solution containing $36.46~\mathrm{g}$ of $\ce{HCl}$ in $1~\mathrm{l}$ $\ce{H2O}$ and right there is another catch.


text:
What is the solvent and what is the solute?
This question is and I am learning about mixtures, solutions, solutes, solvents, and other terms related to these words.
I know that a solute is the substance being mixed with or dissolved into the solvent.
Substances and mixtures: are these solvents and solutes correct?

coffee with milk - solvent:coffee , solute:milk
salt water - solvent:water , solute:salt
air freshener - solvent:air , solute:scent
brass - solvent:zinc , solute:copper

The explicit notion of a solvent/solute pair is that solvent dissolves the solute and that I can evaporate the solvent to get the solute back. There is only one of the answers that fits that description.

@MaxW I agree but say two answers can possibly fit the description (however the second might contain a mixture of solvents and some low-boiling solutes).

In a binary homogeneous mixture, the component present in the larger amount is generally called the "solvent". The component present in the smaller amount is generally the "solute".
In your first example, however, it is more complicated as coffee is already a complex solution (but mainly water). Milk is not a homogeneous mixture. It is a fat/protein emulsion (also mainly water).
The other three are solutions: salt water is a solution of a solid dissolved in a liquid (which is the most familiar type), air freshener (one dispersed) is a solution of one gas in others (but air itself is a mixture), and brass is a solid solution.


text:
Formation of Bonding and Anti-bonding Molecular Orbitals
If two orbitals combine in-phase, a bonding molecular orbital is formed. When they combine out-of-phase, an anti-bonding molecular orbital is formed. For a single molecule, how are both orbitals present at the same time? They are formed differently.

First, remember that orbitals are a more or less theoretical construct, based on mathematical models. With that in mind, remember that an orbital in the strictest quantum-mechanical sense is a wave function of one and exactly one electron. (I am separating α and β spins here.) Since an orbital somehow nicely corresponds to one possible particle state, it may seem logical to postulate a ‘rule of conservation of orbitals’ — and exactly that happens in quantum chemistry.
So if we take one 1s orbital of one hydrogen atom and another 1s orbital of another hydrogen atom, we need to create two new orbitals by linear combination if we choose to do so. The two orbitals we create correspond to σ and σ*, respectively. (In a spintastic sense, we would have taken two orbitals of each hydrogen, one corresponding to α the other to β. We would receive four orbitals: $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}, \unicode[Times]{x3c3}_\unicode[Times]{x3b2}, \unicode[Times]{x3c3}^*_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}^*_\unicode[Times]{x3b2}$. The electrons now occupy the lowest energy orbitals which are $\unicode[Times]{x3c3}_\unicode[Times]{x3b1}$ and $\unicode[Times]{x3c3}_\unicode[Times]{x3b2}$.)

Finally, based on my question on SN2 mechanism, the antibonding orbital was already present. The electrons from the nucleophile filled the antibonding orbital of C-X. That weakened the C-X bond. But since the electrons from Nu fill ABMO of Carbon how does it form a bond? shouldn't it fill the BMO to make a bond?

@AdityaDev And once again, I unfortunately fail to understand the exact question you’re asking … sorry ^^'

1)C-X has both bonding and antibonding orbital

2)Nucleophile's lone pair electrons fills ABMO of C-X bond.

3)This weakens C-X bond. But to form a bond, shouldn't electrons fill in BMO? Nucleophile's electrons is filled in ABMO.

@AdityaDev Just because an orbital is antibonding concerning one bond does not mean it is always antibonding everywhere.

If you have read about Linear Combination of Atomic Orbitals,  it should help you.

If two orbitals combine in-phase, a bonding molecular orbital is formed. When they combine out-of-phase, an antibonding molecular orbital is formed. For a single molecule, how are both orbitals present at the same time? They are formed differently.

What you have said is absolutely right. As per my knowledge, the MO theory was established to explain the paramagnetism of O2 and the violation of octet rule. It is again like the hybridization concept, a hypothetical one, in which they have emphasized more on the wave like nature of the electron. And you may read in the LCAO the rules for combination of Atomic orbitals to give molecular orbitals. One of them states that the number of Molecular orbitals formed should be equal to the number of Atomic orbitals participating.
So I may suggest, first read that up. And again we can't question much in chemistry as the laws are based on the observations, whether hypothetical or not.

That was not my question. Are BMO and ABMO present together?

Yes they are...

Well a lot of chemistry is based hypothetical models that don't really exist. Your sentence "One of them states that the number of Molecular orbitals formed should be equal to the number of Atomic orbitals participating." explains the notion well.  Think of hydrogen. Two hydrogen atoms each have a 1S orbital that can hold two electrons, but the hydrogen molecule just has one bond which has two electrons. So in essence one of the 1S orbitals becomes a bonding orbital over both atoms, and the other orbital becomes and antibonding orbital. Think of H2-. Where would you stick the extra electron?

In the other empty orbital maybe.

"Bad models" - Newton's gravity law is actually wrong, but it works just fine for most of terrestrial experiences. It worked well enough to get men to the moon and back, but it doesn't work good enough for GPS satellites. But the "true" equation is just painful to use in most circumstances, the precision isn't justified, so why bother?

I am not getting the point you are heading to @MaxW

If you are trying to say that there are many laws and theses in this world which would better remained unquestioned then I agree with you.

@AdityaDev You seem to implicitly assume that an AO which takes part in the formation of one MO is "spent" and can't take part in more MOs. This is not so. Whether an empty ABMO actually "exists" is a philosophical question; let's turn to something simple and unambiguous. Think of $sp^3$ carbon; all four of its $sp^3$ orbitals exist at once, and all are different combinations of s and p orbitals. How can it be? Well, just like that.

@IvanNeretin I would like to agree with you. Had thought about it this way too.


text:
How is the degree of unsaturation derived?
The formula for degrees of unsaturation is:
$$1+ \frac{1}{2} \sum ( n_i (v_i-2))$$
$v_i$ = valency of atom
$n_i$ = number of atoms with that valency
Could someone explain how the formula works, and why divalent atoms don't contribute to unsaturation (as evident from the formula).

This is the correct formula:
$$ DU = 1 + \frac{1}{2}\sum _{v=1}^{j} n_{i}(v-2),$$
where
$j$ = maximum valence of any atom in molecule
$v$ = valence of atom(s) 1 to whatever
$n_{v}$ = number of atoms with that valence
Wikipedia gives a good explanation. In simple terms the formula will give the number of double bonds in an alkene. For all alkanes the number would be 0 of course. Rings and ethynes would complicate the use of the formula. Think of using the equation on chemical formulas like $\ce{C_{x}}\ce{H_{y}}$ or $\ce{C_{x}}\ce{H_{y}}\ce{O_{z}}$.


text:
Melting points of NaCl and RbCl
Why is the melting point of $\ce{NaCl}$ more than that of $\ce{RbCl}$ ? Using Fajan's Rule of polarizing power, $\ce{RbCl}$ has more ionic character and thus should have a higher melting point.

