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Why do full shell ions such as chloride form coordinate bonds, while noble gases do not?
Both $\ce{Cl-}$ and noble gases have full outer shells, why does $\ce{Cl-}$ react whereas the noble gases mostly don't?

Taking the general molecular orbital scheme of an octahedral complex — for the sake of discussion, let’s assume it to be $\ce{[FeCl6]^3-}$ — as shown in figure 1, there is an incentive for both the metal and the chlorides to undergo complex formation.

Figure 1: Molecular orbital scheme of an octahedral complex including σ and π interactions from the ligands to the metal. Based on the schemes given by Professor Klüfers in his coordination chemistry course at the LMU Munich.
Complexing a metal is especially favourable for the ligands since they are able to lower their electrons’ energy levels significantly by forming dative σ bonds — and in the cases of halides or other full shell ions, even the electrons in π-symmetric orbitals benefit a little by π forward-bonding.
Therefore, the question should better be rephrased as:

Why are noble gases not known to form coordinate bonds?

Adding an electron to an atom means that all its orbitals are raised in energy: There is additional negative charge repulsing the electrons. This brings the energy levels of the valence electrons closer to those of the metal they are bonding with, and only for similar energy levels is the energy gain from interacting maximised. Note that electronegativity (that inversely correlates rather well with the energy level of the highest occupied orbital) rises from left to right across a period, so the noble gases have highest electronegativity and therefore the lowest valence energy levels. (The correct physical observable would be ionisation energy, but the same holds generally true for that.)
Therefore, the orbitals of noble gases should be substantially lower than those of halides, which are again notably lower than those of oxido/sulphido ligands. This gives a large energy difference between metal and potential ligand orbitals meaning that noble gases would easily be displaced by better ligands such as water.
However, I could imagine without having checked the literature that metals with high oxidation states may be able to be complexed by noble gases as their orbitals are again substantially lowered.

Part of a complex's reactivity is due to its valence, part is due to its charge. The noble gases have a neutral charge in addition to complete valence shells. They really have very little incentive to interact with anything. While $\ce{Cl-}$ does have a full outer shell, it's running around with a negative charge that will allow it to interact with anything else that is also charged, whether positive or negative.

I am not sure about this, but does the greater size of $Cl^-$ compared to Argon also play a role in this?


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isomerism in coordination complexes
Is there any difference between geometrical and cis-trans isomerism in coordination complexes, because in my book it is given that complexes of the form [Ma4bc] does not show cis-trans isomerism but shows geometrical isomerism

Geometrical isomers are compounds where all atoms are connected to the same atoms but their arrangement in space differs. This does include cis-trans isomers but in the context of coordination compounds also fac-mer isomers ($\ce{[MA3B3]}$ cases).
The case of $\ce{[MA4BC]}$ is tricky. In principle, they show the same two possible arrangements that, if $\ce{B}=\ce{C}$ would be termed cis and trans. In lab practice, most chemists would use cis and trans to describe them. In IUPAC theory, that response and terminology may or may not be valid.

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Sugar cube is filled with liquid (coffee), only if corner is sinked for few millimeters
Sorry for quite stupid question, but I could not find answer for it (perhaps my question is wrong).
I noticed one thing about sugar cubes. When I sink one corner of cube in coffee, just for few millimeters, after a few seconds it's already turning brown (filled with coffee, so to say). Another thing, if I sink it for a short period of time and take it out, coffee liquid stops spreading, if I sink it again it continues.
Why this is happening?

This is simply capillary action; it has to do with physics rather than chemistry.

What do you mean by 'the liquid stops spreading'?

@Joel , well coffee is filling cube (sugar cube is becoming brown), when I take it away from coffee it stops filling

Oh right! Look into @Ivan Neretin comment.

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Rocket propellants: accuracy of vapor pressure estimation techniques
I would like to determine the vapore pressures of a couple of substances (rocket propellants) over a significant temperature
range within a ~10% error margin.
For a first test, I tried to approximate the vapor pressure curve of Dinitrogen Tetroxide (N2O4) via the Clausius-Clapeyron (CC) relation, the Lee-Kesler (LK) method and the Ambrose-Walton (AW) method and compare the results to each other and to literature values that are available for N2O4.
Unfortunately my results deviate alot (up to 60%) from the published values.
Attached you find a plot of my attempts and a small table that compares the different methods and their deviation from the literature values.
Don't let the first picture mislead you! As the table and the detailview show, even at low pressures/temperatures the deviation is quite significant!
Is this large error margin really within usual boundaries? What are your experiences with vapor pressure approximation? Could you provide some tips on how to improve my results?

Some Details on my approach
Clausius-Clapeyron:
I chose the boiling temperature at atmospheric pressure as input, but I have no idea which value to
choose for the enthalpy of vaporisation. Many source state the assumption of a constant enthalpy,
but that puzzles me, since for N2O4 it varies by some 16kJ/mol over the temperature range I'm interested in.
Which value is usually taken - the enthalpy corresponding to the values provided as input or the one for the target temperature?
Lee-Kesler / Ambrose-Walton:
As expected both methods produce very similar results and are convenient to use, since they only
rely on the critical pressure, critical temperature and the acentric factor. But MY results are really
disappointing, especially since Wikipedia promises a maximal deviation of 10%.

As a side note:
For different reasons, I don't want to use interpolation of substance specific vapor pressure tables,
but instead I would like to use one of the many vapor pressure equations. It is important that the
approximation results in a closed representation of the vapor pressure curve , therefore I would like
to avoid methods that have points of discontinuity (e.g. Antoine equation?). Since I don't have access
to public databases that cover the substances I'm interested in, approximations that rely on "exotic"
empirical parameters (Wagner, DIPPR, etc.) are also out of the question.

To give a good answer to this question we need to know what compounds you are talking about.

For a first test, Dinitrogen Tetroxide is my main concern.
Later on, I'll have to deal with a range of different substances like:
liquid hydrogen and liquid oxygen

all kinds of hydrazines (UDMH, MMH, N2H4)

Nitric Acid (HNO3)

Hydrogen Peroxide (H2O2)

with N2O4 you should take in account dissociation of  N2O4 into NO2 molecules. Which invalidates assumption used by equations present, as the pressure above the liquid is not defined by vaporised compound only, but also by products of its reversable decomposition (which, to lesser extent, happens in liquid state in addition to auto-ionization =))

@permeakra How serious is the impact of that effect? Do I have to assume that none of the popular equations will output reasonable results for N2O4 or are the results just less realiable and my calculation are within usual boundaries? (I'm a mechanical engineer and my knowledge about chemistry is very limited)

Have you tried doing a least squares fit to the ln p vs T data to determine the three parameters in the Antoine equation consistent with your own data?

@ChesterMiller Not yet, I will try that. Thank you!

@cl10k here is the plot for N2O4 dissociation  (search for "Thermal dissociation)
http://nitrogen.atomistry.com/nitrogen_tetroxide.html || In my opinion none of the standard equations are reliable. They do not take reactions of evaporating liquid  into account. You may approximate your curve by any fitting curve, and maybe get acceptable result due to blind luck and cancellation of errors, but that's that. For than matter, nitric acid produces dimers in gas phase, so the equations would be unreiable for it too. And maybe for water and hydrogen peroxide, less likely for hydrazine derivative.

I would like to determine the vapor pressures of a couple of substances (rocket propellants) over a significant temperature range within a ~10% error margin.

The best way to determine vapor pressures is to measure them.  Is that not an option for you?

It is important that the approximation results in a closed representation of the vapor pressure curve , therefore I would like to avoid methods that have points of discontinuity (e.g. Antoine equation?).

The Antoine equation isn't discontinuous by nature, it's only discontinuous if you use two different parameter sets for a single substance in order to widen the temperature range of applicability of the equation.

What are your experiences with vapor pressure approximation? Could you provide some tips on how to improve my results?

I would recommend the Antoine equation as a first step beyond the Clausius Clapeyron equation.  Unlike the CC equation, the Antoine equation does not assume that the enthalpy of vaporization is independent of temperature.  It also provides a single, closed-form representation of the vapor pressure dependence on temperature.
Unfortunately, the Antoine equation requires compound-specific parameters.
However, many of the other methods you report using, such as the Lee-Kesler and Ambrose-Walton methods, also require compound specific parameters: specifically, the critical temperature and pressure of the compound of interest.  Even the Clausius-Clapeyron equation requires an (assumed constant) enthalpy of vaporization, which is also a compound-specific parameter.
Thus, the Antoine equation is probably the best place to start.  As Wikipedia shows, it can have accuracy better than 10% deviation for some compounds.  Ideally you would measure vapor pressures for your two compounds and then fit the data to the Antoine equation.  If measuring the vapor pressure of your compound is not possible, you could start by examining the Antoine equation predictions for a series of compounds that approximate your unknowns in chemical structure and molecular weight.  However, with this latter approach, I doubt error under 10% will be possible.
Antoine equation parameters for your compounds of interest

UDMH has Antoine parameters available from NIST.
The case of dinitrogen tetroxide is complicated by the equilibrium $\ce{N2O4 <=> 2 NO2}$.  It is unusually difficult to fit vapor pressure data to.
Hydrazine has Antoine parameters available from NIST.
In general NIST has Antoine parameters for a variety of substances.

Thanks for your great effort on answering my question! I have to play a bit with the Antoine equation before I can make any useful comments on your suggestions. I'll return to you later!
PS: I'm don't know much about chemistry, could you maybe comment on cryogenic substances (LH2/LOX) and if vapor pressure approximation for those compounds is any different?

If you found the answer useful, an upvote would be appreciated :-) [And, if in a few days no one has posted a better answer, an answer acceptance as well]

Sure, but you need 15 rep to upvote and I only have 6... :)


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How can anions exist?
Consider an nitrogen atom with 7 protons and 7 electrons. How can an nitrogen anion $\ce{N^-}$ exist? Shouldn't the 7 electrons in valence shell repel the extra one?
What force does hold the extra one electron in the valence shell? There isn't an extra proton in the $\ce{N^-}$ that would hold the extra electron.

Note that nitrogen is, aside of noble gases, one of few elements ( also e.g. Be, Mg, Mn, Zn) with negative electron afinity, therefore energy is released by repulsion of the extra electron and such an anion can exist only for a minimal time period of the order of microseconds to milliseconds.

