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Can I react a Grignard reagent (ethylmagnesium bromide) with water and then ammonia?
I came across this question where I had to convert $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$.
If I need to do so can I just react the given compound with water to get ethane and then react it with $\ce{NH3}$.
If the answer is no then why can't I?
Is there any set of principles in organic chemistry which govern as to which substances can be replaced with what?
Kind of like the reactivity series of metals?
For e.g.: If I react an alkane with a halogen to get a haloalkane and then react this back with hydrogen to get the alkane then I don't think so it would be correct to do so. And I don't think that organic chemistry works that way.
But how do I understand while solving a question whether I am going in the right direction.
Please help.
NB: This is not a homework question its just a question I stumbled across while my preparation for a college entrance test. (edit the title maybe)

A known method of producing primary amines directly from Grignard reagents is reaction with NH2Cl in ether. It makes more sense to convert the original alkyl bromide into the amine using a nitride/reducing agent or even the Gabriel synthesis.

I came across this question where I had to convert
  $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$. If I need to do so can I
  just react the given compound with water [...]

Always be cautious about unwanted, side reactions that may occur. Proton exchanges are one huge class of possible side reactions. Protons transfers occur readily in many circumstances because the protons is small and can be shuffled around without much to get in its way, physically.
So in the case of your hypothetical reaction between a Grignard reagent and water - realize that Grignards are strongly basic and therefore will deprotonate water. Unfortunately protonated Grignards are no good in terms of reactivity. They just fall back apart into hydrocarbons and a basic halide salt.
Further reading regarding Grignards - It is tempting to use Grignards to create carbon-carbon bonds, but anything that is acidic as water or more acidic will be deprotonated by the Grignard before any carbon-carbon bonds are created. When preparing Grignards, glassware must be perfectly dry. Not just wiped dry with a paper towel but heated dry. Grignards also live in ether solvents since ethers are only very weakly acidic.
Grignards may be used to attack aldehydes and ketones in creating carbon-carbon bonds, but realize that if the aldehyde or ketone contains a functional group that is about as acidic as water or more so - then the first equivalent of Grignard you add isn't going to the aldehyde or ketone carbon - it's going to perform a proton-transfer.

Is there any set of principles in organic chemistry which govern as to
  which substances can be replaced with what?

In general, it is good to keep in mind the motif nucleophile attacks electrophile in introductory organic chemistry.
Exceptions to motif: nucleophile attacks electrophile
Other heuristics you can use in predicting what reactions might commence ...

Bond energies. This is mainly useful for studying radical reactions - to which the motif nucleophile attacks electrophile doesn't apply as well. Are we going from weaker to stronger bonds? That's a favorable move, enthalpically.
Polarities. This falls under nucleophile attacks electrophile.
Kinetics. It's good to learn the difference between thermodynamics and kinetics early in organic chemistry. Sometimes chemical entities may be thermodynamically unstable but kinetically stable. Maybe a carbon is particularly electrophilic, but it's too well "hidden" by neighboring groups (steric hindrance). So will a nucleophile attack this carbon? Probably not to a great extent.

The short answer is no, you can't.  Water will hydrolyze a Grignard reagent, destroying its reactivity.  The product of the hydrolysis, according to this web site is an alkane and a basic magnesium bromide:
$$\ce{CH3CH2MgBr + H2O -> CH3CH3 + MgBrOH}$$
So before you ever got to reacting anything with ammonia, your reactants would be lost as ethane gas bubbles out of your reaction.

I agree to this.my other questions?

Alkanes react only with very powerful oxidazing agents, like chlorine or oxygen, because they are very unreactive (due to their weak polarity).

Yes, there are such principles in organic chemistry and they are based most of the times on the reactivity of the substante. (acidity, polarity, aromaticity).

Yes, a Grignard reagent can react with ammonia like the following:
$$\ce{RCH3Mg + NH3 -> R-H + MgNH3OH}$$

I made a minimal modification of your equation but I couldn't actually get what you were trying to say. Could you edit it? And welcome to chemistry.SE!


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Conductivity as a function of acid concentration
I have conducted an experiment measuring the conductivity of both hydrochloric acid and sulfuric acid in solution respectively, with varying concentration. Wikipedia as well as a question on the site suggests a $\sqrt{c}$ relationship is viable. However, my data resembles this:

Clearly not a simple $\sqrt{c}$. It reminds me of the inverse gamma distribution. I have tried looking at papers by Onsager on his conductivity theory, but they're unclear, and he doesn't propose a function. Are there any papers that look at accurately modelling conductivity as a function of concentration which exhibit this form? Obviously for low concentrations (e.g. 1-2 molar), a linear or $\sqrt{c}$ would suffice.

At low concentration, conductivity is proportional to concentration (a linear relationship).
Each ion will have its own unique mobility, as discovered by Kohlrausch. $\ce{H+}$ has the highest mobility. As you can see in your graph the acids have higher conductivities than the salts. $\ce{OH-}$ is also highly mobile. As concentration increases, the linear relationship breaks down for two reasons.
Firstly, in infinitely dilute solution, for the strong electrolytes in the graph, there is complete dissociation into separately solvated ions. However, as concentration increases, a portion of the electrolyte exists as ion pairs.  See for example Equations for Densities and Dissociation Constant of NaCl(aq) at 25°C from “Zero to Saturation” Based on Partial Dissociation J. Electrochem. Soc. 1997 vol. 144, pp. 2380-2384.
Secondly, the mobility of the ions that are solvated is decreased by the fact that they are no longer moving through water, but are instead moving past other ions as well.
To have a maximum in the curves of the question, and to account for the above factors, it is necessary to subtract a term from the linear term.
You need a function of the form:
$$\text{Conductivity} = Ac - Bf(c),$$
where $A$ and $B$ are constants, $c$ is concentration, and $f(c)$ is some function of concentration. Historically, the function $Ac - Bc\sqrt{c}$ was the first and most simple to use.
I would start by trying to fit your data to that function.
More advanced treatments replace $\sqrt{c}$ with $\sqrt{I}$, where I is ionic strength. Further advances involve higher order $I$ in addition to $\sqrt{I}$, such as $I \ln (I)$, $I$, and $I^{3/2}$.
The $I \ln (I)$ and $I$ terms arise upon considering the ions as charged spheres of finite diameter, rather than a simple point charges, as explained in Electrolytic Conductance and Conductances of the Halogen Acids in Water and references cited therein. This is the reference to look at if you want the best function to fit the HCl curve in your graph as it includes numerical coefficients.
For more information see The Conductivity of Liquids by Olin Freeman Tower, which although not the most recent work, has the advantage of being understandable, or Calculating the conductivity of natural waters.
For a very advanced consideration, see "Electrical conductance of electrolyte mixtures of any type" Journal of Solution Chemistry, 1978, vol. 7, pp. 533-548.

I completely agree with the discussion about ion pairing and the decreased mobility. We study charge transport, and similar curves exist - at high concentration, the mobile charges become "frustrated" because of electrostatic interactions.

I also wonder whether, particularly with sulfuric, if there's a density/viscosity effect at work at high concentrations.

@GeoffHutchison viscosity effect is mentioned in this document: http://www.nist.gov/data/nsrds/NSRDS-NBS331.pdf  I'll try to add more to my answer from there.

Your carefully plotted results seem in line with this table from the Foxboro Company. Though there are references that give a simple square-root relationship, they are clearly wrong at high concentrations of many electrolytes.
There is a practical guide to conductivity measurement at Conductivity Theory and Practice, which mentions:

For some samples with high concentrations, the conductance = f (concentration) curve may show a maximum.

Wikipedia on Conductivity states:

Both Kohlrausch's law and the Debye-Hückel-Onsager equation break down as the concentration of the electrolyte increases above a certain value. The reason for this is that as concentration increases the average distance between cation and anion decreases, so that there is more inter-ionic interaction.

So, as you've done, actual conductivity curves are often determined experimentally, and formulae then are derived to fit the curves.

Yes, all of this is clear. The problem is finding a function to fit to the curve. Is there a more sophisticated theory that suggests a function, rather than I having to guess? Because there are many that can fit the curve 'decently.'

There are many tools to derive a formula to fit a curve. E.G. using Excel (http://www.cpp.edu/~seskandari/documents/Curve_Fitting_William_Lee.pdf), or MATLAB, online in Wolfram using it's Fit language (http://reference.wolfram.com/language/tutorial/CurveFitting.html), or this Regression Fitter (http://www.had2know.com/academics/regression-calculator-statistics-best-fit.html)

That's not the problem; I have Mathematica that is capable of fitting any function to a curve, and determining the parameters. My point is that several functions may fit. I'd like a theory which derives or motivates a function, so that my choice of function is justified.

Sorry for the cop-out, but all the literature I've come across is pragmatic, without full numerical analysis for all the competing factors affecting conductivity. Determine the applicable range of concentrations, decide how close the fit must be, and see if you can derive a formula that fits within an acceptable limit. If someone out there has a better theoretical grasp, please answer.

To expound on DavePhD's nice answer, molar conductivity follows Kolrausch's law, $\Lambda_m=\Lambda_m^{\circ} - \kappa \sqrt{c}$ at low concentrations, which means that extensive conductivity follows $\Lambda_m c = c\Lambda_m^{\circ} - \kappa c \sqrt{c}$ in this domain.
As you've discovered this relationship breaks down at high concentration.  In a loose analogy to the virial equation of state for gases, I'd propose fitting your data to a modification of Kolrausch's law, specifically, to an expansion in $\sqrt{c}$:
$\Lambda_m=\Lambda_m^{\circ} - \kappa_1 \sqrt{c} - \kappa_2 (\sqrt{c})^2 - \kappa_3 (\sqrt{c})^3 + ...$
I'm not enough of Debye-Huckel theorist to know if this expansion has the same theoretical justification/derivation in terms of intermolecular forces that the virial equation for gases does.  But, (a) it definitely introduces extra degrees of freedom that you will need to fit your more complex data, and (b) if your data did follow Kolrausch's law (and the data for e.g. sodium chloride doesn't look too far off, at least by eye), your fit would tell you so, meaning you wouldn't have to convert or translate your fit parameters to a new system of units to compare with tabulated values for Kolrausch's law -- your fit would tell you the Kolrausch parameters.
UPDATE: I dug around in the literature a bit.  Wikipedia pointed me towards this 1960s JACS paper.  There they give the Fuoss-Onsager equation, which can be written as:
$$\Lambda_m=\Lambda_m^{\circ} - \beta_1 \sqrt{c} - \beta_2 c - \beta_3 c \ln{c}$$
The $\beta_i$ parameters are expanded in the paper to show dependence on some underlying solvent and solute properties, but that isn't important for fitting compound-specific curves to your conductivity data.  You can find papers on the Fuoss-Onsager equation and cite them if you like.  And this equation still has the advantage that is reduces to Kolrausch's law for $\beta_2 = \beta_3 = 0$.  So if that seems better to you than the expansion in $\sqrt{c}$, try that one.

