Quantifying CO2 generated/released by the soft drink industry and breweries I need some help in developing a method by which to calculate how many tons of "evil" $\ce{CO2}$ are routinely being released into the atmosphere SOLELY due to our beverage addictions, notably beer and soft drinks.  I realize that not all manufacturers and products are consistent in their carbonation levels, and that an average probably has to be applied industry-wide (unless individual companies are required to record and publish such data) per unit measure or monthly/annual volume output.

Is this the proper forum to ask such a question?  If not, suggestions on where else I go?

On a previous post I got part of the answer, relating to an individual 12 oz can, or 1 liter, but I'm wondering if there is an easier way in which I can go about plugging weekly, monthly or annual volumes of beverages into formulas which will yield the results I'm looking for, notably a discussable unit of measurement in tons or ?  I'm willing to do all the numbers work once I'm pointed in the right direction.

I'm not sure if anybody's ever thought about this before, because I've never heard it discussed.  But if, according to some environmentalists, we should all watch our carbon footprint (and I'm not debating that here), then it seems like the most logical place to start would be with our hundreds of millions of daily beverage choices?  Isn't it?

The xkcd What If? blog, by Randall Munroe, addressed a similar question.  He states the result better than I can.

  For most of the history of civilization, there were about 270 parts
  per million of carbon dioxide in the atmosphere. In the last hundred
  years, industrial activity has pushed that number up to 400 parts per
  million.

  One "part per million" of CO2 weighs about 7.8 billion tons.
  A can of soda contains in the neighborhood of 2.2 grams of CO2, so you
  would need about 450 quadrillion cans of soda. That's enough to cover
  the Earth's land with ten layers of cans.

  Randall Munroe - xkcd What If? #88:Soda Sequestration. http://what-if.xkcd.com/88/

For further perspective, each breath of a healthy adult male involves about 500mL of gas exchange.  The air is, on average, 0.04% CO2 prior to inhaling.  Exhaled air contains between 4 and 5 percent carbon dioxide, I'll call it 4.04% for a nice, round 100-fold addition.  The mass of 500mL of air is about 0.6g, and though it will be slightly more dense with the added carbon dioxide, I'll assume that it remains at that mass for simplicity's sake.

So 4% of the 0.6g is CO2 that was added to the atmosphere by a single adult male breath.  That's 0.024g.  About 90 breaths add as much CO2 to the atmosphere as the carbonation in a can of Coke.  Where does the additional carbon in those breaths come from?  The Coca Cola itself, of course.

The sugar in the can of Coke (39g) contains about 16g of carbon atoms.  After being metabolized, that turns into 57.2g of carbon dioxide.  Not considering any of the energy required to harvest, process, or transport the sugar, that's still 26 times a much carbon dioxide going into the atmosphere from the calories in a can of Coke as opposed to its carbonation.

To put it more succinctly, the CO2 output from the carbonation in Coke could be completely offset by reducing overall caloric intake by 6 calories per can.


what is a sugar unit? How a sugar unit is defined?
I am trying to understand the structure of calicheamicins, and it is noted that They consist of an enediyne headpiece and a modified carbohydrate tail consisting of four sugar units and an unusually substituted aromatic ring. I can understand in the picture where sugar unit is referred, but I would like a general definition.
Thank you

Calicheamicins are natural products, which were isolated from the fermentation broth of Micromonospora echinospora subsp. calichensis. These antibiotics exhibited significant activity against Gram-positive and Gram-negative bacteria in vitro. Calicheamicin $\gamma_1$ (Ref.1) demonstrated antitumor activity against P388 leukemia and B16 melanoma in vitro (Ref.2).

The structure of Calicheamicin $\gamma_1$ contains biologically active part of bicycloring system, which is described in the following diagram:

The rest of the structure contains four modified sugar units and one aromatic nucleus. The sugar 1 unit is a 6-deoxy-4-aminohexose unit (that's why it is called a modified sugar). Its anomeric oxygen (hemiacetyl $\ce{OH}$) bound to the $\ce{C}$9 position of the endiyne unit in $\beta$-linkage. The sugar 2 unit is a 2,6-dideoxy-4-thiohexose unit, hence it is also a modified sugar. This linkage is unusual among other sugar linkages in natural origin. The inkage is a hydroxyl amine bond ($\ce{-NH-O-}$) without eliminating $\beta$-$\ce{OH}$ of 2nd sugar unit. According to proposed biosynthetic pathway, these two sugars have been added to endiyne unit in first and second steps of the pathway:

The third step of the biosynthesis is addition of the aromatic nucleus (5-iodo-4-hydroxy-2,3-methoxy-6-methylbenzoate unit) to 4-thial group of sugar 2 as a thioester bond. The forth and fifth additions of other sugar units can be alternative. Nonetheless, one sugar (a 6-deoxy-3-O-methylhexose unit) attached to the free 4-phenolic $\ce{OH}$ group of aromatic nucleus in $\alpha$-linkage. The last sugar unit is attached with its $\alpha$-$\ce{OH}$ of the anomeric center to the free 2-$\ce{OH}$ group of the first sugar in $2 \rightarrow 1$ linkage. The unusual about this sugar is it is a pentose (a 2-deoxy-3-O-methyl-4-N-ethylaninopentose unit) instead of hexose among other three.

You can find a good description of mechanism of action of Calicheamicin $\gamma_1$ in Ref.3.

I hope this is what you want to know about this molecule.

References:

May D. Lee, Theresa S. Dunne, Marshall M. Siegel, Conway C. Chang, George O. Morton, Donald B. Borders, “Calichemicins, a Novel Family of Antitumor Antibiotics. 1. Chemistry and Partial Structure of Calichemicin  $\gamma_1$,” J. Am. Chem. Soc. 1987, 109(11), 3464-3466 (https://doi.org/10.1021/ja00245a050).
W. M. Maiese, M. P. Lechevalier, H. A. Lechevalier, J. Korshalla, N. Kuck, A.Fantini, M.J. Wildey, J. Thomas, M. Greenstein,  “Calicheamicins, a novel family of antitumor antibiotics: taxonomy, fermentation and biological properties,” The Journal of Antibiotics 1989, 42(4), 558-563 (DOI: 10.7164/antibiotics.42.558).
Miklós Simonyi, ”The Concept of Chiral Conformers and its Significance in Molecular Pharmacology,” In Advances in Drug Research: Volume 30; Bernard Testa, Urs A. Meyer, Eds.; Academic Press: San Diego, CA, 1997, pp. 73-111 (ISBN: 0-12-013330-X).


Synthetic route to a spirocyclic amide I have a task to come up with a synthesis of the following molecule (2-azaspiro[5.5]undecan-3-one), but no starting materials have been explicitly specified:

This is my idea. I think that all the steps shown are quite reasonable.Do you have any critics/suggestion?

I think this can be done in 2 steps though I can't find a literature example.

Alkylate the anion of the commercially available cyclohexane carbonitrile (cyclohexyl cyanide) with ethyl acrylate.
Hydrogenate the resulting cyano-ester over Pt to reduce the -CN to -CH2NH2 which with spontaneously cyclise to give the required product.


Acid/Base dissociation constants relationship From what I understand, the concentration of hydronium and hydroxide molecules is constant in pure water (and equals $10^{-7}\ \text{M}$, which is measured experimentally).

What I don't understand is why this remains true for an aqueous solution with an acid or base in it. Shouldn't it affect these concentrations?

When you write the equilibrium reactions for a weak/weak acid/base pairs (one with its conjugate), and get that $K_\mathrm a\cdot K_\mathrm b = [\ce{H+}][\ce{OH- }]$, what's the explanation for it? Why does their multiplication (either as pure water or in a acidic/basic solution) remain constant?

The concentrations do change in the presence of acid or base and are no longer $10^{-7}$M.

Instead, [H3O+][OH-] $= 10^{-14}M^2$

This comes from:

$\ce{2H2O <=> H3O+ + OH-}$

$K = \frac{a(\ce{H3O+})a(\ce{OH-})}{(a(\ce{H2O}))^2}$

Where "a(x)" is activity of species "x".

Then, approximating activity of $\ce{H2O}$ as 1, and the activity of the ions as the concentration of the ions there is a constant $K_w$, such that:

[H3O+][OH-] $= 10^{-14}M^2 = K_w$


Why does the E1cb mechanism proceed in the specific way it does? We have been learning the E1cB mechanism in organic chemistry, but we have not been delving into the motivation as to why this mechanism occurs instead of others. Consider the following mechanism:

I am confused as to why we push the electrons to the left, creating an anion on the oxygen, instead of simply performing an E2 reaction on the right hand side. What is the functional reason for drawing this intermediate? Why does it arise?

-OH is a poor leaving group and so E1 and E2 are not favourable. In the case of E1, the formation of a primary or secondary carbocation is not stable either.

For E1cb, the intermediate enolate is more stable and since the removal of the -OH is not in the rate determining step, it is more favourable than E1.