See I am getting confused. Whether to use Fajan's rule or to go according to the periodic trend. One way we can say as the distance between ions increases bond strength decreases and hence melting point decreases. But then according to Fajan's rule the charge density over RbCl is least and hence covalent character would be minimum and hence melting point should be maximum. :/

@Joel if we take into consideration of Molecular weight, then LiCl should have the highest melting point according to me.

@MaxW or do I just accept this also as an exception?

No. Think about it. Are the NaCl and RbCl "molecules" floating around in empty space?

Another hint. Given that Na+, Rb+ and Cl- ions all have a different size, what would be the effect of that?

I'm not sure here either just throwing it out there, would where the ions are on the scale of reactivity effect how strongly the ions are attracted which could be a melting point variable? Sodium is considered more reactive than lithium.

Not reactivity. Just think of the Na+, Rb+ and Cl- ions as solid charged spheres - like individual legos. How do I get a solid?

Yeah but how reactive they are would contribute to the bond strength and enthalpy?

@MaxW I get the answer by what you have said. But then it disproves this Fajan's rule, doesn't it?

@Adit-Daftary - You can't possibly understand my point if you're still asking about Fajan's rule. I explicitly said to consider that the ions where single charged spheres.  What kind of bond would be between an atoms with a + and - charges??

there would be anIonic bond.

The enthalpy of sublimination of the different compounds could be a factor here. Na has a higher enthalpy of sublimination than Rb.  Li has a higher enthalpy of sublimination than both Na and Rb but due to excessive polarization (Fajans' rules), displays high covalent characteristics resulting in a lower sublimination point. I think. ..

The same may apply to Rb.

No doubt about it: it is due to the ionic radii and thus the distance between ions. The Rb-Cl bond is not as strong because of the distance between them. Na has an E.N. value of 0.9 Paulings while Rb is at 0.8; thus no significant difference in the magnetic forces of attraction.
LiCl has an even lower melting point due to excessive polarization (this is when Farjan's rules come into play), and the result, along with electronegativity again, is still an ionic compound, but with more covalent character.

@Nicholas If you did use Fajan's rule for Li why not for Rb?

Well - look at the rule. Farjan states that small cation size is a contributor to polarization.

See as you know they are ionic compounds and have some force of attraction. To break this energy is needed and the attraction between Na and Cl is more compared to Rb and Cl. Think about it like breaking a box when it is arranged in a lattice structure.
http://www.google.co.in/imgres?imgurl=https://upload.wikimedia.org/wikipedia/commons/e/e9/Sodium-chloride-3D-ionic.png&imgrefurl=https://en.wikipedia.org/wiki/Lattice_energy&h=948&w=1000&tbnid=Gm4SV7MURH7-UM:&docid=bdbhKWpu1E4z3M&hl=en-in&ei=DtkwVvSZCIjH0gTV0a4g&tbm=isch&client=safari&ved=0CCAQMygDMANqFQoTCPTY_9-t5cgCFYijlAod1agLBA

I think there's a bit more to it than that..

Perhaps you could explain why the attractive force between Na and Cl is larger. Also please avoid posting enormous links. If you want to insert a picture you can do that in the editor.

@bon Not with 1 rep, though.


text:
Why does basicity go down along the group 15 hydrides?
As we go down Group 15, the sizes of atoms increase and these are the central atoms in their corresponding hydrides. Shouldn't the large size of central atom (in case of $\ce{Bi}$) actually help the lone pair inside the $\ce{sp^3}$ hybrid orbital to easily be donated and thus increase the basic character down the group?
$\ce{N}$ is small and shouldn't it hold lone pair with much more strength than others, but $\ce{NH3}$ seems to be strong base while $\ce{BiH3}$ is the weakest one.

Going down this group the proton affinity decreases, so basicity decreases:
$$\mathrm{PA}(\ce{X})=\mathrm{D}(\ce{X+-H})-\mathrm{IE}(\ce{X})$$
where $\mathrm{D}$ is the bond disassociation energy, and $\mathrm{IE}$ is the ionization energy.  Down a column this expression is dominated by the $\mathrm{D}(\ce{X^+-H})$ term, and since $\ce{N}$ has the highest bond strength with hydrogen, it is the most basic.
The structural (or hybridization) way to look at it is that the lone pair in $\ce{NH3}$ is very, very different than the lone pair of a similar compound of $\ce{P}$, $\ce{As}$, etc. $\ce{NH3}$ is almost perfectly $\mathrm{sp^3}$ hybridized, but the others have little to no hybridization, so the lone pair is actually held in a orbital that has mainly $\mathrm{s}$ character.  This makes the lone pair far less directional (the charge is distributed over a larger area), making it less reactive and harder to form bonds, so less basic.
Finally, $\ce{N}$ is the only one on the series where the HOMO (which contains the lone pair) is concentrated on the $\ce{N}$. For the rest of the series, the HOMO is very delocalized and significant electron density is held on the $\ce{H}$ atoms, so it will not react well as a base as well.

The very idea of $\ce{sp^3}$ hybridization you are referring to is pretty much not there, starting from $\ce{PH3}$. Look at its bond angles, they are almost exactly $90^\circ$. This means that phosphorus contributes to the $\ce{P-H}$ bonds almost exclusively with its $\ce p$ orbitals, and the lone pair just sits there hidden on $\ce s$ orbital, rather than stick out as $\ce{sp^3}$.
The same is even more true for the following hydrides.
Also, with nitrogen being the most electronegative of all, its partial negative charge in $\ce{NH3}$ attracts $\ce{H+}$, which is not the case for $\ce{PH3}$ and the rest.

As far as I understand the structures , Gr 15 elements do make 4 Sp3 hybrid orbitals. 3 of them make sigma bonds with s orbital of H and one lone pair reside in the 4th sp3 hybrid orbital. You can easily calculate the hybridisation.  In case of NH3 , bp-bp repulsion is maximum but as we go down the group electron density shifts away from the central atom due to its large size and thus bp-bp repulsion is reduced but lp-bp  repulsion is still the same which makes the H-E-H bond angle closer to 90 degrees. (E is Gr15 elements ). But what does any of these have anything to do with my real question

Why does the basic character of hydrides go down the group?

authors: πααρτθ Σαρθι


text:
How can the intense color of potassium permanganate be explained with molecular orbital theory?
In the permanganate ion, manganese is in the $+7$ oxidation state, therefore it is a $\ce{d^0}$ ion. $\ce{d^0}$ and $\ce{d^1^0}$ ions don't absorb visible spectrum radiation because there are no electrons to jump to higher crystal field orbitals (or in the case of $\ce{d^{10}}$ ions, there are no free orbitals for electrons to jump to). I think that the explanation to this question has something to do with the ligand donating electrons into $\ce{Mn^{VII}}$'s empty d orbitals or molecular orbital theory or $\ce{d-d}$ transition or none of them.
Also, chromate is also colored wherein $\ce{Cr^{VI}}$ is a $\ce{d^0}$ ion, therefore its not absorbing radiation due to electrons jumping to higher orbitals in the crystal field. What is causing it to absorb visible radiation?