One reason cations and anions exist is due to the stability of a full or half-full valence shell. The stability from those electronic configurations means that the atom or molecule does not require protons to "hold" the extra electron.
Recall also that nitrogen has three (or five) valence electrons, rather than seven. The 1s shell is full and is not considered part of its valency. The three 2p electrons are the valence electrons although they hybridize with the 2s electrons to produce the trigonal pyramidal structure of ammonia with its lone pair.
The single anion $\ce{N^-}$ could exist, but would not be stable because it puts four electrons in the p shell. The p shell would prefer to have three electrons as it does in the nitrogen atom or no electrons as it does in the $\ce{N^3+}$ cation.

Thank you for your answer! Do I get your answer right when I suppose that the extra electron(s) in the valence shell of an anion are held there by the electrical field of the electrons in the electron shells and also thanks to "right" spatial formation of the electron shells?

An extra electron is held because it is attracted to the nucleus because the particular orbital is not shielded by inner electrons. A second electron is endothermic because the negative charges repel.  Elements with s1 configurations [Group I] have exothermic electron affinities because the s orbitals penetrate closer to the nucleus overcoming electron repulsion. p and d orbitals penetrate less so electron repulsions have a greater effect, electrons are added to each orbital and the 4th electron as in nitrogen must share an orbital and repulsions are greater than the nuclear attraction.

Something to consider:  does N become an anion all by itself?  No.  It gets those electrons from a much less electronegative atom.  When it does that it generates an anion and a cation.  They attract each other.  Potential energy decreases (energy is released) and that "released" energy drives the REACTION.
As I like to say to students:  It's not like a sodium atom is walking down the street and one of its electrons falls out of its pocket. Metals don't just LOSE electrons. That electron has to be TAKEN by a more electronegative atom.

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Hybridisation of carbon atom in cyclopentadienyl anion
How do I figure out the hybridisation of the carbon with the negative charge?
One logic is that it is an $\mathrm{sp^3}$ carbon because there are 3 sigma bonds and 1 lone pair around it.
Another reasoning, though, is that it is $\mathrm{sp^2}$; we don't count the lone pair, because it is in conjugation with the double bonds.
What is the correct explanation?

There is no carbon with negative charge, nor is there a lone pair. All carbons are in fact identical; the negative charge is distributed over the whole ring. So your second option is closer to the reality.

Related answer: http://chemistry.stackexchange.com/a/10936/4325

It entirely depends on what that electron pair is doing.
If the lone pair is hanging out by itself in its own orbital, then the carbon is probably sp3 hybridized, as greater hybridization is typically more stable. This is what you'll see on an alcohol or an amine lone pair. Those lone pairs aren't really interacting with anything else in the molecule, hence the "standard" logic of the rule you invoke.
However, in certain compounds you're able to stabilize the system further by delocalizing the electron pair. In this sort of case, the sp3 hybridized lone pair orbital is not able to participate in the delocalization, and thus will be higher in energy than an electron pair that is in the remaining p orbital of an sp2 hybridized carbon. This is the sort of thing you see with amide groups, where the "lone pair" on the nitrogen is sp2-like, which allows the electron pair to delocalize into the carbonyl, allowing for resonance stabilization.
Note that delocalization is not always going to happen. You can potentially draw up examples where delocalization would decrease the stability of the compound. For example, 1,4-Dihydropyrazines would be anti-aromatic if both lone pairs from each nitrogen were to participate in the ring system. This is highly disfavorable, and the molecule contorts itself to avoid it, putting the nitrogens into sp3 (pyramidal) hybridization, except where it needs to be planar to conjugate with an exocyclic double bond.
In this particular case - the one of the cyclopentadienyl anion - you would be able to form an aromatic system if the lone pair were to delocalize into the ring - much like pyrrole does. The lowered energy obtained from this aromatic delocalization more than makes up for the small amount of energy gained by going from sp2 to sp3, so the carbon should be sp2 hybridized.
The important thing here is not simply to invoke rote-memorized rules to determine hybridization states, but think about what the electrons are doing and what the comparative stability of the various options are. The hybridization will change to be the most stable (lowest energy) state. -- To be perfectly honest, neither sp2 nor sp3 is the "true" answer. The actual   answer to these sorts of questions lies more toward Molecular Orbital Theory, where the molecule is taken as a whole. Per-atom hybridization is a convenient shorthand, but still one which ultimately rests on the basis of the true molecular stability.


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Woodward Fieser rules for dienes - non conjugated double bond as exocyclic?
When looking for exocyclic double bonds in a diene, should a double bond not connected to the conjugated system be considered?
For example, in this compound, should the marked double bond considered as an exocyclic double bond?

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

Should the marked double bond considered as an exocyclic double bond?

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.
If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

The bond you've pointed at isn't in conjugation with the 'triene'
The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.

Thanks for the answer. But according to the paper by Fieser the pointed double bond is exocyclic to the 5 membered ring or right. What I want to know is whether a double bond must be a part of the conjugated system for it to be considered as an exocyclic double bond in the calculation?

If you find a double bond outside a ring (as in the 4 ring system here), it will be considered as exocyclic. So, the pointed double bond is exocyclic to the adjoining cyclopentane ring.

Considering the double bond is in a six-membered ring, I have difficulties seeing it as exocyclic …


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Can the "rate coefficient" of a chemical reaction really be called a "rate constant"?
Can the rate coefficient of a reaction ever be referred to as a constant? To my limited knowledge, it cannot, because it is dependent on factors such as temperature (I'm getting that info from the Arrhenius Equation).
I'm looking for clarification here.

Yes the rate coefficient can be referred to as a "rate constant".
This is in fact a very, very common name for the rate coefficient.  Encyclopedia Brittanica has an entry for rate constant.  Wikipedia has a reaction rate constant page.  Google's N-gram viewer shows that "rate constant" is used more often in the English written corpus by a factor of more than five, relative to "rate coefficient".
The "constant" in rate constant refers to the fact that it does not vary with concentration of the reactants or products.  However, as you note, it could depend on other variables, such as temperature, pressure, catalyst, light, etc.

authors: احمد حسين بندارى


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Why is Mo(VI) more stable than Cr (VI)?
Acids of $\ce{Mo}^\mathrm{VI}$ are more stable than those of $\ce{Cr}^\mathrm{VI}$ in terms of redox reactivity.
In p-block elements, as we go down the groups, the stability of lower oxidation state keeps on increasing. I understand how this can be explained with the inert pair effect.
On the other hand, as we go down the group from $\ce{Cr}$ to $\ce{Mo}$, why does the opposite happen and there isn't any inert pair effect? Shouldn't the lower oxidation state be more stable down the groups in d-block elements? Wouldn't there be less shielding in $\ce{Mo}$ making is less electropositive than $\ce{Cr}$?

Related question with the same answer.

Related: Why is WF6 more likely to be formed over CrF6?

Related generalized question: Why do heavier transition metals show higher oxidation states?

The inert pair effect plays no particular role in the stability or instability of transition metals’ oxidation states. It is due to the large and rising energy difference between a shell’s s- and p-subshells which make the ionisation of the lower electrons more difficult. However, it is equally wrong to say that lower oxidation states be stabilised in lower periods: the most stable halide with the lowest oxidation state a halogen can have is fluoride ($\mathrm{-I}$).
All 3d-metals before iron can be oxidised up to their highest oxidation state while being strong oxidants in that state. This is not so true for the 4d-metals; an indication that the 3d electrons are better attracted to the nucleus. Indeed, the 3d orbitals do not contain radial nodes and their electrons are thus closer to the nucleus. The radial nodes in 4d and higher d-orbitals ‘add distance’ between the nucleus and the corresponding d-electrons (do not copy that wording: it is bad). It is this proximity to the nucleus and the consequential lesser extension of the orbitals into space that draws electrons into them, destabilising the higher oxidation states and stabilising slightly lower ones.
The same effect is true for f-elements; see Nicolau’s answer to a related question on f-element oxidation states’ stabilities.


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Use of statistics in this field
I am currently pursuing a statistics course and I an interested in knowing how can statistics help this field in future. (I am interested because I love chemistry very much) Specifically, are there any open problems in chemistry which have remained as a conjecture or we have not been able to solve due to lack of adequate knowledge of statistics ?

I am answering because I love statistics very much. Statistics is a terrible battle ground and (as of my experiences) has more pitfalls than answers. It is definitely a most valuable tool in the hands of those who know how to handle it. So here is my advise: a) don't expect too much b) be sure how to handle your tools and, c) above all, gain some experience on how to interpret the results.

The tags suggest an answer towards physical chemistry, but the use of statistics in factorial design and principal component analysis, as well as the evolution of these as subjects are extremely useful in industrial process optimisation,  from food to organic chemistry to machine settings. Where this could be focused better, for example,  is the excipient mixture composition of tablets and the solid state characteristics arising from each potential mixture with the active ingredient(s).

@Beerhunter : It would be of more help if you can suggest a link to something of this sort. aren't there any open challenges in chemistry involving mathematics?

From a commercial angle, the challenges change if companies offer a financial incentive as they set deadlines. For open (academic?)  challenges I can't help, but at the risk of being pilloried for putting links in,  look at www.innocentive.com and ninesigma.com for commercial challenges currently under investigation across all sciences. You may have to check every few weeks to get something relating to what you're talking about.


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Why is Sodium acetate called a salt of weak acid and strong base, when Acetic acid acts as a strong acid in Sodium hydroxide soln.?
Why is Sodium acetate called a salt of weak acid and strong base, when Acetic acid behaves as a strong acid (i.e. it shows almost cent percent dissociation) in Sodium hydroxide solution ?

Strength and weakness are relative terms, of course when we compare acetic acid and sodium hydroxide in terms of acidity it is considered as strong acid, but this is not the case when this compare acetic acid and sulfuric acid in the same terms. In other words sodium hydroxide is very much weak acid to a degree that is considered as a strong base.

@M.Ibrahim Please add that as an answer. Thanks! :)

@M.Ibrahim you got my question wrong. Acetic acid acts as a strong acid in sodium hydroxide soln., doesn't it ? and sodium acetate is formed when acetic acid reacts with sodium hydroxide, so doesn't  it behave as strong acid then ?

do you mean the strength of sodium acetate salt ? It is for sure a weaker acid than the parent acetic acid.