Chemicals that react with blood Is there a chemical or chemicals that can be infused in fabric that will react when exposed to blood that would produce a bright or glowing color other than red?

Luminol (IUPAC name: 5-Amino-2,3-dihydrophthalazine-1,4-dione) displays chemiluminescence when it is oxidised, and glows a blue colour. It is used in forensic science to detect iron (in haemoglobin) and also to detect copper in biological material.

The process leading to luminol emitting light is explained in this image.

As for whether luminol can be "infused" in fabric, this is difficult to answer, as you have not explained what you mean by "infused". As it is soluble in water, it will easily be washed out, and is thus not suitable for long-term use. Moreover, it will react with a variety of substances (apart from blood), for example bleach.


How will the pressure of a gas in manometer change if it goes to a place with higher pressure? Imagine a manometer where air pressure is $\pu{0.8 atm}$ and the gas pressure is more. We move it to sea level which the air pressure is $\pu{1 atm}$. How will the gas pressure change? (the temperature is constant)

Assuming an ideal gas with T constant, $$ PV = nRT $$ the entire right-hand term is constant. Therefore$$ P \propto \frac{1}{V} $$ which is Boyle's law. Since the atmospheric pressure goes up, the liquid portion of the manometer will go down on the side open to the atmosphere, and it will go up on the side open to the gas. Since the volume of the gas goes down, its pressure must increase by Boyle's law.


Volume of anion and comparison Recently, I came across a book in which I was solving the exercises and found out that the solution to a question says that the fluoride ion, $\ce{F^-}$, is smaller in size than the hydride ion, $\ce{H-}$.

As far I know, the hydrogen has only one shell in contrast to fluorine. Addition of a electron to hydrogen adds a electron to the 1s orbital and thus to the first shell. Similarly in fluorine, the electron is added to the 2p orbital and therefore to the same shell.

Anions are a bit bigger than their parent atom, and there is some serious electron repulsion, but how can such a repulsion make the hydride ion bigger than the fluoride ion, given that the fluoride ion also faces some repulsion within itself. Are there some other factors - or is the fact which I reported is very wrong?

The quick answer to which ion, $\ce{H-}$ or $\ce{F-}$, is bigger is "it depends". The general reasoning behind the claimed fact the hydride anion is larger than the fluoride anion typically comes down to charge density. In $\ce{H-}$, there are two electrons for the one proton, giving it a charge ratio of 2, while in $\ce{F-}$, there are ten electrons and nine protons for a charge ratio of roughly 1.1. The significantly higher charge ratio of hydride leads to a much more diffuse orbital, and a potentially larger ionic radius.

However, there are a few issues that arise, including:

How do we measure the ionic radius? Ions and atoms don't have a hard boundary; the probability of finding an electron a certain distance from the nucleus drops, but there's no firm boundary.
Ionic radii depend on the nature of the counterion

One potential way to measure the ionic radius is to look at lattice parameters of solids, determined by X-ray crystallography. This does not give much information about what the ions look like in solution, but this is not of huge importance in this case as formal hydride is unlikely to exist in solution. More importantly, X-ray crystallography gives the distance between the two ions, but it does not give the ionic radii. However, we can directly compare the lattice parameters of two solids sharing the same cation with the idea that a larger lattice parameter would be indicative of a larger anion.

For the alkali metal binary salts, the fluoride ion does typically lead to a smaller lattice constant than hydride (e.g. 4.027 and 4.083 Å for lithium salts, respectively). However, for calcium hydride and fluoride, the latter gives a larger orthorhombic cell, implying that fluoride is a larger anion.1 Indeed, for most of the analogous binary hydrides and fluorides in this reference, the fluorides produce larger lattice constants. Using different approaches to measuring ionic radii, including the soft sphere and hard sphere approaches, Lang and Smith succinctly declare that "the ions maintin the following order of size: $$\ce{Li+}<\ce{H-}<\ce{Na+}<\ce{F-}<\ce{K+}<\ce{Cl-}<\ce{Rb+}<\ce{Br-}<\ce{Cs+}<\ce{I-}$$
All the anions are larger than isoelectronic cations, but the Pauling order of ion size is not maintained."2

Therefore, while it may be true that the hydride anion appears larger than the fluoride anion in a few cases, this is not true in general.

1 J. Solid State Chem., 1970, 2 (2), 144

2 Dalton Trans., 2010, 39, 7786


CuI, CuSCN Solubility in Ammonia Solution I am trying to dissolve copper(I) iodide and copper(I) thiocyanate in ammonia.  I am quite sure that they both dissolve in liquid ammonia, and are insoluble in ethanol.

If I were to purchase a 2.0M solution of ammonia in ethanol, like the one here, would the CuI and CuSCN still dissolve in the ammonia solution?

Also, are there any other solvents that you know of for these materials?

The reason these salts are soluble in ammonia (including aqueous ammonia -- so why buy ammonia in EtOH?) is that they form complexes with ammonia:
$$
\ce{CuI}\text{(s)}+2\ce{NH3}\text{(aq)}\to\ce{Cu(NH3)_2^+}\text{(aq)}+\ce{I-}\text{(aq)}.$$
Have fun!


Sodium metal burning off when conducting electrolysis and I don't know why I melted down some table salt and used high voltage welder power source to try and separate the sodium and chlorine it was partially successful because I could smell and see chlorine gas but during the reaction what I think was sodium would float to the top of the molten salt and burst into flame. Is there a way to prevent this?

I was thinking maybe an airtight crucible, switching the anode and cathode or keeping a lower temperature to prevent it vaporizing. Any answer would be very helpful.

First, sodium burns very nicely in air. However, it also burns in chlorine, returning back to the $\ce{NaCl}$ with which you started.

Most welders provide alternating current, meaning the the electrode becomes alternately positive and then negative (50 times per second [50 Hz] in much of the world, and 60 Hz in North America). Therefore, both $\ce{Cl2}$ gas and $\ce{Na}$ metal are produced at each electrode and recombining.

To produce sodium metal you need a direct current supply, such as from a battery, and some means to separate the chlorine bubbling out at the positive (anode) electrode from the sodium metal at the negative (cathode) electrode. A diagram of the Down's cell should help you understand this, and you can view a video of sodium production.

As a demonstration, I've made (for a very short time) small amounts of sodium in an open crucible using a laboratory 10 volt, 6 ampere supply... One sees a small metallic bead forming under the molten salt at one wire and immediately burning when it hits the air, with sodium's characteristic yellow glare.


Why is the hydride ion negative? Recently, I saw a video about sodium hydride and it saw it was made up with $\ce{Na+}$ ions and $\ce{H-}$ ions. My  problem is: aren't all hydrogen ions positive? And in what conditions can $\ce{H-}$ occur?

And in what conditions can $\ce{H^{−}}$ occur?

In any phase. The hydride ion however doesn't stick around for long in water since it's such a strong base. First, hydrogen isn't that electronegative to start with, so tacking on an extra electron isn't helping with its stability. Second, combination with an hydrogen proton, $\ce{H+}$, is highly entropically favorable because the product, $\ce{H2}$, is a gas.

The $pK_a$ of $\ce{H2}$ has been estimated to be ~35 so that speaks a lot to the strength of the hydride ion as a base.


Enthalpy change defined at constant pressure only? Is change in enthalpy defined only at constant pressure?

I know Q-W=del(U)
And Q at constant pressure equals del(H) (enthalpy change).

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

Is del(H) =nCp del(T)?
How?

Is change in enthalpy defined only at constant pressure?

No. The definition of $\Delta H$ is $\Delta H = \Delta U + \Delta (PV)$  Simple as that.

  I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy
  change).

  Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No.  Only if the applied pressure is held constant during the change.

  Is del(H) =nCp del(T)? How?

No.  Only for an ideal gas.  For real gases, this equation is not correct outside the limit of ideal gas behavior.  For an ideal gas, $\Delta U=nCv\Delta T$ and $\Delta (PV)=nR\Delta T$, so $\Delta H=n(C_V+R)\Delta T$.  And then $C_p=(\partial H/\partial T)_P=C_v+R$, so $\Delta H=nC_P\Delta T$.

For a real gas, liquid, or solid,
$$dH=nC_PdT+\left[V-T\left(\frac{\partial V}{dT}\right)_P\right]dP$$
Note that the term in brackets is zero for an ideal gas.


Chemical reaction calculator Is there any kind of program that is able to calculate the end result of a specific chemical reaction? Say I wanted to know what happens when you mix $\ce{HClO + NaOH}$, is there anything that could give me the answer to any question like this? Wolfram alpha works pretty good, but it doesn't give me an answer if it isn't a balanced equation, and I don't know what that means. $\ce{HClO}$ is an acid and $\ce{NaOH}$ is a base, so there has to be some reaction, balanced or not.