Possibly a duplicate of this. However, I feel the other question approaches the issue from a wildly different angle leading to possibly different explanation concepts.

I made a very short answer here: http://chemistry.stackexchange.com/a/32332/16683 However, if you want a MO description, it definitely won't satisfy you. I recommend editing the question (especially the title) to reflect what you are asking for because otherwise it can easily be considered a duplicate.

Crystal field theory is nice and dandy, but people often forgot that d-d bands are typically weak, therefore many times you see an intense colour in complex chemistry, it is charge transfer (CT) band, not a d-d band. Which  should be present in the spectrum on any metal in high oxidation state, anyways. The explanation is simple: the dipole integral for LMCT and MLCT bands are much larger than for d-d bands.

Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$, $\mathrm{4s}$ and $\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as $\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of the same symmetry will mix, so the $\mathrm{t_1}$ ones stay where they are originally. I have colour-coded both the $\mathrm{t_2}$ (red) and $\mathrm{e}$ (blue) orbitals.

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.
Note that the image is qualitative only. The order of energy levels can be different; it is not possible to predict it a priori. The more electronegative a side is, the lower its orbital energies will be. Manganese(VII) is lacking seven electrons so the orbital energies will be very low, will oxygen profits from its high electronegativity. All in all, the energies will be somewhat similar.
We can consider the filled orbitals to be mostly ligand-centred and the empty ones to be mostly metal-centred, in full alignment with the picture of a $\ce{d^0}$ $\ce{Mn^{VII}}$ core and four $\ce{O^2-}$ ligands with electron octets.
The electronic transition that causes the intense purple colour corresponds to an excitation from $\mathrm{t_1}$ to $\mathrm{e}$ — $\mathrm{t_1}$ should always be the highest ligand orbital, while tetrahedric symmetry typically implies that $\mathrm{e}$ lies lower than $\mathrm{t_2}$ on the metal’s side. The higher the central metal’s oxidation state, the lower its orbitals and the closer it is to the ligands’ p-orbitals.
Since the transition is from a ligand-centred to a metal-centred orbital, it is called Ligand to Metal Charge Transfer (LMCT). Oversimplified, one could think of it as a one-electron reduction of manganese by an oxygen ligand to intermediately create $\ce{Mn^{VI}-O^{.}}$.
The transition is obviously spin-allowed (filled to empty orbitals) and the Laporte-rule is invalid in tetrahedric complexes as there is no centre of symmetry. Therefore, the observed colour is very intense.

Manganese is $\pu{[Ar] 3d5 4s2}$. Mn(VII) is $\pu{[Ar]}$ in $\ce{MnO4^-}$. Why are there $24$ electrons in the shown MO of $\ce{MnO4^-}$

@Maurice ‘on the right the twelve degenerate ligand p-orbitals’. These are the 24 electrons present in the twelve 2p orbitals of the oxido ligands. To be even more correct, the four 2s orbitals should also be considered but they are further away and will not have a massive effect.


text:
What kind of optical activity requires chirality?
From "Meso compound" in Wikipedia:

A meso compound or meso isomer is a non-optically active member of a
  set of stereoisomers, at least two of which are optically active. This
  means that despite containing two or more stereogenic centers, the
  molecule is not chiral.

What is the meaning of "optical activity" in this context? In other words, what kind of photo activity requires chirality?

Chirality can be observed by the way a beam of light is refracted after passing it through a solution of the subject compound

Rotation of plane of polarised light, as always.

The meaning of optical activity in this context is the same as usual: the ability to rotate the plane of polarized light. To have it, a molecule must be chiral.
Chirality is the absence of certain type of symmetry. Since most small molecules have that symmetry, many students get the wrong perception that chirality is a property which must be obtained by having some functional group (asymmetric center, etc.), much like having a $\ce{-COOH}$ group makes a compound an acid. This is not quite true. Not only a compound may be chiral without any chiral centers, but it can also be achiral despite having some chiral centers, as are meso compounds.


text:
Shifting of axial and equatorial protons
I know this NMR represents cyclohexanol and I have figured out what each peak represents. If the 3 peaks at 1.9, 1.7,1.5 all represent equatorial protons as they are shifted downfield compared to the axial. How does the peak at 1.3 represent 5 axial protons. Shouldn't they have split signals just like the equatorial protons, due to the electronegativity of the oxygen an its deshielding effect. Or is there a reason why axial protons aren't affected by the electronegativity of the Oxygen.

The distinction between axis and equatorial hydrogens is not going to be visible in an NMR spectrum for cyclohexanol. The conformations of the molecule interconvert too quickly at room temperature to make a distinction in the proton NMR spectrum.
So, in practice, you will only see 4 distinct C-H bond types in the molecule and the splits that will be visible on your spectrum depend on the resolution (which is not particularly high on the spectrum you have). The splits cause by neighbouring hydrogens for the groups between shifts of 1 and 2 overlap even in high resolution spectra and it isn't clear whether what you are seeing this jumble of peaks as 4 separate environments because of the low resolution or whether one of the peaks is an impurity.
Some software packages for structure drawing also generate theoretical NMR plots which can help interpret such spectra.

No, I see 5 peaks not 4 between 1 and 2 ppm. This is a "reference" spectra so it would not contain an impurity which had such a notable peak. From left to right I see 2,2,1,4,1 protons.  here is earlier discussion. http://chemistry.stackexchange.com/questions/39765/nmr-analysis-problem

I think the problem with assigning the spectra is to assume that the molecule is in one chair form or the other. It seems that you have to consider 4 conformations, not 2.

Here is a high resolution spectra for cyclohexanol which has a deuterium in the OH group.  https://www.google.com/search?q=cyclohexanol+nmr&biw=1317&bih=684&tbm=isch&imgil=oOj7TGDEdYJP4M%253A%253ByryowAprRysuPM%253Bhttp%25253A%25252F%25252Fbmrb.wisc.edu%25252Fmetabolomics%25252Fmol_summary%25252Fshow_data.php%25253FmolName%2525253DCyclohexanol%25252526id%2525253Dbmse000431%25252526whichTab%2525253D1&source=iu&pf=m&fir=oOj7TGDEdYJP4M%253A%252CyryowAprRysuPM%252C_&usg=__6Fp2LLVNtmy1ieUIXqezgtkXIrw%3D#imgrc=oOj7TGDEdYJP4M%3A&usg=__6Fp2LLVNtmy1ieUIXqezgtkXIrw%3D

‘The distinction between axial and equatorial hydrogens is not going to be visible in an NMR spectrum for cyclohexanol’ — it’s right there in OP’s spectrum. If you deem that spectrum incorrect, I demand you show corresponding 2D spectra that prove OP’s assignments (which are confirmed by SDBS) be incorrect.

