text:
What is Drago's rule? Does it really exist?
My textbooks states the Drago's rule in inorganic chemistry as follows:

The more electronegative atom prefers the orbital having more $\mathrm{p}$ character, and lone-pairs or less electronegative elements prefer such orbitals, as have more s character.
Now, according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}.$
So, in $\mathrm{sp^3d}$ more electronegative elements like fluorine go
to axial position, but such elements go to equatorial position in $\mathrm{sp^3d^2}$ or $\mathrm{sp^3d^3}$.
Also, as a result of Drago's effect when atoms of group $14,$ $15$ and$16$ (belonging to third period and above) are attached to atoms having electronegativity less than $2.5$ then it has valence electrons in pure s-orbitals. And apparently no hybridization effect is observed in such cases. That is why in $\ce{NH3}$ the bond angles are $107.8^\circ$ but for $\ce{PH3},$ $\ce{AsH3},$ $\ce{SbH3},$ $\ce{BiH3}$ the bond angles are very
close to $90^\circ$.

I couldn't make much sense out of the above rule. My questions are as follows:

What does the following part mean?

according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

Does it mean that the hybridized orbitals split in this way? If so, why should they split?

Why would the more electronegative atom prefer the orbital with more $\mathrm{p}$ character?

Even if Drago's rule holds, why is the statement in the paragraph beginning with "Also, as a result of Drago's effect when atoms …" true? I could not understand why the same won't occur for atoms with electronegativity $>2.5$.

I could not find much information regarding Drago's rule on the internet. Not even on Wikipedia. One of my acquaintances (who is a chemist) said that no such rule exists. But I didn't get much time to discuss the matter with him.
So does such a rule really exist? And, in that case, how can we answer the above two questions given as bullets  using the existing theories in chemistry?

Sounds rather like https://en.wikipedia.org/wiki/Bent's_rule

Related:http://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h?noredirect=1&lq=1

Also related:http://chemistry.stackexchange.com/questions/14791/hybridization-in-ph3

@Doraemonドラえもん If the term is textbook specific, you might want to also post the name of the textbook & authors (and edition, if any). There's a chance that that information will help track down the usage. (Also, is this an English language textbook? "Drago's rule" might be a translation error - again, knowing the textbook details may help.)

@R.M. It is not actually a standard textbook which is available worldwide. It is a study material prepared by a coaching institute. I don't think posting the name of the institute will be of much use here. Anyway, I don't think the propositions of the rule are wrong. Some of it is indeed explained by Bent's rule as pointed out by Mithoron. But I could not find the explanation regarding the orbital splitting as proposed by the so-called "Drago's rule". I hope someone can verify whether such a splitting really takes place or not.

@R.M. And yeah, the textbook/study-material is written in English. But I don't think the name of the rule should be much of a problem. The main problem is checking whether all the claims made by the rule are true or not.

That sounds like an extension of Bent's rule shoehorning in some d orbitals. That kind of hybridisation doesn't make much sense as we have often discussed here. I would not waste time on understanding this rule.

@Martin-マーチン "That kind of hybridisation doesn't make much sense as we have often discussed here." Can you give some reference or links for this ? Where has it been discussed before ?

@Doraemonドラえもん In like every question where using of d orbitals in p-block was mentioned? Like http://chemistry.stackexchange.com/questions/32695/is-clh7-possible or http://chemistry.stackexchange.com/a/49844/9961 or...

This might go back to Professor Russell Stephen Drago (1928-1997) https://webster.chemistry.msstate.edu/prof/drago.htm and his textbook of Physical Methods in Inorganic Chemistry from 1965. @EdV any ideas (Drago was born in Turners Falls and received his B.S. from UMass Amherst).

@Karsten Theis Well, I vaguely remember being in the room when Russ Drago and Alan Katritzky visited our (Jim Winefordner’s) lab at the University of Florida at Gainesville: https://chemistry.meta.stackexchange.com/a/4757/79678. I never talked with Drago, but I do not doubt that the rule is named after him.

Going through what you posted, I think ‘Drago’s rule’ (which I never encountered, either at school or at university) gives good predictions but uses a largely terrible set of arguments.
For any atomic system with more than one electron, quantum chemistry predicts the energies of the $\mathrm{s}$ and $\mathrm{p}$ subshells of a shell to differ — while for hydrogen-like systems (one-electron systems) all subshells of a given shell have the same energy. The $\mathrm{s}$ subshell of a certain shell in multi-electron systems has a lower energy than the $\mathrm{p}$ subshell, which in turn has a (far) lower energy than the $\mathrm{d}$ subshell and so on and so forth. Since the $\mathrm{s}$ subshell is determined to be more stable by this method, it is not surprising that the first two electrons of a given shell are added to said $\mathrm{s}$ subshell; the next six electrons are added to $\mathrm{p}$ subshells whereafter a new shell is used.
This immediately explains that it is most stable for any atom to treat the $\mathrm{s}$ electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the $\mathrm{s}$ electrons to be energetically favourable with respect to any configuration that does a priori. (You can consider that part of the quoted statement to be correct.) Therefore, the a priori predicted most stable configurations of certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 $\mathrm{p}$ orbitals to form bonds; keep one lone pair in an $\mathrm{s}$ type orbital. (Predicted bond angle: $90^\circ$.)

tetracoordination (no additional lone pairs): since the former is not possible, hybridise $\mathrm{s}$ and $\mathrm{p}$ orbitals to form $\mathrm{sp^3}$ hybrid orbitals. (Predicted bond angle: $109.5^\circ$.) Actually, in MO terms you should consider the four ligands to supply four group orbitals whose symmetry always fits one of the three central atom’s orbitals, resulting in one lower bond energy and one higher bond energy corresponding to triple-degenerate bonds. The $\mathrm{sp^3}$ description is mathematically equivalent, though.

penta- and hexacoordination; tetracoordination with additioal lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an $\mathrm{s}$ orbital if possible. Use remaining $\mathrm{p}$ orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^\circ$ going from 2e2c bonds to 4e3c bonds; $180^\circ$ between a pair of coordinating atoms contributing to the same 4e3c bond.)

Using the aforementioned 4e3c bonds nicely explains why electronegative atoms prefer to occupy these positions. You should always remember that 4e3c bonds can be described by the following resonance structures (drawn for $\ce{ClF3}$):
$$\ce{F-Cl^+\bond{...}F- <-> F^-\bond{...}Cl^+-F}$$
In this resonance, each outer atom taking part in a 4e3c bond has a formal average charge of $-0.5$; it is generally better to assign charges to electronegative atoms.

The theories I have outlined so far do not adequately predict the bond angles of water, ammonia and related compounds. Indeed, these compounds are remarkably complicated although they seem simple. The gist, however, is that a bond angle of $90^\circ$, while being most favourable on a pure orbital basis, introduces steric strain between the outer atoms which approach each other too closely if the central atom is relatively small. To combat this destabilisation, the bond angle is extended by mixing $\mathrm{s}$-contribution into the bonding orbitals — which in turn means that the lone pair will be less $\mathrm{s}$-like. The ideal bond angle for dicoordinated systems based on sterics is $180^\circ$, showing that the electronic contribution ($90^\circ$) is more important than the steric one ($180^\circ$). Larger central atoms than oxygen and nitrogen, such as phosphorus, antimony, selenium, etc., allow for smaller bond angles since the bond lengths are larger increasing the spacing between the ligands.
The reason for the special behaviour of nitrogen and oxygen is not their high electronegativity but their small size. Chlorine, which has an electronegativity similar to nitrogen, behaves as predicted for large atoms.

It has been commented that you should not invoke $\mathrm{d}$ orbital participation for main group elements. Remember what I said in the second paragraph: a shell’s $\mathrm{d}$ subshell has a higher energy than the next shell’s s subshell. Thus, before any same-shell $\mathrm{d}$ subshell is added to hybridisation, the next shell’s $\mathrm{s}$ subshell should be. Since hybridisations involving two different shells’ $\mathrm{s}$ subshells are not observed, we should reject the possibility of using $\mathrm{d}$ orbitals for hybridisation. Therefore, we can also flat-out reject the central two paragraphs of your quotation.

"...you should not invoke d orbital participation for main group elements". Does that mean all hybridizations like $\mathrm{sp^3d}$ ,$\mathrm {sp^3d^2}$ and $\mathrm{sp^3d^3}$ are invalid for main group elements ? And which elements do you consider as "main group" elements ?

And, in that case, how do you explain the hybridization of compounds like $\ce{PCl5}$ ? If it is not $\ce{sp3d}$ then what is it ? How can you explain its hybridization using only $s$ and $p$ orbitals ?

@dora http://chemistry.stackexchange.com/q/52179/4945 hybridisation simply isn't a good enough concept for these cases.

' Since hybridisations involving two different shells’ s subshells are not observed, ...' No one ever observed any hybridisation. But try mixing 1s and 2s (etc), they have the same symmetry...

@Doraemonドラえもん $\ce{PCl5}$ is $\mathrm{sp^2}$ hybridised. The two axial chlorine atoms form a 4-electron-3-centre bond with phosphorus.

PCl5 is sp3d hybridized

@ShubhangWalavalkar That is incorrect. It is an old theory proposed, I believe, by Pauling that has never managed to die despite the overwhelming evidence presented against it possibly because it was supported by a Nobel laureate. Please see this answer as a starting point. If that doesn’t satisfy you, there are many other posts on this network that can enlighten you.

so according to you 3 center 4 electron bond changes bond angle of 90 degree( 2 center 2 electron ) to 180 ????

@Samardeepsingh Molecules typically don't 'go to 4e3c bonds from 2e2c bonds'. So there is no 'change of bond angle'. Rather, a molecule will either adopt a configuration with two 2e2c bonds or one 4e3c bond, depending on electronic factors. One is predicted to give a 90° bond angle, the other is predicted to give a 180° bond angle. Which is which is up to the chemist to determine.