This is a fundamentally very difficult question. Your example of of $\ce{HClO + NaOH}$ hides several assumptions, e.g., time frame and aqueous solution, which really means some number of $\ce{HClO}$, some number of $\ce{NaOH}$, and presumably a huge number of $\ce{H2O}$. When mixing all three components a wide variety of reactions take place, e.g., disproportionation of water ($\ce{H2O + H2O -> H3O^{+} + OH^{-}}$), solvation of $\ce{NaOH}$ etc.

In general, a full exploration of all possible reactions and their barriers have to be assessed in order to make a prediction. For instance, $\ce{HClO + H2O -> HCl + O2H2}$ is a possible reaction, or $\ce{NaOH + 2 HClO -> NaCl + H2O2}$ even in the absence of extra water. The number of possible recombinations of the input materials generally increases exponentially with the number of atoms (e.g., the number of possible isomers of an n-alkane).


If α-glucose is less stable than β-glucose, why is glucose almost always α in its compounds? Maltose, saccharose, turanose, trehalose, trehalulose, amylopectin, amylose - they are all α. Why? Also, why is cellulose β?

I think it has to do with configurations and conformations of sugars and biological needs.

A sugar with a six-membered ring is known as a pyranose, the simplest compound containing such a ring. Similarly, sugars with five-membered rings are designated furanoses.

The cyclization of a monosaccharide renders the former carbonyl carbon asymmetric. The resulting pair of diastereomers are known as anomers. In the α anomer, the OH substituent to the anomeric carbon is on the opposite side of the sugar ring from the CH2OH group at the chiral center that designates the D or L configuration (C5 in hexoses).

  The other anomer is known as the β form. Sugars accurately described
  as cyclic polyhydroxy hemiacetals or hemiketals, because the linear
  form exists in very minute quantities (although β forms comprises
  larger percentages in mutarotation mixtures and under physiological
  conditions)

Hexoses and pentoses may each assume pyranose or furanose forms. The equilibrium composition of a particular monosaccharide depends somewhat on conditions but mostly on the identity of the monosaccharide.

The pyranose ring, like the cyclohexane ring, may assume a boat or a chair conformation.

The relative stabilities of these various conformations depend on the stereochemical interactions between the substituents on the ring. The boat conformer crowds the substituents on its “bow” and “stern” and eclipses those along its sides, so that in cyclohexane it is ~25 kJ /mol less stable than the chair conformer. The ring substituents on the chair conformer fall into two geometrical classes: the rather close-fitting axial groups that extend parallel to the ring’s threefold rotational axis and the staggered, and therefore minimally encumbered, equatorial groups.

Since the axial and equatorial groups on a cyclohexane ring are conformationally interconvertible, a given ring has two alternative chair forms; the one that predominates usually has the lesser crowding among its axial substituents.

  The conformational situation of a group directly affects its chemical
  reactivity. For example, equatorial OH groups on pyranoses esterify
  more readily than do axial OH groups. Note that β -D-glucose is the
  only D-aldohexose that can simultaneously have all five non-H
  substituents in the equatorial position. Perhaps this is why glucose
  is the most abundant naturally occurring monosaccharide.

Glycogen and starch ingested in the diet are hydrolysed by α -amylases and glycosidases, enzymes in saliva and the intestine that break (α1n4) glycosidic bonds between glucose units. Most animals cannot utilise cellulose as a fuel source, because they lack an enzyme to hydrolyse the (β1n4) linkages.

In cellulose the glucose residues have the β configuration, whereas in amylose the glucose is in the α configuration. The glucose residues in cellulose are linked by (β1n4) glycosidic bonds, in contrast to the (α1n4) bonds of amylose. This difference gives cellulose and amylose very different structures and physical properties.

The most stable three-dimensional structure for the (α 1n4)-linked chains of starch and glycogen is a tightly coiled helix, stabilized by interchain hydrogen. For cellulose, the most stable conformation is that in which each chair is turned 180° relative to its neighbours, yielding a straight, extended chain. (In contrast, the average plane of each residue along the amylose chain forms a 60° angle with the average plane of the preceding residue, so the helical structure has six residues per turn). All ⎯OH groups are available for hydrogen bonding with neighbouring chains. With several chains lying side by side, a stabilizing network of interchain and intrachain hydrogen bonds produces straight, stable supramolecular fibers of great tensile strength.

Steric factors and hydrogen Bonding influence homopolysaccharide folding

Polysaccharide molecules can be represented as a series of rigid pyranose rings connected by an oxygen atom bridging two carbon atoms (the glycosidic bond).

  There is, in principle, free rotation about both C-O bonds linking the
  residues however, rotation about each bond is limited by steric
  hindrance by substituents. The three dimensional structures of these
  molecules can be described in terms of the dihedral angles, ϕ  and Ψ,
  about the glycosidic bond:

The bulkiness of the pyranose ring and its substituents, and electronic effects at the anomeric carbon, place constraints on the angles ϕ and Ψ; thus certain conformations are much more stable than others.

Summarising this, the abundance of the forms of sugars has a lot do with factors affecting conformations such as boat and chair – actually this the reason why , equatorial OH groups on pyranoses readily react compared to axial ones, (sterics, rotation about the C-O bond – this is limited in cellulose as it forms almost linear chains and a strong network of H bonds between adjacent molecules limiting flexibility), and other biological factors such as availability of enzymes to break glycosidic bonds connecting the molecules (as noted animals don’t have the ability to break (β1,4) glycosidic bonds).

Hope this helps

References

Lehninger Principles of Biochemistry

Biochemistry Voet and Voet


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

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

Ultimately, the octet rule is justified by the fact that in many cases an $n s^2 n p^6$ electron configuration is the most energetically favorable one. Recall that electrons are arranged in different electron shells around the nucleus. Each shell corresponds to a principal quantum number $n$ and can hold only a certain number of electrons, namely up to $2n^2$. Each shell is composed of subshells each of which corresponds to an azimuthal quantum number $l$.

Now recall also that the binding energy of the electron decreases with both $n$ and $l$, and thus, once $n s$ and $n p$ orbitals of an $n$-th shell are filled, in many cases it is indeed energetically unfavorable to fill the subsequent orbitals: $n d$ if (there are any) or $(n+1)s$. The energy gain from binding an electron to an atom might not be enough to compensate a cost of separating it from a different atom.

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

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

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


Why does tetrachlorocuprate(II) form in aqueous solution even though water should be the stronger ligand? In aqueous solution, $\ce{Cu^{2+}}$ forms the $\ce{[Cu(H2O)6]^2+}$ complex. Given that water is a stronger ligand than $\ce{Cl-}$, though, why does why does the  $\ce{[CuCl4]^2-}$ complex form upon addition of chloride ions?

[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.

This is incorrect. Water ligands may be stronger field ligands in the spectrochemical series but that does not in any way affect the strength of the ligand–metal bond or the kinetics and thermodynamics of certain complexes, their formation and their breakdown.

Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.

The thermodynamic energy differences between different complexes are often relatively small and typically the kinetic barrier for ligand exchange is low, too. Thus, in many cases which complex is formed depends strongly on the concentrations of individual ions. The mere fact that complexes without aqua ligands can and are formed shows that water is not that good a ligand after all. In fact, due to the positive charge of the central metal having an anionic ligand bind is often preferred due to simple electrostatic interactions.

To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.


Does the mechanism of the Schmidt reaction involve a nitrene as an intermediate? Per the mechanism described here the reaction does not involve nitrene as an intermediate.(EDIT-1---I initially thought the written mechanism described so, but as Zhe points out,I was wrong in my observation.)

But, can anybody shed some light on why the reaction does not involve nitrene; though it possibly can?

Nowhere in that link is there a nitrene formed. All steps are concerted, that is, you see that alkyl shift happens at the same time as dinitrogen fragmentation.

So indeed, nitrenes are not intermediates. I suspect that mechanistic studies have been done. You might consider looking at this paper.


Finding/Calculating Thermal Conductivity of Magnesite Our research group is having a very difficult time locating anything about the thermal conductivity of $\ce{MgCO3}$. It is a pretty commonly mineral, so we were pretty surprised by this. Is any way to derive thermal conductivity from other properties of magnesium carbonate? (Magnesite)

In the dissertation of K. Kratz (2006) the following values of thermal diffusivity along the crystallographic axes of magnesite are listed:

$D_{[001]} = 6.55 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[100]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[010]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$.

Ch. Clauser and E. Huenges, Thermal Conductivity of Rocks and Minerals lists the following values for the thermal conductivity of magnesite in the temperature range from 25-100°C:

$\lambda_{\perp} = 7.32 \pm 0.57 ~\mathrm{W m^{-1} K^{-1}}$
$\lambda_{\mid} = 7.86 \pm 0.17 ~\mathrm{W m^{-1} K^{-1}}$


What are the common names of this two types of plastic? I need to do some plastic molding design. I want to know the names/types of plastic which are commonly/widely used in industries (please see the pictures below).