On another note, this does not answer OP’s question.

The 4 conformations (2 chair & 2 boat) of the molecule interconvert too quickly at room temperature to make a distinction in the shown proton NMR spectrum.
From SDBS.
(1) Enter Cyclohexanol
(2) Select full math & H NMR
(3) Search
(4) Select 1H NMR : 400 MHz in CDCl3 on left side
One of the possible 4 conformations of the molecule is shown below.

Assign......#.....Shift(ppm)

  A       1      3.578
  B       1      2.84
  C       2      1.882
  D       2      1.728
  E       1      1.542
  F       4   1.30 to 1.22
  G       1      1.16

So the OP was wrong that the 3 peaks at 1.9, 1.7,1.5 all represent equatorial protons. There are 2-2-1 protons at those peaks.

This is an answer to a question but not to the question the OP asked.

@jan have you considered the possibility that the OP makes incorrect assumptions?

@matt_black I have considered that possibility but dismissed it after the assignments were confirmed by SDBS. (SDBS No. 581HSP-40-238)

I corrected the proposed solution. Using SBDS image I had it wrong that the  protons were on two carbon atoms before.


text:
Why does superglue lose strength at high temperatures?
I'm not a chemist. When I superglue my fingers together, by accident, I put them under warm water to peel off the superglue. Why does superglue become less "sticky" at higher temperatures? Is it the water or the temperature?

Well....... You could always find out yourself...... By supergluing your fingers together and running cold water over them to see whether it helps. ^^

Use acetone to wash superglue off your fingers. It works great ;)

It is the temperature because as you probably know all kinds of substances get denatured at high temperatures. Another reason could the that the rate of reaction increases after running hot water, therefore taking no effort to take it off. Hope the answer helped!

Do you actually have any evidence to suggest this is the case. There are also many substances which don't denature at the sort of temperatures we are talking about. Answers based on unqualified speculation are discouraged on chemistry.SE.

The glue is sticky because it is uncured. Water is part of the reactants in the chemical curing. Raising the temperature makes the reaction faster. So you both put the glue in contact with a copious supply of water and raised the temperature which make the curing much faster.

It's normally suspended in methylethyl ketone or something similar.

But why does raising the temperature make the superglue easier to peel off my fingers. Does it lose some strength or start to breakdown above a certain temperature?

It would seem that the glue had not completely cured before you used the water. The water caused the glue to cure fast, before it had time to completely bond with  your skin. Water will not dissolve the cured glue.


text:
Extent of dehydration by concentrated sulphuric acid
I know that conc $\ce{H2SO4}$ is a dehydrating agent. And the most easiest mechanism for example would be this:

But I want to know how to find the extent of dehydration of it.
I came across this question:

And the answer is:

 d

My thinking:
(a) $\ce{O}$ is EN and will pull $\ce{-OH}$ towards itself inductively, thus decreasing the electron density on $\ce{-OH}$, and decreasing its power to pull $\ce{H}$ from concentrated $\ce{H2SO4}$.
Similarly for (b) and (c) and I think due to inductivity (c) will have more chances than (b).
As for the (d) I think $\ce{-CH3}$ group is an electron pusher so it should increase the electron density, and hence reducing the effect on $\ce{-OH}$ group .
1) But my doubt is: is this the only reason for the extent of dehydration by conc $\ce{H2SO4}$ on (d) to be more than (c)?
2) I don't think there would be much difference if it tautomerises to enol form.

The extent of dehydration would relate to how much hydrate the starting material could potentialy form with the sulfuric acid?

Could you please tell in more detail?

You should think about the relative stabilities of the products. Which has the most stable dehydration product and why?

I'll give you two more hints. (1) You must think in 3D. The ring on the molecule isn't flat. (2) a,b and c have OH moved relative  to C=O group. The OH would have the most effect "meta" or "ortho" to the C=O, so b is very likely out off the bat.

@MaxW thank you, it kind of  helped. But I want to know why d over c? Is the reason what I have already written or something else. If I'm wrong, then please tell me where I'm wrong or I'm missing here.

The most stable Alkene is the most highly substituted Alkene. (d) gives a tri-substituted alkene, all others give the less disubstituted alkenes. Most stable Alkene (d) therefore more of it is formed! (Edited thanks to Bon for highlighting my error)
Also a factor to keep in mind is conjugation. (a), (b) and (d) will have conjugated double-bonds, while (c) will not. This conjugation also adds stability to the system. But the major stability contributor is the higher substituted alkene in (d).

@bon only saw your comment after posting my answer...credit to you...

No, credit goes to you because you posted an answer ;). However, you meant tri-substituted, didn’t you?

indeed I did..tri substituted it is, others are disubstituted..

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


text:
Ionization energy and Electron Affnity
Which of the following statements correctly interprets the relationship between ionization energy, atomic radius  Electron affinity when comparing two atoms.
The correct choice as per my instructor is that when atomic radius increase ionization energy decrease and electron affinity becomes less positive.
I do understand that as the atomic radius increase it becomes easier to remove and an electron due to shielding from multiple orbitals. Hence Ionization energy is decreased as the atomic radius increase. Additionally, electron affinity is an exothermic process.I am not sure how  electron affinity (addition of an electron) related to ionization energy and atomic radius. Please clarify. Thank you.

EA is an endothermic process. If I'm not wrong, EA is conventionally defined to be the negative of $\Delta H$ for the process $\ce{X(g) + e- -> X-(g)}$. So, using your logic: larger radius > incoming electron is more shielded > incoming electron is less attracted to the nucleus > $\Delta H$ less negative > EA less positive.

There are two definitions of EA out there, and they differ by their sign. One defines positive EAs as exothermic, the other negative. Yeah because logic.

For example ,when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released).
Electron affinity is defined as the change in energy (in kJ/mol) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.
When an electron is generally added to a neutral atom, i.e first electron affinity, energy is released; thus, the first electron affinities are negative. However, when an electron is added to a negative ion, i.e. second electron affinity, more energy is required. Thus, more energy is released to add the electron into an ion because the negative ion has to force the electron to go into its electron orbital; thus, the second electron affinities are positive.
First Electron Affinity (negative energy because energy released):
$\ce{X(g) + e- -> X- (g)}$
Second Electron Affinity (positive energy because energy needed is more than gained):
$\ce{X- (g) + e- -> X^2- (g)}$
This is because you are practically trying to forcefully add an electron to it.
Coming to your question: Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.
One might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table.
As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately.)
The electron affinity is a measure of the attraction between the incoming electron and the nucleus — the stronger the attraction, the more energy is released. The factors which affect this attraction are exactly the same as those relating to ionization energies — nuclear charge, distance and screening.
The increased nuclear charge as you go down the group is offset by extra screening electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about.
So,
as atomic radius increases, the nuclear attraction on outtermost electron decreases — this leads to decrease in IE — but the shielding also increases — less energy is released when an electron is added to the outside orbital — decreasing the EA.