@M.Ibrahim I did not mean the acidic strength of sodium acetate salt.

So, What did you mean ?

Your definition of strong acid is wrong.
Simplistically speaking, it is defined with respect to dissociation in water, not in NaOH.
Acetic acid only dissociates partially in water, so it is a weak acid. It doesn't matter whether it dissociates completely in NaOH or not. That's irrelevant.

The dissociation is affected by the PH of the medium, water is more acidic than NaOH. It is not necessary to use the standard definition to compare these characteristics especially qualitatively as here.

No, sorry to be blunt, you don't understand the question at all.

Good question.
You are right. In sodium hydroxide solution, you should not call sodium acetate a salt of strong base and weak acid. If so then the pH should be even more greater than what we expect. However, the pH should be less than expected, though not appreciable. It is because the ionization of NaOH will compete with the ionization of sodium acetate as well as with the hydrolysis of acetate ion.
However, defining of strong or weak acids or bases is done in water not in sodium hydroxide solution.
Ref: www.adichemistry.com

Yeah, that's exactly what I meant to say, but unfortunately very few people understood that, and what is more surprising is that the ones who understood the least spoke out the most !!


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How do I determine which liquids are immiscible with one another?
I am trying to work out which type of ferrofluid I can use to form a weak magnetic seal for a container of liquid photopolymer resin. Is there a mathematical approach to determine what substances will form a solution, and which will remain immiscible, or will I have to buy each ferrofluid and try each one?

The mathematical approach requires insanely hardcore quantum chemistry and is not all that reliable anyway. You'd better stick to the old good "polar mixes with polar" rule.

Ah yeah I was afraid that would be the case. All good though, I'll stick with that. Cheers.

generally, there are three major classes of (mostly) mutually unmixable liquids: nonpolar hydrocarbons, polars like water and lower alcohols and metals (like mercury). I heard that perfluorocarbons are unmixable with hydrocarbons, but can't support the claim on spot.

That helps heaps, thanks. Now I just need to work out which to make my ferrofluid, and which to make my resin. I have a choice of water or oil for both so I guess I need to weigh up the advantages and disadvantages of each.

@permeakra Yeah, perfluorated stuff is another "class".

http://chemistry.stackexchange.com/questions/641/liquid-which-neither-dissolves-in-water-nor-oil

If you want high accuracy for solubility, UNIFAC is a good way to go. The basic idea behind it is to break the molecules down into groups then generate correlations for how each of the groups interact. The weighted sum of all the group contributions gives you the total molecular interactions. From there you can get solubility.
The Dortmund Data Bank maintains a comprehensive list of all the group coefficients and offers a program to get the various properties, like solubility. Unfortunately, the free version is pretty limited. You an only access a handful of fluids.
If you want to put in a bit more work, someone has put an Excel version on the internet. It doesn't explicitly do solubility, but basically if the activity coefficients are very far from 1, you can assume they are insoluble.
Furthermore, I'm not sure what the ferrofluids will do. I would guess that the concentrations would be low enough that you could just calculate solubility using the surfactant. Otherwise, you might have to search the literature for how to handle magnetic nanoparticles with UNIFAC.


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Why does CO2 have higher boiling point than CO?
The intermolecular forces between $\ce{CO2}$ molecules are dispersion forces, while the forces between $\ce{CO}$ molecules are mostly dipole-dipole attraction forces. So, why does $\ce{CO2}$ have a higher boiling point than that of $\ce{CO}$?

related http://chemistry.stackexchange.com/questions/42946/can-nonpolar-molecules-exhibit-dipole-dipole-forces

I don't think it answers my question.

There are many other points on the scale from dipole-dipole interaction to dispersion forces. $\ce{CO}$ is a very weak dipole; on the other hand, $\ce{CO2}$ is a pretty strong quadrupole.

the forces between COCO molecules are mostly dipole-dipole attraction forces.  || Orientation of CO molecules in solid is random with weak preference of head-tail. This suggests that dispersion interactions dominates in CO solid.

Also the MW of $\ce{CO2}$ is about 57% greater than that of $\ce{CO}$ meaning that it will take considerably more energy to raise the kinetic energy of the heavier molecule to where it has the necessary surface escape velocity.

CO$_2$  does cannot exist at a liquid under standard conditions, but CO can. CO$_2$undergoes sublimation and only exists in liquid form at high pressures so it is inaccurate to compare the boiling points if they are not under the same pressure.

$\ce{CO2}$ has more electrons than $\ce{CO}$. This means that it has a much larger electron cloud as compared to $\ce{CO}$, so its more easily polarised and thus, the ease of forming instataneous dipole-induced dipole bonds increases. Even though $\ce{CO}$ is a polar molecule and it forms permanent dipole-permanent dipole bonds, in this case the id-id bonds are stronger.
P.s. Just a $17 \space year \space old$ A Level Chem student here, I might be wrong

I believe you are on the right track here. Additionally to your thought, CO2 has a well developed quadrupole moment.

The larger the small covalent molecule, the greater the intermolecular bonds, hence higher boiling / melting point.
$\ce{CO2}$ has 3 atoms involved in the molecule and is therefore larger than $\ce{O2}$ that has 2 atoms. Hence, $\ce{CO2}$ has a higher boiling / melting point compared to $\ce{O2}$. (Exception to this is water molecules.)


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The nomenclature of mineral acids
My basic knowledge about nomenclature is: A mineral acid is called hypo when it has one less oxygen than the original one,
a pyro acid when 2 molecules of acid add up with release of $\ce{H2O}$
a meta acid when one molecule releases $\ce{H2O}$ and
a per acid when it has one extra $\ce{O}$ in the structure connected with a peroxide bond.
Question 1
I was taught in my class that this structure:

is correct because when we are drawing structures of a hypo-,per-,meta- or pyro- oxyacid, one must try to maximise the covalency. Why is that so?
Also, usually in pyro- acid we find a $\ce{M-O-M}$ bond where $\ce{M}$ is the central atom(s). Like $\ce{P-O-P}$ in pyrophosphoric acid as show below:

Then shouldn't the structure of pyrosulfurous acid be like this:
 Why is the structure not similar to pyrophosphoric acid?
Question 2
We know that in hypo-acids, there is one less $\ce{O}$ in the empirical formula of the acid i.e. Hypophosphoric acid should have one less $\ce{O}$ than in $\ce{H3PO4}$ i.e. it should be something like $\ce{(H3PO3)_$n$}$. If we put $n=2$ we have $\ce{H6P2O6}$.

But why is its formula $\ce{H4P2O6}$?
I might lack some information about it or maybe I've understood the concept wrong. Please help.

related http://chemistry.stackexchange.com/questions/40506/what-does-the-prefix-meta-mean-in-inorganic-chemistry http://chemistry.stackexchange.com/questions/43502/logic-in-amount-of-h-and-o-in-acid

@Mithoron None of them answers my questions. None, I repeat, None of the questions explain why do I need to maximise the covalency when I'm drawing a structure. And why is the structure of pyrosulphurous acid not the one I think it should be. And the formula for Hypophosphoric acid is different. My questions is not "How to name a mineral acid? " or "Why is meta ,hypo or pyro used? ". I couldn't find anything more than I already know

Compounds simply are as they are and have to be named somehow.

I know. What would you name the structure that I've made? It has the same formula H2S2O5. Is that structure not possible? If yes, then why?

The English names are not very systematic, they just give hints.  In principle you have to either memorize the name-structure correspondence, or know "all" acids of sulfur, so based on the name you can guess what the structure it corresponds to.
In addition, the hypo having one oxygen less is just a shortcut. What the names refer to is the oxidation state of central atom from lowest to highest.
Broadest range of acids are for chlorine: hypchlorous - chlorous - chloric - perchloric.
Answer 1: Please refer to the Wikipedia page on sulphur oxo-acids
.
The $\ce{H2S2O5}$ acid drawn is hypothetical one, but it's fine to have a name for it. Note, that sulfur is in low oxidation state.
In the pyrophosphoric acid, the phosphorus is in high oxidation state and has extra oxygen in between them.
For the summary formula $\ce{H2S2O3}$ you try to mix these properties, putting "extra" oxygen between sulfurs in low oxidation state. This won't work. The acid you should get when you attempt to obtain this one will be the thiosulfuric acid, where one sulfur mimics oxygen as $\ce{S^2-}$.
Answer 2: Chemistry of phosphorus is distinctly different from sulfur.
Phosphoric acid and Phosphorus acid for the acids with one phosphorus atom.
Your formula $\ce{(H3PO3)_{$n$}}$ with $n=2$ is not really justified, the name corresponds just to the formula on hypophosphoric acid.

Answer 1 - Covalency is maximized in any molecule formation because bonding electron pair is considered to be in noble state while lone pair always distorts the symmetry or it causes strain in molecular structure.
Answer 2 - Often we use hypo-, per-, etc. to show various oxidation state of the central atom. Well in your case, in hypo-, one $\ce{-OH}$ is removed.Or you can say that there is one less oxygen than the -ite.
For clarification in nomenclature you can see the IUPAC 1971 pdf.

Then shouldn't the structure of Pyrosulfurous acid be something like:

Yes.
See page 472 of Advanced Inorganic Chemistry, Volume 1
by Satya Prakash and see page 514 of Inorganic Chemistry by J. E. House
Furthermore, in NOMENCLATURE OF INORGANIC CHEMISTRY
IUPAC Recommendations 2005 at page 130 it is stated

$\ce{[(HO)(O)2SS(O)OH]}$ disulfurous acid

is the acceptable name but adds the footnote:

This common name presents a problem because the unsymmetrical structure is not the structure which would otherwise be associated with the ‘diacid’ construction
  (disulfurous acid would systematically be $\ce{[HO(O)SOS(O)OH]}$). The use of an additive name eliminates this potential confusion, but the problem with the use of disulfurous acid as a parent name persists in the naming of organic derivatives.

and on page 129:

$\ce{H4P2O6}$  = $\ce{[(HO)2P(O)P(O)(OH)2]}$   hypodiphosphoric acid

All these names are considered "common names" rather than following any system by the IUPAC recommendations


text:
Can a Birch Reduction be done using gaseous anhydrous ammonia?
Instead of generating anhydrous ammonia as a liquid using either pressure or cryogenic freezing, I would like to avoid the hassle entirely by using anhydrous ammonia generated as a gas. Is there a way to use a gaseous form of ammonia for the birch reduction that doesn't generate sodium amide? I'm a hobbyist chemist so I don't have access to a ton of glassware or specialty equipment.
And before someone asks me if I'm making meth- No. I like my freedom and this is just a hobby of mine. Mostly why I'm trying the birch is because I have heard the color of the solvated electron is quite beautiful. Chemistry is a wonderful science and I wish simply showing an interest in it wasn't considered so unusual!