I think the question would be why there are higher-order terms.

The easy answer is because the assumptions of Kolrausch's law break down at high concentrations...but I gather you (and the OP?) are asking why that is?  That is a very different question than "what is a justifiable way to fit my experimental data".  I worry that the OP wants a one-equation, globally applicable summary of the theory of the liquid state.  That's an impossibly tall order.

I can't speak for the OP, but I suspect even a sense of the leading corrections (e.g., in Dave's answer) would be helpful. Clearly you can find an empirical fit, but the question would be what the higher order terms mean physically.

I updated my answer to explicitly refer to Fuoss-Onsager.

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Does the added component (intended to disturb equilibrium) always ends up being more than the original amount, in the new equilibrium?
The title can be a little unclear. But what I am basically saying is say we add a reactant to a reaction to push the reaction forward toward the product side, that reactant which was increased in amount/ concentration will surely less. Part of it used to produce more products to gain new equilibrium. But will it always be more than the original amount in the equilibrium before?
Without knowing much of advanced math, I just experimented with some simple fractions; thus far I haven't seen anything that violate this rule.

Are you considering Le Chatelier's principle ? And do you mean "increased in amount/concentration will surely become less" ?

Yes la chatier. What I meant was the component of a reaction that is being increased in amount to push the reaction in either one of the two directions.

Is the amount of component always gonna to be more at the new equilibrium than was originally.

Okay , got it now . Its a very good question and the answer is Not Always . I am giving you a hint to think about : "What happens if there are only a solid and a gas in the reactant side ? Write down the eqn and calc Eq. Const expression for each case " . If you get it from here well and good . If not , I will answer the question .

Did you get it then?

Sorry I couldn't get it.

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Will all thermodynamic equations have a linear dependence with moles?
I'm taking chemical thermodynamics this semester, and currently seeing things like the Maxwell's Relations.
My professor never uses explicitly the variable "n" and always works with molar quantities like $\bar{V}$ and stuff like that.
I think this simplifies the algebra a great deal, and can be useful if you have the temptation to get a "molar quantity". However, I don't like it right now.
I think theres nothing that can assure me that if I were to work with the variable $n$ in my derivations, I won't somehow get a quadratic dependence of my thermodynamic equation with $n$, or logarithmic, who knows.
But it seems that there is always a linear dependence with $n$. This makes physical sense to me, however, how can I prove this to myself? Maybe this is more of a question of mathematics, but it has scientific implications as well.
Thanks.

Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.
Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is:
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$
We choose $\lambda = 1/N$, so
$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,
$ N \,S(e, v, 1) = S(E,V,N)$
So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):
$S(E,V,N) =$ $whatever$
If you work with molar properties you will find just:
$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:
$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$
So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.
Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.
EDITED:
After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of.

One exception to the general rule that equations will be linearly dependent on moles is if surface tension is to be modeled.  With surface tension, we have to specify the shape of a system as well as the extent.  The fundamental equation for a single-component system including surface tension would be:
$$ d\bar{G} = \bar{V}dP - \bar{S}dT + \gamma d\bar{A} $$
where $\gamma$ is the surface tension and $d\bar{A}$ is a differential change in surface area (i.e. $\bar{A}$ is surface area per mole).
If the shape of the system is constant (imagine say a sphere of water that grows bigger and bigger as we add more water molecules, but still stays perfectly spherical), then $A$ is proportional to $N^{2/3}$, so $\bar{A}$ is proportional to $N^{-1/3}$.  The exponent of 2/3 is not one, therefore the free energy is not linear in the number of moles.
These kinds of cases are rarely studied in chemical thermodynamics at the undergraduate level, however.


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Is carbohydrate alcohol inebriant?
First of all, though I think that my question is related to this society, if you think reverse, tell me to delete the question:
I would like to know if carbohydrate alcohol is inebriant; Can anyone reference me to some valid source?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. I think I've seen you here before, though. Anyways, I think your Q is on-topic here, but better dealt with in bio.SE. I leave it for community to decide whether it's off-topic or not.

Related: http://biology.stackexchange.com/questions/13547/can-sugar-alcohols-and-glycerol-in-particular-cause-intoxication

Thanks dear for everything! however, it is my first night in this society!

By "carbohydrate alcohol", do you mean alcohol obtained by fermentation? You can also make alcohol by acid-catalyzed hydration of ethylene. It is worth pointing out that a pure chemical, such as ethyl alcohol, will behave the same regardless of how it was produced.

I'm assuming you mean "do the alcohol groups in carbohydrates such as sugars cause inebriation".
If that is what you mean then the answer is no. Inebriation is a function of some biological effects that are very specific to small alcohols not just to the OH group in a molecule (I suspect that methanol causes inebriation as well as ethanol, but given that it also kills you this is not something i'm willing to test).
Some of the biological effects are based on specific interactions with proteins in nerves, others are gross physical effects (e.g. dizziness is a product of changes to the density of fluid present in the balance organs in your inner ear when the alcohol concentration in the blood is high). The neuromodulation effects are very specific to the size and shape of the whole molecule not just the OH group and the dizziness is a function of the density of ethanol and its ability to circulate in the blood (larger alcohols and other OH-containing chemicals won't do this).
The body has a whole bunch of mechanisms to handle and manage sugars (as they are a primary way of providing fuel to cells). It would be surprising indeed if animals were inebriated by such compounds as they have evolved sophisticated ways to use them as a key part of their metabolism.


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Salicylic acid - polarity and solubility
Really confused about this, since there is carbon and two other types of atoms: should be polar, right? However, some people online say it is non-polar. Also, is it soluble in water?

I'm going to bed, but some hints, before I come back to answer (if no one else does): You should think of the shape, as well as the atoms themselves. Types of atoms aren't the only important factors in predicting polarity. An example is Carbon dioxide. And if a substance is polar, it'll assumingly dissolve in water, but that's a rule of thumb.

The shape tells me it is polar: not symmetrical at all. Also, since there is carbon and hydrogen PLUS oxygen, it should be polar. However, I don't think it dissolves in water well, so maybe non-polar?

Well it has polar groups - carboxylic and phenolic but benzene ring reduces its polarity. One could say carbon atoms win 7:3 with oxygens, but it's still not bad :) PubChem say it's solubility is about 2g/L, but this site gives more info - it dissolves much better in hot water and its solubility in methanol, which is quite polar, is good, and in benzene (non-polar) bad. Generally it seems to be somewhere in the middle between strongly polar and strongly non-polar compounds.

So would you say it has dipole dipole forces or not?

Yes, also hydrogen bonding.


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Purifying washing soda
I bought a box of A&H Washing Soda, the packaging of which implied that it was pure washing soda (sodium carbonate).
On dissolving it in osmotically-purified water, however, I found that the water took on a brown tinge, which would seem to indicate a contamination or adulteration of some type. Also, the washing soda tended to form hard chips or clumps, which I have not seen in previous experiments doing the same thing.
I suspect there may be some kind of organic compound being used to adulterate the washing soda.
How can I determine if there is an organic adulterant and remove it?

In addition to the answer given by Curt: It might be interesting to see whether the same effect can be observed using destilled water, rain water or any (tab) water that didn't go through the RO purifier.

If there is an organic contaminant in the $\ce{Na2CO3}$, you could try to rinse the sodium carbonate with an organic solvent in which $\ce{Na2CO3}$ is insoluble.  Based on this paper, acetone and methanol would qualify.  Since those solvents, especially methanol, are toxic, I'm not recommending either one of those.  Based on the data for methanol and acetone, I'm guessing absolute ethanol would also be incapable of dissolving sodium carbonate, so it would probably the best (relatively) non-toxic solvent for extraction.  If you can't get absolute ethanol due to tax or beverage laws, you could probably used denatured ethyl alcohol instead.  It usually has small percentages of methanol, acetone, and/or isopropyl alcohol added to make it non-potable.
After you wash the sodium carbonate in this solvent, you should air dry it for several days, and then dry it in an explosion-proof drying oven at 100 - 200 °C.  Be very careful when combining flammable solvents with heated environments!
The problem could also be with the water you used, although you did say you used osmotically purified water.  The reason I think the water may be the culprit is that if the washing soda contained a brown contaminant, I would think it render the powder visibly brown.  If no brownness is evident until adding the soda to the water, it makes me think the high pH from the washing soda is precipitating minerals from the water, in particular iron.

It's also possible that the brownness is from an inorganic contaminant in the washing soda.  This would be difficult to remove with organic solvents and would require a totally different purification procedure.  Recrystallization maybe?

Life is short. I would recommend getting sodium carbonate from an alternate source. It probably comes in different grades - avoid technical or practical grade and favor reagent grade if available. If the problem persists, the contamination would appear to be from the water.


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How to rationalise the difference in the melting points of acids and alcohols with inter molecular forces?
Acids: arachidic acid, palmitic acid, lauric acid, etc.
Alcohols: butan-1-ol, pentan-1-ol, etc.
The melting points of acids with $\ce{-COOH}$ at the end compared to melting points of alcohols, $\ce{-OH}$ at the end, is higher. I need to relate this to intermolecular forces. I was going to say that because of many more electrons in the acids, there are greater London dispersion forces and also more polarity (dipole dipole forces) because of the $\ce{COOH}$ at one end. Is this correct?

The ultimate goal of melting-point reasoning is to explain how intermolecular forces lead to better packing of molecules in the solid phase.
Both alcohols and organic acids have hydrogen bonds, but organic acids have stronger hydrogen bonds due to the withdrawing carbonyl group, $\ce{COO-}$. This causes the acids to interact stronger with each other. More importantly, organic acids form dimers (usually) with their hydrogen bonds (acetic acid dimer, for example). For this reason, the packing of the acids in the solid phase is significantly better than that of alcohol; hence the higher melting point. Acetic acid: $16~^\circ\mathrm{C}$ vs. $-114~^\circ\mathrm{C}$ for ethanol.