These rules go back to a publication by Russel S. Drago criticizing the use of the VSEPR model in Introductory Chemistry. As an alternative, he suggests a set of heuristics. First, if there are no lone pairs, use essentially the same rules as VSEPR. How to treat lone pairs is described below:

Here we shall consider central atoms with eight or
less electrons about it. When there are lone pairs on the
central atom, each lone pair counts as a group in the
above scheme if the central atom is a second row element.
If the central element is a third, fourth, etc., row atom
and if the groups attached to the central atom are oxygen
or a halogen, the lone pair also counts as a group. If the
groups attached are less electronegative than bromine
(that is, if almost anything other than a halogen or oxygen
is attached), the lone pair occupies an unhybridized orbital
and does not count as a group. When the s orbital is
used to accommodate a lone pair, only p orbitals remain
for bonding, and the geometry involves an arrangement of the groups at about 90° angles to each other. Some examples
are listed in Table 2.

(Table 2 states that $\ce{H2O, NH3, and PCl3}$ are tetrahedral whereas $\ce{PH3 and H2S}$ use p-orbitals to bond and have ninety degree angles). The claim is that the rule of thumb is better than VSEPR, and not off by more than 4 degrees from the observed structures.
He goes on to say that chemists, when they encounter a molecule they don't know, would draw a Lewis structure and compare it to those of molecules they know rather than trying to predict a structure on paper from first principles. Another quote:

It is deceptive to lead a freshman
into believing anything else. The rules presented above
summarize these analogies and enable you to predict the
geometries of many molecules.

Source: A criticism of the valence shell electron pair repulsion model as a teaching device, RS Drago - Journal of Chemical Education, 1973


text:
Hydrolysis products of chlorides of group 15
$\ce{NCl3}$ on hydrolysis produces $\ce{NH4OH}$ and $\ce{HOCl}$.
$\ce{PCl_3}$ when hydrolyzed produces $\ce{P(OH)3}$ i.e. $\ce{H3PO3}$.
$\ce{AsCl_3}$ when hydrolyzed produces $\ce{As(OH)3}$.
However when $\ce{SbCl3}$ or $\ce{BiCl3}$ are hydrolyzed they produce $\ce{SbOCl}$ and $\ce{BiOCl}$ respectively.
Why are the products so widely varying ? How can we explain the reactions theoretically ?

The first one is because nitrogen is more electronegative than chlorine while all the others are electropositive with respect to chlorine. The antimony and bismuth reactions are probably due to some stability of those compounds that I, unfortunately, don’t know about.

Nitrogen trichloride

Nitrogen trichloride is quite unstable and hydrolyse easily:
$$\ce{NCl3 + 3 H2O → NH3 + 3 HOCl~~~~~~~~(1)}$$
$$\ce{NH3 + H2O -> NH4OH~~~~~~~(2)}$$
$$\ce{NCl3 + 4H2O -> NH4OH + HOCl~~~~~~~(1) +(2)}$$

Phosphorus trichloride

$$\ce{PCl3 + 3H2O → H3PO3 + 3HCl}$$

Arsenic trichloride

$$\ce{AsCl3 + 3 H2O → As(OH)3 + 3 HCl~~~~~~~~(1)}$$
This is the first and main reaction of hydrolysis of arsenic chloride to form arsenous acid. But other species (arsenite ions) like $\ce{[AsO(OH)2]−}$ and $\ce{[AsO2(OH)]^2−}$ also exist in solution and are the conjugated bases.
$$\ce{AsCl3 + 4H2O ⇄ H[As(OH)4] + 3HCl~~~~~~~(2)}$$
$$\ce{AsCl3 + 2H2O ⇄ HAsO2 + 3HCl~~~~~~~~~(3)}$$
And at last aresnic trichloride fully hydrolyse to form arsenic(III) oxide
$$\ce{2AsCl3 + 3H2O → As2O3 + 6HCl~~~~~~~~~~(4)}$$

Antimony trichloride

$$\ce{SbCl3 + H2O → SbOCl + 2HCl~~~~~~~~(1)}$$
With more water it forms $\ce{Sb4O5Cl2}$ which on heating to 460° forms $\ce{Sb8O11Cl2}$.
$$\ce{SbCl3 + H2O ⇄ Sb(OH)Cl2 + HCl~~~~~~~(2)}$$
$$\ce{4SbCl3 + 5H2O ->[50 C] Sb4O5Cl2 + 10HCl~~~~~~~(3)}$$
Final reaction to form antimony trioxide.
$$\ce{2SbCl3 + 3H2O ->[\Delta] Sb2O3 + 6HCl~~~~~~~~~~(4)}$$
Note that speculation on the composition of antimony oxychloride has been raised. Some say it is a mixture of antimony trioxide and antimony trichloride.

Bismuth tricloride

$$\ce{BiCl3 + H2O -> BiOCl + 2HCl~~~~~~~~(1)}$$
Main reaction to form bismuth oxychloride. Further hydrolysis yield and intermediate monohydrate $\ce{BiCl3.H2O}$.
$$\ce{BiCl3 + H2O(vap) ->[50 C] BiCl3.H2O~~~~~~~~(2)}$$
$$\ce{$$BiCl3 + 2H2O → Bi(OH)2Cl + 2HCl~~~~~~~~~(3)}$$
Explaination
Now, you may been be wondering that hydrolysis of nitrogen and phosphorus yield one product but that of arsenic, antimony and bismuth yield 3 or 4 products. This is because of presence of vacant d or f orbitals where electrons can reside easily. For this reason, $\ce{Sb^3+, Bi^3+}$ solvate in aqueous solution to form antimonyl or bismuthyl ($\ce{SbO+, BiO+}$) and other species like $\ce{[Bi6O4(OH)4]^6+, [Bi6(OH)12]^6+ ,  [Bi(H2O)9]^3+}$ and similar antimony ions. Other factors also play role like relativistic effects, lanthanoid contraction, inert pair effect etc.

source: Stack Exchange

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

sort: votes


text:
Clarification needed regarding description of molecules without hybridization using pure p orbitals
I need some clarification regarding this answer given by the user @Jan.

@Jan says:
Therefore, the a priori predicted most stable configurations of
  certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 p orbitals to form bonds; keep one lone pair in an s type orbital. (Predicted bond angle:
  $90^\circ$)

I could not understand the meaning of the first bullet point clearly.
I cannot agree that in mono, di and tri coordination pure $p$ orbitals are always used. Pure $p$ orbitals have bond angles of $90^{0}$. In that case, how can we explain the structure of compounds like $\ce{BF3}$ or ethene where $\mathrm{sp^2}$ hybridization is said to exist and bond angles are $120^{0}$ ?
Or did @Jan imply that we need to use pure $p$ orbitals in mono, di and tri coordination in case no hybridization takes place ? If so, how can we understand when hybridization will take place and when it will not? Please give some examples.

In brief: the hybridization model fails badly for hypervalent species. All have large contributions from charge-shift bonding, which leads to the actual valence electron population in the region around the central atom being $\leq 8$ in all cases. I'll post an answer with more detail when I can.

Yes, those bullet points are only very brief xD They can fail quickly; $\ce{BF3}$ does indeed fail the first bullet point because I did not consider phrasing it in a way to include boron. By the way, I’m leaning towards voting to close this as too broad since the two concerns you raise should be answered in two very different ways (and they are not really conceptionally related, except for both being simplified). This is coming at the danger that the pentacoordination part may be closed as a duplicate of one of the many questions here dealing with the topic.

@Jan Umm, before you vote to close I would ask you whether the question can be edited. Should I divide the question into two or more parts ? BTW your answer to my previous question was not really clear and the bullet points are really too brief.

@Doraemonドラえもん Yeah sure you can edit! In fact, the intended way to address closed (or: ‘on hold’) questions is to edit them to make them fit again. In this case, I would suggest asking about each of the bullet points in a separate question, i.e. editing this one down to one of the two and putting the other into another question ^^

@Jan Done. Please check if my questions are making sense, individually. http://chemistry.stackexchange.com/questions/64673/how-to-explain-shape-of-molecules-if-hybridization-involving-d-orbitals-in-main