The wiki mentions Birch-like reactions in THF and liquid amines. I guess, that mixture of dry tetramethylethyleneamine and some primary/secondary amine should work. That said, it would require an argon line and I wouldn't do it without a fume hood. However. Some aromatic compounds demonstrate quite significant affinity to electron, and can form stable in solution carbanions. This suggests, that at least for naphtalene and heavier aromatics something can be worked in dry THF with proton donor added by drops. The products, however, may be unexpected.

I saw that on Wikipedia but I don't have the equipment available to safely try that as far as I can figure. I found this PDF (http://goo.gl/MT9Q8i) which identifies a specialized product sold by the SiGNa company which allows ammonia free "birch" reductions which takes place in a solution of tetrahydrofuran and the special sodium they symbolized as Na-SG(I). The reaction they detail takes place at 0 deg C and produces reduced phenanthrene at 60% yield. However, since I'm not actually attempting to reduce something, simply learn and see the classical birch, that is not helpful.

@permeakra Can you identify what products would be produced by the dry THF reaction added dropwise? I'm quite curious.

Naphtalene (and some other aromatics) produce radical-anions in THF with sodium. Such radical-anions are strong reducing agents, but also may be protonated. The radical produced may pair, dispropotionate or reduce further.

Lower molecular weight alkyl amines can be used in place of anhydrous ammonia for a birch reduction, if your just wanting to observe the blue colour this might be a more practicle alternative.

I'm not sure but could hexamine (hexamethylenetetraamine) or EDTA be used in place of ammonia?

A word of caution: pure ammonia, in either its gaseous or liquified form is extremely dangerous to handle, especially in the context of your 'hobby' where, in your own words you "don't have access to a ton of glassware or specialty equipment." Any ammonia that escapes from your reaction and isn't ventilated properly will cause severe burns to anything it comes into contact with, including completely trashing the lining of your nose and lungs etc. In an attempt to relieve your curiosity, a photograph of a birch reduction showing the characteristic blue is shown below:

That said, the question is an interesting one in the context of laboratory chemistry.

Is there a way to use a gaseous form of ammonia for the birch reduction that doesn't generate sodium amide?

Yes and No. The most common way of performing these reactions on a small scale in a lab context is to essentially pass the gas over your reaction as illustrated below:

Practically, you're not actually making liquid ammonia and then transferring it to your reaction, but rather your passing it over your reaction flask which is fitted with a condenser which will cause in-situ liquefaction of the gaseous ammonia. Even a simple dry ice condenser is sufficient to do this, and the formation of the active reagent is easily observed by the appearance of the characteristic blue/green colour.  Any gas not liquified is then passed to the back of the fume cupboard, or, ideally, bubbled through a solution of something to neutralise it and hence stop it being released.

Thank you for that answer. So if I were to use a cold finger I could condense the gas immediately before the reaction, and that would only take some cheap dry ice? Also your picture of the dissolved reducing metal/ ammonia solution shows a solution with a ton of free e-; it seems almost black instead of blue.
Also I am well aware of the dangers of ammonia. I have a respirator which is rated for ammonia that I purchased online in preparation for this reaction. I also made an airtight setup for the reaction using ptfe tape and stainless steel pipes for the connections.

Yes. A cold finger with dry ice/acetone is sufficient to cause the ammonia to condense into the reaction flask. One issue you will have if you;re generating the ammonia yourself, however, is that you're unlikely to generate a useful pressure of ammonia. You may also have issues keeping it sufficiently dry.

As long as the ammonia is pretty dry/ almost dry I should be able to simply use an excess of sodium or lithium  metal right?

As I have no idea what you're trying to achieve by doing this , I have no idea. If all you want is to observe the blue colour then yes, I imagine you'll get away with not having things dry

I wasn't really trying to achieve anything specifically. https://www.youtube.com/watch?v=I_DD2_7QkRk Is where I got the idea to do this as a fun yet complicated project. I am taking far more safety measures than that guy though.

What is the formula for calculating the reaction pressure of a reaction? Something similar to pressure(A + B -> AB) = pres(a) + pressure (b) + pressure (ab)?

You'd need a pressure gauge/flow regulator

Ya, but I was trying to calculate the pressure for an idealized reaction assuming an ideal gas. I'm not going to rely solely on the math when the consequence is an explosion of ammonia everywhere lol.

Let us continue this discussion in chat.

@EmilyM just to add some numbers to NotNicolaou's excellent answer, the boiling point of ammonia is -33 C. A dry ice/acetone (or isopropanol) bath is -78 C, so sufficiently cold to condense ammonia.


text:
Enzyme-Kinetics. Michaelis-Menten Curve
A question regarding the Michaelis-Menten curve. For those of you who are well-versed in chemical kinetics(or even better, enzyme-kinetics) you most likely have come across this curve before. It is a hyperbolic curve, eventually reaching a plateau, where the Initial velocity(reaction rate) of the reaction is given as a function of the substrate concentration. Befofe I go on to say what has been bothering me I just want to remind that once the curve has reached a plateau this is known as the maximal initial velocity and it is caused by enzyme saturation by the substrate.
Now what has been truly bewildering me is the fact that when these experiments are done the enzyme concentration, which is always kept constant, is always much much lower than the substrate concentration(even when the substrate is at it's lowest concentrations). I'm talking about Enzymes having nanomolar concentrations while the substrate is five or six orders of magnitude higher. Surely the amount of substrate is already more than enough to saturate the enzyme from the get-go with the smallest concentrations. How is this not so?? When I saw the ratio of enzyme molecules to substrates I figured it is way more than enough to saturate the enzymes thus as a result the Michaelis-Menten curve wouldn't make sense and that the curve should start of as a plateau at the maximal initial velocity and just stay that way forever. I hope this wasn't too confusing.
Thanks for any attempts at an explanation

Michaelis-Menten kinetics are derived in detail in the Wikipedia article. https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics

Part of your confusion lies in the all-too common belief that it's the ratio of ligand to binding partner (e.g. substrate and enzyme) which determines the amount of binding. That's not the case. Instead, it's the absolute concentration of free ligand which determines how much is bound.
Let's take a microscopic view, and look at a single enzyme molecule. The enzyme doesn't know or care about what the other enzymes are doing, it only knows about its own local environment. To bind a substrate molecule, it needs to encounter (collide with) a substrate molecule in the correct orientation. How frequently it does that is directly proportional to the amount of free, bind-able substrate molecules there are in solution. For bound enzymes, (ignoring catalytic turn over for the moment), transitioning to the unbound state is a first-order process, independent of the concentration of anything else in the solution. (The total macroscopic rate of unbinding events depends on the number of bound complexes in solution, but for an individual bound enzyme it's concentration independent.) So for a single enzyme molecule, the fraction of the time it spends in the bound state is only dependent on the concentration of free ligand, so the time average of a large number of molecules is going to reflect that same distribution, by the ergodic hypothesis.
If binding is difficult but release is easy, then you're going to need a large amount of substrate in order to push the rate of binding to be comparable to the rate of release. If binding is easy but release takes a long time, you won't need as much substrate. That's what the KD of a binder (and roughly KM of an enzyme) measures. It's the concentration of the free ligand which makes the effective on-rate an the off-rate identical, and thus results in enzymes binding about half the time (or equivalently, at any one time half the enzymes are in the bound state).
So it doesn't matter if you have 100 molecules of substrate for every molecule of enzyme - if you're dilute enough, the concentration of substrate will be too low, the rate of enzyme/substrate encounters will fall off, and any enzyme-substrate complexes which form will fall apart long before some other one can form.
Enzyme/substrate (and ligand/binder) ratios are important, but mainly because it's the concentration of free substrate, rather than the concentration of total substrate, that's involved in the calculation. As your enzyme binds substrate, it depletes the pool of free substrate. If your enzyme concentration is large in comparison to your concentration of substrate, substrate going to the enzyme-bound state could cause a significant drop in free substrate concentrations, leading the nominal substrate concentration to be different from the actual free substrate concentration. (This is also part of why people primarily work with initial rates - the amount of free substrate drops over time due to catalysis.)
Caveat: KM is not quite the binding affinity of an enzyme for the substrate. In addition to the standard off rate of dissociation, there's also the "off rate" due to catalysis. There's also the lifetime of catalytic intermediates and the product-bound state which comes into play. With typical Michaelis-Menten assumptions - single step catalysis which is very slow compared to binding, with rapid product dissociation - the difference is minimal, but for many enzyme systems those assumptions don't hold.

A few points:

Michaelis-Menten kinetics usually implies a biological as opposed to a chemical reaction.
Enzymes are generally a big molecule behaving as catalyst. Envisioning an enzyme as a molecular machine helps to understand its function.
Enzymes create an intermediate reaction path.
Without the enzyme the dominant reaction is as follows:
$$\ce{S -> P}$$
With the enzyme the dominant reaction is:
$$\ce{E + S <-> ES -> E + P}$$
The enzyme is never consumed; however, it is part of the substrate-enzyme complex.
Either the two steps in the enzyme reaction can be the rate limiting step.

If the formation of the enzyme-substrate complex is limited say by diffusion then the enzyme will not saturate.
When the substrate becomes depleted, which is usually the case in biological reactions, the reaction will naturally slow down.


text:
Is there a method to separate ions from water other than by double displacement precipitation?
I'm working on a problem for school where we have to detect Fe3+, Ca2+, Cl-, and SO42- ions in water, and we have been given AgNO3, KSCN, BaCl2, and a HC2H3O2/Na2C2O4 mixture (all in solution) to do so. We have already determined the ion content by precipitating them out using successive double displacement reactions, but we also have to do it using another method. Is there another way to separate or detect ions in solution?