Thanks! If I say that there are more london dispersion forces as well because of higher number of electrons would that also be correct?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].
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@NoobCoder Here we are comparing two classes of compounds: acids and alcohols, not individual molecules, so to compare the number of electrons is probably a vague comparison. Besides, suppose we're comparing acetic acid and ethanol, the acetic acid would have more electrons than ethanol because it has an extra oxygen. The extra electrons may increase that polarizability a bit for acetic acid, but it would help if the acetic acid molecules don't come close together. It is the dimerization that helps the molecules pack closer together and increase the melting point.

Huy Nguyen has already addressed the importance of dimers through hydrogen bonds in the solid state of alkanoic acids.
If you want to correlate the chain length of alkanoic acids of the form $\mathrm{{C_{n}H_{{2n+1}}-COOH}}$ with their melting points, a look at the data might be helpful:
\begin{array}{lrr}
\mathrm{(trivial)\ name} & \mathrm{n} & \mathrm{mp}\\
\mathrm{\tilde{} acid} &  & \mathrm{[°C]}\\
\hline
\mathrm{acetic} & 1 & 17\\
\mathrm{propionic} & 2 & -21\\
\mathrm{butyric} & 3 & -5\\
\mathrm{valeric} & 4 & -32\\
\mathrm{caproic} & 5 & -3.4\\
\mathrm{enathic} & 6 & -7.5\\
\mathrm{caprylic} & 7 & 16.7\\
\mathrm{pelargonic} & 8 & 12.5\\
\mathrm{capric} & 9 & 31.6\\
\mathrm{undecanoic} & 10 & 28.6\\
\mathrm{lauric} & 11 & 48.3\\
\mathrm{tridecanoic} & 12 & 41.5\\
\mathrm{myristic} & 13 & 54.4\\
\mathrm{pentadecanoic} & 14 & 31.6\\
\mathrm{palmitic} & 15 & 28.6\\
\mathrm{margarinic} & 16 & 48.3\\
\mathrm{stearic} & 17 & 41.5\\
\mathrm{nonadecanoic} & 18 & 54.4\\
\mathrm{arachic} & 19 & 75.5\\
\mathrm{behemic} & 21 & 80\\
\mathrm{lignoceric} & 23 & 84.2\\
\mathrm{montanic} & 27 & 90.9\\
\end{array}
(Source: German and english wikipedia)
Plotting melting points vs length of the alkyl substituent gives the following result:


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Safety in handling of concentrated ammonium fluoride
I recently saw a protocol for extracting organic materials that are bound in the silica matrix formed by diatoms.  The extraction protocol used 10 molar ammonium fluoride ($\ce{NH4F}$).
I briefly considered using the protocol and then endeavoring to remove the ammonium fluoride (and water) by lyophilization or with a rotary evaporator.
However, I am concerned about the safety of doing this.  My first worry is that any glass equipment or vessels in the lyophilizer or rotary evaporator systems would be damaged by the fluoride.  My second worry is that even if the equipment used did not contain any glass, dangerous solutions of hydrogen fluoride (i.e. hydrofluoric acid) could be formed in the cold trap of the equipment.
My first question is, am I right that it is a bad idea to evaporate ammonium fluoride in glass equipment?
My second (but related) question is, even if glass is not used, what are the safety risks of evaporating ammonium fluoride under vacuum?  Imagining that a glass-free apparatus could be found, is there a way to do the evaporation safely?

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.
The rate of etching at 25 degrees C is proportional to :
$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$
according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.
The relevant equilibrium constants at 25 degrees C and ionic strength of 1M are
$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$
$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$
From the second equilibrium, it can be seen that if the fluoride concentration is 10M as in the question, the more active $\ce{HF2-}$ will be present at a concentration of about 100 times $[\ce{HF}]$.
Multiplying the two equilibrium equations together, and substituting $[\ce{F-}] = 10M$:
$$\frac{[\ce{H+}]}{[\ce{HF2-}]} = 1.4 \times 10^{-6} $$
Therefore at pH = $\log (1.4 \times 10^{-6}) = 5.9$, there will be about 1M $\ce{HF2-}$, diminishing by a factor of 10 for each pH unit increase.
Experimentally, concentrated Ammonium Fluoride has a pH of 7.2 according to http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf (which also says to avoid glass).
At pH 7.2, [$\ce{HF2-}$] is 0.05M.
In conclusion, there will be a significant concentration of the most active etching species, $\ce{HF2-}$, present in 10M ammonium fluoride and I would be concerned about glassware.

Note that I don't have first hand experience on this - I never used ammonium fluoride to release organic material from diatomeous earth, but I frequently used tetrabutylammonium fluoride (TBAF) in dichloromethane THF to cleave silyl ethers.

Cleaving silyl ethers with TBAF

To a solution of 3.8 mmol of a cyclopentylsilylether in 50 mL THF at 0 °C under argon, 4.5 mL of a TBAF solution (1.0 M in THF) were added. The mixture was stirred for 10 min, allowed to come to RT and the stirred for another 10 min.
  Then, 100 mL diethyl ether were added.

The organic layer was extracted twice with each 20 mL water and then dried over sodium sulfate.

To my knowledge, release of $\ce{HF}$ from $\ce{NH4F}$ (we are not talking about $\ce{NH4HF2 =\ NH4F*HF}$ here) is only to be expected under the following two conditions:

acidification
heating $\ce{NH4F}$ up to its melting point - sublimation is accompanied by decomposition.

Thanks Klaus.  Was TBAF incompatible with glass or was using it with glass OK?  I recognize that the story with ammonium fluoride might be different, but am just trying to gather as much information as I can.  (And yes, I'm talking about the neutral salt not the acid salt.)

@CurtF. TBAF was perfectly fine with standard lab glassware.

@CurtF. I wrongly remembered dichloromethane as the solvent. If actually was THF. I added the procedure from an old manuscript.

I think that because ammonium is a much stronger acid than tetrabutylammonium, in ammonium fluoride solution there is a much greater opportunity for F- to get a proton and create the active etching species HF and HF2- .  Especially in THF, there is no possibility of F- getting a proton from tetrabutylammonium.

@DavePhD Good point! That makes sense!

authors: مجاهد الكليبي


text:
Finding molarities of individual reagents given the chemical equation and total concentration in a container
I am given a table that gives me the different total pressures at certain times. The decomposition of phosphine is the reaction:
$$\ce{4 PH3(g) <=> P4(g) + 6 H2(g)}$$
The time/pressure values are:
\begin{array}{rr}
\mathrm{time\ [min]} & \mathrm{pressure\ [Torr]} \\
\hline
0 & 100 \\
40 & 150\\
80 & 167\\
100 & 172\\
\end{array}
This reaction takes place at 950K. I can then find the molarity of the total gas mixture at each time using $$\frac{n}{V}=\frac{P}{RT}$$
I need to know how to get the pressure of the $\ce{P4}$ for each individual time. How would one go about doing this?

The assumption you need to make is that at time zero, only $\ce{PH3}$ is present.  So relative to the moles at time zero, or $N_0$, you should find $\Delta N = N(t) - N_0$, the total new moles formed by the reaction since it began.
The reaction converts 4 moles of gas to 7.
So pressure increases mean the reaction is moving to the right.  Every mole of pressure increase must be $\frac{6}{7}$ due to $\ce{H_2}$ and $\frac{1}{7}$ due to $\ce{P4}$ formation.
So just find the total new moles of gas and multiply by $\frac{1}{7}$.

So for time equal to 0 the concentration of P4 is 0 as well, correct?

That's the assumption I would make.


text:
Conc. vs Time Graph for Second Order Reaction
From my understanding plotting the conc. over time for a 0-order reaction yield a negative linear slope. Plotting conc vs time for a first order reaction intuitively yields an exponential type slope. however, my teacher never covered the looks of a conc vs time graph for a second order reaction. What may this look like. In conjunction, what does it look like when we graph its rate over concentration?

Let's take a single-component second-order reaction:
$$\ce{2A -> B}$$
If it is a 2nd-order reaction, the rate will be proportional to the concentration of $A$, squared:
$$\frac{dA}{dt} = -kA^2$$
That equation can be integrated:
$$\frac{dA}{A^2} == -k dt$$
$$-(A^{-1}-A_0^{-1}) = -k(t-t_0)$$
And solving for $A$:
$$A = \frac{1}{k(t-t_0) + A_0^{-1}}$$
That is the equation of a hyberbola.  It goes to zero more slowly than a the expression for a first-order reaction, i.e. an exponential decay.  This makes sense because as $A$ gets more and more diluted as the reaction occurs, collisions between $A$ molecules become rarer and rarer.
See this web page for more information.


text:
Is the dissolution of sodium acetate trihydate endothermic?
Sodium acetate trihydrate dissolves in water to its constituent ions:
$$\ce{NaOAc.3H2O (s) ->[H2O] Na+ (aq) + OAc- (aq) + 3H2O (l)}$$
The crystallisation of sodium acetate from a supersaturated solution is well-known to be exothermic. Since dissolution is the reverse of crystallisation, should the enthalpy of dissolution be positive (i.e. an endothermic process)?

I don't particularly like calling the melting of a hydrate salt dissolving.