The most common types of plastics in use today are polyethylene (PE) and polystyrene (PS). Each of these comes in several variants.

It is difficult to tell just from a photograph exactly what plastics are used in the products shown.

Also, various additives are often added to plastic to enhance certain properties such as high-impact, chemical resistance, recyclability and flexibility.

The type of plastic is sometimes marked on the item itself to help with recycling, typically in the form of an extruded imprint, or you could try to contact the manufacturer.

Both PE and PS are suitable for moulding and extrusion. You can also do some simple physical and/or chemical testing to determine what the plastics contain, although this will involve destroying at least a part of the item. Low density polyethylene (LDPE) typically has a lower melting point $(105-130^{o}C)$ than polystyrene $(200-240^{o}C)$.

LDPE is also softer and slightly rubbery whereas polystyrene is typically more rigid & crackly when crushed. Without physically touching the items to get a proper feel for what they are made of, as a first guess, I would say that the colourful containers such as plastic bins and laundry baskets in your first photo look like some sort of polyethylene, whereas the clear plastic packaging in the second photo looks more like polystyrene.


Why is Sodium Chloride White? Why is solid sodium chloride white - and for that matter, why are the majority of salts white in solid form?

Sodium chloride is colorless because the electrons are tightly bond tho the sodium cation and the chloride anion. Visible light has not enough energy to excite the electrons to higher energy levels. So there's no absorption or emission of visible light of a certain wavelength and therefore no color.
I wouldn't say that the majority of salts is colorless, since the more complex the composition of a salt is the more likely it is not colorless.


Which allotrope of phosphorus is kinetically the most stable? Which allotrope of phosphorus is kinetically the most stable?

I ruled out white and yellow as being quite reactive.

Red and black allotropes are both polymeric and comparitively less reactive.

Black looks the most stable one thermodynamically. But kinetically, I have no idea whether it should be red or black.
The answer provided to me is Red.

(Relevant chat discussion)

I think the key to this question lies in the understanding of the difference between thermodynamic and kinetic stability. In this post, I find Thomij's answer the most rigorous and enlightening.

From this you should learn, that the thermodynamically most stable allotrope corresponds to the global minimum on the potential energy surface, while a kinetically stable allotrope, i.e. hindered through an activation barrier, corresponds to local minimum. From this it can easily be seen, that the thermodynamically most stable allotrope is also the kinetically most stable allotrope. This is due to the fact, that the barriers for interconversion of the different allotropes always have to be higher for the most stable one.

In the following two dimensional case you see that the thermodynamically most stable, i.e. the global minimum, is surrounded by two local minima. It must always be true, that $E_\mathrm{a}(1\to2)<E_\mathrm{a}(2\to1)$ if $1$ is a local and $2$ is a global minimum. The same applies to the other case. If this was not true, then $2$ cannot be the global minimum and hence not the thermodynamically most stable modification.

Having established that, it is therefore true, that black phosphorus is the thermodynamically and kinetically most stable form, see Hollemann-Wiberg, p. 680.

However, looking at a different reaction, the case might not be so simple any more. This is even true for different temperatures, as the following graphic will show (quoted from the linked book).


What are the magnetic properties of [Ni(S2C2Ph2)2]^z's 3 oxidation states? I was doing some research on Wikipedia and I came across the article on non-innocent ligands. It talked about the chemical $$[\ce{Ni(S_2C_2Ph_2)_2}]^z$$ as an example. It mentioned that the transition metal complex was diamagnetic when it was neutral(z=0). I decided to figure out how many unpaired electrons the central nickel atom would have at various formal oxidation states. I found the following:

$$
\begin{array}{ccc}
\ce{Ni(II)}&(z=-2)&=2~\text{unpaired}\\
\ce{Ni(III)}&(z=-1)&=3~\text{unpaired}\\
\ce{Ni(IV)}&(z=0)&=4~\text{unpaired}\\
\end{array}$$
I don't understand why the complex is diamagnetic when its charge is 0. The radical ligands would be coupled with each other anti-ferromagneticly but there would still be 4 unpaired electrons in the central nickel atom. Wouldn't they make the complex paramagnetic? Could someone please explain to me why it is diamagneic and tell me whether the other 2 oxidation states are para or dia-magnetic.

$\ce{Ni^{2+}}$ is a $d^8$ metal. In a square planar orbital splitting, all $8$ electrons are paired, hence it is diamagnetic. Even when the complex is neutral, the electron loss is at the ligands (by definition, non-innocent), and they are antiferromagnetically coupled, which means that their spins "cancel" each other. Since there are two, the net moment is zero.

All complexes are $\ce{Ni(II)}$, so assuming all complexes are square planar, all $d$ electrons are paired up and the magnetic moment from these is 0. Both the $z=-2$ (no radicals) and $z=0$ (antiferromagnetic coupling) area diamagnetic. The $z=-1$ complex is paramagnetic. There is a single unpaired electron in one of the ligands that can't couple to the other one, where all of them are paired up.


Is chlorine the central atom in NOCl? We've been told that the in the structure of $\ce{NOCl}$ the central atom is $\ce{Cl}$ (rather than $\ce{N}$) due to some exception.

Looking up on internet I could not find anything to support this. I would like to know if it really exists in such a state due to some exception. If yes, what is that exception.

Wikipedia gives a structure with a central nitrogen atom:

The nitrogen atom (blue, using the standard color scheme) can form all three covalent bonds without building up an excess positive or negative charge, and so is favored in the central position through which all three bonds pass.


Possible Synthesis of a Polymer Following a question I have asked previously, what I intend to inquire this time is quite similar as well, except, with a polymer.

So previously I asked whether such monomer PCDO is possible to synthesize.

That originated from this similar, but different polymer poly-PCDO. It is true that I also don't think this PCDO from my previous question is poly-PCDO after monomerization (I'm still not so sure though). But the PCDO molecule pop out when I was attempting to make this original poly-PCDO a monomer.

I think the synthesis of the polymer in this structure would be easier to theorize and explain.

One can dehydrate divinyl ether with benzene (or perhaps phenol) and lithonate by condensation polymer.

$$\ce{n(C4H6O + C6H6O + HLiCO3) -> (C11H9LiO4)_{n} + nH2O + nH2}$$

Due to my limited knowledge in this area, I can only assume the phenyl group would most likely bond with lithonate being bonded to the ether already. Would someone explain to me how this will or will not work?

Under acidic conditions, a part of the divinylether will probably just undergo cationic polymerization:


Carboxylate "cage" concept My professor emphasizes the concept of the carboxylate "cage" or the idea that electrons participate in resonance within a carboxylate group such as in the benzoate anion but the electrons in the carboxylate group cannot leave this "cage" - i.e. participate in resonance with the phenyl substituent.

Is this idea valid?

I've already poked a hole into it by pointing out that electron donation to the "cage" is possible - i.e. consider carbonic acid with one hydroxyl group replaced by oxygen connected to a phenyl group. Valid resonance structures can be drawn in which electron density from the oxygen in the ester linkage participates in resonance with the electrons "in" the carboxylate group. So donation into the "cage" is possible. Now, this begs the question - is withdrawal from the cage possible as well? He insists no. Now, he might be hiding some edge cases for pedagogical purposes, but I don't care - if there is an exception - I want to know about it (because chemistry always gets interesting at the exceptions).

Let's start by deciding what compounds we're talking about here. You reference an example by your instructor of benzoate, which is the conjugate base of benzoic acid.

What your instructor is suggesting is correct for benzoate. The most important resonance structures for benzoate are the two equivalent structures that place negative charge on the oxygen atoms. It is also possible to draw a resonance structure where electrons are removed from the ring and both oxygens carry a negative charge.

The "carboxylate cage" essentially means that there is no way to delocalize the negative charge of the carboxylate into the aromatic ring. The only way to move electrons into the ring is from the C-O pi bond, which leads to (what I would consider) an unreasonable resonance structure, as it renders an oxygen both positively charged AND non-octet.

Carbonic acids are either a subset of carboxylic acids or a closely related functional group to carboxylic acids. Depending on what property you are interested in, they may be useful to compare or the difference might be too great. In this case, starting from the conjugate base of phenyl carbonic acid (as drawn in the question), there will be no way to delocalize the charge into the aromatic ring.


How can you predict a double displacement reaction with no precipitate? Is it possible to predict a double displacement reaction between two ions where the products remain in aqueous solution? I see how you can predict precipitation reactions if you know whether the product(s) is/are soluble, but that doesn't apply to all possible double displacement reactions.

If there is no precipitate, then there is nothing to drive the reaction. Both possible products would be soluble, and so there is no net reaction (it's just the reactant ions in solution).


Regioselectivity of deprotonation of 1,4-dicarbonyl In the intramolecular aldol condensation of hexane-2,4-dione (above), why does the base abstract one of the $\ce{-CH3}$ protons instead of one of the $\ce{-CH2}$ protons? Both sets of protons are α to a carbonyl group and should therefore be acidic.