There's no need to separate, only detect...

Can you describe the limits of your problem in more detail? Do you have to actually complete the task with the equipment available to you or do you just have to suggest a way you could do it? If you have access to other stuff, there are many options.

@ericksonla We have to complete the task, and we don't have access to any other chemicals unless we acquire them ourselves.

authors: Кукулевський Марк


text:
Oxidation state of sulfur in H2S2O5
How can I find the oxidation state of sulfur in $\ce{H2S2O5}$ using its structure?
This is the structure of $\ce{H2S2O5}$:

Consider every bond and which way it is polarized.

S-S bond is not polarized. Both sulphur will have different oxidation states

How can I find the oxidation of sulphur @Mithoron using the structure? Are there any rules to be followed?

It looks like homework... Hmm,  http://chemistry.stackexchange.com/questions/35343/oxidation-of-carbons is analogical

@Mithoron: if I take $x$ as the oxidation state of sulphur, $2+2x-10=0$, $x=4$. But Wikipedia says 5 and 3.

Then don't take $x$ as the oxidation state of sulphur. I mean, you seem to dislike that answer with 3 and 5. Why, really?

Use the states of oxygen and hydrogen to know the oxidation state of sulphur.

While technically correct and nicely painted, this answer provides little explanation as to its conclusion. How do we know that O is -2 and H is +1, really? Are they always this way? Well, good like trying to apply that to $\ce{H2O2}$.

Electronegativity. What is the possible oxidation states for oxygen and hydrogen?
Oxygen is the more electronegative element of the compound. So, his charge will be -2, and the hydrogen is bonded with the oxygen so his charge will be +1. The sulphur charge is consequence.


text:
Is Neutron radiation considered electromagnetic wave?
I'm confused because gamma rays are classified as an electromagnetic wave. But why is neutron radiation not considered electromagnetic wave? I know Alpha and Beta, as well as positron and proton, is not considered electromagnetic wave because it can be deflected by magnetic or electric field (because they are + or -).
But a neutron is neutral, so why is it not electromagnetic wave? Can it be deflected too by electric and magnetic field? Is that why? Or is there another reason?

This probably belongs on physics.SE. I'm not real sure how to flag it to be migrated though.

Charge is only one of the properties which characterise a particle. Charged species cannot be photons (electromagnetic radiation), but that does not imply neutral species must be photons.

True, $\alpha$ and $\beta$ are not considered electromagnetic waves. But this is not because they are charged and can be deflected. This is because they are not electromagnetic waves. They consist of certain particles, and so does neutron radiation.

Electromagnetic radiation consists of electromagnetic waves (oscillations of the electric and magnetic field), whereas corpuscular radiation consists of actual particles. Certainly, wave–particle duality can make the distinction between waves and particles fuzzy. However, electromagnetic radiation (whether described as wave or as photons) is massless (its invariant mass is zero) and it travels at the speed of light. In contrast, the particles of corpuscular radiation have a non-zero mass and therefore a velocity below the speed of light. The particles of corpuscular radiation can be charged positively (e.g. α particles) or negatively (e.g. β− particles), or they can have no charge at all.
Neutron radiation is a corpuscular radiation because it consists of particles (i.e. neutrons). Neutrons are subatomic particles with no charge and a mass of $m=1.674\,927\,471(21) \times 10^{-27}\ \mathrm{kg}$ [source], which is slightly larger than the mass of a proton.
The kinetic energy of free neutrons can vary significantly. For example, the so-called fast neutrons, which are released during nuclear fission, have energies above $E_\text{kin} = 1\ \mathrm{MeV} \approx 1.6 \times 10^{-12}\ \mathrm{J}$. The thermal neutrons, which are used to initiate nuclear fission in moderated reactors, have energies below $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$.
Since, in classical mechanics, kinetic energy $E_\text{kin}$ is
$$E_\text{kin} = \tfrac{1}{2}mv^2$$
where $m$ is mass and $v$ is velocity,
the velocity of a neutron with a kinetic energy of $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$ may be estimated as
$$\begin{align}
v &= \sqrt {2\frac{E}{m}}\\[6pt]
&= \sqrt {2\frac{1.6 \times 10^{-20}\ \mathrm{J}}{1.674\,927\,471 \times 10^{-27}\ \mathrm{kg}}}\\[6pt]
&= 4.4\times 10^3\ \mathrm{m\ s^{-1}}
\end{align}$$
which is clearly less than the speed of light.

Electromagnetic waves are oscillating electric and magnetic fields, oriented perpendicularly to each other and their direction of motion. According to quantum mechanics, the energy of an EM wave is quantized, and each quantum is called a photon (from which the idea of light as a particle is formed). If you look at an EMR spectrum, you will see all the different sort of EM waves categorized by their wavelengths—radio waves, microwaves, infrared, visible light, UV, x-rays, and gamma rays.
I assume you’re grouping gamma waves with alpha and beta particles because of nuclear decay, and thus assume that gamma waves are particles, just like alpha and beta particles are. But they aren’t exactly the same. Alpha and beta particles (and positrons and the like) are not EM waves not because they are charged and capable of being deflected by a magnetic field, but because they are simply particles with mass and volume—they are not pure energy as a photon is. That is what a gamma ray ‘particle’ is—a photon, or quantum of light energy. It is a particulate manifestation of an EM wave in the highest range of frequency, and is devoid of mass and volume.
That said, neutron radiation is not an EM wave because

It does not correspond to a wave on the EMR spectrum, and thus is not a photon in its particulate interpretation.
It is a particle that has mass and volume, just like any macroscopic object you’re familiar with. It is not massless and dimensionless as a photon is. It is not pure energy.

It is a particle that has mass and volume, just like any macroscopic object you’re familiar with - I don't think a neutron is like any macroscopic object I am familiar with. Although it has a mass, and a volume (sort of) it doesn't behave like anything macroscopic.


text:
What is the correct Lewis structure of NH3O?

Choose the most appropriate Lewis structure for $\ce{NH3O}$ among the two:
 and

The oxygen in the second case has 3 nonbonding pairs around it.
I chose the first as an answer, yet the second seems to be the correct one. Can anyone please explain why?

The second structure does exist in the crystalline phase. I know crystals that conceive the second structure. One such crystal could be Hydroxylammonium 5-oxo-4,5-dihydrotetrazol-1-ide azane oxide. It can be downloaded from CSD (https://www.ccdc.cam.ac.uk/structures/?), with identifier YEKJEF (CSD reference). Also, check, the reference, D.Fischer, T.M.Klapotke, J.Stierstorfer, Propellants,Explos.,Pyrotech. (2012), 37, 156, doi:10.1002/prep.201100108

Technically, both the structures $\ce{H2N-OH}$ and $\ce{H3N\bond{->}O}$ may exist. However, in reality hydrogen atom is rather prone to migration and the second structure is not favorable. So, for a compound with composition $\ce{NH3O}$, the correct structure would be $\ce{H2N-OH}$. The structure of second type is stable for compound $\ce{NOF3}$ and may be observed for amine oxides like $\ce{(C2H5)3N\bond{->}O}$ or pyridine N-oxide.
Why $\ce{H2N-OH}$  and not $\ce{H3N\bond{->}O}$? One  can come with several explanations for that, but I would like to focus on one: charge distribution. In $\ce{H3N\bond{->}O}$, the nitrogen has to carry a formal positive (and actually very real) charge, while for $\ce{H2N-OH}$, the structure has no formal charges. Still, according to some sources[1] up to 20% of hydroxylamine in water exists as ammonia oxide $\ce{H3N\bond{->}O}$, probably, due to stabilization by hydrogen bonds.
Somewhat similar uncertainty may be found for phosphorus compounds (like hypophosphorous acid) and sulfur compounds (like sulfinic acids).

Kirby, A. J.; Davies, J. E.; Fox, D. J.; Hodgson, D. R. W.; Goeta, A. E.; Lima, M. F.; Priebe, J. P.; Santaballa, J. A.; Nome, F. Ammonia oxide makes up some 20% of an aqueous solution of hydroxylamine. Chem. Commun. 2010, 46 (8), 1302. DOI: 10.1039/b923742a.

So basically the first structure is the most probable one and the teacher is wrong?

@HasanSaad Pretty much. Point to this article if needed.  http://scitation.aip.org/content/aip/journal/jcp/22/2/10.1063/1.1740030


text:
Compare the work done during dissolution of carbides

1 mole of each $\ce{CaC2}$, $\ce{Mg2C3}$, $\ce{Al4C3}$ react with excess water in separate open flask. Work done during dissolution shows the order

$\ce{CaC2}=\ce{Mg2C3}<\ce{Al4C3}$
$\ce{CaC2}=\ce{Mg2C3}=\ce{Al4C3}$
$\ce{Mg2C3}<\ce{CaC2}<\ce{Al4C3}$
$\ce{Mg2C3}<\ce{Al4C3}<\ce{CaC2}$

Calcium carbide gives ethyne on reaction with water. One mole of calcium carbide reacts with $2$ moles of water.
Magnesium carbide gives prop-1-yne on reaction with water. One mole of magnesium carbide reacts with $4$ moles of water.
Aluminum carbide gives methane on reaction with water and one mole of aluminium carbide reacts with $12$ moles of water.
Work done $=P\Delta V$. But since the amount of water is large compared to the amount of carbide, will the work done differ?
(Answer given is $1$)

somewhat related: http://chemistry.stackexchange.com/questions/34099/carbides-of-different-elements

The relevant equations are:
$\ce{Mg2C3(s) + 4 H2O (aq) -> C3H4 (g) + 2Mg(OH)2 (aq)}$
$\ce{Al4C3 (s) + 12H2O(aq) -> 3CH4(g) + 4Al(OH)3(aq)  }$
$\ce{CaC2(s) + 2H2O(aq) -> C2H2(g) + Ca(OH)2(aq)}$
As you can see, most number of gaseous moles are released in the 2nd reaction while equal number of gaseous moles are released in the first and third reaction, therefore maximum work is done in the first reaction under same conditions of temperature and pressure.
Note that:
In open vessel reaction, gases expand against a constant external atmospheric pressure, thereby doing work on the surroundings , thus the work done $= P_{\text{atm}}\Delta V$ is independent of the gas's pressure and only dependent on $\Delta V$
Also, recall that Avogadro's law states:

For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

Hence, here,  $\Delta V$ only depends on the moles of the gas.