The process of dissolving sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ in water is endothermic. The molar enthalpy of solution at $T=25\ \mathrm{^\circ C}$ is $\Delta_\text{sol}H^\circ=19.66\ \mathrm{kJ\ mol^{-1}}$.*
Accordingly, since crystallization is the reverse process of dissolution, the crystallization of sodium acetate trihydrate from aqueous solutions is exothermic. For practical purposes, the enthalpy of solution with a reverse sign is taken as enthalpy of crystallization.
The enthalpy of solution mainly depends on two energy contributions: lattice energy and hydration energy. Lattice energy is the energy released when the crystal lattice of an ionic compounds is formed. Conversely, energy equal to the lattice energy has to be supplied to break up the crystal lattice; i.e. this process is endothermic. Hydration energy is released when water molecules hydrate the ions; i.e. when water molecules surround the ions and new attractions form between water molecules and ions. This process is exothermic.
Therefore, whether the process of dissolving a salt in water is exothermic or endothermic depends on the relative sizes of the lattice energy and the hydration energy. If the lattice energy is greater than the hydration energy, the process of dissolving the salt in water is endothermic. Conversely, if the lattice energy is smaller than the hydration energy, the process of dissolving the salt in water is exothermic.
For anhydrous sodium acetate $(\ce{NaC2H3O2})$, the lattice energy is actually smaller than the hydration energy. The molar enthalpy of solution is negative; at $T=25\ \mathrm{^\circ C}$ it is $\Delta_\text{sol}H^\circ=-17.32\ \mathrm{kJ\ mol^{-1}}$.* Thus, the process of dissolving anhydrous sodium acetate in water is exothermic.
However, hydration of ions can also occur in crystalline solids. Sodium acetate trihydrate $(\ce{NaC2H3O2.3H2O})$ contains three water molecules for each $\ce{NaC2H3O2}$ unit in the crystal. The corresponding amount of hydration energy is already released when crystalline anhydrous sodium acetate absorbs water and is converted to crystalline sodium acetate trihydrate. Therefore, the additional amount of hydration energy that is released when sodium acetate trihydrate is dissolved in water and completely hydrated is smaller than the total hydration energy that is released when anhydrous sodium acetate is dissolved in water. Hence, the resulting value for the molar enthalpy of solution is larger for sodium acetate trihydrate than for anhydrous sodium acetate.
* “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


text:
Ksp of calcium hydroxide
What would the Ksp of calcium hydroxide be at around room temp (66 Fahrenheit) and 1 atm of pressure? I know the Ksp value varies with temperature, but is there a notable difference from 20 degrees C to 25 degrees C?

According to IUPAC Solubility Data Series Volume 52, Alkaline Earth Hydroxides in Water and Aqueous Solutions (1992), the best fitting to experimental solubility values of calcium hydroxide in water was obtained with a three-parameter equation:
$\ln\left({b/{\text{mol kg}}^{-1}}\right) = 86.1534 - 3492.14/({T/{\text{K}}}) - 13.7494\,\ln({T/{\text{K}}})$
where $b$ is molality and $T$ is temperature.
The results have a standard uncertainty of
$\sigma \left( b \right) = 1.7 \times {10^{ - 4}}\;{\text{mol k}}{{\text{g}}^{ - 1}}$
over the temperature range from 273.15 K to 623.15 K.

For
${T_1} = 20\;^\circ\text{C} = 293.15\;{\text{K}}$
the three-parameter equation reads
$\begin{aligned}
  \ln\left({{b_1}/{\text{mol kg}^{-1}}}\right) =  & \;86.1534 - 3492.14/({T_1}/{\text{K)}} - 13.7494\,\ln({T_1}/{\text{K)}} \\
   =  & \;86.1534 - 3492.14/(293.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(293.15\;{\text{K}}/{\text{K)}} \\
   =  & \; {-}3.8651 \\
  {b_1}/{\text{mol kg}^{-1}} =  & \;0.02096 \\
  {b_1} =  & \;0.02096\;{\text{mol kg}^{-1}} \\
\end{aligned} $
In the same way for
${T_2} = 25\;^\circ\text{C} = 298.15\;{\text{K}}$
the three-parameter equation reads
$\begin{aligned}
  \ln\left({{b_2}/{\text{mol kg}^{-1}}}\right) =  & \;86.1534 - 3492.14/({T_2}/{\text{K)}} - 13.7494\,\ln({T_2}/{\text{K)}} \\
   =  & \;86.1534 - 3492.14/(298.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(298.15\;{\text{K}}/{\text{K)}} \\
   =  & \; {-}3.8978 \\
  {b_2}/{\text{mol kg}^{-1}} =  & \;0.02029 \\
  {b_2} =  & \;0.02029\;{\text{mol kg}^{-1}} \\
\end{aligned} $
Thus, taking into account the above-mentioned uncertainty, the calculated solubility of calcium hydroxide in water is
${b_1} = \left( {{\text{0}}{\text{.02096}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T_1} = 20\;^\circ\text{C}$
and
${b_2} = \left( {{\text{0}}{\text{.02029}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T_2} = 25\;^\circ\text{C}$.

Interesting related read: William B. Euler, Louis J. Kirschenbaum, and Ben Ruekberg; J. Chem. Educ., 2000, 77 (8), p 1039   (mirror).

Thank you. Ah, so the $T/K$ value was just to get rid of the units.

@Anonymous Yes. Generally, such numerical value equations are deprecated; the use of quantity equations is normally preferred and is strongly recommended. However, logarithms of units are ugly; thus, it is easier to eliminate the units and use a numerical value equation in this case. Anyway, this equation is not based on any chemical theory; it is just an arbitrary equation that sufficiently describes the experimental data.


text:
How to determine enzyme inhibitor efficacy?
At what percentage reduction or increase in Km and Vmax is an enzyme inhibitor regarded as effective?- is any percentage increase or reduction regarded as being a clinically useful level of inhibition, or would it have to be over a certain limit?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

If the inhibitor is a competitive inhibitor, to quantify inhibition the $K_m$  of the inhibitor ($K_I$) is determined.
$K_I$ is compared to the $K_m$ of the substrate.
Then, the degree of inhibition is a function the inhibitor and substrate concentrations.
$V_{max}$ is unaffected by a competitive inhibitor.
In the presence of an inhibitor the kinetics are modified such that the enzyme has an effect $K_m$ as:
$K_m(1 + [I]/K_I)$


text:
Nuclear Testing and Carbon Dating
Will the nuclear testing done during the past decades or if a nuclear war happened in the present day, affect the accuracy of carbon dating in the future? (Hypothetically, we haven't developed a better dating method)
I mean you could just adjust the dating to the amount of Carbon-14 if there will be a sudden change, but will the fossils found in around and near the blast sites be dated differently from the unaffected areas?
I'm sorry not really good in science, I don't know if this should be in Physics STEX and I might be wrong in my understandings. Thanks in advance!

The usual path to C14 is the N14(n,p)C14 reaction.  Most of the N14 will be in the air, so the C14 will disperse. One would have to run some numbers of total neutrons generated in an all-out exchange to determine the additional C14 generated above normal background.  And, as @nordmaj indicates below, the historic C14 concentrations are calibrated using, e.g., tree ring series.

If the amount of extra C14 was unknown this might become a problem. If the change in atmospheric C14 was recorded and the records survive into the future, it would be easy to adjust the dates. At present we have good data on C14 concentration over time for various locations around the world. I don't know the spatial resolution though.


text:
On biological detergent breaking down organic material
This is again related to the crime story I'm working on.
I would like to know how much biological detergent (in kilograms) and water (in liters) and how long it would take to dissolve $60 \,\rm{Kg}$ of organic material at $50^\circ$ Celsius with standard store-bought bio-detergent. Just an estimate of the time required.
Bonus questions:

Would it dissolve hair?
Do I need a special container as to not commit the same mistake Jesse had in Breaking Bad?

http://chemistry.stackexchange.com/questions/4877/is-it-actually-possible-to-dispose-of-a-body-with-hydrofluoric-acid

In Breaking Bad there was hydrofluoric acid and even there they would need more of it than what was shown, and you want to dissolve body with stuff for washing dishes?

http://chemistry.stackexchange.com/questions/3949/dissolving-organic-tissues gives additional directions, but no calculation of times.

@Mithoron Detergents for dishwashing machines (household and lab) contain significant amounts of sodium hydroxide  and therefore might help in the cleanup after the chopping.

@Klaus Warzecha But those "biological" use rather enzymes it seems.

@Mithoron You're probably right. That's why we never had them in the OC labs. Most of the time, there wasn't any automatic dishwasher at all, just the good, old tank with isopropanol and potassium hydroxide.

In the household, hair and fat residues in the pipes are typically dissolved using strongly alkaine drain cleaners.
In addition to the hints and directions given in the commentaries, namely this and that you might want to consider, at least in part and for the dramatic effect, a biological solution:
In the preparation of skeletons for museum displays, adult insects and larvae of the bacon beetle (Dermestes lardarius) are frequently used. After gutting the carcass (I'll leave the gory details to the novellist), plan a couple of weeks for them.

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


text:
How to predict the 31P NMR spectra?
What can I predict about the $\ce{^{31}P}$ spectra of $\ce{TiP2O7}$?
To get the required information about the crystal structure of the compound

Here is a screenshot from the ICSD website:

Since there is only one oxidation state (+5) for $\ce{P}$, there will be only one peak. And from the Wyckoff symbol, 4c, the intensity of the signal is 4, i.e. 4 $\ce{P}$ atoms in the unit cell.
I can't go any further to complete the task and extract all the relevant information. I'm using SIMPSON software to simulate the spectra. Can somebody walk me step by step through this?

No, you cannot derive that there is only one peak because phosphorus only has one oxidation state. If there are four atoms in a unit cell, there can be up to four peaks of phosphorous out there, depending on whether they translate to each other or not.

If the question you were asked to answer was literally "what can you predict about the $\ce{^31P}$ NMR spectrum of $\ce{TiP2O7}$ (from the crystal structure), then I believe you have vastly overcomplicated the problem.
This is a question about solid-state NMR (as $\ce{TiP2O7}$ is a solid).
In solid-state NMR of crystalline compounds, the chemical shift depends on the exact arrangement of atoms within the crystal unit cell. The number of signals (for each element) depends on how many non-equivalent atoms of a particular element there are within one unit cell.
From the screenshot, it says that "formula units per cell" = 4. As the formula is $\ce{TiP2O7}$, this indicates there are 4*2 = 8 different phosphorus atoms in the unit cell.
So the answer is:
You expect to observe 8 phosphorus signals in the $\ce{^31P}$ NMR spectrum.
There is no need to simulate the NMR spectrum with simpson. In fact, it is not possible to simulate the spectrum using simpson with only the crystal unit cell parameters. In order to set up a simulation in simpson, you actually have to supply the chemical shift and the interactions between the sites. You can calculate the interaction between the sites from the distances specified in the unit cell, but there is no information about chemical shifts!


text:
Why would this reaction occur? Won't it be unfavourable?
I was just reading about how Nitrogen and Oxygen combine to form Nitric oxide.

My question is, why would this reaction even occur? Isn't the valency of Nitrogen 3- and Oxygen 2-, and therefore this reaction seems to form a product that still hasn't achieved the octet electronic configuration.
Could I please have some help?
Thank you in advance.
Edit: Or even $\ce{N2O}$ or $\ce{NO2}$ for that matter?

NO is one of most common stable radicals.