I'm confused because I've seen similar reactions where the proton is taken from the $\ce{-CH2}$ group.

Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very slightly in acidities, and in general, whether one deprotonates at the more or less substituted carbon can be controlled by reaction conditions.

The mechanism you are drawing is the first step of what is known as an aldol condensation. Because all the steps in this reaction are easily reversible, simple aldol condensations with $\ce{OH-}$ are typically under thermodynamic control, meaning that the most stable product is preferentially formed.

Therefore, let's consider what would happen if we deprotonated the $\ce{-CH2}$ group. The deprotonation itself is definitely a possibility, but what happens after that is quite implausible:

Once you form the enolate, the next step in an aldol condensation is the enolate attacking the other carbonyl group. In this case, it would lead to a three-membered ring, where there is a lot of angle strain and the substituents eclipse each other. The dehydration to form the α,β-unsaturated ketone is even less likely, because the cyclopropene ring thus formed would possess incredible angle strain.

On the other hand, deprotonation of the terminal $\ce{-CH3}$ group eventually leads to the formation of a five-membered ring after nucleophilic attack on the other carbonyl group. This is much more stable, and you can eliminate water via an E1cb-type mechanism to get a cyclopentenone.

So, you could certainly deprotonate the $\ce{-CH2}$ position. It is just that, after it gets deprotonated, nothing very good can come out of it. This deprotonation is said to be unproductive.


Is it actually possible to dispose of a body with hydrofluoric acid? In the TV show "Breaking Bad", Walter White frequently gets rid of people who get in his way by submerging them in a plastic container full of hydrofluoric acid. This, at least in the TV show, completely dissolves the body leaving nothing but a red sludge behind at the end.

Is it actually possible to dispose of a body with hydrofluoric acid?

If hydrofluoric acid wouldn't work, are there any acids corrosive enough to achieve the stated effect from the show?

Hydrofluoric acid is toxic and corrosive, but actually isn't that strong of an acid compared to other hydrohalic acids; the fluorine has a very good orbital overlap with hydrogen and is also not very polarizable, therefore it resists donating its proton, unlike other hydrohalic acids which are good proton donators. It will break down some tissues, but it will take a relatively long time and won't turn the entire body into stuff that can be rinsed down the drain. Hydrochloric acid is a much stronger acid, and as it has several uses from pH-balancing pool water to preparing concrete surfaces, it's available by the gallon from any hardware store. However, it isn't very good at dissolving bodies either; while it will eventually work by breaking down the connective tissues, it will make a huge stink and take several days to dissolve certain types of tissues and bones.

The standard body-dissolving chemical is lye aka sodium hydroxide. The main source is drain clog remover because most drain clogs are formed by hair and other bio-gunk that accumulates naturally when humans shower, exfoliate etc. It works, even though the body's overall chemistry is slightly to the basic side of neutral (about 7.35-7.4) because the hydroxide anion is a strong proton acceptor. That means that it strips hydrogen atoms off of organic molecules to form water (alkaline hydrolysis, aka saponification), and as a result, those organic molecules are turned into simpler molecules with lower melting points (triglycerides are turned into fatty acids, saturated fats are dehydrogenated to form unsaturated fats, alkanes become alcohols, etc). Sodium hydroxide is also a ready source of the sodium ion; sodium salts are always water-soluble (at least I can't think of a single one that isn't). The resulting compounds are thus either liquids or water-soluble alcohols and salts, which flush down the drain. What's left is the brittle, insoluble calcium "shell" of the skeleton; if hydrolyzed by sodium hydroxide, the resulting calcium hydroxide ("slaked lime") won't dissolve completely but is relatively easy to clean up.


"Generally Octahedral" - Meaning? What does this question mean by generally octahedral? Can this molecule adopt other shapes?

Also I have my suspicions about the Br-S-F bond angle being a perfect 180 degrees. Wouldn't vdW repulsions distort the bond angle and make it ~180 degrees? The way I see it, if we place the bromine axial, wouldn't its sheer size push down on the fluorines in the equatorial plane and thus distort the 180 degree bond angle?

On the other hand, now that I think of it, we only have one bromine pushing down on all the fluorines with an equal force and the net force should cancel out, making the 180 degree bond angle a theoretically perfect 180 degree bond angle.

Are there any ways to verify this other than jumping through hoops in one's mind? I tried making Marvin Sketch render the molecule but it wasn't useful (or I wasn't using it correctly).

Also, another choice I was doubting is choice 1. Are you sure there are 8 F-S-F bond angles < 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles < 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?

The term "generally octahedral" refers to a distorted octahedral arrangement of the ligands referring to the central atom (as ron already stated). In this case the distortion is due to the massively different bond lengths.

In addition to ron's answer I did a little calculation at BP86/cc-pVDZ. I did not use any symmetry constraints at first and the final geometry deviated from ideal $C_\mathrm{4v}$ geometry only slightly due to rounding errors. After symmetry constraint reoptimisation, the following parameters may be found:

\begin{array}{lrl}
\text{parameter}                & \text{value}&\\\hline
\mathbf{d}(\ce{S-Br})           & 2.039 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{ax}})       & 1.638 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{eq}})       & 1.650 &\mathrm{\mathring{A}}\\\hline
\angle(\ce{F_{ax}-S-Br})        & 180.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-Br})        &  90.7 &{}^{\circ}\\
\angle(\ce{F_{ax}-S-F_{eq}})    &  89.3 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F'_{eq}})   &  90.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F''_{eq}})  & 178.7 &{}^{\circ}\\\hline
\angle_\mathrm{D}(\ce{F_{eq}-S-F'_{eq}-F''_{eq}})
                                & 178.7 &{}^{\circ}\\\hline
\end{array}

I was unable to determine a interconversion barrier related to the $\angle(\ce{F_{eq}-S-Br})$ bond angle change. I was also not able to find a different stable isomer.

In solid state it is most likely, that the geometric parameters will change due to packing effects and/or more internal entropy.

Addressing the edit:

  Are you sure there are 8 F-S-F bond angles < 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles < 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?

Yes, the bromine pushes down the fluorines of the equatorial plane resulting in $\angle(\ce{F_{ax}-S-F_{eq}})=89.3^{\circ}$ bond angle. Due to this, the sulfur is out of plane with the equatorial fluorine atoms. Since the planes expanded by $\ce{Br-S-F_{eq}}$ and $\ce{Br-S-F'_{eq}}$ have to be at $90^\circ$ due to symmetry constraints, this is creating also a slightly smaller angle than $90^\circ$ for the equatorial fluorines. In this very case it can be found to be $\angle(\ce{F_{eq}-S-F'_{eq}})=89.992^{\circ}$, which was rounded by the program to fit the one decimal place restriction. The application of this is just trigonometry and you can derive it as a simple thought experiment - if you like.


Le Chatelier’s principle when pressure of gases increased with the same stoichiometric coefficient $$\ce{C (s) + H2O (g) <=> CO2 (g) + H2 (g)}$$
What will happen to the equilibrium when the pressure of gases is increased? What will happen to the yield of $\ce{H2}$?

Since pressure increased means the volume is decreased, equilibrium will shift to the direction with the least number of moles. But since the stoichiometric coeffients are all the same, am I right to say that it will not shift?
But I doubt my answer as the question states - "pressure of gases increased". The reactants include solid carbon, meaning I do not count that in. Does this mean that equilibrium shifts to the right?

Your thinking is on the right track. However, what you really need to look out for is number of gas moles.

In order to understand this, think in terms of the equilibrium constant. The equilibrium constant for this reaction is:
$$K_\text{p}=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Remember, since carbon is a solid, changing the volume/pressure is not going to change anything about it, so it is not in the equilibrium constant.

Since we are introducing a supposed change into the system, we have to write its reaction quotient. The reaction quotient, $Q$, is just a way to determine which way the reaction will have to shift in order to return to equilibrium. We write it in pretty much the same way:
$$Q=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Even though it looks the same, remember that we are calculating this after introducing a change. So what we really plug into this equation is the following, where $x$ is the factor by which we increase pressure (suppose that $P_{\text{CO}_2}$, $P_{\text{H}_2}$, and $P_{\text{H}_2\text{O}}$ are the equilibrium pressures):
$$Q=\frac{(P_{\text{CO}_2}\cdot x)(P_{\text{H}_2}\cdot x)}{(P_{\text{H}_2\text{O}}\cdot x)}$$
$$Q=\frac{x^2P_{\text{CO}_2}P_{\text{H}_2}}{xP_{\text{H}_2\text{O}}}$$
Since we are increasing pressure, $x$ is obviously positive. Thus, $Q>K_\text{p}$.

Due to Le Chatelier's principle, since we introduced a stress into the system, the system will shift away from that stress. $Q$ and $K_\text{p}$ are a good way of mathematically representing this stress. $Q$ is essentially a representation of the system's current state, and $K_\text{p}$ is the representation of the system's ideal state.