Good answer! Though I think it's worth pointing out the question is slightly ambiguous. "Work done" by whom? Surroundings or system? Because the signs will be opposite for each, and the order will reverse itself. Fortunately though, none of the options for this question are like that, so we're safe.

@GaurangTandon we generally talk wrt system in chemistry...

Of course, we can assume that.


text:
Why does ionization increase from Li to Be?
If ionization energy decreases from $\ce{N}$ to $\ce{O}$ due to the pairing of electrons (causing electric repulsion and greater potential energy) in the $\mathrm{2p}$ orbital in the $\ce{O}$ atom, then why doesn’t ionization energy decrease from $\ce{Li}$ to $\ce{Be}$ as electrons are paired in the $\mathrm{2s}$ orbital?

The s-orbitals have a much greater degree of penetration than the p-orbitals, so they experience the nuclear charge more strongly.  This electrostatic attraction is sufficiently greater than the electrostatic repulsion between the two electrons that the ionisation energy increases.
Because the p-orbitals do not penetrate through the s-orbitals to the nucleus as much there is less electrostatic attraction.  By placing a second electron in a p-orbital the repulsion is now more significant compared to the attractive force, and the ionisation energy is less than expected.
However, each new element also increases the nuclear charge, which in turn increases the attraction that each electron has to the nucleus.  When the next p-orbital becomes doubly occupied this stronger attraction once again dominates, increasing the ionisation energy.


text:
Free energy of an ideal gas
I want to calculate the Gibbs free energy for an ideal monoatomic gas, i.e. $$U=\frac32nRT.$$
Free energy is defined as $G(T,p)=H-TS$ and using the definition of enthalpy $$G(T,p)=U+pV-TS=U+nRT-TS.$$
To calculate the entropy I use the total differential of entropy given by
$$\mathrm{d}S=\left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T
  +\left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p.$$
Using the Maxwell relation
$$\left(\frac{\partial S}{\partial p}\right)_T =
-\left(\frac{\partial V}{\partial T}\right)_p,$$
and for an ideal gas with $V=\frac{nRT}{p}$ we hence get
$$\left(\frac{\partial S}{\partial p}\right)_T=-\frac{nR}{p}.$$
Further
$$\left(\frac{\partial S}{\partial T}\right)_p =
\frac{c_p}{T}=\frac{\frac52nR}{T},$$
 so all in all
$$\mathrm{d}S=\frac{\frac52nR}{T}\mathrm{d}T-\frac{nR}{p}\mathrm{d}p.$$
How do I calculate $S$ from here, because integrating would leave me with $\Delta S$ which I cannot use?
The point of reference should be standard conditions ($1~\mathrm{atm}$, $1~\mathrm{mol}$, $298~\mathrm{K}$, ...)

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. I have updated your post with some nicer typesetting. If you want to know more, please have a look here and here. We prefer to not use markup in the title field, even if it is tempting, see here for details.

Why can't you use $\Delta S$?

Well, because $G(T,p)=U+nRT-TS$ and I cannot just plug in the change in entropy there...

Why do you need to have an absolute value of G.  Isn't it sufficient to have G relative to some reference state?

Are there no questions or comments about the answer I provided to this question?  I obtained the result in two different ways, and got the same answer both ways.  Do people feel that it is not correct?

Let $H^0$ be the heat of formation of the gas at $298~\mathrm{K}$ and $1\ \mathrm{atm}$. and let $G^0$ be the free energy of formation of the gas at these same conditions.  The enthalpy of the gas at temperature $T$ (any pressure) is then:
$$H(T)=H^0+C_p(T-298)$$  First let's get the free energy at temperature T and 1 atm.  At constant pressure, the change in free energy with respect to temperature can be calculated from:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
If we integrate this equation by parts, we obtain:
$$\frac{G(T,1~\mathrm{atm})}{T}-\frac{G^0}{298}=\frac{H(T)}{T}-\frac{H^0}{298}-\int_{298}^T{\frac{C_p}{T}dT}$$
Some algebraic manipulation of this equation yields:
$$G(T,1~\mathrm{atm})=G^0+\frac{(G^0-H^0)}{298}(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
If we define $S^0=\frac{(H^0-G_0)}{298}$, this equation becomes:
$$G(T,1~\mathrm{atm})=G^0-S^0(T-298)+C_p(T-298)-TC_p\ln(T/298)$$
Adding the free energy change between $1~\mathrm{atm}$ and pressure $P$ at temperature $T$ yields:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[C_p\ln(T/298)-R\ln(P/P_0)]\tag{1}$$
where $P_0=1~\mathrm{atm}.$  The term in brackets in this equation is the same as $S(T,P)-S^0$.  If we substitute this into the previous equation, we obtain:
$$G(T,P)=G^0-S^0(T-298)+C_p(T-298)-T[S(T,P)-S^0]$$
If we combine terms in this relationship, we obtain:
$$G(T,P)-G^0=C_p(T-298)-(TS(T,P)-298S^0)$$
The term $C_p(T-298)$ is $\Delta H$, and the term $(TS(T,P)-298S^0)$ is $\Delta (TS)$.  So, as expected, the equation is the same as $$G(T,P)-G^0=\Delta H-\Delta (TS)$$
where
$$S=S^0+C_p\ln(T/298)-R\ln(P/P_0)]\tag{2}$$
Either Eqn. 1 or Eqn. 2 can be applied, depending on preference.

authors: ΧΑΡΑΛΑΜΠΟΣ ΚΑΡΑΝΤΩΝΗΣ


text:
Diesel additive spill. Odor elimination
I am not chemist, however I am interested in science and know some basics.
So I have real life problem to solve and I hope there will be some people interested to help me or at least understand more how chemistry works. Anyway I would appreciate any answer.
So I have unfortunate spill of diesel addtive VIF (czech republic product). Composition is like this from the site(some translations maybe wrong):
Name - % in mixture
2-ethylhexyl nitrate - 50-100
solvent naphtha light - <5
solvent naphtha hard - <5
2,6-Di-tert-butylphenol - <5
1,2,4-Trimethylbenzene - <2.5
1,3,5-trimethylbenzen - <2.5
So I spilled this stuff in my car's trunk. Don't ask how it was, this is a fact and I cannot revert it back. It was a canister with it, so I spilled around 2 litres of it. I noticed it only after a day.
I took out the rug from the trunk and it was full of this liquid. The rug is textile. I tried to take it out with water and actually it was kind of eager to go out with water. It was lot of it, and I didn't want to do it manually. I cleaned my hands with soap and no smell was left on my hands after washing them, so I assumed I can clean the rug in the washing machine. Which was probably the worst mistake.
So instead of it being cleaned, I have

Rug that smell like the worse mechanical/technical smell in the world, however with no stains.
Washing machine which not only smells worse, but it gives this smell to anything that was washed in the machine and the smell is kind of concentrated.

So I have several questions on what can give this smell? Which component exactly? I wrote to the manufacturer with this questions, but they only told me I can try some solvent like technical petrol. But it was not an answer to the question "how to get rid of the smell"
So after reading some threads, everybody advising activated charcoal as best deodorizer.
I bought a kilo of it, and put around 200 grams to the washing machine for like 20 hours, and after that I didn't notice any change in the smell.
So the questions is: what is the formula for absorption with activated carbon? How to calculate how much of it I need to deodorize well? Will it deodorize at all? Maybe activate carbon pores are too small for the molecules or etc?

The compounds listed are all fairly 'greasy', that is, likely immiscible with water. The best way to remove them would be to wash them off with something also 'greasy' such as hexane/heptane/xylenes etc which you may/may not be able to to pick up in a DIY store depending on where you live.

Ok, I can try it with a rug, but how about washing machine? It looks like it's already on the materials of the washing machine. I cannot see anywhere where I can wash it off. So is it possible to just remove the smell? Or it's impossible?

The inside of a washing machine is often calcified, and lime soap hangs in every corner. I guess the smell might be in there. Clean stainless steel surfaces don't smell. Buy a packet of citric acid, put it in the machine, and let it run at 90dC. Afterwards perhaps try again the charcoal, also at 90dC. One more run with washing powder (with bleach), again at 90, and I'll bet you'll find the situation improved.

@Karl you are too optimistic about a packet of citric acid, IMHO I'd aim at about 100 g/ every liter of the working volume.

@permeakra I was thinking about 100 gramms for the whole machine. That's about twice the amount recommended on the packet of citric acid i have in my basement for this. What's your reasoning?

Can I suggest not adding Charcoal to your washing machine....

@NotNicolaou why not?

@Karl Technically, the more concentrated solution of chelation agent is used, the faster and more complete is the dissolution. And here an absolutely complete dissolution of lime is required.

guys and what about quantity of activated charcoal? Actually before charcoal I tried baking soda and vinegar, but it didn't help the smell. Maybe it was not right quantities? Just for education, can somebody give me a link or explain we how to calculate quantities of absorb substances needed for eliminate the smell? Just the science of it, i am curious to know how to calculate in future without help of internet.

Just a crazy idea, you can actually try to oxidize the hydrocarbons to carboxylic acids which get washed away with water. You could also try sulphonating the phenolic compounds with sulfuric acid which can be washed away with water. The only idea that comes to my mind is to convert those greasy substances to highly polar substances which can be washed away with water. Other than that, the only other way to do away with these non-polar substances is to use a non-polar solvent such as xylene.

Sulfuric acid in a washing machine?!? That's nuts. Adding a nonpolar solvent like xylene is nuts too. There are rubber hoses and gaskets in a washing machine that weren't designed to be used with such solvents.

Warning: Do not put any flammable organic solvents like technical petrol, hexane, heptane, xylenes into your washing machine. These substances form explosive mixtures with air and can quite easily lead to severe explosions in a washing machine. I would try go get rid of the smell by repeatedly running washing programmes with 90° C and ordinary washing powder (without clothes of course). If you do not notice any improvement between the rounds, the odour might come from substances dissolved in rubber or plastic parts of your washing machine and you might be lost.

@aventurin This was the case, the smell was "integrated" to the sealing rubber. After I changed it - the smell almost disappeared.