Btw, welcome to the community of chemistry.SE! This might be interesting.

@MARamezani So the octet rule does not always hold? I assume that should be in really rare cases.

@ShashankRammoorthy Actually yes. Octet isn't something microscopic world has to always follow. They've got their own criminals you know! :) Wikipedia takes short notes on the exceptions, but I think it covers all of them. Or, you could google something like "octet exceptions", or, a better way is to search chemistry.SE!

Think about it this way.  If a molecule of nitrogen and a molecule of oxygen were to collide with enough kinetic energy to break the bonds that hold the molecules together, and they collide at such an angle that allows their orbitals to overlap and form new bonds, then $\ce{NO}$ can be made.  The octet rule is a rule.  Whenever you see the word rule in chemistry you should think that it can be broken, some rules are broken often.
The reaction is endothermic and is not spontaneous but by random chance, some $\ce{NO}$ can form in a mixture of nitrogen and oxygen.

So NO should be unstable, right? Nitrogen's valency is not satisfied.

You're right: $\ce{NO}$ is a free radical and the reaction between $\ce{N2}$ and $\ce{O2}$ can be achieved - supposed that enough energy is provided.
This may happen through an electrical arc discharge in the lab or, in the mesosphere (50-85km above earth) through ionizing cosmic rays. Note however that nitric oxide react with oxygen to eventually furnish nitrogen dioxide $\ce{2 NO + O2 <=> 2 NO2}$.


text:
Calculating concentration and volume in acid metal reactions
In the reaction:
$$\ce{3 Zn + 2 H3PO4 -> Zn3(PO4)2 + 3H2}$$
I know the initial mass of $\ce{Zn}$ to be 13.08 g and the volume of phosphoric acid to be 0.5 $\mathrm{dm^3}$. How do I work out the final concentration of $\ce{Zn(PO4)2}$ and the volume of $\ce{H2}$ gas evolved?
Once I work out the moles of Zn, are the moles of $\ce{Zn(PO4)2}$ just $\ce{Zn}$ moles/3?

Your conclusion on the molar ratio of zinc phosphate and zinc is right and you'll figure out the rest too!

Your equation uses molar ratios, rather than weights or volumes.
If you know the weight of zinc used, you know the number of moles. From this, you can calculate the number of moles of zinc phosphate generated.
Supposed that the phosphoric acid is sufficiently concentrated to react all zinc, the number of moles calculated above together with the volume of the phosphoric acid used allows to calculate the concentration of zinc phosphate. Note: Do not omit units in your calculations!
As far as the hydrogen is concerned: The number of moles produced can be calculated using the amount of zinc used. Additional knowledge of the molar volume of gases under standard conditions is helpful to figure out the volume.


text:
Why does the "gap" between atomic radii get smaller down the periodic table?
The question asks: why is the difference in atomic radii between $\ce{K}$ and $\ce{Na}$ larger than the difference in atomic radii between $\ce{Cs}$ and $\ce{Rb}$? I do have some kind of notion that the gaps between shells decrease as you go down the periodic table, but why would this be the case?

The first thing to remember is that the "shell" model is an over-simplification of quantum chemistry. A better picture is that the electrons represent "clouds" (i.e., electron density).
Your question basically boils down to "why don't 5th and 6th row atoms get much larger?"
Quantum mechanics can give a relatively easy formula for the "average" radius of a given hydrogen orbital
$$\langle r \rangle = a_0 n^2 \left( 1 + \frac{1}{2} \left[ 1- \frac{l(l+1)}{n^2} \right] \right)$$
Here, $a_0$ is the Bohr radius, the expected radius of the hydrogen atom, and $n$ is the principal quantum number and $l$ is the angular quantum number (e.g., l=0 for "s" orbitals and n=5 and 6 for Rb and Cs).
It's pretty obvious that as n gets large for s orbitals (l=0), that formula approximates:
$$\langle r \rangle \approx \frac{3 a_0}{2} n^2$$
(The left side becomes smaller and smaller, so the coefficient gets closer and closer to 3/2.)
So for a hydrogen atom (one electron), the 5s and 6s orbitals in fact get very large.
Your question is about atoms. So we have to consider many electrons. Basically, this is another representation of the lanthanide contraction caused by relativistic effects.
Indeed, in that last link, there's a formula for the atomic radius after relativistic corrections.

For atomic numbers 11 and 19, the relativistic correction is 0.99 (i.e., no real change to the radius). For atomic numbers 37 and 55, it's 0.963 and 0.916, respectively. So ~9% of the "ideal" radius is lost with $\ce{Cs}$.

Now, I'll admit I am hardly authoritative on this, but it seems to largely explain the effect. Rb->Cs would correspond to the 2nd->3rd row transition metals and that small increase in size is usually explained by the lanthanide contraction.


text:
What's the relation between the octet rule and quantum exclusion and how is it related to oxidation state?
What in the quantum nature of electrons gives rise to the octet rule? How is it related to the fact that an element cannot have an oxidation state higher/lower than the complementary to 8?

When I almost finished writing the answer I found almost identical question on Physics.SE with the accepted answer providing a very similar explanation.

What in the quantum nature of electrons gives rise to the octet rule?

Ultimately, the octet rule is justified by the fact that in many cases an $n s^2 n p^6$ electron configuration is the most energetically favorable one. Recall that electrons are arranged in different electron shells around the nucleus. Each shell corresponds to a principal quantum number $n$ and can hold only a certain number of electrons, namely up to $2n^2$. Each shell is composed of subshells each of which corresponds to an azimuthal quantum number $l$.
Now recall also that the binding energy of the electron decreases with both $n$ and $l$, and thus, once $n s$ and $n p$ orbitals of an $n$-th shell are filled, in many cases it is indeed energetically unfavorable to fill the subsequent orbitals: $n d$ if (there are any) or $(n+1)s$. The energy gain from binding an electron to an atom might not be enough to compensate a cost of separating it from a different atom.

How is it related to the fact that an element cannot have an oxidation
  state higher/lower than the complementary to 8?

Suppose an atom has $x$ electrons in the outermost shell, where with the exception of transition metals $x \leq 8$. In the limiting cases the atom will either gain $8-x$ electrons to completely fill $ns$ and $np$ subshells or lose all $x$ electrons. In the first case we will end up with the atom in $-(8-x)$ oxidation state while in the second case with the atom in $+x$ oxidation state. But these are only the limiting cases, while in principle atom can gain few electrons and loose few others at the same time. Besides, atoms of the same type can evenly share electrons with each without satisfying the octet rule without gaining or loosing electrons. Thus, usually elements have few possible oxidation states, though, in the above mentioned limiting $[-(8-x), +x]$ range.

Excerpt from Wikipedia's list of oxidation states of the elements.

I'm able to give an answer only for He, Ne and Ar and the answer has to do with the magnetic dipole moment of the involved electrons, known also as the electrons intrinsic spin.
The electrons intrinsic spin is somehow abstract while the magnetic dipole moment is well imaginable as a bar magnet. And exactly this feature gives rise to the octet rule as well as to Pauli's exclusion principle.
The argumentation goes the way that exactly 2 respectively 8 (and only 2 and 8) bar magnets could be in equilibration respectively in a stable state. For two bar magnets this is not a big deal. The bar  magnets will be faced north by south poles and following this picture one get for He one the shell with two electrons with opposite spin.
How the magnetic dipole moments of the 8 electrons of Ne would be distributed? The answer lays in an octahedral structure (The second picture shows the same of 8 symmetrical distributed axis or areas or volumes.):

Imagine for every of the four red corners a bar magnet with north pole to the centre and for every of the for yellow corners a bar magnet with south pole to the centre. That is the reason why Ne as well as Ar are stable as they are.
So far the magnetic dipole moment has not been considered (or only indirectly through the intrinsic spin), however it has to be.

authors: ألكساندر المطيري


text:
Possible to grow optical calcite?
I notice that most optical calcite for sale seems to be from natural (mined) sources.
Also, I know in World War 2, mining optical calcite was a strategic objective.
Is there some reason why high-grade optical calcite cannot be lab grown?

The crystal structure is trigonal, so has very low symmetry and high chance of twinning or other line defects forming during, say, Czochralski type growth methods.  My guess would be the good optical quality stuff sat and annealed in situ for a long time to clear most of the twins/dislocations out.
Support for this comes from A.J. Gratz et al., Geochimica et Cosmochimica Acta 57 491-495 (1993) where they observed, with AFM, calcite growth occuring primarily on screw dislocations.

This paper from 1974 seems relevant:

Optically clear single crystals of calcite up to 2 cm in length have been pulled from a Li2CO3 flux in the range of 50–75 wt% CaCO3 under a pressure of 1–15 atm of carbon dioxide. The necessary phase equilibria for this growth was experimentally verified. The best growth parameters were a pull rate of 0.25 mm/hr with a rotation rate of 28 rpm. Seed orientation and perfection are critically important.


text:
How can you predict relative Rf values of cis-trans isomers in thin layer chromatography?
This is a "homework" question.
I'm writing up a lab in which we synthesized 1,4-diphenyl-1,3-butadiene. We analyzed the products using TLC with 4:1 Hexane:Ethyl Acetate as the solvent.

One of the isomers has a larger Rf value than the other (0.83 and 0.95).

I'm having trouble understanding how to figure out which of the isomers is which on the TLC. I know more polar compounds are going to have a lower Rf, like cinnamaldehyde.
My best guess is that, although the all the double bonds are conjugated, the phenyl rings have a higher electron density than the rest of the molecule. Since the trans,trans isomer is essentially symmetrical, it has little or no molecular dipole. The cis,trans isomer has a small molecular dipole because it is asymmetric. Therefore the cis-trans isomer has more affinity for the stationary phase (giving it the Rf value of 0.83).
Am I on the right track here?

Your analysis seems logical.

Well, apparently I was wrong.
In fact, the trans,trans isomer is adsorbed more strongly by the stationary phase because it is entirely planar. The cis,trans-isomer cannot adopt a planar conformation because of steric hindrance between the two phenyl rings. The non-coplanar geometry lowers the efficiency of interactions with the stationary phase, making it more mobile.
So the cis,trans isomer has a higher Rf, and the trans,trans isomer has a lower Rf.


text:
Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent?
The question is the same as the title states. In my textbook,[NCERT Chemistry I for Class 12 , pg no 217.] the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state has half-filled $\mathrm{t_{2g}}$ orbitals; thus it is a good reducing agent.
On the other hand $\ce{Mn^{3+}}$ gets converted to $\ce{Mn^{2+}}$ as its +2 oxidation state has a half-filled d-subshell.