Yes, what I wrote on the topic in the previous question was very, very brief and while writing it I was not thinking of boron — an element that usually calls for exceptions due to it only having three valence electrons.
However nonetheless, the first thing you should do for any compound is to consider all atoms in ground state configuration and see whether you can form the required bonds from occupied orbitals in that state. Let’s take the example compounds $\ce{BF3, NF3, AsH3, C2H4}$ (tricoordinate) $\ce{CO2, H2O, H2Se, BeCl2, CH2}$ (dicoordinate) and $\ce{HCl, N2, CO}$ (monocoordinate) for the discussion. Any structural data referenced herein is taken from the corresponding Wikipedia articles.
Ideal cases
I consider the examples $\ce{AsH3, H2Se}$ and $\ce{HCl}$ to be ideal cases. Taking each atom in its ground state ($\ce{As}:{ [\ce{Ar}]}\,\mathrm{4s^2\,4p^3}; \ce{Se}{: [\ce{Ar}]}\,\mathrm{4s^2\,4p^4}; \ce{Cl}{: [\ce{Ne}]}\,\mathrm{3s^2\,3p^5}; \ce{H}{: \mathrm{1s^1}}$), arsenic has three half-filled p orbitas, selenium has two and chlorine has one. Each of these elements (except hydrogen) has a single lone pair in an s orbital and if there are further lone pairs, they are in p orbitals (arsenic has no more). Thus, we can construct $\sigma(\ce{s-p})$ bonds until all valences are satisfied.
Moving on to bond angles: well, $\ce{HCl}$ has no problems as there are only two atoms. But $\ce{H2Se}$ and $\ce{AsH3}$ also have hardly any problems since the central atoms are so large that the comparatively smaller ligands can be attached at almost perfect $90^\circ$ angles without significant steric repulsion. The bond angles repored ($\ce{AsH3}{: 91.8^\circ}; \ce{H2Se}{: 91^\circ}$) are in adequate agreement.
Ideal cases with steric distortion
The next set of examples along the route you should be going are $\ce{NF3}$ and $\ce{H2O}$. These compounds fulfil the general criteria outlined in the previous section. However, their central atoms are decidedly smaller. Thus, perfect $90^\circ$ angles are unobtainable as there would be a steric clash of the ligand atoms. The ligand atoms move apart to prevent steric repulsion; this is achieved by mixing in s-contribution into the bonding orbitals. $25~\%$ s-contribution would correspond to a tetrahedral bond angle of $109.5^\circ$ and $\mathrm{sp^3}$ hybrid orbials. The table below compares the bond angle and length data of the lighter homologues discussed here and the heavier ones discussed above (consider the fluorines in $\ce{NF3}$ to be pseudohydrogens).
\begin{array}{lcccc}\hline
\text{element} & \ce{N} & \ce{As} & \ce{O} & \ce{Se} \\ \hline
d(\ce{X-H})/\mathrm{pm} & 137 & 151.9 & 95.84 & 146 \\
\angle(\ce{H-X-H})/^\circ & 102.5 & 91.8 & 104.5 & 91 \\ \hline\end{array}
$\ce{CH2}$, carbene, is a dangerous case. As far as I know, no definite structures are reported (although Wikipedia mentions bond angles determined by EPR) since it is too reactive and since it is susceptible to dimerisation to form ethene. It also comes in three distinct forms: a singlet form, a linear triplet and a bent triplet. The most ideal structure of these three is the singlet carbene, which also fits nicely into this subsection.
For a singlet carbene, you would also start off from carbon’s ground state $[\ce{He}]\,\mathrm{2s^2\,2p^2}$. The two p orbitals each form a $\sigma(\ce{s-p})$ bond to the hydrogen atoms. The two hydrogens, however, need to move away from each other to ease steric repulsion; thus a little bit of s-contribution is mixed in for a final bond angle of $102^\circ$. A third p orbial remains unoccupied — note the similariy to $\ce{H2O}$ which has everything carbene has plus two electrons in a p-type lone pair.
Practically ideal cases which naturally lead to multiple bond
Herein, we shall discuss $\ce{N2}$ and $\ce{CO}$. For both, we can apply the method outlined above; I’ll start with the simpler $\ce{N2}$ case. We initially have one s lone pair and three, singly occupied p orbitals on each nitrogen. One of these forms a standard $\sigma(\ce{p-p})$ bond with the other nitrogen. Now, thankfully, the other p orbitals are nicely aligned and fit perfectly for generating two $\pi(\ce{p-p})$ bonds. Together, that gives us three bonds or a triple bond, composed of one σ and two π bonds.
The carbon monoxide case is essentially the same except that carbon would have one empty p orbital and oxygen would have one lone pair in a p orbital; only one π bond can be formed in the way outlined above. However, the unoccupied carbon orbital and the fully populated oxygen orbital can align likewise and form what could formally be described as a second, dative π bond. That renders nitrogen and carbon monoxide isoelectronic. Note, however, that the simplification introduced herein is not able to explain the big differences between $\ce{CO}$ and $\ce{N2}$ that arise from their actually rather different molecular orbital schemes.
Cases, in which an s type lone pair is impossible
I’ve cleared away all the simple cases. If you go through the list, the only ones that remain are $\ce{BF3, C2H4}$ and $\ce{BeCl2}$ — although we might as well include linear and bent triplet $\ce{CH2}$ in here. These all have in common that assuming an s type lone pair would not allow for sufficient electrons available for bonding.
If that is the case, the simple answer always suggests hybridisation (athough the MO description does not require it and nonlocalised molecular orbitals would not support it). At least one p orbital must be mixed in to allow for the generation of enough σ bonds; yet you should always try and make the p contribution of the hybrid orbitals as small as possible (in simplified terms) as that means the s lone pair ‘loses’ less energy en route. (Beware, the terms and images I used are inaccurate!) Any orbitals, populated or not, that are not required for σ bonds remain as p type ones. For $\ce{CO2}$ and linear triplet $\ce{CH2}$, that gives us an $\mathrm{sp}$ configuration of the central carbon with two unused p orbitals; in $\ce{C2H4}$, bent triplet $\ce{CH2}$ and $\ce{BF3}$, the central atoms are $\mathrm{sp^2}$ configured with one additional p orbital.
If possible — i.e. in $\ce{CO2}$ and $\ce{C2H4}$ — π bonds should then be formed. Each of the oxygens in carbon dioxide has one p-type semipopulated p orbital available for a π bond as does the second carbon of $\ce{C2H4}$. The carbenes cannot form π bonds due to the lack of partners, and thus they remain as diradicals. The linear one has two degenerate p-type radical orbitals, the bent one almost has an $\mathrm{sp^2}$ type radical orbital and one p type. (In fact, bent triplet carbene has a larger bond angle than the $\mathrm{sp^2}$ predicted $120^\circ$. This is in line with removing part of the s contribution of the radicals and supplying that to the bonding orbitals. Nobody expects you to correctly analyse the structures of carbenes, though.)
Beryllium
You may have noticed that that element had gone missing at the end of the previous section. This is because it is the most special of all cases. All other alkaline earth metals behave as metals (form dications and practically ionic structures) while beryllium still features notable non-metal properties.
I think that the best description of $\ce{BeCl2}$ is via a four-electron-three-centre bond akin to $\ce{XeF2}$. That again requires an unhybridised central atom. The lowest of the three relevant MOs of the 4e3c bond would be formed of the fully bonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ combination; the second bonding MO would then be $\mathrm{p}(\ce{Cl}) + \mathrm{p}(\ce{Cl})$ where the nodal plane traverses $\ce{Be}$ — mixing in its 2p orbital probably increases the bonding character — and a final antibonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ orbital wherein the phase of beryllium’s orbital is of opposite sign with respect to the chlorine orbitals’. If that is the case, it should possibly be discussed with the ‘hypervalent’ cases of the other question.
I could be terribly wrong, though. Thankfully, nobody will request you to predict the MO scheme of $\ce{BeCl2}$ until at least the final bachelor’s examination.

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


text:
How to explain shape of molecules in penta and hexa coordination if hybridization involving d-orbitals (in main block) is considered incorrect?
In this answer @Jan says:

penta- and hexacoordination; tetracoordination with additional lone
  pairs and related: attempt to form as many normal bonds with p
  orbitals as possible; keep one lone pair in an s orbital if possible.
  Use remaining p orbitals to construct four-electron-3-centre bonds to
  the remaining atoms. (Predicted bond angles: diverse. $90^{∘}$ going
  from 2e-2c bonds to 4e-3c bonds; $180^{∘}$ between a pair of coordinating
  atoms contributing to the same 4e-3c bond.)

Let us consider $\ce{PCl5}$. According to several books the hybridization in this compound is $\mathrm{sp^3d}$. But @Jan says that pure $d$ orbitals never take part in hybridization in case of main group elements. Okay. So, if we assume that there is no hybridization in $\ce{PCl5}$ then as Jan says we form "as many normal bonds" with $\ce{Cl}$ as possible using $p$ orbitals. Since, $\ce{P}$ has 3 p orbitals we use them to form mutually perpendicular bonds. Suppose they are the top, bottom and leftmost bonds as in the following diagram:

After this, we are still left with two $\ce{Cl}$ atoms and only one $s$ orbital. How can we place two $\ce{Cl}$ atoms in only one pure $s$ orbital ? This part makes no sense to me.
Now, let us consider the $\ce{ClF3}$ example that @Jan gives. If we follow Jan's method we keep one lone pair in a pure $s$ orbital , then the shape can never be trigonal bipyramidal like this:

This is because $s$ orbital is spherical in nature and does not look like a dumbbell shaped $p$ orbital.
Also, could someone please give me some examples in cases of "penta- and hexacoordination; tetracoordination with additional lone pairs", where $3c-4e$ bonds are formed using remaining $p$ orbitals when no hybridization occurs (as mentioned by @Jan) ?

I think they’re looking good =)

This is very broad as written, and it also reads more like a critique of another person's answer to a question than as a question itself.

@ToddMinehardt Well, Jan himself/herself told me that this question is fine. And it simply asks for clarification. No one is criticizing anyone or anyone's answer here.

@Jan Do you also think that is very broad as Todd says? Should I edit it further ? (In my opinion if I trim this question further it will lose its essence )

@ToddMinehardt Imho, the question is fine. The critique is on scientific level imho; they question the (very simplified) concept I gave in another answer asking for expansion. It was too much to include in the other answer but I think it’s good here.

In fact, I even think this can serve as a canonical dupe target for ‘My book says compound X uses d orbitals; here I read it doesn’t. Why/how?’

One possibility, PCl3^2+ Cl- Cl-, P uses sp2 hybridization.

Sounds like it's overflow from the other question - I trust your assessment, withdraw close vote. Thanks for clarifying.

This has been asked (and explained) in one form or the other multiple times: http://chemistry.stackexchange.com/q/29142/4945 http://chemistry.stackexchange.com/q/18427/4945 http://chemistry.stackexchange.com/q/28692/4945 http://chemistry.stackexchange.com/q/6785/4945 Simply let go of the hybridisation concept, forget the VSEPR model and try to form a MO scheme with ligand group orbitals, the lone pairs will fall in place. An while I disagree with the name 'hypervalent' the wikipedia article is quite well explained. (cc @jan )


text:
Stoichiometry and POAC Contradiction
Lets say we have the reaction
$$\ce{H2 + O2 -> H2O}$$
And we are given 10 moles of $\ce{O2}$ and are asked that
How many moles of hydrogen gas do we need to complete the reaction ?
Stoichiometry gives the answer as 10 moles but by applying Principle Of Atom Conservation,  we get
$$\text{Moles of }\ce{H2}  =2\cdot\text{moles of }\ce{O2}$$
Why is the POAC giving wrong results?

What is the abbreviation POAC?