To get the smell out of your washing machine and rug you have to do two things
1.) Get the source of the smell into the water and
2.) Keep it in the water for flushing.
To get it into the water I suggest using dish soap. Dish soap is far more design to handle the greasy materials you listed and will dissolve them better than regular laundry detergent.  The draw back to this is that dish soap will form foam quite readily which laundry detergent is formulated not to do. If you have a top loading washer this can cause suds to over flow your washer.  You may obfuscate this issue by monitoring and adding rubbing alcohol as needed to suppress the suds. If front loading washer you do not need to monitor but it will take a long time, as these machines can sense suds and will wait for the suds to die down before restarting. I also suggest using LA's Awesome as it is a very effective degreaser and uses limonene (orange oil) as an emulsifier.
To keep the smell in the water I suggest activated charcoal  which can be bought for use in aquarium filters.  This will absorb the petro-chemicals and carry them away.  I suggest you use activated charcoal after a first attempt with the prior suggest method of soap and degreaser as charcoal can coat your drier affecting the next load or two, but washing a rag can help clean it out.
One of the comments mentions scaling as a possible issue.  LA's awesome has chealating agents and this should be of no issue.  If you choose to etch some scale I would suggest vinegar for this over citric acid.


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Why do we have an alkyl shift rather than epoxidation in a pinacol reaction?
Why do we have an alkyl shift rather than epoxidation in a pinacol reaction?
I don't understand why the hydroxyl group doesn't attack the positive charge.

Excellent question.
When the first Pinacol rearrangements were being studied, there was a suggestion that the product of the reaction was indeed the epoxide (migrations/shifts weren't well known reactions at this point in time, and methods of characterisaton were limited). The reaction that was being attempted is shown below (source: Wikipedia), however at this point there was even dispute over the structure of acetone, further adding to the complications.

One of the first reports of a Pinacol rearrangement

Today, we of course know that Pinacol and semi-Pinacol rearrangements occur when vicinal diols are treated with protic acid (or Lewis acids) to give the carbonyl product as a result of a [1,2]- alkyl shift.

Source: Strategic Applications Of Named Reactions. Wiley.

Looking at the mechanism (below), we first have a rate-determining loss of water via protonation-elimination. This generates the all important cationic intermediate.
The simple answer to your question is that rearrangement occurs faster than epoxidation (which is never observed to any significant extent), this is intuitive if we consider that:

Under the acidic conditions, the alcohol required to form the epoxide is fully protonated, and hence not a very good nucleophile (in agreement with R-OH generally being a poor nucleophile except in basic conditions where we get the anionic alkoxide RO(-)
The migration generates a more stabilised carbocation by the influence of the oxygen LP donating into the carbocation to form the resulting ketone.

Source: Strategic Applications Of Named Reactions. Wiley.

As an aside, if you'd kept reading through Clayden by another page or so, you would have discovered that epoxides also undergo Pinacol-type rearrangements, so realistically even if the epoxide did form, it would simply re-open under the acidic/Lewis-acidic conditions before undergoing a rearrangement reaction.

Source: Organic Chemistry, 1 ed. Oxford University Press (Warren & Clayden)

great answer, thanks! But why does the OH group stabilize the positve charge , i always thought an OH group on an alkyl is electron withdrawing?

It's not so much that the oxygen is 'stabilising' the positive charge, its that the lone pair of the electrons are kicking down to form the pi-bond, forming the ketone/aldehyde. I'll try and dig up a study of the kinetics, but IIRC, once the water has left, the [1,2] shift and subsequent formation of the pi bond are incredibly rapid (i.e. the cation adjacent to the oxygen is very short lived indeed)

Hanna, it is often said that oxygen is electron-withdrawing by induction and electron-donating by resonance. Here the resonance of the carbocation to the carbonyl is a much more significant factor than the induction.


text:
What factors effect the energy change in a reaction?
The reaction of zinc metal and copper sulfate solution produces copper metal and zinc sulfate solution. What factors affect how much energy is given out for this reaction?
So far I have:The mass of zinc, Pressure of surrounding area, Concentration of copper sulfate, Surface area of the solid reactants (i.e zinc), Volume of copper sulfate, Light intensity, pH of the copper sulfate.
If you can think of anymore or provide a link where I can find the information it would be much appreciated!

Some of your parameters do not affect anything, and some affect only the speed of reaction, but not its energy output.

Do you have an  idea on what does affect energy change?

In terms of enthalpy, $\ce{\Delta H}$, in chemical reactions, it's always change in bonds. By definition, chemical reactions are those which alter bonds between atoms and molecules, and breaking old bonds/forming new bonds involves energy. For example, combustion reactions give off heat because they form products with strong bonds (the $\ce{C=O}$ double bond, for example; strong bonds release a lot of energy and are thus more stable).
In reality, the work that can be done by a chemical reaction $\ce{\Delta G}$ (Gibbs Free Energy) also relates to a change in the entropy of the system. For example, if a reaction produced a lot of gas, the gas could expand and also do work, in addition to the energy calculated from enthalpy.


text:
Soluble or insoluble?
In the picture attached here, is the compound in water soluble or insoluble?
I get confused while identifying solubility of a solute in a solvent when a situation like this arrises. The crystals of the solid have become liquid, but they give a certain transluscent white colour to the solvent (water in this case). So in this picture, is the solute soluble in water?

A true solution must be transparent.

It is impossible to determine just from the image. All solutes have a limited solubility, so you can always reach a point where the solution is saturated. Also soluble is a fuzzy concept. Is soluble having 1 gram, or 0.1 gram, or 1 milligram of a solute being dissolved in 100ml of water?


text:
Why is the periodic table periodic?
I am not a chemist, but I am interested in Science in a general sense. Can anybody explain why the periodic table is periodic in nature? I would appreciate links for further reading.

Could you be more elaborate on what sense of "periodic" interests you? What kind of table did you expect that would be able to contain 118 elements? Or are you interested in its history?

@Ϻ.Λ.Ʀ. why exactly it is that elements with a set number of extra electrons have similar property? What properties are particularly interesting/significant or surprising? Also, the history of how these results were discovered is interesting to me. Thanks for your help.

http://chemistry.stackexchange.com/questions/4529/what-is-a-period-of-the-periodic-table, http://chemistry.stackexchange.com/questions/2843/periodic-trends

Elements interact with the rest of the world through their electrons. How those electrons interact with other atoms (or electromagnetic radiation) determines how that atom behaves.
You see this with isotopes. Different elements are distinguished from each other by the number of protons in their nucleus. Changing the number of neutrons has little effect on their properties. Why is this? It's not the protons themselves which change the properties, instead it's how the change in total positive charge of the nucleus changes how many electrons the atom can hold and how they behave that changes the properties. Different numbers of electrons - and specifically different numbers of electrons with respect to the net charge on the atom - change the properties of the atom. Changing the number of neutrons changes the nucleus, but has little effect on how the electrons behave, so has little effect on the chemistry of the atom.
So what does that have to do with "periodicity" of the periodic table? Well, there's a structure  to how the electrons are "stored" in an atom. These are orbitals. For a bunch of quantum mechanical reasons that are rather advanced for this question, there are repeating patterns in these orbitals. Most particularly, in the l quantum number which determines the shape of the orbital. Each period corresponds (roughly) to the n quantum number. For each n, there are a set of ls which go along. (Because of tradition, orbitals with different l quantum numbers get letter labels: s, p, d, f, ...)
This is why you see periodic trends. Having exactly filled s and p orbitals is the most stable, and each n has its own set of s and p. So each time you get a filled s and p, you get an inert noble gas. For halogens (like chlorine and bromine), you need just one extra electron to fill the s and p orbitals, so they really like taking a single electron from other elements. In contrast, the alkali metals (like sodium and potassium) can most easily get to a completely filled s and p orbital state by losing a single electron, resulting in their propensity for a positive charge. It's how close the atoms are to filled s and p orbitals - regardless of their n quantum number - and what it takes to get there which provides the bulk of the "periodicity" of the periodic table.
It turns out that the s and p orbitals are the most "exposed", so what's going on there contributes the majority of the behavioral change between elements. Elements that differ only in how the electrons in their d orbitals are situated have much less property differences (this is why most transition metals have similar properties). Elements which differ only in f orbitals (like the Lanthanides) are even more alike. There certainly are differences here, but they tend to rely on more subtle effects than those that cause the alkali/chalcogen/halogen/noble gas differences.
Note that this sort of orbital-based periodicity breaks down as you get lower in the periodic table. That's because you actually start to get relativistic effects, and how the orbitals behave starts to change. Periodic trends are just that - trends - and aren't absolute.

The answer can be really simple indeed. The elements are organized according to their atomic number in ascending order, and if you analize the properties of the elements in a period (row) or in a group (column) you'll see that the properties of those elements follow a "pattern". For example generally speaking the atomic radius in a group increases with the increase of the atomic number and insted ionization energy decreases. This has exceptions, of course, but this regularity helps to have a fast idea about the properties of an element


text:
Why HF is said to be formed from H-1s and F-2pz overlap?
According to Wikipedia, HF is formed by the interaction between hydrogen's $1\mathrm{s}$ orbital and fluoride’s $2\mathrm{p}_z$ orbital:

In hydrogen fluoride HF overlap between the H 1s and F 2s orbitals is
  allowed by symmetry but the difference in energy between the two
  atomic orbitals prevents them from interacting to create a molecular
  orbital. Overlap between the H 1s and F 2pz orbitals is also symmetry
  allowed, and these two atomic orbitals have a small energy separation.
  Thus, they interact, leading to creation of σ and σ* MOs and a
  molecule with a bond order of 1.

So:

If there is a too large energy separation between H-1s and F-2s, isn't the energy separation between H-1s and F-2p even larger?
Can it be experimentally verified that the bonding is indeed via the F-2p and not the F-2s?

No. The F2p is much closer in energy to the H1s than the F2s is. You have to remember that fluorine is a highly electronegative element and as a result all the orbital energies are lower than that in hydrogen.

@orthocresol why high electronegativity results in lower energy orbitals?

Is the lower energy of highly electronegative element due to the closeness of the orbital to the nucleus?