Now my questions are:

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$?

Why can't the half-filled $\mathrm{d}$-orbital argument be used for $\ce{Cr}$?

$\ce{Mn^3+}$ could be oxidized to $\ce{Mn^4+}$, which would have a half-filled $\mathrm{t_{2g}}$ set. As oxidation and reduction both can take place in aqueous medium, why wouldn't $\mathrm{t_{2g}^3}$ or $\mathrm{d^3}$ configuration be more stable than $\mathrm{d^5}$ configuration in aqueous medium?
A similar observation is made in case of $\ce{Fe^2+}$ and $\ce{Cr^2+}$, where it observed that $\ce{Cr^2+}$ is more powerful reducing agent than $\ce{Fe^2+}$ (due to reasons explained above).
If possible, I would like to see the crystal field splitting calculations that show this to be the case.

since you have talked about aqueous solution there is term enthalpy which dominates overall change. the difference is hydration enthalpy!

Duplicates don't necessarily have to be done in chronological order, see e.g. the answers at https://chemistry.meta.stackexchange.com/q/4611/16683, but anyway I don't plan to pronounce on which one should be closed as the other (not right now, at least; that can be a decision for later). My main point was that the questions are very similar.

http://ncert.nic.in/textbook/pdf/lech108.pdf link for the textbook

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?

Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$.  In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons.  I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons.  $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons.
You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.

Why can't the half-filled d-orbital argument be used for Cr?

Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell.  $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.

I was asking why not $\ce{Cr^+}$.

And actually, $\ce {Cr^2+} $ also has 3 $\ce {t_2g}$ electrons.

would Cr+ be d5 or s1d4?

i don't know why it would be s1d4 as d5 would be completely half filled and s is in the outer shell

from slide 16 here: http://alpha.chem.umb.edu/chemistry/ch371/documents/2.Chromiumacetate_001.pdf it seems like Cr(I) is s1d4

Won't Cr5+ be at d5 as well? How do we use that argument only for Mn3+->Mn2+ ?

Why not Mn3+->Mn4+? Then this would become a half filled t2g , right? Why can't we apply this argument?

Answer in incomplete


text:
Bipyridyl complex isomers
What type of stereoisomers do these two complex compounds represent?

These are examples of enantiomerism or optical isomerism. If you imagine a mirror placed between the two images you will see that they are mirror images of each other. If you try to rotate them to match up (easiest to do with a model, either physical or virtual) you will find that they are non-superimposable and this is what gives them their isomeric properties. The two isomers rotate plane polarised light in opposite directions (this is where the name, optical isomerism, comes from).
The $\Delta$ and $\Lambda$ labels distinguish the two isomers by their absolute stereochemical configuration.
An decent explanation of this type of isomerism can be found here.

But what type? Central, axial, planar or helicity chirality?

@Marko As you mention, there are multiple types. In this case, it's stereoisomers around a central point (the metal center).

@Marko This is an example o helical chirality.

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


text:
Replacement for optical calcite (Glan Thompson prism)?
I notice that Glan Thompson prisms seem to be the preferred optics for doing polarimetry and have been for at least 75 years. One instrument maker says that polaroid films can be used but they tend to degrade and warp over time.
It seems strange to me that no better substitute has been found for Glan Thompson prisms given their expense. I know at one point Polaroid was saying that Sodium Nitrate crystals (which can be grown) could potentially be used, but this effort seems to have gone nowhere.
Why has no cheaper alternative been found? Aren't there growable crystals that have the birefringence necessary?

There are other polarizers out there, depending on wavelength, power, etc.  For example, Thor Labs offers a fairly broad range (including calcite Glan Thompsons).

I have a pair of Glan Thompson polarizers (Karl Lambrecht) that cost a pretty penny many years ago. I also have several other (less expensive) pairs of calcite polarizers. Nothing to date can match high quality calcite in terms of low scattering, low fluorescence and low extinction ratio. In some cases, with the highest quality calcite devices, it is possible to find 'sweet spots' having $10^{-10}$ extinction ratio. (The best I ever got, after hours in a dark room, was about $8 x 10^{-9}$). In contrast, the plastic sheet polarizers are nice and convenient, but nowhere near good enough for serious polarimetry work. Sodium nitrate crystals might have excellent optical properties (I do not know anything about them), but would they need to be protected from water vapor in the air? After all, potassium nitrate is used in gunpowder and I suspect one reason is that it 'keeps the powder dry'!

Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!


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How actually do we decide acidity and basicity order?
In a book, I read two quite contradictory statements.:

$\ce{HIO > HBrO > HClO}$ wrt acidity while

$\ce{HF > HCl > HBr}$.

Aren't they contradictory? If not can anyone please explain the accurate reasoning?

HF is a weaker acid than HCl and HBr. The sign of inequality must be reverse in the second statement.

I'm not sure about the first order, but I'll try to explain the reason for the second one. The acidic strength increases as $$\ce{HF<HCl<HBr}$$
One reason is the bond strength between the hydrogen and the halogen atom. Since the bond between $\ce F$ and $\ce H$ is very strong, the molecule does not easily lose $\ce H^+$. On the other hand, as the size of the halogen atom increases, the overlapping of the orbitals between halogen and hydrogen becomes less efficient. Thus, $\ce {Cl}$ and $\ce {Br}$ can more easily lose $\ce {H^+}$ than $\ce {F}$. Hence, the acidic strength increases as size of halogen atom increases.
Second reason is the stability of formation of conjugate base. When an acid loses $\ce {H^+}$ ion, it forms what is called as conjugate base. Similarly, when a base loses $\ce {OH^-}$ ion, it forms a conjugate acid. In case of $\ce {HF}$, we have $$\ce{HF <=> H^+ +F- }$$
Here, $\ce {F^-}$ is the conjugate base. Now, due to small size of $\ce {F}$, it is difficult for $\ce {F-}$ to hold negative charge. There are strong repulsions.
Now, for $\ce {HBr}$, we have $$\ce{HBr <=> H^+ +Br^-}$$
$\ce {Br}$ has a greater atomic radius. It is also less electronegative than $\ce {F}$. Hence, the repulsions are less in $\ce {Br-}$. Thus, $\ce {Br^-}$ is quite stable. The more stable is the conjugate base, the more readily the acid loses $\ce {H^+}$ ion, and so, more is its acidic strength.
Check this link. Maybe you'll find it useful.

There should be an equilibrium sign in that reaction. I don't know how to put it. :p

Use <=> for equilibrium arrows $\ce{<=>}$. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

https://en.wikipedia.org/wiki/Hydrogen_fluoride (also @Karan Singh)

$$\ce{HOCl>HOBr>HOI}$$
$\mathrm{p}K_a(\ce{HOCl})=7.5$,  $\mathrm{p}K_a(\ce{HOBr})=8.6$,  $\mathrm{p}K_a(\ce{HOI})=10.6$
The $\ce{H-O}$ bond  in those oxo-acides ionizes more readily when the oxygen atom is bonded to a more electronegative atom.

The conjugate base $\ce{OCl-}$ is more stable than $\ce{OBr-}$. Because $\ce{Cl}$ being more electronegative, is better able to accommodate the negative charge.
The conjugate base $\ce{OBr-}$ is more stable than $\ce{OI-}$. Because $\ce{Br}$ being more electronegative, is better able to accommodate the negative charge.

$$\ce{HI>HBr>HCl\gg HF}$$
$\mathrm{p}K_a(\ce{HF})=3.1$, $\mathrm{p}K_a(\ce{HCl})=-6.0$, $\mathrm{p}K_a(\ce{HBr})=-9.0$, $\mathrm{p}K_a(\ce{HI})=-9.5$
$\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ are all strong acids, whereas $\ce{HF}$ is a weak acid.
As halogen size increases (as we move down the periodic table), $\ce{H-X}$  bond strength decreases and acid strength increases (a weaker bond produces a stronger acid, and vice versa).

While in mhchem mode the subscript has to be encased in curly brackets, otherwise it does not work. The proper typesetting is also \mathrm{p}K_a(\ce{...}).

As an add-on to this excellent answer: acidity and basicity are experimentally determined. The order of acidity and basicity is based on countless experiments that establish. We are left as scientists to explain why the order is such as it is. The how is the easier part - experiment!

As Tejas said the sign of inequality in the second statement should be opposite. This is because in HF there is extensive hydrogen bonding which tends to reduce its acidic character. The reason being the high electro negativity of Fluorine. The extent of hydrogen bonding would decrease down the group due to decrease in electro negativity.
A similar explanation may also be extended to reason the first one.
Your book is definitely incorrect.

Hey Thanks! But i am still not getting it precisely. Can you pls explain it more elaborately :) ?


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Why conductivity of electrolyte decreases on dilution, but molar conductivity increases?
I learnt that on dilution of an electrolytic solution, it's conductivity decreases. However, there is an increase in the molar conductivity of the solution. The book explained the reason by comparing the definition of the two terms. However, it gave a hypothetical example which contradicts the statement it earlier stated. Maybe there is a printing error, but it has got me confused now.
Can someone please explain why the conductivity of an electrolytic solution decreases but molar conductivity increases on dilution, maybe with the  help of a good hypothetical example?

http://www.sciencehq.com/chemistry/effect-of-dilution.html

Also http://chemistry.stackexchange.com/questions/6292/effect-of-concentration-on-molar-conductance/6294#6294 answers your question

Because too many cooks spoil the soup, maybe?

I've just been doing an experiment on this, and my understanding's not that great, but as I understand it the conductivity decreases because you have a lower concentration of charge carriers in the solution. The molar conductivity increases however, because as the charged ions get further apart they interact and slow each other down less. Essentially each ion becomes more able to move through the solution, but the increase in conductivity from this effect is outweighed by the decrease due to there being fewer ions in a given volume.
Personally I find it easier to think about the other way around. As you increase the concentration of the solution you get more charge carrying ions, so the conductivity increases. But because the ions get closer together they begin to interact and slow each other down so the conductivity per mole of the ions is reduced.
I hope you still find this useful and that I managed to explain what I meant

I agree with the answer, but what will be the effect in resistance of a fairly concentrated solution of a strong electrolyte (say 2N $\ce{NaCl}$)? Will the resistance increase because of some unionized molecules of solute in high concentration?