You applied the POAC incorrectly . Even your stoichiometric answer is incorrect.
First Balance the reaction
$$\ce{2H2 + O2    -> 2H2O}$$
We get that one mole of $\ce{O2}$ requires 2 moles of $\ce{H2}$.
Thus 10 moles $\ce{O2}$ need 20 moles of $\ce{H2}$.

@Zhe Looks good now

@Harsh Sharma You can refer your RCM for more

FYI: Formatting formulas


text:
Why does NO2 react with HCl ? What is the mechanism of the reaction?
I found the following reaction given in a certain book:
 $$\ce{2NO2 + 4HCl -> 2NOCl + Cl2 + H2O}$$
I cannot think of any reason why an acidic oxide like $\ce{NO2}$ will react with an acid like $\ce{HCl}$.
So, why will they react? And, what will be the mechanism of the reaction?

It is a redox reaction. One compound is a strong oxidant; another can reduce it, albeit not very willingly. So it goes.

What do you mean acidic oxide?

@NMJD acidic oxides are oxides that on hydrolysis gives acid and on reacting with base gives salt. Basic salt on the other hand gives base on hydrolysis.

Your equation is not balanced.

@NilayGhosh - I'm aware. I wanted user39139 to answer the question; some students misunderstand "acid oxide" and assume it should act as an acid in the way of HNO3 or HCl. I wanted to see if that was at the root of this question, since otherwise the answer just lies in redox potentials (which has already been suggested). If this misunderstanding persists, that answer might not help OP. I suppose to be clearer I should have said, "Can you define what you mean by 'acid oxide'"

Yes, it is a typical redox where nitrogen dioxide is an oxidising agent and hydrochloric is a mild reducing agent as Ivan noted.
The reaction proceed in two phases:
$$
\begin{align}
\ce{NO2 + 2HCl ->[\Delta] NO + Cl2 + H2O}   & \tag{R1}\\
\end{align}
$$
Now some nitrogen monoxide react with water to form nitrosyl chloride.
$$
\begin{align}
\ce{Cl2 + 2NO -> 2NOCl}   & \tag{R2}\\
\end{align}
$$
Finally, we get our main reaction
$$
\begin{align}
\ce{2NO2 + 4HCl -> 2NO + 2Cl2 + 2H2O}   & \tag{R1*2}\\
\ce{2NO2 + 4HCl -> (2NO + Cl2) + Cl2 + 2H2O}\\
\ce{2NO2 + 4HCl -> 2NOCl + 2H2O + Cl2}  & \tag{R3}\\
\end{align}
$$

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


text:
How can I draw metallocenes in ChemDraw?
How can I properly draw the structure of metallocenes (e.g. ferrocene) and similar sandwich compounds in ChemDraw?

In particular, two problems cannot be solved with the usual workflow when building a molecule in ChemDraw:

How can I draw a bond starting from the centre of a ring structure (e.g. cylopentadienyl or benzene) if there is no atom?
How can I connect the centre of another ring structure to an existing molecule with the correct bond length and angle?

This question came up among the editorial staff of the chemistry section of the German Wikipedia. I have answered this question there myself, and I think it might be useful to share the solution to this problem here.

In the first step, separately draw the individual ring structures. ChemDraw comes with an template library of predrawn structures, including some suitable cyclopentadienyl rings.

You can place the ring structures anywhere in your document.

Select one ring structure. On the Structure menu, click Add Multi-Center Attachment. This adds a multi-center node as attachment point to the structure.

Repeat this process for the other ring structure. Asterisks indicate the created multi-center nodes.

You can use the multi-center attachment points to draw the coordination bonds. Note that bonds connecting to delocalized systems should be drawn as plain bonds; it is not acceptable to use dashed bonds. When one bond crosses another, if necessary, you can click Bring to Front or Send to Back on the Object menu to indicate which bond is in front. The bond which is behind will have a small section cut out.

Click to select the end (i.e. the terminal atom) of the coordination bond in one structure, and then shift+click to select the end of the coordination bond in the second structure. On the Object menu, click Join.

The two structures are joined so that they share an atom. Note that the second structure is moved accordingly.

Finally, you can add any substituents as required.
(Also note that for unsubstituted ferrocene in the gas phase, the cyclopentadienyl rings are actually eclipsed.)


text:
Why is Helmholtz free energy often defined as the useful work attainable at constant volume if constant volume implies no expansion?
From Wikipedia:
"In thermodynamic systems where the temperature and volume are held constant, the measure of useful work attainable is the Helmholtz free energy function"
https://en.wikipedia.org/wiki/Work_(thermodynamics)#Free_energy_and_exergy
By definition, constant volume means that no PV work (expansion work) is done by the system. Because of this, is it not confusing to define Helmholtz free energy as a measure of the maximum useful/non-expansion work attainable at constant volume and temperature rather than just the maximum work attainable at constant volume and temperature, since all work done at constant volume is inherently useful/non-PV work? Adding the "useful" makes it seem like the useful work is separate from work done in general for a constant volume process. Am I missing something?

I don't understand. Isn't that the same thing?

No.  You are not missing anything.  There are other kinds of work besides PV work, such as electrical or work in driving a shaft within a constant volume system.

What I mean is that adding the "useful" is confusing because all work done at constant volume is inherently useful because constant volume means no PV work.

I see your issue. But you can't just change the definition of "work" which is not inherently useful in general. Even though it is in this case, we might add the qualifier "useful" to disambiguate and be as clear as possible. But I agree, it's redundant given the context.


text:
Why is phosphorous acid more acidic than phosphoric acid?
Phosphoric acid has basicity of 3 i.e it can loose 3 $\ce{H+}$ while phosphorous acid has basicity of 2. Why is phosphorous acid more acidic than phosphoric acid? Acidity refers to the ability to liberate protons. Phosphoric acid liberates more protons than phosphorous acid (as the basicity of phosphoric acid is 3 and that of phosphorous acid is 2). Why then is phosphorous acid more acidic than phosphoric acid?

Consider that proticity is not the only factor governing acidity. For example, hydrochloric acid is considered to be strongly acidic, but it only has the capacity to donate one proton. At the same time, carbonic acid is considered to be weakly acidic, though it has two protons available for donation.

@T.Kent does carbonic acid really exists ? ^^

I'm pretty sure it is because phosphorous acid is more polar than phosphoric acid. This is because in the phosphoric acid there are more OHs and an O, so it balances out more towards the center.
In phosphorous acid, there is a H directly bonded to the central phosphorus. Hydrogen doesn't pull as much on electrons as oxygens do; hence, that part of the molecule becomes more positive, causing a stronger dipole moment than that in phosphoric acid.

You’re using a wrong definition of acidity. Acidity and its adjective acidic do not mean ‘number of protons liberated eventually.’ Instead, acidity is well defined as the equilibrium constant $K_\mathrm{a}$ of the corresponding first deprotonation reaction $(1)$:
$$\begin{align}\ce{H_nA &<=> H_{($n-1$)}A- + H+}\tag{1}\\[0.6em]
K_\mathrm{a} &= \frac{[\ce{H_{($n-1$)}A-}][\ce{H+}]}{[\ce{H_nA}]}\tag{2}\end{align}$$
For phosphoric acid, the corresponding deprotonation is reaction $(3)$ and the equilibrium constant is as written; for phosphorous acid, see $(4)$.
$$\begin{align}\ce{H3PO4 &<=> H2PO4- + H+} &&K_\mathrm{a} = 7.11 \times 10^{-3}\tag{3}\\[0.8em]
\ce{H3PO3 &<=> H2PO3- + H+} && K_\mathrm{a} = 5.01 \times 10^{-2}\tag{4}\end{align}$$
Phosphorous acid has the higher acidity constant corresponding to a lower $\mathrm{p}K_\mathrm{a}$ value and is thus more acidic than phosphoric acid.


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

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

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

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

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

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

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

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

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

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


text:
Why magnesium has the lowest melting and boiling point in its group?

It is written in the third point ( shown in the picture) that presence of d orbitals results in stronger metallic bond.
1) We check the strength of metallic bonding to compare the melting and boiling point in case of metals. According to the general trend, melting and boiling point decreases down in the group( in s block) because the strength of metallic bonding decreases.

2) Presence of d orbitals decrease the size of atom thus increasing the electrostatic attraction between the delocalized electrons in the metal lattice and the cation and as a result, increasing metallic bonding. But magnesium IS IN A DIFFERENT PERIOD.
Can the ineffective shielding by d electrons be so much effective in decreasing the size of cation?
3) We didn't use any of this d orbital argument in comparing melting and boiling point of alkali metals. It decreased down the group following the expected trend.
How should  we explain this fact ?

Explanation based on d-orbitals is rather questionable. Do a search on SE for "magnesium melting point", this question has been asked numerous times.

@Nilay You have >3,000 rep, so you can already vote to close as a dupe by yourself. (Click on "close" next to "edit".)

I read some of the answers posted earlier on this topic, including the one that Nilay Ghosh mentioned but I don't really find them convincing. What should I do if I don't find the earlier answers convincing?

http://chemistry.stackexchange.com/questions/27143/why-does-magnesium-have-the-lowest-melting-point-of-all-earth-alkalis

@Blue Put a bounty on the existing question.


text:
Preventing build-up of Diethyl Ether Peroxides
I've never stored Diethyl Ether before but am thinking of producing minor amounts myself and storing said Ether inside of glass/PTFE bottles.
I am aware of possible pressure build-up due to vapors and storage temperature, but what always puts me off is the formation of explosive peroxides.
Now as the above Wikipedia article mentions:

Diethyl ether hydroperoxide decomposes in the presence of sodium hydroxide and $\ce{Fe^2+}$ -containing salts.

This is obviously stating that I can stop the peroxide formation simply by adding in some $\ce{NaOH}$ which I have an abundance on hand.  However that above information is regarding hydroperoxides. The less nasty of the two peroxides.
Ethylidene peroxides are the explosive ones that detonates by simply touching it. This is what scares me. There's no information regarding the stopping of formation on the page, and frankly my google searches for "Safe storage of Diethyl Ether" and "Stop Diethyl Ether Peroxide formation" seem to come up with "put some iron wire in it" or Potassium Hydroxide.
So my question is: what is the most reliable, safe, and long-term solution to prevent building up of both of Diethyl Ether's peroxides, especially the Ethylidene peroxide, with the least amount of contamination possible.