So, how do we reach this ideal state? It's simple; we want $Q$ to equal $K_\text{p}$. Obviously this will happen through a change in pressure. Since $Q$ is too big, we need to decrease the numerator in order to make it equal to $K_\text{p}$. The numerator represents the pressure of the products, so the reaction will shift away from the products, using up more products and creating more reactants, until it reaches equilibrium.

Hope that helps.


Oxidation state of ions in a unit cell Given that $\ce{Al2O3}$ is face-centered cubic (cubic-closest packed) with respect to oxide anions, I am instructed to find the oxidation state of each aluminum ion "fragment" in the unit cell.

Let's suppose that the picture above is of $\ce{Al2O3}$ and not $\ce{NaCl}$. The larger oxide ions are green; the smaller aluminum cations are purple.

There are a total of 4 oxide ions in a unit cell, and a total of 3 aluminum ions in the unit cell.

Given that 4 oxide ions impart a charge of -8, and that a unit cell must be neutral, then each of the "three" aluminum ions must have an oxidation state of +8/3 in the unit cell, correct?

I cannot find any textbook discussing the oxidation states of atoms in unit cells ... so I am at a loss at whether my work is correct or not.

This is a really weird question.

Let's start with the general case for a FCC lattice, you'd count up the oxygen atoms per unit cell. In the $\ce{NaCl}$ example, as noted in the comment, there are four $\ce{Cl-}$ per unit cell (6 on the faces, 8 on the corners). Then you'd count up the number of Al atoms per unit cell and divide up the charge as you note above.

In the case of "fractional" oxidation state, it implies that there's a mixture of oxidation states.

Let's say, hypothetically, you had 3 cations and need a +8 charge to balance out the anions. That tells me there are two +3 cations and a +3 cation somewhere. The average is +8/3 but I need to assign formal charges as integers.

But...

Normal $\ce{Al2O3}$ is not an FCC lattice. It's triclinic.

Here, you would do the same thing - count up the number of Al and O atoms per unit cell, considering the fractional volume of the atoms inside the unit cell. It's a bit trickier, since the volumes aren't as obvious as the FCC case.

Instead

This question is about $\gamma$-alumina

Notice that the oxygen atoms actually form an FCC lattice. I have an oxygen at each corner of the unit cell, and one in the middle of each face. So that's 4 oxygen atoms per unit cell = +8 charge from the anions.

How many Al atoms are there?

There are 5 Al atoms completely in the unit cell, plus one per edge (12, each counting 1/4 = 3) so a total of 8 Al atoms!

So that's an average oxidation state of +1 per Al...


Boiling point and vapor pressure Does a liquid having higher vapor pressure have a lower boiling point?
I think so because I read in wikipedia that, "A substance with a high vapor pressure at normal temperatures is often referred to as volatile." A volatile liquid has lower boiling point.

Not necessarily. The vapor pressure of a liquid (or solid) is the pressure of the vapor in equilibrium with its condensed phase in a closed system. The boiling point of a liquid at a given pressure is the temperature at which the vapor pressure is equal to the given pressure. This means the vapor pressure is a function of the temperature,
$ P^* = f(T) $, and the boiling point is a function of the pressure, $ T_{bp} = f(P) $.

So your question is a bit vague. In general, yes, liquids with higher vapor pressures will have lower boiling points. In any first year chemical engineering course, students are introduced to the Antoine Equation, an emperical relation between vapor pressure and temperature. Plotting the Antoine equation for a set of liquids lets one visualize the relationship between vapor pressure and temperature very quickly; these plots are called Cox charts. For example, examine the Cox chart below:

Now look at the relationship between ethanol and benzene. At 50 F, benzene has a higher vapor pressure than ethanol. At pressures below ~10 psi, the general trend holds and benzene has a lower boiling point. However, the lines for benzene and ethanol cross at ~10 psi, so above ~10 psi benzene actually has a higher boiling point than ethanol.

So in general, yes, liquids with higher vapor pressures will have lower boiling points. However, the question is vague and exceptions do exist.


The effect of air pressure on the melting point What is the effect of air pressure on the melting point of a substance?
a) No effect
b) The melting point increases when the air pressure increases
c) The melting point decreases when the air pressure increases

The boiling point has inverse relation with vapor pressure of the liquid and positive relation with atmosphere (air) pressure. The melting point has positive relation with vapor pressure of the substance, but what about its relation with air pressure?

For most substances, higher pressure (or air pressure, in your case) will cause the melting temperature to go up. To think about it intuitively, imagine that you have a certain solid. Melting it would increase the volume of that substance because liquids take more space than solids. If you increase the pressure, it becomes harder for that transformation to occur. It's harder to change into a state that needs more volume if you are under pressure! So, you need more energy, hence temperature, to melt.

Notice that this relation is actually reverse in water. Because ice takes more volume than liquid water, it's actually easier to melt it at high pressure and it requires less heat. Try taking an ice cube and apply pressure on it with a knife, see how it liquifies at the contact point.


Hydration of lithium salts Why are lithium salts heavily hydrated? I am not able to understand this because I feel salts of larger cations like Caesium should be more hydrated because of the larger size, they can able to hold more water molecules.

What is the flaw in my reasoning?

The larger $\ce{Cs+}$ ion may interact with more water molecules at once than $\ce{Li+}$, but each interaction is weaker. The prevailing interaction between water molecules and ions are ion-permanent dipole interactions (or just ion-dipole), which are proportional in strength to the amount of charge on the ion and the partial charge on the dipole, and inversely proportional to some power of the distance between the centres of the charges (via the shell theorem). Hence, the closer the water molecules get to the ion, the stronger the interaction. Since $\ce{Cs+}$ is very large, the charge centres don't come so close, and the interaction is relatively weak. $\ce{Li+}$ is much smaller, so there is strong solvation. For sufficiently small, highly charged ions, the first solvation shell comes so close to the ion that it allows covalent bonding to take place.

You may find this succinctly described as $\ce{Cs+}$ having a lower charge density than $\ce{Li+}$, which is about the same thing.


Dissociation of water into H+ and OH-: Does the law of mass action hold at nanoscale? Suppose water under neutral conditions is confined in a virtual spherical nanocontainer with a radius of 25 nm. To calculate the number of hydronium ions, one uses water dissociation constant which is derived from the law of mass action. In this case, considering the temperature at 25°C I come up with a number which is smaller than one.

A logical suggestion here is probability. Is there any limitation to the law of mass action, i.e. does it break down at the nanoscale level and quantum mechanics and thus probability kick in? If yes, is there any boundary to specify things with respect to the size of the system?

There is definitely a breakdown of mass action at very low numbers of molecules.  This can happen either because concentrations are tiny, or because the system of interest is tiny.

In either case, your idea to turn to probability is an excellent one, and is how these systems are modeled.  But the reason doesn't necessarily have to do with quantum mechanics.  It's just a consequence that matter is not really a continuum that can have any concentration, but is in fact granular.  At the scale of atoms, there can either be an atom, or not be an atom.  You can't have 0.5 of an atom.

The right boundary at which mass action breaks down depends a bit on the particular system, in particular, if reactions are occurring that depend on H+.  In the less complicated case of no reactions depending on $\ce{H+}$, then the boundary between mass action being valid and not is actually fairly easy to calculate. If you know the average or expected value of $\ce{H+}$ ions in your water sphere (which you do, since you calculated it) the actually observed number of ions will follow a Poisson distrubtion.  If the mean number of atoms is $N$, then the standard deviation of atoms is $\sqrt{N}$, so the relative uncertainty in the mean number of atoms is proportional to $\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}$.  According to your calculation, $N$ is less than one, so the relative uncertainty is greater than 100%!  But if you enlarged the sphere so the expected number of ions was $N=10$, then the uncertainty would be about 32%, still pretty bad.  If you enlarged the sphere so $N=10000$, then the relative uncertainty is only 1%, meaning that mass action predictions for this system are pretty close to accurate!  Getting $N$ tht high requires a sphere about 10,000-times bigger in volume, which means the radius would have to be about 38 times bigger.  But keep in mind this result is only true for the water dissociation problem.  So each chemical system has its own scale where mass action breaks down.

For systems where reactions are occurring that depend on $\ce{H+}$, this simple logic breaks down.  Complex probabilistic simulations need to be done.  Depending on the details of the reactions, the chemical master equation or the chemical Langevin equation would have to be used.

One time I calculated the concentration of frogs in an aquarium was about 1 zeptomolar.  I don't think we could assume frogs followed the law of mass action!


Is supercriticality dependent on partial presssure, or total pressure? Supercritical CO2 is commonly used as a solvent in organic chemistry (notably in industrial processes like decaffeination). I presume that in such cases, the CO2 is typically pure, or very close to it.