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Which gas makes Orange juice bottles swell?
I forgot a bottle of 1L Orange Juice (100% orange juice with pulp) in the fridge for about a week, the bottle was previously open so air got inside as well.
After a week the bottle got inflated and when I opened it I could hear the gas coming out. No odours, orange juice still the same colour (not sure if the same taste, I threw it away).
Which gas was produced and how did it happen? The air that got in seems to have played a part on it (these bottles stay in the supermarket closed for a long time without getting inflated).

What temperature was the bottle when you opened it and heard the gas coming out? To be more precise, did you take the bottle out of the fridge, leave it closed inside a warm room for a while, and then try opening the bottle?

No, the bottle was already inflated inside the fridge but if I keep it outside it inflates faster.

CO2, produced from sugar by yeast that came in through the air. Although i thought orange juice would be sour enough to keep yeast from growing.

I can't imagine the gas was anything except $\ce{CO2}$ formed by the action of microorganisms contaminating the orange juice.  They degrade the organic components of the juice to power their metabolisms, and produce carbon dioxide as a byproduct.
A common contaminant of pasteurized fruit juice is a genus of bacteria called Alicyclobacillus, but this paper says that Alicyclobacillus bacteria do not cause bottle swelling and do cause a medicinal smell.
By searching Google for ["bottle swelling" "orange juice"], I found an English-language abstract of a thesis from China which implies that the culprit could be a yeast.


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Why not to consider hydrogen and oxygen moles to determine an empirical formula
A question says:
Find the empirical formula of an organic compound from the following composition:
34.62% C, 3.88% H, 61.50% O.
The answer is $\ce{C3H4O4}$.
It was found by using the mass (percentages) divided by the molar mass of each element.
But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why?
in this case it will be $\ce{CH3O3}$
On the other hand, in a similar question in the text book, they consider the moles of hydrogen and oxygen in calculations by multiplying by 2 after getting the moles.
Could anybody explain?

The answer is $\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element.

Exactly! That's the way to do it.

But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\ce{CH3O3}$

No, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:

It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.
In summary, the "natural appearance" of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question.


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Electrolysis in aqueous solution – which equations to use to predict product at each electrode?
When trying to work out what will be produced at each electrode during electrolysis, I'm confused as to which equations to use for hydrogen and oxygen.
For instance, an electrolytic cell containing aqueous nickel(II) bromide, with a nickel cathode and a platinum anode.
At the cathode:
$$\ce{Ni^2+ + 2e- -> Ni}$$
I have read that this does not occur because $\ce{Ni^2+}$ is a worse oxidising agent than $\ce{H+}$, which makes sense because $\ce{H+}$ is lower down on the left side of the electrochemical series meaning its reduction is more feasible.
Water is therefore reduced at the cathode:
$$\ce{2H+ + 2e- -> H2}\quad E^\circ=0.0\ \mathrm V$$
or:
$$\ce{2H2O + 2e- -> H2 + 2OH-}\quad E^\circ=-0.83\ \mathrm V$$
The second equation, however, has water higher up in the electrochemical series. This would suggest nickel should actually  be reduced in preference to water.
How do I know which of the two "water-reduction" equations to use?
This is even less transparent for the anode, where bromide should be oxidised to bromine, but there are also two equations showing production of oxygen: one below and one above bromide in the series:
$$\begin{alignat}{2}
\ce{O2 + 4H+ + 4e- &-> 2H2O}\quad &E^\circ&=+1.23\ \mathrm V\\[6pt]
\ce{O2 + 2H2O + 4e- &-> 4OH-}\quad &E^\circ&=+0.40\ \mathrm V
\end{alignat}$$
I don't understand which of these equations should be used and how you're supposed to know which to use.

The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.
The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:
$$E = E^\circ  + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$
For the given reaction
$$\ce{2 H+ + 2e- <=> H2}$$
the Nernst equation reads
$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$
Since
$${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$$
the potential $E$ depends on $\text{pH}$:
$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$
Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.

Thank you for this answer, though it is a little over my head. It still doesn't answer the question of which equation to use when trying to report what process is happening at each electrode. Which half-equation would be most suitable for explaining the production of oxygen or hydrogen at each electrode?

@Raf Since the potential of water depends on pH, you may use your first equation at E = 0.00 V for an acidic solution and the second equation at E = −0.83 V for an alkaline solution, respectively. Or better still, you calculate the actual potential E for the actual pH of the solution (using the formula given in my answer). (The oxidation reaction is analogous.)

Thank you Loong. I've realised what I was looking for was an answer to my question without using the Nernst equation, but it just can't be that simple.

Actually we generally say that in a given mixture, which substance is going to reduce 1st does not depend upon concentration. But there are a few exceptions to this fact,like this one. Also, you might remember that electrolysis of aqueous sodium chloride generally gives chlorine gas but if excess overpotential is applied then oxygen is liberated instead


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What is the meaning of "superionic"?
I see the term "superionic" applied to high pressure water in articles like Giant planets may host superionic water, but I don't understand what the term really means.
How is "superionic" distinguished from simply "ionic"?
Why does the article refer to hydrogen "atoms", which would imply that they are neutral, as opposed to protons or hydronium or some other actual ion?

Written by an intern? You will probably have to read the original articles by Laurence E. Fried: Phys. Rev. Lett. 2005, 94, 217801 and Phys. Rev. Lett. 2005, 94, 125508

@KlausWarzecha Those articles helped some.  Seems that there are H2O, H3O+ and OH- species that are all very shortlived and protons are moving much more than the oxygens.  But still I don't comprehend any defining characterists that would let me know when to use the term "superionic".  For superconductor or superfluid there is a clear defining property, but I don't seen anything like that for superionic.

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

@Mithoron I still don't understand

http://www.nature.com/nmat/journal/v10/n9/full/nmat3066.html - it seems superionic means possesing extremely high ionic conductivity

Have you seen this one? The thing seems to be highly theoretical.

How is ionic clearly distinguished from covalent?

Yes I have already seen it @M.A.R, but it rather seemed to be lowly theoretical (or highly practical) contrary to your own observation of the thing seeming to be lowly practical.

Superionic water is water under extreme heat and temperature with the properties of both solids and liquids.
It is where the water only exists as a mixture of $\ce{H+}$ and $\ce{OH-}$ ions. At high pressures, the oxygen crystallises, leaving the hydrogen ions free - thus making superionic water highly conductive.
It's crystal structure is a face-centred cubic lattice, and it is thought that Uranus and Neptune hold a layer of superionic water.


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Which is More Stable with Boranes: Amines or Phosphines?
Given borane's ($\ce{BH3}$) LUMO, which of these Lewis bases would have a more stabilizing interaction?  In acid bases that means the higher HOMO and lower LUMO equals a stronger dative bond I believe.
So which would have a stronger interaction (greater net stabilization) with borane, trimethylphosphine or trimethylamine:
$$\ce{P(CH3)3 ~vs~ N(CH3)3}$$
Phosphorus is less EN and it's electrons reside in a higher energy 3p orbital, but doesn't it compensate for this by placing the electron pair an orbital with more "s" character? Also carbon is more EN than phosphorous drawing some strain.
I would say trimethylphosphine has the HOMO just because of it's higher orbitals and lower overall EN making donating a pair of electrons more favorable.
Am I right?

@Dave You mean an orbital with more s character, not a character, right?

If so I suppose you could check using bond angles ... although there is another issue at hand -the relative sizes of P and N

OK, first off, let's comment on the stability question. When making a Lewis adduct (and rationalizing with the frontier orbital approach), you want the resulting complex to be stable. That requires two possibilities:

The HOMO of the Lewis Base ($\ce{NH3}$ or $\ce{PH3}$ here) is low (i.e., hard to remove an electron).
The coupling between the acid and base orbitals is large.

I want those electrons to be stabilized. Either (or both) approaches can work. So to correct your statement, you don't want a "high HOMO" or "low LUMO."
The catch, of course, is that a large splitting between the Lewis acid LUMO and Lewis base HOMO can lead to low coupling (poor orbital overlap).
Now, I went and ran some quick B3LYP/6-31G(d) calculations on both $\ce{N(CH3)3}$ and $\ce{P(CH3)3}$ because I couldn't find any quick comparisons in the textbooks I use.
The amine has a computed orbital energy of -5.64 eV, and the phosphine has a computed orbital energy of -5.98 eV. (Standard disclaimers about orbital energies being non-physical apply.)
This would suggest that the phosphine would form a stronger bond with a borane.
There's supporting evidence here: "Ab initio molecular orbital study of the substituent effect on ammonia and phosphine–borane complexes" Journal of Molecular Structure: THEOCHEM, (2004) 709(1–3), pp 103-107.

The complexation energies of $\ce{H3BXH_{$3−n$}F_{$n$}}~(\ce{X~$=$~N, P}; n=0–3)$ and the proton affinities of $\ce{XH_{$3–n$}F_{$n$}}$ compounds have been investigated at the G2(MP2) level of theory. The G2(MP2) results show that the phosphine complexes are more stable than the corresponding ammonia ones. ... The NBO partitioning scheme shows that the stability of the phosphine complexes was related to the hyperconjugation effect.

Basically, the phosphine complexes seem more stable because there is some back-donation of electrons from the borane into the phosphine. It's not just a simple frontier orbital picture - more orbital interactions come into play.

For some reason, that I don't know, you have to enclose one character (non-numeric) sub- and superscripts in the mhchem environment, i.e. $\ce{F_n}$ produces $\ce{F_n}$, while $\ce{F_{n}}$ produces $\ce{F_{n}}$.


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Density Functional Theory with Generalized Gradient Approxmations (GGA) - What Happens if the Electron Density Shows a "Kink" at the Ion Position?
My question may be stupid, so please correct me if you find anything which is obviously erroneous.
In the following I will place a question mark (?) besides points/steps I consider doubtful.
My observations:

In density functional theory (DFT) (at zero temperature), the energy of the system $E$ is broken up as $E = \int_{V} d v \left[ e_{K}(n(\vec{r})) + e_{H}(n(\vec{r})) + e_{XC}(n(\vec{r})) \right]$, where $V$ is the system size, $e$ the energy density per unit volume, $n(\vec{r})$ the electronic density, subscript "$K$" for "kinetic", "$H$" for "Hartree" and "$XC$" for "exchange-correlation".
In the generalized gradient approximation (gga), $e$ also depends on $\nabla n(\vec{r})$. Thus write $e(n(\vec{r}), \nabla n(\vec{r}))$.