Where does the wikipedia article say etyhlidene peroxide is not reduced with Fe2+?

@Karl Well for me, the real question was, where does the article say it IS reduced? Nowhere.

You state that iron salts like Fe2O3 should help. This is incorrect, since this contains basically Fe3+ only, which cannot reduce the peroxides (that's the manner in which the peroxides are rendered safe). Moreover, Fe2O3 is basically insoluble. In my experience, the canonical reagent for the reduction is FeSO4.

@TAR86 Thankyou for the correction, far out I'm making some errors today! I did see somewhere that Ferrous Sulphate was the way to go but then again, I made this question for a reason. I'll edit it out.

There's no information regarding the stopping of formation [of ethylidene peroxide] on the page...

The required information is given implicitly. The wikipedia page states

Diethyl ether peroxide, also known as ethylidene peroxide, [...] is a polymerization product of diethyl ether hydroperoxide

With other words:
If you prevent the formation of diethylether hydroperoxides by

storing diethylether in a brown bottle over sodium hydroxide

you will automatically prevent the formation of subsequent products, such as the ethylidene peroxide.
Note that there is a different approach, which is recommended in terms of workplace safety:
Substitute diethylether with methyl tert-butyl ether (MTBE), which does not form peroxides.

Ah, I didn't see the polymerization piece.... My bad. :P

Why the hydroxide?

@Karl If I remember correctly, the hydroxide deprotonates the hydroperoxide and renders it insoluble.

What if we just used Ethanol?

@frayment What do you mean? Substitute diethylether with ethanol?

@KlausWarzecha Sorry for the vagueness. Instead of adding in Sodium Hydroxide, will Ethanol act as a deprotonating substance or atleast have the same effect? In other words, is it the same reason I see people adding Ethanol that your telling me to add Sodium Hydroxide?

@frayment No, that wouldn't work. The hydroxide will convert the hydroperoxide to an insoluble precipitate. Moreover, you might run into layer separation problems with water if your diethyl ether does contain (too much) ethanol. In summary, I'd suggest to use MTBE nowadays, if possible.

Sure, the hydroperoxide, if already present, precipitates. Ok, that means  you don't have it in your work if you take ether from that storage bottle, which is of course good. Just have to be careful with the dry NaOH/NaOOEther mixture when the bottle gets empty.

@Karl True. That's why the use use of test strips, which you mentioned in your answer, is crucial.

Do they show anything when the hydroperoxide has precipitated? I guess it's more of an alternative: test strips or NaOH.

@Karl Honestly, I don't know. But better safe than sorry.

The ethylidene peroxide is the polymerisation (polycondensation actually) product of the hydroperoxide. Firstly that only occurs when the solution gets concentrated, i.e. when the ether and residual water evaporates. Secondly, it is also destroyed by Fe(II) just as the hydroperoxide, and thirdly, you must store ether away from light anyway, and then you should never get peroxides. Have your peroxide test strips at hand and check from time to time, and IF they give a positive result, add the reducing agent.
https://en.wikipedia.org/wiki/Diethyl_ether_peroxide

We are medical lab doing occasional water:ether extractions. We have always had the practice of adding deionized water to diethyl ether, and drawing ether from the top of this stock. This apparently removed some minor impurities in the diethyl ether stock able to dissolve in water. I was taught this is a safer way to work with ether, but have never seen this mentioned in documentation from any supplier, such Sigma or any publication or educational text. We only have small amounts and try to not keep it around very long, so I have never been able to determine what the water addition does regarding peroxide formation over longer storage (years).


text:
Type of analysis for water
What other analysis suitable for water as general? Right now, I have turbidity, pH, Total Soluble Solid and color.

Hardness
TC Total Carbon

TIC Total Inorganic Carbon
TOC Total Organic Carbon
VOC Volatile Organic Carbon

COD Chemical Oxygen Demand
BOD Biochemical Oxygen Demand

@anonymous My pleasure. If the anwer was really helpful, please consider to accept it.

Also trace metal concentrations, disinfectant byproducts, fluoride, nitrite, nitrate.  I'm kind of presuming drinking water with these though.

@airhuff Absolutely! I only listed  some parameters for which wikipedia pages with further explanation exist.


text:
Storage of the Halogens in PTFE Bottles
I'm looking to buy some Iodine for various reactions, and noticed that the person who sells it has it in a glass jar with a plastic lid of unknown type.
I'm looking up chemical resistance charts and it says that Iodine, Chlorine and Bromine both work absolutely fine when placed inside of a PTFE bottle, assuming that the Chlorine is not in a gaseous form.
As for Fluorine, one website states that it attacks the plastic severely, whereas the other states that the PTFE is resistant until around the 100°C mark.
These sites are both here and here.
Now when I go look at people asking the question, "How should I store elemental Iodine?", everyone seems to respond along the lines of keeping it in a glass jar with a ground glass joint or PTFE/Teflon lined cap.
However since my research suggests that PTFE alone is capable of storing Iodine, is this a viable option? I would love to know from past experiences, because no-one seems to recommend PTFE on its own on any site I checked. Of course however I will also be storing said PTFE container inside of another non-reactive material container.
I understand that I'm probably better off just storing it in glass containers but I have PTFE bottles lying around and would like to use them if at all possible.
Since this question revolves around the Halogens themselves, could the same be potentially done for Bromine and Chlorine?

http://chemistry.stackexchange.com/questions/34190/is-there-any-chemical-that-can-destroy-ptfe-or-teflon?s=5|0.9921 http://chemistry.stackexchange.com/questions/6183/does-a-sulfuric-acid-nitric-acid-mixture-dissolve-ptfe-ie-teflon?s=9|0.5349 Teflon is very resistant

Though PTFE is not attacked by the halogens, they may discolor PTFE (particularly iodine, and at elevated temperature), since the plastic is somewhat permeable to halogens (end even to water).

Hmm... I might be better off just buying a glass jar with a Teflon-only lid instead. Thanks for the reference, some good source material for later storage!


text:
Which characteristics of the compound affect the number of theoretical plates in a GC experiment?
Some days ago, I tried for the first time a chromatography machine.
The data I collected are the following:
$$
\begin{array}{lrrr}
\hline
\text{Compound}                 & t_\mathrm{R}/\pu{min} & w/\pu{min} & N     \\
\hline
\text{Benzene}                  & 2.325                 & 0.047      & 39153 \\
\text{Toluene}                  & 2.712                 & 0.067      & 26214 \\
\text{Chlorobenzene}            & 3.690                  & 0.099      & 22228 \\
\textit{p}\text{-Chlorotoluene} & 4.332                 & 0.106      & 26723 \\
\hline
\end{array}
$$
I know the relationship between retention time $t_\mathrm{R},$ peak width $w$ and theoretical plates $N.$
Known the different properties, boiling points, polarities, or shapes of the molecules involved, is there a way to predict which molecule will yield an higher number of theoretical plates?
Which characteristics of the molecules affect the different number of plates for each molecule?

The number of theoretical plates can be calculated as:

$\mathrm{N_t = \frac {H}{HETP}}$
$\mathrm{N_t =}$ number of theoretical plates
  $\mathrm{H =}$ total height (or usually length) of the column [$\mathrm{m}$]
  $\mathrm{HETP =}$ Height Equivalent to a Theoretical Plate [$\mathrm{m}$]

Where $\mathrm{HETP =}$ can be calculated from the Van Deemter equation:

$\mathrm{HETP = A + \frac {B}{u} + (C_s + C_m) * u}$
$\mathrm{A =}$ Eddy-diffusion parameter, related to channeling through a non-ideal packing [$\mathrm{m}$]
  $\mathrm{B =}$ diffusion coefficient of the eluting particles in the longitudinal direction, resulting in dispersion [$\mathrm{m^2 s^{−1}}$]
  $\mathrm{C =}$ Resistance to mass transfer coefficient of the analyte between mobile and stationary phase [$\mathrm{s}$]
  $\mathrm{u =}$ Linear Velocity [$\mathrm{m s^{−1}}$]

So, the characteristics specific to the analyte molecule that determine the number of theoretical plates are B and, unfortunately from the standpoint of developing an ab initio model, C.
Various models have been developed, however.  Below is a list of references to some articles of interest in this field.
[ 1 (abstract)]"The problem of plate modeling: Theoretical and computational results", Computer Methods in Applied Mechanics and Engineering, Volume 100, Issue 2, October 1992, Pages 249–273
[ 2 ]"Plate Models in Chromatography: Analysis and Implication for Scale Up", Chromatograpy,Springer-Verlag
[ 3 (abstract)]"Neural Network Modeling of Structured Packing Height Equivalent to a Theoretical Plate", Ind. Eng. Chem. Res., 2000, 39 (5), pp 1520–1525

I think it may be worth noting explicitly in this answer that the column chemistry is a primary determinant of it HETP and of N.  GC usually uses nonpolar stationary phases such that vapor pressure is a primary determinant of these parameters.  Would the same be true if an e.g. an aqueous stationary phase was used to analyze ammonia gas?

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


text:
Why do chemistry and physics have different sign convention in thermodynamics?
In physics class, we write the first law of thermodynamics as $\mathrm dU =\mathrm dQ - \mathrm dW$ and in the physical chemistry class, we write the same law as $\mathrm dU =\mathrm dQ + \mathrm dW$.
The reason being the sign convention is different in both the cases.
In physics we take work done by the system as positive and in chemistry work done on the system.
I realize that this does not cause any change in the actual law of nature but I just want to know why we have different sign conventions. Wouldn't just one convention make life easier?
Is there a historical reason? Or is this is to differentiate between subjects?