Implying that there be only one definition of electron affinities; 2) implying that all first electron affinities have the same sign. Enough for $-1$

I'm not telling that there is only one defination, but this is the most common one at least to me. Yes, there are exceptions, but we are talking on general or most common basis here.


text:
Influence of neighboring fluorine and hydroxyl on carbocation
Which of the 2-fluorobutan-3-yl and butan-2-ol-3-yl carbocations will be more stable? I know that fluorine has a greater electron-withdrawing inductive effect, but what about the resonance effect of those?
$\hspace{35 mm}$

Firstly, they don't have a +I effect? Also, resonance doesn't play a part in these carbocations. Try drawing resonance structures to stabilise the carbocations. You can't. But, to answer your original question, all you need to do is to look at the comparative reactivities of fluorobenzene and phenol. Which one is more reactive? So which one shows a greater +M effect?

They are stabilized by Neighbouring group participation (anchimeric assistance) not mesomeric effect.

I heavily edited your question, you can change it if you wish.

@Mith If the heteroatom is attached to the positive carbon, then that happens. But not if it is one carbon away. http://i.imgur.com/TSqejdS.png

@orthocresol Sorry, your link is wrong, what they call Ngp is simple mesomeric stabilisation.

More precisely, during reaction which proceeds via this kind of intermediate anchimeric assistance can be called synartetic acceleration. http://goldbook.iupac.org/N04100.html

@Mith OK, you're right. The way I drew it above is just how I was taught it. I guess, loosely speaking it is still NGP since the presence of the heteroatom increases the rate of an SN1 occurring there, but by IUPAC definition the presence of conjugation isn't allowed.

The answer is actually quite straightforward. Fluorine is more electronegative than oxygen, thus it is less willing to share its lone pair with the carbon bearing the positive formal charge. Thus, the $\ce {-OH}$ group provides greater stabilisation. In fact, the donation of the lone pair by the $\ce {O}$ generates a familiar moiety, an epoxide. The donation by $\ce {F}$ is almost negligible since the positive formal charge on $\ce {F}$ would be highly unfavoured.

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


text:
Solid state, closed packed structures, BCC lattice
How many next nearest neighborhood respectively each potassium has in bcc lattice?

I am just not able to proceed, I know how to find first nearest neighbours but don't know how to find second nearest neighbors

Well, second nearest neighbors are the neighbors of the first nearest neighbors, not counting those that are already first nearest. You get there by looking at one unit cell of bcc and imagining the next unit cells out and which atom(s) in those cells are closest.

BCC is the very case when the distinction between "first nearest" and "second nearest" does not work really well. However, a line still can be drawn between the two. How many first nearest neighbors did you find, to begin with?

You're overthinking the problem. I'll give you a hint. I paint one particular "molecule" of potassium red. How many atoms are in a "molecule" of potassium metal? How many nearest neighbors does my red "molecule" have?

I find 8 first nearest atoms

You need to remember that the BCC is an abstraction within the "infinite" bulk solid. All the atoms are the same in K metal and EVERY atom is at the center of a BCC. Thinking about the "edges" of a chunk of K metal will get you into trouble.

@JonCuster Your comment is somewhat misleading. Let's consider a square planar lattice with parameter 1. The first nearest neighbors (all 4 of them) will be at distance 1. Now let's look at the first neighbors of first neighbors, excluding the atom at origin itself. Some of them will be at distance 2, and some at $\sqrt2$. Surely not all of them are the next nearest neighbors.

Yes not all of them are next nearest but it gives some idea about which atom is second nearest

Second nearest neighbors are the neighbors of the first neighbors. So for BCC let's consider the atom at the body centre, for this atom the atom at the corner are nearest and for the atoms at the corners the atom at body centres of other cubes are nearest. Little imagination(there are 6 body centred atoms surrounding the atom we are considering) and counting gives the answer as six.


text:
When using the Arrhenius Equation, beyond what range would the values of k, A , and Ea be extreme
In the Arrhenius Equation
$$k = A \mathrm e^\frac{-E_\mathrm a}{RT}$$
with:

$k$ defined as the rate constant
$A$ as the pre- exponential factor
$E_\mathrm a$ as the activation energy (in$\ \mathrm{\frac{kJ}{mol}}$)
$R$ as the universal gas constant (in$\ \mathrm{\frac{J}{mol\ K}}$)
$T$ as temperature (in $\mathrm K$)

beyond what range would the values of $k$, $A$, and $E_\mathrm a$ seem unreasonable for any given order or type of reaction?  Specifically is there a simple way to tell if a reaction is as fast as an explosion or as slow as paint drying for verifying if kinetics calculations seem reasonable.

To clarify, are you looking for general rules of thumb regarding the speed/type of a reaction given an activation energy or pre-exponential factor?

If using a general rule takes significantly less effort than computing a mathematically rigorous bound on reaction speed and/or plotting a reaction curve then yes....


text:
Mathematically, how does the exchange integral for a closed-shell system reduce to zero?
I am looking at the Hartree-Fock method of approximation for a closed-shell two-electron system.
I have the basis functions
$$
\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(s_1) \\
\chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(s_2)
$$
where $\vec{x}_i$ is the combined spatial and spin coordinates. The opposite spins are illustrated by the $\alpha$ and $\beta$ labels.
The above gives the total wave function to be the following Slater determinant
$$\Psi(\vec{r}_1,\vec{r}_2,s_1,s_2) =
\begin{vmatrix}
\psi_1(\vec{r}_1)\alpha(s_1) & \psi_1(\vec{r}_1)\beta(s_1) \\
\psi_1(\vec{r}_2)\alpha(s_2) & \psi_1(\vec{r}_2)\beta(s_2)
\end{vmatrix}
$$
Let's bypass a lot of math, and just look at the exchange integral, where the spin and spatial parts have been separated:
$$
\int ds_1 \alpha^*(s_1)\beta(s_1) \cdot \int ds_2 \beta^*(s_2)\alpha(s_2) \cdot \int d\vec{r}_1 \int d\vec{r}_2 \psi_1^*(\vec{r}_1) \psi_1^*(\vec{r}_2) \dfrac{1}{r_{12}} \psi_1(\vec{r}_2) \psi_1(\vec{r}_1)
$$
So the spin integrals do not survive in this case, due to the closed shells, but I am not sure if I see why. Is there some orthonormality for the spin states at play here?

I corrected the typesetting and an obvious mistake (missing complex conjugation) without changing the actual meaning. Note specifically that for vectors with subscripts one better include only the label and not the subscript inside \vec{}, i.e. $\vec{r}_{1}$ (\vec{r}_{1}) usually looks better than $\vec{r_1}$ (\vec{r_1}).