You fell into a wrong analogy with the picture of energetically far apart. Let me break down your picture into something visible:
Imagine the orbitals of fluorine at different heights of a mountain (I’ll use the Zugspitze, because it feels like home). 1s is at the very peak of the mountain (yeah, I’m turning this around just for simplicity). The Eibsee (c. $1000~\mathrm{m}$ over sea level) would correspond to the 2s-orbital. Garmisch-Partenkirchen, the town in the valley before the mountain, ($700~\mathrm{m}$) would be the 2p orbital.
We need to place the 1s-orbital of hydrogen somewhere so that the height difference to the Eibsee is too large for an interaction. One possibility would be the Höllentalanger hut at around $1300~\mathrm{m}$. That would render the energy difference to both fluorine’s 2s and 2p too large. But in fact, hydrogen’s 1s is better placed at the height of Oberau, $650~\mathrm{m}$. That is way too far away from the 2s but rather close to the 2p.
This is confirmed by energy diagrams of $\ce{HF}$ like that presented in this lecture (in German). (Just search for HF on the page, it’s not far down. Ignore the text around it and look at the scheme.)

Now you may be asking yourself why the hydrogen’s 1s energy is so high compared to fluorine’s orbitals. Compare ionisation energies, which are typically understood as the energy difference of the highest orbital to ‘an electron in vacuum’. For hydrogen, this energy is $1312~\mathrm{kJ/mol}$ while for fluorine this is $1681~\mathrm{kJ/mol}$. The reasoning behind it is fluorine’s very high electronegativity, attracting the electrons towards its nucleus. Fluorine’s nucleus has a $+9$ charge which is much better at drawing electrons in than hydrogen’s $+1$ (even though it needs to draw nine electrons in rather than one). This stabilises all of fluorine’s orbitals; and those which are ‘closer to the nucleus’ (s-orbitals) most.

I am not well-educated in modern physical chemistry methods, but I heard that orbitals were visualised. If that is the case, it won’t be long until we can visualise the $\ce{HF}$ orbitals, and indeed show that the bonding orbital (highest energy σ type) will have a nodal plane through fluorine.

No, there is larger energy separation between H$_{1s}$ and F$_{2s}$. To see this lets lie a lot and say that orbital energies are proper energies. So thinking classically, the energy is due to coulomb interaction between the nuclei and the electron, electrons always have the same charge, but fluorine nuclei have much more charge than hydrogen. That means that (in this picture) the electron in a H$_{1s}$ would have greater energy than one in F$_{1s}$ (the second one much greater in absolute value). An F$_{2p}$ electron "feels" like there is not as much charge in the nuclei (because the coulumb repulsion with inner electrons), so its energy is closer to the H$_{1s}$ energy.

By the way, I have to say it, I do not find much sense in the argument about the closeness of orbital energy to assert about the overlaps.

No, it can't be measured because orbitals does not exist. All we can do is perform molecular calculations to get an approximation to the wave function. Then force the separation of this function in hydrogen like orbitals (other functions with whom we are familiar or obsessed) through projections to make this kind of interpretations.

I'm fairly sure there is a way to show that the interaction between two AOs depends inversely on the energy difference - I think it might be done via perturbation theory but I don't know the details. It's probably in Albright, Burdett, & Whangbo's Orbital Interactions in Chemistry somewhere.

@orthocresol I bet you are referring to a Møller–Plesset like treatment for the correlation energy (take a look at https://en.wikipedia.org/wiki/Møller–Plesset_perturbation_theory). But I'm not sure about what you mean by "interaction between two AOs". As they are abstract mathematical construction strictly speaking they can not interact. Electrons can interact and in the framework used in the question, I think that spatial overlap is important. Consider a molecule where similar orbitals are separated by many bonds, we won't use them to generate a molecular orbital.

If you state something like "let's lie a lot" I kindly ask you to state exactly what you are lying about. In this case, why are orbital energies not proper energies. I guess you will state that orbitals do not exist, but I guess it is necessary that you back up this statement. As it stands here, it reads as just an opinion. You also must explain why you are using a model that you believe is wrong.

@Martin-マーチン, My assertion that these are not proper energies is due that they are not eigenvalues of Hamiltonian of the system, but eigenvalue of the Fock operator. I did not add this kind of comments in the answer I thought that it was far too technical when considering the question.

@Martin-マーチン , I just try to go along the question. The usage of this kind of mindset are sometimes useful to solve some problems. The "why" they are useful (considering that they are wrong), that is, why they turn to be (sometimes) acceptable representation of reality is many times harder than learning more more advanced descriptions. Note that I do not use the word "model" because they really aren't, as they are not clearly established.

Wrong way round- search "orbital potential energy" and you'll know what I mean.
Hydrogen 1s is much higher in energy than F 2s and slightly higher then F 2p; that's why the bonding MO is centred mostly on fluorine and the unfilled LUMO is mostly on hydrogen.


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Conversion of 1,2-dione to a 7 membered oxepane?
Has anyone heard of forming a seven membered ring from the rearrangement of a 1,2-dione? What I am talking about specifically is benzil rearrangement:

Could anyone give me guidance for how on earth something like this would form?

Is it in basic medium ?

It's a protic solvent and under illumination so hydroxyl radicals can definitely form. Does that make it possible?

SciFinder has no examples of this transformation.

Is this an exercise from a book or is it your experiment? In which case, how did you confirm the structure of the product?

This is an experiment and the mass spectrometer suggests that this is the best match but it is just so strange because I can't find anything on sci finder either

@Coherent what mass spec data do you have that suggests this product? Since this is an isomer of the starting material, I'd be hesitant about assigning this structure. I'd probably be hesitant about assigning a structure for any reaction with only mass spec data.

By this late date Coherent has probably resolved the structure of his product (9). From the limited information provided, it appears that a photolysis of benzil (1) was conducted in hydroxylic solvent (methanol?). Let's assume for the sake of argument that the product has the structure that Coherent has proposed. Coherent asks, how can it (9) be formed? First of all, hydroxyl radicals need not be invoked. Structure 2 represents an excited state of benzil that may cyclize to diradical 3 whose resonance structure is diradical 4. Rearomatization of 4 provides enol 5, tautomerization of which affords 2-phenylbenzofuran-3-one (6). Now the reorganization 4 --> 5 need not be intramolecular as indicated (red hydrogen) but intermolecular or, less likely, via the solvent acting as a source of hydrogen atoms (methanol C-H bond DHo = 96 kcal/mol. Probably too strong a C-H bond to deliver a hydrogen atom to stabilized radicals).
Benzofuranone 6 is a known compound. If benzofuranone 6 is truly formed in the reaction sequence,  it should independently afford the final product 9 upon irradiation conditions. Cleavage of benzofuranone 6 by the Norrish type I mechanism leads to diradical 7, which may recyclize to 8 and subsequently rearomatize to 9. [I am not overly excited about going from a 5-membered to a 7-membered ring, but then again who knows.]
Finally, compound 9 should be able to be synthesized from O-benzylsalicylic acid by intramolecular acylation.


text:
Can only gases by diatomic?
I know that the elements H, N, F, O, I, Cl, and Br are found naturally as diatomic gases. But can any other elements be found as diatomic gases, or even solids or liquids.
Why or why not?

Actually at STP bromine is a liquid and iodine is a solid.

Any molecule composed of exactly two atoms is diatomic, but I will answer this question assuming that you mean elemental diatomic molecules. Only H, N, O, F and Cl are diatomic elemental gases under STP. Br and I are diatomic liquid and solid respectively, under STP. Oxygen remains diatomic in liquid state, while nitrogen and all halogens(F, Cl, Br, I) are diatomic in all three classical physical states. Gaseous sulfur can be diatomic at or below standard pressure. Structurally similar to diatomic nitrogen, diphosphorus is an unstable allotrope of phosphorus. Diatomic carbon is a gas that exists under very high temperature. There may also be other diatomic allotropes.
Some elements form stable diatomic molecules because of the octet rule (or duplet for hydrogen). Generally a bond can share up to three pairs of electrons, so elements which are three or less electrons short of a stable electron configuration can pair up and share up to three electrons to become stable. Beyond that, elements will find other ways to become stable.
However, chemical stability isn't governed only by the octet rule, there are other factors such as geometry, bond length and nucleophilicity/electrophilicity that affect molecular stability. Therefore, some diatomic allotropes are not the most stable allotropes of the elements. Examples from above are disulfur, diphosphorus and dicarbon.


text:
Will washing child seat belt straps with baking soda and hydrogen peroxide make them unsafe to use?
My friend is afraid to use the child seat for her next child.  She was told washing them with the solution weakened the fabric,  The straps weren't submerged, but they did get throughly wet.

By peroxide, you mean hydrogen peroxide.

Yes, hydrogen peroxide.

@Emily Can you identify the material of the straps, for example glossy polyester (PES) or matt polyamide (symbol: PA and a number)? Polyester is usually resistant to hydrogen peroxide solutions, whereas polyamide will deteriorate.

I have seen some companies that refurbish child seats, perhaps a new strap kit would put your friend's mind at ease.
If the straps were not soaked for hours in concentrated H2O2 and I saw no damage and could not cause them to break with my strength (healthy adult) I would have used it with my kids.  The baking powder or soda should have no effect.
We bought a used seat for the next size up and have no idea of how it was washed.  Would do the same again.
Stay at home, spend the coin, take out a loan or live with the risk, those are the choices. There are other risks in life like not being sure that the excavator operator is of sound mind and will not swing the bucket into your car at the stop sign, does this stop you leaving the house?
EDIT:
It is possible the dangers of washing are circulated by the seat refurbishing and manufacturing industries to increase sales and not safety.  A car seatbelt can last for decades and sees a lot more use and has no washing warnings.

This is not enough evidence that hydrogen peroxide treatment is safe. Also the discussion about attitude to risk is irrelevant and in fact for most parents anything else than zero risk for their children is unacceptable if they can avoid it.

@K_P your attitude is illogical.  There is NEVER zero risk, getting pregnant has attendant risks, adults and parents especially become adept at weighing risks.  Most parents will agree that they are less neurotic with subsequent children, would you have me believe this is due to them becoming irrational about risk or perhaps an understanding that they were over cautious when less experienced.  Also note my answer covered the risk eliminators profile first to avoid comments like yours, I'll keep trying to educate in a balanced manner in spite of it.