The standardized definition according to ISO 80000-5 Thermodynamics of internal energy $U$ for a closed thermodynamic system is $\Delta U=Q+W$, where $Q$ is amount of heat transferred to the system and $W$ is work done on the system, provided that no chemical reactions occur.

Personally I would describe the version with the "-" as the engineering convention. When you are building engines (as were the people who invented thermodynamics) the process you care about putting heat in and getting work out, and you define both those quantities as positive. Pure physics thermodynamics often (mostly?) uses the "+" version.

Even in engineering, there are two conventions... Mechanical engineers think work leaving the system is positive work, chemical engineers think work done on the system is positive... Other engineers just fake it.

It has got to do with IUPAC's ruling then.

This is not a simple physics versus chemistry distinction. I taught Physics for 25 years and saw many examples of either usage in multiple textbooks. In fact, at some point in my tenure, the AP Physics committee swapped conventions on the equation sheet for the AP Exam.
Just my take here: I've always attributed the work-done-by-the-system camp as being more prone to be used by engineering types who want to know "what the system can do for us" in practical applications. On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from without.

"Ask not what your system can do for you..." :-P Also, for what it's worth, only the second version makes sense to me. In the first version you're considering positive heat to be something that is supplied by the surroundings, and positive work to be something that is supplied by the system. That's obviously inconsistent, and by fixing the inconsistency you get the second equation. Just my (correct) opinion... ;)

@Mehrdad Agreed. That's the same argument I gave my students.

As I said above, only mechanical engineers are greedy and care about what the system can do for them :D

@B.Kelly Is that why there is no difference in sign convention of heat?? Since heat released by system is a loss of energy that could have done work for engineers and heat loss is decrease in system energy for theoretician it is negative.  While heat gain is positive . For engineers it is input energy and for experimenter ìt is increase in system energy

There is no difference and the equation will yield the same answer. The difference is by deciding the point of reference for the work done.
Where we have $W = \pm P\Delta{V}$, you just have to decide where do you look from. Am I the system and work is done on me, or am I outside and doing work on the system? anyhow, once you choose your point of reference you should stick to it and do all your calculations according to it. I guess, as in most of these cases it's historical.
Another example is different notation, chemists use $A$ to symbolize the free Helmholtz energy while in physics it's $T$.

Seems like using T for Helmholtz energy would be a nightmare, T is already an overloaded symbol even just sticking to chemists notation (temperature, kinetic energy, certain tensor operators).

I've never seen "T" being used for Helmholtz energy. I think you mean "F".

Physicists are interested in getting work done by the system. In physics we always look for what system is giving to us (surrounding). So, in physics we study work done by the system.
Whereas chemists are interested to know what work have the surroundings done on the system, in order to get the reaction completed. So, in chemistry we study work done on the system.

Chemistry made the switch over to the IUPAC convention back in the early 90's which dictates the first law as the sum of $Q$ and $W$ and in turn requires $W$ to be defined as the negative integral of $P(V)\mathrm{d}V$. See the recommendations of the International Union of Pure and Applied Chemistry.
As pointed out, physicists use $Q-W$ but define $W$ as the positive integral. Same final results though.

This is because....
In physics , we are concerned about the surroundings
And in chemistry, we concerned about the system.
That's why ,
In physics , work done by the system is positive because change in energy (due to the work done) of the surroundings is positive ! (Energy is released from the system to the surroundings)
Whereas In chemistry , work done by the system is negative because the system loses some amount of energy to do work !
That's it!

Take it simple; it's nothing but a reference in different field for different uses that changes the formulae. If system is doing work on surrounding, U will decrease if W is taken positive – and vice versa.


text:
About the equilibrium of a flowing reaction
Imagine to have a tube (to make a reactor simple) inside which a chemical reactions take place and reactants are constantly fed so overall can be considered in steady-state conditions.
Just to pick one, I assume the reaction to be exothermic.
Now in my mind there are the following two options, one of them is not clear.

Assuming that the tube is cooled and kept to a constant temperature and that the reaction is not controlled by kinetics, I think that at the end of the tube (if long enough) it can be considered in thermodynamic equilibrium.
If the tube is adiabatic then a distribution of temperature inside should arise. For each temperature there exists an equilibrium constant, and thus a different composition. In this case, would there be thermodynamic equilibrium at each single step (infinitesimal) along the tube? Again, this is considering an equilibrium-controlled reaction.

What is the actual question here? Is it are these descriptions correct?

Sure.  If the tube is long enough, the system will reach a final temperature and composition.  You can even precisely calculate it by considering the same reaction mixture in an adiabatic batch reactor, and waiting until the system has equilibrated.

@bon The actual question is related to the second point, if it is correct or not!

@ChesterMiller yes, but how about the second point, each step of the reaction can be considered as in equilibrium? If so it is correct, there exist an optimal temperature to maximize the calories conversion

What you describe is a 'flow reactor' commonly  used to measure rate constants in gas phase reactions by placing the detector at different places along the tube. Question (1) yes; eventually equilibrium is reached at some point at a point distant from the source inlet. (2) If the temperature is not controlled the rate constant will change as the gases travel along the tube and equilibrium reached sooner if the temperature is raised. No reliable rate constant can now be obtained.

@porphyrin Thanks, I have a question more on this type of reactor. Back to your answer, what do you mean by no reliable rate constant can be obtained? The other doubt is related to Le Chatelier's Principle application, in a closed volume at equilibrium, removing the products increases the reactants conversion (to counteract the change, reaction shifted to the right); in a flow system how this principle apply, since reactants and products continuously move? Is the reaction always shifted to products?

You can't get more than 100% conversion, and that would correspond to some specific final temperature.  You can then evaluate the equilibrium constant at that hypothetical limiting temperature, and use it to determine what the conversion actually would be if that temperature prevailed.  It will certainly be less than 100%.  So the actual final temperature will be less than for 100% conversion.  With regard to @porphyrin 's comment on a flow reactor, this type of reactor is widely used in industry to carry out chemical reactions in continuous processes.

What I meant wrt rate constants was that you start at zero conversion at the input point and distance of detector along tube corresponds to time at constant flow. Thus, if the temp changes along the tube, the rate constant changes (Arrhenius ) also so a different amount of  reaction has occurred at a given distance than expected and so rate constant computed will be in error.


text:
Why does a hard-boiled egg have more entropy than a normal egg?
The question "Why does a hard-boiled egg have more entropy than a normal egg?" was recently asked on one of my tests.
I said that entropy will decrease as a hard boiled egg has molecules in a more organised way than an raw egg, but I was marked wrong with a remark that entropy will increase because the proteins and other molecules are decomposed into simpler substances due to heat.
I did ask why I was not correct but what I got was the same explanation along with the decomposition reactions.
I don't feel this explanation is satisfying.
For example, do not 10 objects on a table organised properly have more entropy than 5 objects thrown anywhere on the table?
The definition of entropy is the measure of disorder in the system. So clearly in my above example the 10 objects are more organised and thus less disordered than the 5 objects in the second case and same can be said for the main question.
Can somebody please explain to me why a hard-boiled egg has more entropy than an un-boiled egg (I like if the explanation has some math)?

See similar question answered in Physics.SE here.

Here is an explanation I found on the internet that works for me:
The protein in egg whites is albumen, which is tightly folded in an uncooked egg.  The balls will slide past each other, so it is a liquid.  The entropy is low because of the precise folding.
After cooking, the protein molecules are unfolded and tangled, forming a solid. The entropy of the egg whites is actually higher.

But why it is higher ?

Are you familiar with the term configurational entropy?

Yes after reading https://en.wikipedia.org/wiki/Configuration_entropy .

But then does not that mean more things = more entropy regardless of how things are organised ?

It means that the polymer chains have more quantum states readily  available to them in the final state than in the initial, well-organized, folded configuration.

The protein in egg white is a polymer (poly peptide a polymer) and this pokymer has a unique conformational structure. This polymeric structure of protein get decomposed into  many smaller peptides due to hydrolysis of amide linkage of polypeptude chain in protein. So overall we can say one bigger molecule gives many smaller molecules and this will increase the entropy.

I’ve come across several claims that a (hard) boiled egg has greater entropy than an unboiled egg. None of them cite any peer-reviewed research to back up their claim. While it is possible that entropy increases when you boil an egg, it is also possible that it does not. It is even possible that it stays the same.
I did my own literature search to see if I could find any publication that answered the question. What I found left the question very much up in the air. So I contacted Greg Weiss, a scientist who has done research on a related topic: ‘unboiling’ an egg. Dr. Weiss was kind enough to discuss the topic with me via email.

tl:dr
Whether a hard-boiled egg has greater entropy all boils down to the question of how structured are the assemblies found in aggregated ovalbumin. Furthermore, we can't just consider the assemblies found in the ovalbumin aggregate since we're talking about assemblies composed of all the types of protein in the overall albumen. (BTW, it's important to note that heated egg white doesn't just form any type of aggregate, it forms a gel, which presumably implies additional structure.)
For those who are interested in the details, here is the full email thread.


text:
Hydrophobicity of Polysaccharides
Are polysaccharides generally considered to be hydrophobic and/or hydrophilic? My reasoning is that although they have polar groups/electronegative atoms, the polymeric nature of polysaccharides excludes water so I'm leaning towards them being hydrophobic.

They are typically full of hydroxy functions and thus hydrophilic. However, that does not mean ‘soluble’ automatically.

What do you mean by 'the polymeric nature of polysaccharides excludes water'?

Also the OP has shown some thought here and it's a reasonably written question so I don't think this is homework.

Are polysaccharides generally considered to be hydrophobic and/or hydrophilic?

They are generally considered to be hydrophilic unless they are highly modified (e.g. to siloxanes).

My reasoning is that although they have polar groups/electronegative atoms, the polymeric nature of polysaccharides excludes water so I'm leaning towards them being hydrophobic.