Are you sure, by the way, that the second spin orbital has a different spatial part $\psi_2$ and not the same $\psi_1$ as the first spin orbital? I mean, usually for a closed shell case, the restricted formalism is used in which spin orbitals come in pairs with the same spatial part: $\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(s_1)$ and $\chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(s_2)$.

Ohh, you're right. Is that due to that both electrons are occupied in the same orbital, and are thus restricted to the same spatial domain?

Yes, the usual restricted picture for a closed shell system insists on electron "occupying" the same spatial orbital. I don't think it matters for the very question though, since the exchange integral vanishes due to the defining properties of the spin functions irregardless of this restriction.

Let's do the math as explicit as possible. So, the exchange integral is defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{r}_1 \mathrm{d} \vec{r}_2 \, ,
$$
where I used a more appropriate label for the spin coordinate ($m_{s}$) as well as distinguished between summation over discrete variables $m_{s1}$ and $m_{s2}$ and integration over continuous $\vec{r}_1$ and $\vec{r}_2$ ones.
Then, for the restricted spin orbitals,
$$
  \chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(m_{s1}) \, , \\
  \chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(m_{s2}) \, ,
$$
where $\alpha$ and $\beta$ are the so-called "spin up" and "spin down" spin functions defined as follows,
$$
    \alpha(m_{s}) =
    \begin{cases}
        1, & m_{s} = +1/2 \\
        0, & m_{s} = -1/2
    \end{cases} \, ,
    \quad
    \beta(m_{s}) =
    \begin{cases}
        0, & m_{s} = +1/2 \\
        1, & m_{s} = -1/2
    \end{cases} \, ,
$$
we get then
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1) \alpha(m_{s1})
\bar{\psi}_1(\vec{r}_2) \beta(m_{s2})
r_{12}^{-1}
\psi_1(\vec{r}_1) \beta(m_{s1})
\psi_1(\vec{r}_2) \alpha(m_{s2}) \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1)
\bar{\psi}_1(\vec{r}_2)
r_{12}^{-1}
\psi_1(\vec{r}_1)
\psi_1(\vec{r}_2) \, ,
$$
where we used the fact that spin functions $\alpha$ and $\beta$ are real-valued, so that $\bar{\alpha} = \alpha$ and $\bar{\beta} = \beta$.
At this point we could concentrate our attention exclusively on the coefficient in front of the double integral above, since we could show that it is equal to zero, so that the whole expression vanishes irregardless of the value of the integral. So, the coefficient can be written as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) \\
=
\alpha(-1/2) \beta(-1/2) \beta(-1/2) \alpha(-1/2)
+
\alpha(-1/2) \beta(+1/2) \beta(-1/2) \alpha(+1/2) \\
+
\alpha(+1/2) \beta(-1/2) \beta(+1/2) \alpha(-1/2)
+
\alpha(+1/2) \beta(+1/2) \beta(+1/2) \alpha(+1/2) \\
= 0 + 0 + 0 + 0 \\
= 0 \, ,
$$
where all the terms vanish due to at least either $\alpha(-1/2) = 0$ or $\beta(+1/2) = 0$, so that we indeed get
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
=
0 \, .
$$

It can be noted that there is a shortcut in proving that the coefficient above is zero that uses the defining properties of spin functions. First, we could rearrange spin functions as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s1})
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\beta(m_{s2}) \alpha(m_{s2}) \, .
$$
Secondly we note that by the very definition of $\alpha$ and $\beta$ presented above,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \alpha(m_{s}) \beta(m_{s})
  =
  \alpha(-1/2) \beta(-1/2) + \alpha(+1/2) \beta(+1/2)
  =
  0 \cdot 1 + 1 \cdot 0
  =
  0 \, ,
$$
which is already enough to establish that the whole exchange integral vanishes. The same is, of course, true for the second factor as well,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \beta(m_{s}) \alpha(m_{s})
  =
  \beta(-1/2) \alpha(-1/2)
  +
  \beta(+1/2) \alpha(+1/2)
  =
  1 \cdot 0 + 0 \cdot 1
  =
  0 \, .
$$
It is also quite customary to use Dirac bracket notation for such expressions over spin orbitals, so that one could write these findings concisely as
$$
  \langle \alpha \lvert \beta \rangle
  =
  \langle \beta \lvert \alpha \rangle
  =
  0 \, ,
$$
where $\langle \alpha \lvert \beta \rangle$, for instance, is defined as follows,
$$
  \langle \alpha \lvert \beta \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\alpha}(m_{s}) \beta(m_{s}) \, .
$$
In general, for two spin functions $\gamma_1$ and $\gamma_2$, expression $\langle \gamma_1 \lvert \gamma_2 \rangle$ defined as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\gamma_1}(m_{s}) \gamma_2(m_{s}) \, ,
$$
is zero if the functions correspond to different spin states and one otherwise which can be expressed as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  =
  \delta_{\gamma_1 \gamma_2} \, .
$$

Finally, I would like to make a short comment on "integrating" over discrete spin coordinates rather then summing up over it. One could indeed define
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \int \bar{\gamma_1}(m_s) \gamma_2(m_s) \mathrm{d} m_s \, ,
$$
and then think of integration over discrete $m_s$ variable being reduced to summation. But the integration here is just a "symbolic shorthand" since, strictly speaking, we have a summation over discrete $m_s$ variable from the get go. We could just symbolically write it down as an integration as well, if we think it looks more cute for some reason.
In exact same way the exchange integral can also be symbolically defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\iint
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{x}_1 \mathrm{d} \vec{x}_2 \, ,
$$
where the "integration" is informally done over joint spin-spatial coordinates of the electrons. And again, strictly speaking, a summation is done over the spin coordinates and an integration over spatial ones, as it is written explicitly in the very beginning of the answer.

Actually, I think \left| and \right| works best.

@orthocresol, MathJax appearance is browser dependent: what looks nicer in one, might looking worse in another. But let me check.

@orthocresol, mmm... Wait! Where exactly do you want to use \left| and \right|?

I see... which looks good to you? http://meta.chemistry.stackexchange.com/a/2989/16683 My reasoning was, the braket package in $\LaTeX$ defines it that way too.

Basically all the brackets. But I'm not too sure how to do it for <a|b>; it'll probably return an error since the \left isn't closed by a \right.

Just to sum up what you have done. You state that $\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}1) \alpha(m{s1})
\bar{\psi}2(\vec{r}2) \beta(m{s2})
r{12}^{-1}
\psi_2(\vec{r}1) \beta(m{s1})
\psi_1(\vec{r}2) \alpha(m{s2})$ is the sum of all possible spin configurations of the two-electron system. Your conclusion is that "at least either $\alpha(-1/2) = 0$ or \beta(+1/2) = 0". Maybe I am dimwitted, but I don't understand why. Why must at least one of these be zero? I feel the Pauli principle somehow is related.

And by the way, I believe you forgot the operator in the first term of the sum of the spin configurations.