Assumptions:
1. spherical symmetry (closed-shell)
By spherical symmetry $e_{XC}(n(\vec{r}), \nabla n(\vec{r})) = e(n(r), \nabla n(r))$. Of course in writing this we are restricting the variation in density $\delta n$ in the radial direction only. I'm doubtful about the validity of this step.
Anyway, proceed as usual. The exchange-correlation potential is defined as the functional derivative of $e_{XC}$ w.r.t. $\delta n$. Now please point out if there are any mistakes in the following steps:
$$\begin{eqnarray} \delta \int dv e_{XC}(n(r), n^{\prime}(r)) &=& \int dv \left[ \frac{\partial e_{XC}}{\partial n(r)} \delta n(r) + \frac{\partial e_{XC}}{\partial n^{\prime}(r)} \delta n^{\prime}(r) \right] \end{eqnarray}$$,
where $n^{\prime}(r) = \partial_r n(r)$. The first term is done; consider the second term. $\delta n^{\prime} (r) = \partial_r \delta n(r)$. Thus we integrate the second term on the right hand side by parts, and obtain
$4 \pi \int_0^{r_0} dr \partial_r \left[ r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \delta n(r) \right] - 4 \pi \int_0^{r_0} dr \left[ \partial_r \left( r^2 \frac{e_{XC}}{n^{\prime}}\right) \delta n(r) \right]$. The first term is the boundary term; ignore it for the time being. The second term seems to give the correction to the exchange correlation potential due to the introduction of gradient-dependence. Note that the second term is equivalent to
$\int_0^{r_0} dr 4 \pi r^2 \left[ - \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right)\right]$, the square-bracketed term seems to give the exchange-correlation potential (?).
Now $- \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right) = - \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} - \partial_r \frac{\partial e_{XC}}{\partial n^{\prime}}$. What I worry about is the first term, which seems to diverge at $r=0$ (but it shouldn't; or is it cancelled by a piece in the second term?)
A expression for $e_{XC}$ may be the becke88 functional (exchange only) (PRA, 38, 3098). $$ \begin{eqnarray}e^{Becke}_{X} &=& e^{LDA}_{X}(n(r)) - \beta \sum_{\sigma} \int n_{\sigma}^{4/3} \frac{x_{\sigma}^2}{1+6 \beta x_{\sigma} \sinh^{-1}x_{\sigma}} d\vec{r} \end{eqnarray}$$, where $\sigma$ is the spin-index (For closed-shell system $n_{+} = n_{-} = n/2$ (?)), $\beta$ is a parameter (=0.0042 a.u.), and $x_{\sigma} = | \nabla n_{\sigma}|/n_{\sigma}^{4/3}$. In spherically symmetric system, can we change $| \nabla n_{\sigma}|$ to $|n^{\prime}(r)|$?. Now I am not sure about the following claims

$\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ is non-zero at the origin. (?) The reason is that, while it seems that $n^{\prime}(0)$ should be zero, it may not be. Consider helium. When solving the schroedinger eq. for the 1s state under arbitrary potential which tends to $2/r$ at $r=0$, the slope of the radial wave function (excluding the $r$-factor from the volume element) is non-zero (Consider for example the radial wave function of the hydrogen ground state, which behaves as $e^{-r}$). This means $n^{\prime}(0) \neq 0$. But then  $\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ will be non-zero at the origin and thus $- \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} $ will tend to infinity at $r=0$.

This should not be the case because the hartree (i.e. "classical" contribution by all electrons) contribution to the potential is small as $r \rightarrow 0$, and the exchange contribution should cancel the "self-action" in the hartree part.
Now, sorry for the long post. Note that I might have done something very wrong, such that the whole argument does not stand. Thanks for your time.

There is too much going on here. Please enumerate your questions (all the question marks), so that they can be individually addressed. You might also want to collect your assumptions, so that the circumstances of their validity can be addressed.

There is too much going on here, you are mixing assumptions with questions and derivations. Please give us a step by step question so that we could help.

authors: Ŋɩĸɘsʜ Sɩŋʛʜ


text:
Boiling and melting point of trans- and cis-but-2-ene
The boiling point of trans-but-2-ene is lower than that of its cis isomer but the melting point of the former is higher than the later. Why is it not following the same order? Is there any factor of structural symmetry coming playing a role here?

Yes . You are right that structural symmetry comes into play .
Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment .  The dipole moment enables electronic interactions which hold molecules together .  This shows some general factors of boiling point . Also the below link to the google book covers the structural aspect of a molecule towards b.p on pg47.
For melting points , let me quote Joseph Hornback from Chapter 2.6 Organic Chemistry pg46 (google books) :

The shape of a compound is very important and has a dramatic effect on its melting point . More symmetrical molecules pack into the crystal lattice better , allowing closer approach and larger attractive forces , resulting in higher melting points .

Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.(source)
Trans 2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point .
For further reading I suggest you visit 'Molecular Shape' over here.

What about the boiling point ? Why is it not following the same pattern ? Can you include that in the answer too please. Thanks

@DhrubaBanerjee The second sentence refers to the b.p , also pg46,47 of the google book can be viewed for details . Anyway , I will edit to add a few more references .


text:
Do multielectron atoms have nodes?
For hydrogen, other than in the 1s state, the electron wavefunctions have radial and/or angular nodes where the electron probability density is zero.
In helium or further atoms with more than one electron, considering correlation of electrons, what happens to these nodes?  Are there still radial and angular nodes where electron probability density is zero?
For example, if helium is in a 2s2s state, is there anywhere that an electron cannot be?

If you haven't already seen it, this answer of mine might be of interest to you. Maybe I'm overlooking something but this question might be rather difficult to answer. For one thing it will depend on which "kind" of wavefunction you mean. If you are talking about the many-particle wavefunctions then I think it is very difficult to say something definite about the nodal structure. But if you are talking about one-particle states, like Hartree-Fock or Kohn-Sham orbitals, then much more can be said.

For Kohn-Sham orbitals the sperical symmetry of an atom should always permit us to seperate the orbital into a spherical harmonic (whose nodes we are familiar with) and a radial component, which should have $n-l-1$ nodes.

Technically, wavefunction nodes for atomic orbitals only exist in solutions for the non-relativistic Schrödinger equation. Addition of relativistic terms means every region has a non-zero wavefunction value for every orbital; what used to be nodes then becomes "quasi-nodes".

@NicolauSakerNeto As far as I know there exists a nodal line theorem for the radial component of the Dirac equation that gives the number of nodes for $r \in (0, \infty)$. It can be shown, that the number of nodes for the Large Component is $n - l - 1$ (the same as for the Schroedinger equation) and that the Small Component has either the same number of nodes as the Large Component or 1 additional node.

@Philipp I guess to make my question more specific, if helium is in a 2s2s state, is there anywhere that an electron cannot be?

@DavePhD I'll refer to my linked answer above: For multielectron atoms the general form of Courants nodal line theorem applies. Interestingly, one of its consequences is that the ground state wavefunction will never have a node. For excited states (such as your helium example) the matter is different: Here Courants theorem doesn't give a definite answer only an upper bound, so it might or might not contain nodes. And I think this is the answer to your question, you can't make a general statement and it might depend on the multielectron system at hand whether there are nodes or not.

@Philipp Clearly you have far greater knowledge than I on the physical foundation of atomic theory, so I'll gladly defer to your expertise. I do remember however reading something about relativistic orbitals being nodeless in a book or article, but that was two and a half years ago, so I can't remember it too well and I'm not sure I'd be able to pull up the source. Sorry I couldn't be of any use.

@NicolauSakerNeto I think you do me too much honor here :) My knowledge about relativistic quantum mechanics is actually very limited. It is only that a friend of mine wrote his master thesis about this subject and within this thesis I found a passage about the nodes of the radial Dirac equation. It might well be that relativistic orbitals in general do not have nodes. Also my statement was concerned with a spherically symmetric system and the resulting 1D radial Dirac equation. Maybe the source you mentioned considered other cases than this rather specific one.

@NicolauSakerNeto "according to Dirac’s relativistic treatment of an electron, which has now been extended to many-electron atoms, atomic orbitals possess no nodes whatsoever." http://www.chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf Foundations of Chemistry 1: 297–305, 1999

@NicolauSakerNeto Thanks for pointing out the relativistic effect.  I'm trying to read more about it starting with "Relativistic quantum chemistry: The electrons and the nodes" http://pubs.acs.org/doi/abs/10.1021/ed045p558 and "Contour diagrams for relativistic orbitals" http://pubs.acs.org/doi/abs/10.1021/ed046p678

Nice finds. The last article is actually what I read a couple of years ago! "Pseudo-node" is in fact the term I meant to say before. I suppose that technically means you have your answer, though the non-relativistic picture is still of interest.

My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.
A more non-trivial question is, whether the total electron density can have nodes for the ground state. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.
The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state has to be occupied in the ground state.

Indeed as I stated in one of my comments to the original question: You are right in thinking that the ground state of an atom should not show any nodes. There is even a strict proof for this: According to Courant's nodal line theorem the eigenfunction corresponding to the lowest eigenvalue cannot have any nodes (except the trivial one at infinity).

There is a proof of node for the triplet-S state of helium in What Do We Know About Wavefunction Nodes? Recent Advances in Computational Chemistry vol. 2.  (alternative link to preprint version):
The wavefunction can be written as a function of only 3 coordinates: the distance ($r_1$) of the first electron from the nucleus, the distance ($r_2$) of the second electron from the nucleus, and the distance ($r_{12}$) between the electrons.
$\Psi = \Psi(r_1, r_2, r_{12})$
by the Pauli exclusion principle:
$\Psi(r_1, r_2, r_{12}) = -\Psi(r_2, r_1, r_{12})$
Then, suppose both electrons are the same distance ($r$) from the nucleus, but not necessarily at the same point:
$\Psi(r, r, r_{12}) = -\Psi(r, r, r_{12})$
Therefore,
$\Psi(r, r, r_{12}) = 0$
In other words, there is a node defined by $r_1 = r_2$.
The two electrons can never be the same distance from the nucleus.
The reference goes on to say that in 6-dimensional space, the node is a 5-dimensional hypersurface.