There is no rule that polymers exclude water. Linear polymers can either stretch out in water (when the interactions with water are favorable) or clump up (if the interactions with itself or another polymer is favorable.
One notable exception is cellulose. The straight chains of cellulose make perfect interactions (involving both hydrogen bonding and dispersion forces), yielding a non-water soluble fiber. Cotton is made of cellulose fibers. The cotton fibers are highly hydrophilic (they are not water repellent) because the surface of the fiber is studded with hydroxyl groups.
Perhaps surprisingly, methyl cellulose (where all hydroxyl group are modified into methyl ether groups) is moderately soluble in cold water. The methyl groups prevent the packing of straight chains to cellulose fibers, and the polymer contains sufficiently hydrophilic groups to be water soluble (similar to polyethylene glycol, which also is an ether-containing polymer).

Maybe worth mentioning that water soluble starch can be dried to the point that the strands hydrogen bond so tightly to each other that water is excluded and it is extremely difficult to rehydrate and dissolve.


text:
Why is sodium cyanide an inorganic compound?
I have been told that sodium cyanide is an inorganic compound during a lab experiment but I don't know the reason for this.
Shouldn't sodium cyanide be an organic compound because it has a carbon atom bonded to nitrogen?
The Wikipedia article  on cyanide also says that it is an inorganic compound but I fail to find why it is so.
How do we differentiate between an organic compound and an inorganic compound?

We don't. It is a historical concept.

Related: http://chemistry.stackexchange.com/questions/22195

Related: http://chemistry.stackexchange.com/questions/6690/what-is-the-definition-of-organic-compounds

So many questions about inorganic/organic division. Nature is impartible, and one shouldn't expect strict borderlines between artificially divided sub fields of chemistry.

Sodium cyanide is classified as an inorganic compound for historic reasons, much like the entire organic-inorganic classification system is historical in reasoning.
The basic initial idea was that forming organic compounds required a vis vitalis, a life force, and could therefore only occur in living systems. On the other side, inorganic compounds were easily synthesised in vitro and have been synthesised even before alchemy was renamed to chemistry. These included most ionic compounds then known.
Naturally, sodium cyanaide is ionic. Most notably, early chemists would have been able to precipitate cyanide from other cyanide sources, redissolve it and add a sodium source (most likely sodium hydroxide) to regain sodium cyanide; thereby classifying it as inorganic.
In 1828, Friedrich Wöhler disproved the vis vitalis assumption and laid the foundation for synthetic organic chemistry. Yet the distinction between inorganic and organic chemistry was still upheld. Rather than defining it by synthesisability, it was said that any molecule containing carbon be organic — or one of many other similar definitions. This could exclude the cyanide salts but does not have to. The only salts and carbon-containing molecules that are always and consistently classified as inorganic are carbon monoxide, carbon dioxide and (hydrogen)carbonates.
By now, the distinction between organic and inorganic chemistry has lost most of its relevance and is only there for historic reasons. In principle, the distinction is that organic chemists are more interested in the carbon-containing compounds while inorganic chemists lean towards other elements more. But then there are numerous in-between fields that don’t really fit anywhere such as organometallic chemistry, bioinorganic chemistry etc.
Therefore, treat the organic/inorganic distinction as a historical residue and don’t read anything into it.


text:
Meaning of m2 in the symbol for an isotope of an element
What does the notation $^{197\mathrm{m}2}\ce{Pb}$ mean? Specifically, the '$\mathrm{m}$2' part.
I've found this and it appears to have something to do with charge distribution. The original notation was found here.

Related: How do I correctly typeset metastable radionuclide symbols?

As per this Wikipedia page on isotopes:

The letter $m$ is sometimes appended after the mass number to indicate a nuclear isomer, a metastable or energetically-excited nuclear state (as opposed to the lowest-energy ground state), for example $\ce{^{180\mathrm{m}}_73Ta}$ (tantalum-180m).

Note that the lower subscript to the left of the chemical symbol is the atomic number of the element (in the example above, 73 is the atomic number for tantalum) and is omitted in your example.
However, your question specifically is about the "m2" part. I'm quoting the Wikipedia page on nuclear isomers here:

For isotopes with more than one metastable isomer, "indices" are placed after the designation, and the labeling becomes m1, m2, m3, and so on.

In your case, 197 is the mass number for this lead isotope, and m2 is the second excited metastable state, with a half-life of 1.15 $\mu$s.


text:
Can ML algorithms be applied to predict the products of a chemical reaction given a set of reactants?
How would the training data look like ? Would it be a set of reactions ? What would be a good algorithm to run on the data ?
Wolfram Alpha does have an interface for chemistry, but does not seem really robust. The other databases for chemical reactions seem to work on a static database.
The ultimate goal is to predict the products given a set of reactants. Any suggestions will be helpful !

Yeah. I initially went to chemistry stack exchange but I thought they might not be familiar with ML. If you suggest, I will post it on chemistry stack exchange.

Your question has nothing to do with ML. You need to ask for whether it makes sense to use the predictors. ML is not magic, if your inputs are non-sense, your model is also bad.

Adding on top of Emre's answer, product formation depends on certain reactive groups say for e.g., $-OH$ (Alcohol) having more chance of releasing a Hydrogen than $-CHO$ (Aldehyde) or a Methyl $-CH_{3}$. Instead of using reactants as categorical variables you should use reactive groups on that, as variables.
For example if you take a vector of say (Chlorine, Alcoholic_group, Amine) to be coded for Para-Amino Phenol, it will be $(0,1,1)$ as chlorine is absent and for $CCl_{4}$, it will be $(1,0,0)$. But remember the first vector will be same for O-Amino Phenol as they are positional isomers. But both have different properties and reactivities. So categorical is kind of wasted here.
In my more deeper and frank opinion, these values should be continuous. You should be kind of looking for a value to quantify these each dimension of the vector coded. Then you can use classifiers from this vectors to find the probability of reaction happening or for product formation as @Emre mentioned.
So, take a simple, small set of molecules and make a vector having fewer dimensions and test it with categorical and continuous values(Give a value to each dimension based on your chemistry gut opinion). That is the better way to find whether your hypothesis/assumptions are reasonable.
EDIT 1:
My answer basically answers whether reaction is feasible or not but does not answer which products are returned. I am sorry for that. But I hope it kind of points to what you can achieve with ML if you want to take your first step.
Hope this helps.

+1 That's precisely the kind of domain expertise I meant. I think the predisposition you refer to can be encoded with a regularizer. Can you quantify how likely one product is to be released relative to another in such cases?

@Emre Yes that can be achieved, but requires quantifying the groups based on reactivity + availability etc. for the molecules involved which I think is a tough job. I don't even know whether there is any quantifiable term for reactive groups :( . That is why I asked him him/her to use chemistry gut knowledge to mark some molecules and test it on a set to see the class probability.

Have a look at this paper
http://pubs.acs.org/journal/doi/full/10.1021/acscentsci.6b00219
(Added to beat character limit)

Let the question run its course here before reposting there. Without knowing much chemistry, I would approach it as a quantized regression problem (i.e., pick a loss function that does not penalize rounding errors). The input is a list of (chemical, coefficient) tuples, and the output is likewise. The order does not matter. The chemicals can be represented as categoricals, and you should also created derived categorical variables for the elements in each chemical. In other words, the input is simply the list of coefficients for the reactants, plus indicators for the derived variables. The goal then is to estimate the coefficients of the resulting products; zero would mean the product is not present.


text:
Why does the light catalyzed radical addition of tetrachloromethane to an alkene produce only a single regioisomer?
This reaction always occurs in the following, regiospecific way:
$$\ce{CCl4 + R-CH=CH2 ->[$h\nu$]  R-CHCl-CH2-CCl3}$$
The same should apparently also happen when there is a carbon that has two substituents (e.g. methyl groups) and yield only one major product:
$$\ce{CCl4 + (CH3)2C=CH2 ->[$h\nu$]  (CH3)2CCl-CH2-CCl3}$$
I believe the initiation step goes like this:
$$\ce{CCl4 ->[$h\nu$] CCl3^. + Cl^.}$$
Can't both of these radicals, that form in the initiation step, react with the double bond and form a secondary or tertiary radical carbon? Why does only $\ce{CCl3^.}$ apparently attack the double bond?

Is there any other information missing from this problem or in the wording of it?  I would have expected both of the radicals to react significantly, although I would expect the CCl3 radical to be a less stable intermediate than the Cl radical.

Is it not just a steric effect making the $\ce{CCl3\cdot}$ react more slowly and so the $\ce{Cl\cdot}$ wins out at the more crowded site?

While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:
$$\begin{align}
\ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\
 \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\
\end{align}
$$
Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$ are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:
$$ \small
\begin{array}{lcc}
\hline
\text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\
\hline
\ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\
\ce{.Cl} & \text{3.4} & \text{1.1} \\
\ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\
\ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\
\ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\
\ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\
\hline
_{^* \text{units are } \mathrm{kcal\ mol^{-1}}}
\end{array}
$$
As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:
$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$
$$\ce{.CCl3 + CH2=CH-R ->  Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$
$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 ->  Cl3C-CH2-CHCl-R + .CCl3}$$
The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.
$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$

$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947).
$^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979).
$^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034.
$^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).


text:
Separating boric acid from sodium chloride
I want to create some boric acid with borax in the following reaction:
$$\ce{Na2B4O7·10H2O + 2 HCl → 4 BH3O3 + 2 NaCl + 5 H2O}$$
However as you can see, the leftover solution will contain a mixture of boric acid, sodium chloride and water. Both chemicals are soluble in cold water so neither will want to immediately crash out of solution.
My question is, how can I remove the excess water and sodium chloride to leave myself with some powdered/dry boric acid crystals?
I know I can probably just use some liquid ammonia, but then the boric acid is only slightly soluble in it leaving much of it behind. I could perhaps use glycerin also, however I am limited to how much I currently have.

I guess ordinary ethanol dissolves boric acid pretty well.