But suppose you had a mixture of, e.g., 50% CO2 and 50% nitrogen. Pure CO2 becomes supercritical at approximately 31C and 73atm; particularly in terms of its solvent properties, would the CO2 in that mixture also demonstrate supercritical properties at 73atm total pressure, or would it be necessary to double the total pressure, producing 73atm partial pressure of CO2? Or is the interaction more complex than that, with mixture of CO2 and N2 having a joint critical point different from either single gas?

There is no such thing as supercritical properties. By looking at a gas under given pressure and temperature, you'll never be able to tell is it supercritical or not.

If you try to lower temperature or pressure and see what happens, then yes, the interaction is more complex, and a mixture of two compounds does indeed have a joint critical point different from those of both pure compounds; moreover, the very meaning of a critical point becomes different and more complicated.


Why does the water dissociation constant Kw remains constant? At $\pu{25 ^\circ C}$ at equilibrium the value of $K_\mathrm{w}$ is $\pu{1\times 10^{-14} M}$ and the concentration of $\ce{H+}$ and $\ce{OH-}$ is equal i.e. $\pu{1\times 10^{-7} M}$.
Now if we add $\ce{HCl}$ to it the $\ce{HCl}$ will form $\ce{H3O+}$ and so the concentration of $\ce{H+}$ increases and so the equilibrium shifts towards the left to counteract the changes. As a result the concentration of $\ce{OH-}$ decreases but why does still the $K_\mathrm{w}$ stays the same. If we add too much $\ce{HCl}$ such as 1 mole then we know it will completely dissociate and would form $\ce{H3O+}$ whose concentration would be $\pu{1 M}$.
So now what will happen to $K_\mathrm{w}$? Does it still remains constant but why? Does still the concentration of $\ce{H+}$ can not rise above $1\times 10^{-14}$ but why?

Your question reflects your knowledge in chemistry so I'll try to keep this explanation simple as possible.
Water conduct electricity because it contains $\ce{H+}$ and $\ce{OH-}$ ions by auto-ionization:
$$\ce{H2O <=> H+ + OH- \tag1}$$
It has been defined that this ionization is a constant at $\pu{25 ^\circ C}$. Thus by definition:
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \tag2$$
Accordingly, by definition for pure water, $ [\ce{H+}]=[\ce{OH-}]= 1.0 \times 10^{-7}$. And then, there is a widely accepted principle in chemistry called Le Chatelier’s principle:

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Accordingly, increasing the concentration of reactants (species on left hand side of the reaction) will drive the reaction to the right (more products), while increasing the concentration of products (species on right hand side of the reaction) will drive the reaction to the left (more reactants). At constant temperature, even though concentrations of species (e.g., here they are $\ce{H+}$ and $\ce{OH-}$ in equation $(1)$) at new  equilibrium state is changed, the equilibrium constant remains same. In other word, the equilibrium constant is depends only on temperature.
According to Le Chatelier’s principle, in in equation $(1)$, if you add more $\ce{H+}$ or $\ce{OH-}$ (RHS), the equilibrium would be adjusted to reduce that amount by making more water. But since this addition is at constant temperature, $K_\mathrm{w}$ remains constant.


How should a triple bond be depicted in skeletal structures? Draw the skeletal structure for the following compound:
  $$\ce{HOCH2CH2OCH2C#CCH2CH(CH3)2}$$

I can't figure this out.

I can draw it up until the triple bond. My professor marked off because the triple bond was not drawn correctly.

So, is the triple bond depicted any differently from the other bonds in the molecule? If so, why?

My guess, is that you forgot to take the linear disposition of the carbon-carbon triple bonds into consideration.

So the bond lines cutting through four carbon atoms will be linear, and not zig-zagged as in the rest of the molecule (the latter implies tetrahedral geometry accorded by the $sp^3$ carbon-carbon bonds)

The first diagram may feel more "graceful" (the term I used when I defended my incorrect depiction of the triple bond to my teacher when he asked me a similar question a while back), but it is incorrect.


Confused with oxidation numbers If -2 is a oxidation number and 2- is the charge, then what is - by itself, is it a oxidation number or charge? As in $\ce{Ag+}$.

Also do you see $\ce{Cu + 2Ag+ -> Cu^{2+} + 2Ag}$ as a redox reaction?

Because I don't see any increase or increase in the oxidation number of any element in this equation. All I see is charges

According to the IUPAC gold book, oxidation state is:

  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;...

Let's consider your reaction $Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag$ The superscripts represent charges. Reaction equations typically don't include oxidation states. In the reactants, $Cu$ has no charge and thus an oxidation state of 0 (rule 1). $Ag^+$ has a charge of +1 and thus an oxidation state of +1 (rule 2). In the products, $Cu^{2+}$ has a charge of +2 and thus an oxidation state of +2 (rule 2). $Ag$ has no charge and thus an oxidation state of 0 (rule 1).

In molecules, oxidation state can be thought of the charge of an atom when all bonds are treated as ionic. For example, in $H_2O$, Both $O-H$ bonds are polar and, since $O$ is more electronegative, all bonding electrons can be considered to belong to $O$. Thus, $O$ has 2 bonding pairs and 2 lone pairs for a total of 8. $O$ has 6 protons so it's oxidation state here is $6-8=-2$. Each $H$ has no electrons and 1 proton for an oxidation state of +1. The total oxidation state of the molecule is $2(+1)+-2=0$, which matches its charge.


Does RedP+HI reduce all carbon functional groups to alkane? I've seen the related questions 1 and 2, they're not duplicates

So. when introducing Red P+HI reduction, our teacher said that it is the strongest reducing agent, and it can reduce any functional group i.e. all of these: carboxylic acid,  acid anhydride, ether, amide, acid halide, carbonyl, and nitrile

However, today I stumbled upon a question in my book involving a step in the synthesis of ibuprofen, where they did not reduce the carboxylic acid group even after using RedP+HI. I searched a lot on the internet, and finally came across this paper which confirms this:

paper - notice they've only reduced the alcohol in step (ii)

This now makes me wonder if what I was taught is actually correct. Which brings me to the question:

  Does RedP+HI really reduce all carbon functional groups to alkane?

Nope, in general it doesn't. You are correct. See also for example:

A. Aramini, M. R. Sablone, G. Bianchini, A. Amore, M. Fanì, P. Perrone, A. Dolce, M. Allegretti, Tetrahedron 2009, 65 2015–2021. DOI: 10.1016/j.tet.2009.01.005:

M. Dobmeier, J. M. Herrmann, D. Lenoir, B. König, Beilstein J. Org. Chem. 2012, 8, 330–336. DOI: 10.3762/bjoc.8.36 Open access PDF. (This also includes your one as a reference, too.):

If it did, it would be a pretty poor reagent since it would have no functional group tolerance, so not very useful. However, it is not unlikely that under special conditions it might be able to reduce carboxylic groups as well.


What is the physical significance of molecular partition function? What is the physical significance of molecular partition function?

I reason, that the molecular partition function

\begin{align}
q &= \sum_i \exp(-\beta\varepsilon_i),&
\beta &= \frac{1}{k_\mathrm{B}T}
\end{align}

is found in the expression for the populations of each configuration, derived from the Boltzmann distribution:

$$p_i = \frac{\exp(-\beta\varepsilon_i)}{\sum_i \exp(-\beta\varepsilon_i)}$$

The partition function $q=\sum_i\exp(-E_i/k_BT)$ in your question can be regarded as the effective number of levels accessible to the molecule at a given temperature. It also means that in the equilibrium distribution the partition functions tells us how the systems are partitioned or divided up among the different energy levels.

In determining the form of the Boltzmann distribution the total number of 'systems' is N and these are divided up as $n_1$ systems have energy $E_1$, $n_2$ have energy $E_2$ etc. subject to conditions $\sum_i n_i=N$ and $\sum_in_iE_i=E$ where E is the total energy. It is found overwhelmingly that the chance that $n_i$ states have energy $E_i$ is given by $n_i = N\exp(-E_i/k_BT)/q$

In practice the partition function acts as a normalisation term in the calculation all sorts of thermodynamic properties, for example the average (internal) energy of N molecules with quantum energy levels $E_i$ is

$$ U=N\frac{\sum E_i e^{-E_i/k_BT}}{\sum e^{-E_i/k_BT} }$$

and the calculating the entropy $S=-Nk_BT\sum p_i \ln(p_i)$ where $p_i=n_i/N =\exp(-E_i/k_BT)/q$  is the probability that the ith state is occupied.

[ If the states are degenerate, with degeneracy $g_i$, then the partition function is $q=\sum_i g_i\exp(-E_i/k_BT)$.]


Oxidation states of oxygen I'm given a question:

  Oxidation number of $\ce{O}$ in $\ce{BaO2}$ is $x$ and in $\ce{OF2}$ is $y$; then the value of $x+y$
  is what?

Now my main question is that if $\ce{F}$ has $-1$ valency in $\ce{OF2}$ then $\ce{O}$ must have valency of $+2$. But is that possible since $\ce{O}$ mainly has oxidation states of $-1$ & $-2$ or am I making any mistake?