@AndersMB, true, $r_{12}^{-1}$ was missing from the first term. Now to the actual business. It is not me who state the definition of exchange integral, it's the quantum theory. With respect to $\alpha(−1/2)$ as well as $\beta(+1/2)$ being bot zero, they are both zero by definition. And all the terms vanish due to either $\alpha(−1/2) = 0$ (1st term) or $\beta(+1/2) = 0$ (4th term), or both (2nd and 3rd terms). In the later case just one of these being equal to zero is enough in principle for the whole term to vanish. That's what I mean by "at least".

@orthocresol, does it look better now? :)

It most certainly does, meow `(.•.)´

@Wildcat, I am a little confused as to that $\alpha ()$ and $\beta$ represent in the equation. $\alpha$ and $\beta$ is usually a shorthand for +1/2 and -1/2, respectively. Is it so here as well? Is the point then that since $\alpha$ is defined to be spin +1/2, then it cannot be spin -1/2, and vice vera for $\beta$?

@AndersMB, no, $\alpha$ and $\beta$ are not just "labels" as you called them in the body of the question, they are special piecewise functions defined appropriately to describe the spin state of an electron. I will add some information on that in a minute.

Thank you for clarifying. My understanding of basic terminology is not sufficient, it seems.

@AndersMB, which book do you use, by the way?

2nd edition of Leach's "Molecular Modelling: Principles and Applications".

@AndersMB, there you go then, page 35, lines 10-12: "The spin part defines the electron spin and is labelled $\alpha$ or $\beta$. These spin functions have the value $0$ or $1$ depending on the quantum number $m_s$ of the electron. Thus $\alpha(\frac{1}{2})=1$, $\alpha(-\frac{1}{2})=0$, $\beta(\frac{1}{2})=0$, $\beta(-\frac{1}{2})=1$.


text:
Calculating enthalpy of dissolution

In a polystyrene cup calorimeter, $4.3\ \mathrm g$ of ammonium nitrate, $\ce{NH4NO}$, was added to $60.0\ \mathrm g$ of water and stirred to dissolve the solid completely. The initial temperature dropped from $22.0\ \mathrm{^\circ C}$ to a final temperature of $16.9\ \mathrm{^\circ C}$.
Calculate the enthalpy change in $\mathrm{kJ\ mol^{-1}}$ for this dissolution process, as represented by the chemical equation below:
$$\ce{NH4NO3(s) -> NH4NO3(aq)}$$
Assume that the calorimeter does not absorb any heat, that the density of the solution is the same as that of water $(1\ \mathrm{g\ ml^{-1}})$ and that the specific heat capacity of the solution is also the same as that of water $(4.18\ \mathrm{J\ g^{-1}\ K^{-1}})$.
$$M(\ce{NH4NO3})=80.05\ \mathrm{g\ mol^{-1}}$$

My attempt:
$$Q=mcT = 64.3\ \mathrm g \times 4.18\ \mathrm{J\ g^{-1}\ K^{-1}} \times -5.1\ \mathrm K = -1.37\ \mathrm{kJ}$$
But this doesn't seem right because I haven't used all the information given in the question, such as the molar mass of ammonium nitrate.

So you have just worked out how much energy was taken in by the system in the dissolution process, 1.37 kJ. That's for 4.3g of ammonium nitrate. You are asked for kJ mol-1, so you need to find out how many moles of ammonium nitrate , that's why you have the molar mass of ammonium nitrate

Hi, just some notes on good practice: 1) You should write $\Delta T$ instead of $T$; the $\Delta$ sign represents a change in a quantity, in this case, temperature. 2) I think it is usually a good idea to put parentheses around your quantities, especially the negative one. Anyway, regarding the answer itself, @Leeser has already given you a nice hint, so see if you can work it out.

The question requires you to put your answer in kJ $mol^{-1}$. This means kilo joules per mole. So you have calculated the energy in kJ correctly, to get it per mole you need to first calculate the number of moles of ammonium nitrate you used.
Use the equation: $$n=\frac{m}{M_r}$$
n = moles
m = mass in grams
$M_r$ = molar mass
The mole calculation turn out to be 0.05371642723.
Then its just a matter of division, which I calculated to be -25.50 kJ $mol^{-1}$ (2.d.p).
That answer is using your rounded answer for the kJ, so the answer may be off. When i do calculations, i round right at the end to get the most accurate answer.

not quite correct, you need to get the answer in kJ mol-1, hence divide by number of moles of ammonium nitrate in 4.3g

ah yes i missed the mol-1. what an idiot. sorry OP

As you have realised, this answer as it stands is incorrect. I suggest you edit the extra information into your answer, but not before giving OP some time to try and see whether they can work it out themselves.


text:
Questions about Organic solvents insoluble in water
ــPlease introduce some organic solvent insoluble in water densities less than water.
At least ten solventــ


text:
What does "at%" refer to in sol-gel method?
I'm reading on the preparation of ZnO thin films for UV stimulated emission and I found a method called Sol-Gel. During this I found a paragraph saying, " the ration Ag/Zn in the sol is 3 at%......". so I want to know what at% refer to.

Same question on Physics SE - cross posting is generally frowned on since questions can be migrated to the better SE...

Atomic percentage. It can be converted to weight percentage (wt%) too.


text:
Error propagation Titration
I hope I am right in this section.
I am unsure with error propagation. When calculation the error in a titration, many errors has to be taken into account:
Error in Glassware/ Error in Balance/ Error in Burette etc.
I learned that the absolute and relative error have only 1 significant figure and that the total amount is rounded to the decimal place of the error.
Therefore   5.34532g ± 0.001428g  would be 5.345g ± 0.001g
The relative error 0.001g/5.345g = 0.00018709 = 0.0002
If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?
Wouldn't be rounding the error just in the end make more sense?
Many thanks in advance

Related: Molar concentration calculation with absolute uncertainty

If you (or any future reader) want to go for a deeper dive, not necessarily for the original problem, but for something in the future, check this out: EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012. Just search on the web. This is just for future reference, since @MaxW gave a nice answer years ago.

I learned that the absolute and relative error have only 1 significant figure

That really isn't  true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant.
You then simplified to 5.345g ± 0.001g which is fine. My personal preference would be more like 5.345g ± 0.0014g since ± 0.00067g would round to ± 0.001g as well. So I'd prefer one more digit in the ± "precision" than the measurement itself.

If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?

Yes absolutely. Carry "extra" significant figures until the final step.

Wouldn't be rounding the error just in the end make more sense?

That is the preferred method.
We typically use a calculator now to do the calculations. You can set the calculator to show 3 significant figures but it carries more internally.
There are various statistical techniques which allow all the measurements used in the calculations to have random errors injected and then the  calculation is simulated many times to see how the errors propagate through the calculations.