Here is a potential solution that shouldn't even require heating. Note that I have not tried this myself.
Boric acid is soluble in lower alcohols like ethyl alcohol and isopropyl alcohol.
The solubility of sodium chloride in ethyl alcohol at room temperature is about 0.05%$^1$.
Given these properties, I suggest dissolving the dried boric acid / sodium chloride in a minimal amount of cold 95% denatured ethyl alcohol or 99% isopropyl alcohol (available at most pharmacies).
The boric acid should dissolve readily and the sodium chloride should not, so you can just filter it and let it dry to a fairly sodium chloride free boric acid solid.
To make the sodium chloride even less soluble you could cool the solution in a freezer prior to filtering.  If you notice any significant increase in solids that means boric acid is coming out of solution and you'll want to abort this step.
1) Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005

Is there any steps I can add to this solution by heating/distilling or rather anything to get a pure solution of boric acid? I'm going for a sodium chloride-free environment here. :)

If you don't mind sacrificing your recovery of boric acid a bit, and any purification procedure will likely result in decreased recovery, here is what I would do: When all but a small amount of alcohol has evaporated, pour off the remaining alcohol then give the solid a quick rinse in a small amount of ice-water, pour off all the water that you can and dry.  As the boric acid crystalizes out of solution, the crystal formation should exclude most of the NaCl, so that most of the NaCl will be in the alcohol and at the surface of the solid. NaCl is very soluble in water, so most should rinse off.

Hmmm... I'm sorry but I'm no good at these crystallization attempts... I'm giving it a go, but is there any other way I can do this? :(

Could you please share the link for the specific research paper from the "Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005"? I tried but am not able to find it on Google Scholar or Google search.

You could consider using ion exchange resin, there are resins intended for use in the nuclear industry for the removal of boron (borate / boric acid) from water. A common method in light water reactors (PWRs) is to control the reactivity by adjusting the boron content of the coolant.
While you can imagine it is easy to add boric acid / lithium borate to the water, a method of removing it is also needed. One method which is common is to use an ion exchange resin. It is interesting that the acid strength of boric acid increases when it is combined with a polyalcohol which can chelate to the boron.
What I would imagine going for is a weak acid resin such as IRA 67 which has been modified with surgar groups which are able to increase the binding of boric acid / borate. There is a resin which is marketed for this purpose (IRA743) which could be used to remove boron from sodium chloride.
You would then need to elute the resin, I would suggest using a mineral acid solution to release the boric acid followed by treatment with sodium hydroxide to restore the resin back to being the free base form.


text:
Stability of transition and post-transition metal alkoxides
Many transition metal alkoxides are seen in chemistry, for instance those of $\ce{Ti, Sn}$ and $\ce{Al}$, and in particular one of those of aluminium has its own role in the MPV reduction reaction.
I was wondering: how stable these $\ce{M(OR)_n}$ structures are?
To simplify the question:

Are they susceptible to acid/basic attacks (which could possibly hydrolyze them, maybe)?
Being them metal-alkoxide compounds, do they present similar reactivity as alkali metal oxides (eg: use in Williamson's synthesis of ethers)?
Would it be possible to obtain an alkali metal alkoxide from a transition metal alkoxide, hence synthesizing those compounds without using alkali metals?
When used in a reaction as a catalyst (eg: MPV or Oppenauer reactions), a metal alkoxide can be recovered and reused?

Regarding point 3, you don't necessarily have to use the metal itself - e.g. adding NaH to an alcohol will get you the sodium alkoxide. But I get what you mean.

You definitely can't do MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM).
I will answer some of these questions in line after I read through some references.

Alkali metal, main group, and transition metal (TM) alkoxides all have significantly different chemistry. Early- and Late-TM alkoxides also have different behaviour, so we should split this family into at least four categories. All of them are basic and will decompose via hydrolysis if dissolved in water; however, this type of chemistry is not usually carried out in presence of water.
In TM chemistry, alkoxide or phenoxide are mostly used as ancillary ligands, meaning they do not usually participate in the chemistry. Although they are liable to be hydrolyzed under protic conditions, alkoxides are actually known for being especially stable TM ligands because they do not undergo β-hydride elimination (BHE) (1), which is one of the most common decomposition pathways for TM alkyls. For instance; you often cannot make a stable t-butyl TM complex because the β-hydride will be transferred to the transition metal, liberating 2-methylpropene unless the metal is totally coordinatively saturated. An important example of a BHE-stable TM t-butoxide is the following Mo(IV) Schrock catalyst, which is stable:
$\hskip2.4in$
It should be noted, though, that the Schrock catalyst doesn't actually have any β-hydrides. Later, I'll show an example of a TM-alkoxide that does.
An important feature of alkoxides as ligands for TM chemistry is that they are electronically malleable. Since an alkoxide has 5 electrons it can behave either as an σ-donating X ligand or as a σ- and π-donating LX ligand (and formally as L2X). Presumably, this electronic malleability has something to do with their stability to BHE. An alkoxide's π-donor character might stop the metal center from liberating a coordination site to accept the β-hydride. For more information on BHE-stability, see (2), as cited by Schrock in (1).
Since early-TMs are more electropositive, the bonding character of their alkoxides is more ionic, although not to the same degree as that of alkali metal alkoxides. Alkoxides are very good bridging ligands, and if they are not sufficiently bulky, alkoxide complexes often polymerize. With a sufficiently bulky monomeric early-TM alkoxide, the main decomposition pathway is formation of the oxo complex. For example; the decomposition of a metal diolate shown in the next image was observed by Gable et al. (3) Their Hartree-Frock calculations implicated an empty metal MO at rhenium in the M=O bond formation, as shown in the MO picture in the next image.
$\hskip0.2in$
Late-TM alkoxides are expected to have weaker M-O bonds; and not as many have been isolated, but a new study (4) has examined these M-X (where X = O, N) bonds.
A really nice example illustrating a few important TM-alkoxide concepts is tantalum(V) ethoxide, Ta2OEt10. See wikipedia: https://en.wikipedia.org/wiki/Tantalum(V)_ethoxide
$\hskip1.5in$
Ta2OEt10 is a stable TM-alkoxide with true β-hydrides, bridging oxygens, and it's readily hydrolyzed when exposed to protons. It can be synthesized via salt metathesis with sodium ethoxide, a reaction that has equilibirium lying in the opposite direction of the alkali metal alkoxide formation you proposed.
In MPV reduction (and OPP oxidation), Al(O-i-Pr)3 is usually used as the catalyst. Aluminium is a metal, but it is a main group metal, and its chemistry is somewhere between that of boron, a semimetal, and that of early transition metals, except, of course, that it does not have a partially filled d shell. Al(O-i-Pr)3 is an electron deficient Lewis acid, comparable in chemistry to boronate esters, but much more reactive. The proposed mechanism for MPV/OPP [R]/[O] goes something like this:
$\hskip0.2in$
It is a true catalytic reaction, so the Al(O-i-Pr)3 is indeed regenerated, but I don't know if it's a routine part of the procedure to recover it. Al(OR)3's are not, however, soluble in IPA, but alcohols and ketones usually are, so you could probably precipitate Al(O-i-Pr)3 in IPA providing you have a way to seperate your ketone and your secondary alcohol afterwards.
Interestingly, Ti(O-i-Pr)4, is analogous to Al(OR)3, but its reactivity totally different, since titanium is a true TM and not just a Lewis acid main group metal like aluminium. Ti(O-i-Pr)4 is the catalyst for sharpless epoxidation, so this might be a good reaction to look at as an example of early-TM alkoxide reactivity.
To answer your questions in list form:

Pretty much all alkoxides are susceptible to acidic attacks, but, being fairly strong bases themselves, they are probably stable to weaker bases
Although they will react with acids, they will not necessarily react with electrophiles unless they are stronger electrophiles than the metal center; however, early-TMs are fairly strong electrophiles. I don't think people use TM alkoxides for ether synthesis though, and you always have to keep in mind d orbitals and the unique rules of TM chemistry when you're discussing TM complexes.
As mentioned above, and in my comment; "You can't have MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM). As Schrock says in (1), "the relatively high electrophilicity of an early transition metal makes dissociation of alkoxide ion, at least in a neutral species, unlikely".
If you can find a suitable means of seperation like a solvent that will precipiate your metal alkoxide, you should be able to seperate it, although some catalysts will be deactivated after a certain number of cycles.

(1) R. R. Schrock, Polyhedron, Vol. 14., No. 22, pp. 3177-3195, 1995
(2) D. C. Bradley, R. C. Mehrotra and D. P. Gaur, Metal Alkoxides. Academic Press, New York, 1978
(3) K. P. Gable, J. J. J. Juliette. J. Am. Chem. soc, 1996, 118, 2625-2633
(4) J. R. Fulton, A. W. Holland, D. J. Fox, R. G. Bergman. Acc. Chem. Res., 2002, Jan; 35(1): 44-56


text:
Aldolisation reaction and chalcone formation
I realized a reaction of mixed aldolisation between acetone and 2-nitrobenzaldehyde, using a solution of 10% NaOH in ethanol.
I obtained a mixture of products at the end of the reaction. The problem is when I added the ethyl acetate I saw a blue liquid mixed with brown solid.  After a night of this reaction I found a lot of white crystals mixed with the blue liquid.
Can someone explain the presence of this blue liquid with the brown solid, knowing that the crystals are soluble in ethanol?

Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.
Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

I hope this helps.

In the aldol reaction, we would expect to come up with the following product:

However, this reaction is bound to happen poorly. This is due to the presence of the nitro group, which has so much electron-withdrawing power that you can consider it to be twice as powerful as a carbonyl group (although the nitro group is not as powerful here due to its attachment to the benzene ring). Hence the nitro group then subsequently attacks the ketone in many intramolecular reactions, an example is shown here:

Since there are so many aromatic compounds that are fully conjugated forming, these should all contribute to the blue colour that you are getting.
I believe the brown solid is the result of some polymerization of the ketone with either the aldol or the nitro group.

I also have to add that the product I drew happens to be an activated form of pyridine, which is susceptible to electrophilic substitution as well. Hence, electrophiles such as another nitro group can easily attack the pyridine to form polymers.