Your instinct about not rounding early is good. Before I get to that, I wanted to address one thing in another answer:
"> I learned that the absolute and relative error have only 1 significant figure
That really isn't true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant."
Theoretically possible, but pretty unlikely. To the best of my knowledge, even the National Institute of Standard and Technology never give more than 2 digits of uncertainty because the number of measurements required to give a 3rd digit any meaning is unreasonably large.
The 1 significant figure in error "rule" is pretty good. There are two exceptions. 1) If the uncertainty is very well known (because it was calculated based on a large number of measurements), two sig figs can often be justified. 2) If the first digit in the uncertainty is a 1, two sig figs are usually justified. Think of it this way: Rounding ±0.16 to ±0.2 changes the uncertainty by 25%, while rounding ±0.86 to ±0.9 changes the uncertainty by only 5%. In many cases, the uncertainty is known to better than 25%. It's rarer to know it to 5%.
So, one or two sig figs are justified in the answer, but what does that mean for rounding? A good guideline is to keep at least one more digit than is significant until the last step. There's nothing really wrong with keeping even more, they just are unlikely to make a difference in the final (rounded) result.


text:
NaOH solution not stable
$\ce{NaOH}$ solution can react with $\ce{CO2}$ to form sodium carbonate.
Is there any other reason why $\ce{NaOH}$ solutions for accurate titrations cannot be stored for long (or the titre has to be updated every time). Wouldn't filling the vessel with argon fix the problem.

This one reason is quite enough. As to filling the vessel with argon, it will surely solve this problem and create many more.

NaOH is hygroscopic, ie it absorbs water from the atmosphere. So over time your solution will become more dilute so can't be used over a period of time. Therefore it cannot be used as a primary standard. Usually you determine the molarity of your NaOH solution just prior to its use. If you carry out your titration under argon or other dry atmosphere, sure use it but not hta feasible!

Thank you Leeser, Is NaOH solution  still hygroscopic in solution as it is already dissolved in water or  it does not matter.

Your questions has two a two part answer.
(1) How would you get the $\ce{NaOH}$ out of a sealed bottle?
(2) A concentrated stock solution can be made for $\ce{NaOH}$. In the top of the bottle you use a drying tube stuffed with solid $\ce{NaOH}$ to remove $\ce{CO2}$ from air. On the bottom you have an outlet from which you pour the $\ce{NaOH}$. The gist is that $\ce{Na2CO3}$ is the least soluble carbonate. So most of the $\ce{CO3^2-}$ in the original solution participates. So if you let the solution sit for a couple of days it will be very stable.
The point with the stock solution as outlined is that $\ce{Na2CO3}$ "sticks" to inside of the container and doesn't flow out with solution. (You don't shake bottle up like you do with tomato juice that you drink to resuspend the solids.)
So for (1) it is high near impossible to make a $\ce{NaOH}$ solution without some $\ce{CO3^2-}$ in it. The $\ce{NaOH}$ itself has $\ce{CO3^2-}$ unless it is specially made, sealed in argon and so on.
So your stock solution is concentrated. I boil some distilled water to get rid of $\ce{CO2}$ and seal it till it cools. I then take $10~\mathrm{ml}$ of my stock $\ce{NaOH}$ solution and dilute to $250~\mathrm{ml}$ (or whatever ratios) to get the desired concentration of $\ce{NaOH}$. (That I can check with KHP titration.)

$\ce{NaOH}$ also attacks glass and quartz containers. Before the advent of plastic bottles, that was an additional reason to prepare a solution just before use. BTW, I've had to deal with glass bottles in which solid $\ce{NaOH}$ pellets absorbed enough moisture to fuse the cap onto the bottle!


text:
How many valence electrons does a copper atom have?
The valence electron configuration of $\ce{Cu}$ is $\ce{4s^1 3d^{10}}$. According to this the answer is $11$, but shouldn't it be $1$ since the d shell is already filled up?

Yes, copper only has 1 valence electron. Remember: valence electrons only include the electrons in the highest energy (n) shell.

How does it form Cu(II) then?


text:
Relationship between coordination numbers and stoichiometry
In fluorite ($\ce{CaF2}$), the coordination number of $\ce{Ca}$ is $8$, and that of $\ce{F}$ is $4$. Is there any logic behind the ratio of coordination numbers being the same as the stoichiometry of the ions?

I think it isn't a coincidence that the ratio of cations to anions is 1:2, and the coordination is 8:4. You can have other $\ce{AB2}$ structures, like rutile where the coordination is 6:3. Also, please improve your grammar.

You didn't explained why that happens?

You can, to some extent, rationalise it by looking at the unit cell. For a highly regular structure like that of fluorite, if you reduce the coordination number of Ca to 7, for example, then the ratio of Ca:F will no longer be 1:2. However, I do admit that that is not quite the best "proof".

I just want to know why the ratio of coordination numbers is same as that of ratio of stoichiometric coefficients

Is explanation related to the formula of compound

Surely there is a logic behind that, and a pretty simple one. Let's take a crystal and count all $\ce{Ca}$ atoms in it. We'll get some huge number N (possibly about as huge as $N_A$). Now how many fluorine atoms are there? Each $\ce F$ is a neighbor of some $\ce{Ca}$, so if we just count all neighbors of all $\ce{Ca}$, we won't miss anything... and will end up with 8N. But wait, each $\ce F$ is a neighbor of four $\ce{Ca}$, so it was counted 4 times! We divide by 4 to compensate for that and happily arrive at 2N, which is consistent with the formula $\ce{CaF2}$.
So yes, it is no coincidence that the ratio of coordination numbers is also the ratio of stoichiometric coefficients.


text:
Joule–Thomson effect: why does temperature decrease during expansion
When a gas under a high pressure in a piston (isolated from surrounding is allowed to pass through a valve), why does its temperature decrease once during expansion?
I haven't still studied thermodynamics but would like to know why exactly the temperature decreases.
Is it because under high pressure the heat was concentrated under small volume once expanded the heat is distributed?
Once the gas molecules are allowed to expand through the valve its kinetic energy increases. Where is the source of energy coming from, is it from its internal energy?

I answered this a while ago, and wrote out all the maths take a look: http://chemistry.stackexchange.com/questions/61517/reason-for-negative-joule-thomson-coefficient-of-helium-and-hydrogen-at-ntp-cond?answertab=active#tab-top

In order to expand, the molecules may do work, such as pushing against a piston, pushing against the container (as in water rockets), spinning a turbine, etc. Energy is transferred from molecular motion to another form, so the temperature drops.
In free expansion, however,

the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but the temperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.

[from Wikipedia].

If a real gas undergoes a free expansion, energy must be supplied in order to overcome the attractive forces which are not neglected in this case, this energy comes from the internal energy of the molecules. As a result, the molecules temperature decreases. What about the ideal gas, where does the energy come from in order to pass through the valve?.

No, no energy is applied from outside: the kinetic energy of the molecules is the source. See the demo at https://www.youtube.com/watch?v=tEFHkcx2cz0

What i have realised from the idea once the gas(blue) expands its temperature drops, because it uses some of its internal energy (kinetic in this case) thus its temperature drops.

What about real gases?