The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegative atom acquire a positive charge.
Depending on this hypothesis oxygen have 5 oxidation states.

In all the oxides,oxygen has an oxidation state of $-2$. Eg. $\ce{CO2,CO}$
In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of $-1$. For example, consider $\ce{H2O2}$, here $\ce{H}$ is less electronegative so it will acquire a charge of $+1$ and to balance the $2$ positive charge of 2 H-atoms,each oxygen atom will acquire a charge of $-1$.
In all superoxides ($\ce{KO2,CsO2,RbO2}$), oxygen has an oxidation state of $-\frac{1}{2}$,this is because $\ce{K,Cs,Rb}$, being elements of the first group and less electronegative than oxygen acquire a charge of $+1$, to balance it, each oxygen atom acquires a charge of  $-\frac{1}{2}$.
In one of the exceptions $\ce{OF2}$, the fluorine being more electronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 fluorine atoms oxygen acquires a charge of $+2$.
As last, there is $\ce{O2F2}$, similarly here to balance the $-2$ charge on 2 $\ce{F}$-atoms each oxygen atom acquire a charge of $+1$.


Why is SN2 favored greatly over SN2' in this reaction? Secondary allylic halides can be produced in high yield with the following Mitsunobu-type reaction:[1]

The yield being $94~\%$.
The last step in the mechanism is:

Why is the $\mathrm{S_N2}$ reaction favored so much over the $\mathrm{S_N2'}$ reaction?

Reference: Magid, R. M.; Fruchey, O. S.; Johnson, W. L.; Allen, T. G. Hexachloroacetone/triphenylphosphine: a mild reagent for the regioselective and stereospecific production of allylic chlorides from the alcohols. J. Org. Chem. 1979, 44 (3), 359–363. DOI: 10.1021/jo01317a011.

While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.

Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects.  The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there.  The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.

Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.


How to derive the pressure dependency for the Gibbs free energy? $$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$
The above is an exact differential. Firstly what does this mean physically? (I hear about the fact that a change in G is independent of the "path" but I don't understand this argument because I don't know what is meant by this path. Also how would something be dependent on the path?)

It can also be shown that $\mathrm{d}G$ can be written:
$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T $$

So by comparing coefficients:
$$\left(\frac {\partial G}{\partial p}\right)_T=V$$
It is useful to integrate this to see how $G$ varies with $p$:
$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$
$V$ can be expressed in terms of $p$ using the equation $V=\frac{n\mathcal{R}T}p$ Where $n\mathcal{R}T$ are constants. The above equation becomes:
$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$
Integrating the LHS of the above equation is the crux of my question (I don't understand how that can be integrated). I am not sure whether the following step is correct mathematically, please explain why this is correct/incorrect. I cancel out the $\mathrm{d}p$'s on the LHS:
$$\int_{p_1}^{p_2}\mathrm{d}G=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$
Then the result is:
$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
Or...
$$\Delta G=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
This result is correct but I don't understand the left hand side. I've been told that the reason it's possible to write "$G(p_2)-G(p_1)$" or "$\Delta G$" is because $G$ is a state function/exact differential but, again, I don't understand this. If $G$ was not a state function/exact differential how would the result change?

This question has a lot of parts! I hope I manage to answer most but probably edits will be required.

$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$

What does this mean physically?
This is simply an expression that shows the contributions to the change in the Gibb's free energy, which varies with both pressure and temperature (i.e., $G$ is a function of $p$ and $T$). However, it is possible to separate the exact differential of $\mathrm{d}G$ into two terms using the mathematical expression of partial derivatives, first term is the variation of $G$ with pressure at constant temperature, and the second term is the variation of $G$ with temperature at constant pressure.

What does it mean for $\Delta G$ to be independent of the path? (same question: what does it mean for $\Delta G$ to be a state function?)
You can calculate a change in $G$, or equivalently, $\Delta G$, for any reaction. So for example for a reaction like a combustion process for glucose

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g),}$$

you can calculate $\Delta G$ for this reaction at any particular temperature/pressure that you would like. And in fact, if you don't do the combustion directly but somehow do it via completely dissociated gaseous atoms

$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 C~(g) + 18 O~(g) + 12 H~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$

You will still end up with the same $\Delta G$ for the overall reaction. Obviously if you calculate the $\Delta G$ for each individual step/arrow in the gaseous atoms case it will be very different, but they will combine to give the same overall $\Delta G$ as the direct combustion. In other words, $\Delta G$ won't be able to tell you whether you have done the combustion directly or done it laboriously via gaseous atoms, it just tells you what is the difference between the two states, first state being glucose + oxygen, second state being carbon dioxide + water. $\Delta G$ for glucose that you have eaten and combined with oxygen you breathe in and breathe out carbon dioxide + water would still the same $\Delta G$ as for combustion, even though now it is all done through cells, Kreb's cycle and a lot of biochemistry.

How would something be dependent on the path?
As you can see from my example above, the amount of effort you need to make $$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$ happen can differ a lot, and for each of the cases, what you "get back" from doing that reaction is different, such as heat released when burning glucose vs keeping your cellular biochemistry going with respiring glucose. In physics/physical chemistry, this is formally "work done". Work is dependent on path.

I am a little unsure whether you went from

$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T \tag1$$

to

$$\left(\frac {\partial G}{\partial p}\right)_T=V \tag2$$

in the right way, so I am going to go through it again.

At constant $T$, $\mathrm{d}T=0$. So $(1)$ becomes

$$ \mathrm{d}G=V\mathrm{d}p \tag3$$

which after rearranging is equivalent to $(2)$.

The integration then should go from $(3)$ to give

$$ \int_{p_1}^{p_2}\mathrm{d}G=\int_{p_1}^{p_2}V\mathrm{d}p \tag4$$

The reason why what you wrote look a little weird

  $$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$

is that you usually don't just cancel partial differentials like you did. Whereas $(3)$ is an exact differential, which you integrate over in $(4)$, and that is OK. Unfortunately I am not sure whether what you wrote is wrong, it is just something I would not write myself.

The rest of the derivation is fine.

Final question. I believe you are fine with mathematically going from $\int_{p_1}^{p_2}\mathrm{d}G$ to $G(p_2)-G(p_1)$, however, if you think about this result physically, $\Delta G$ does not depend on any other $p_3, p_4, p_5...p_n$ that the system might pass through on the way from $p_1$ to $p_2$. If $G$ is not a state function, then you would expect to have to consider at the very least all these different points of $p$ that you might have gone through, and different ways of doing the reaction passing through different $p$ would change $G$ even if the initial and final $p$ are the same. In that case, an integral like what we have written here would be a very poor way of finding $\Delta G$ because we are missing the contribution from all the intervening $p_n$, and we would have to use some different mathematical tools. In fact, I believe we would have failed to derive the master equation $(1)$ much earlier on, so the rest of the derivation would simply not follow.


When is there hybridization and when is there no hybridization? When is there hybridization and when is there no hybridization? Do all structures have hybridization or only some specific structures have it and why?

As Ivan already mentioned, hybridisation is not a physical phenomenon. Physically, the most accepted theory at the moment is that you have nuclei and the electrons roam freely in between them, forming standing waves (orbitals). These standing waves generally spread across the entire molecule unless it is a standing wave of a core electron.

However, it is non-trivial for us chemists to come up with a proper description of these molecular orbitals. Hence, we use a certain set of assumptions. For example, the LCAO approach for molecular orbitals used in some calculations assumes a set of hydrogen-like orbitals on every atom and linear combines them to create the most stable set of molecular orbitals. Since that is still rather hard to do, chemists have also derived a pen-and-paper only approach that does this on a per-atom basis (called hybridisation) and then checks what kind of orbitals would form which bonds — so that’s two steps of linear combinations following each other.

But all of this is only our thought experiment to better describe the physical picture of a molecule. Nature does not need hybridisation or linear combinations. Nature just ‘does it’ and comes up with a final solution. So whenever hybridisation ‘is involved’, it is always man thinking about how to look at the molecule.

Once you have taken that for granted, the question may remain:

  But when do I need to take hybridisation into account for orbital considerations?

There’s a set of simple rules:

Elements of the third and higher periods will remain mostly unhybridised unless they require hybridisation for any specific issue ($\ce{SiH4}$ cannot be explained without $\mathrm{sp^3}$ hybridisation on silicon).
Halogens and noble gases will never be hybridised unless they are hepta-/ octacoordinated.
Generally assume a hybridisation for all elements of the second period. Use the one that best explains the geometry. For oxygen, assume at least one unhybridised p-orbital.
For a nitrogen with three single bonds, remember that the hybridisation will always switch between $\mathrm{sp^3}$ and $\mathrm{sp^2 + p}$. This is the nitrogen inversion.
Hydrogen needs no hybridisation.


In which pH range are silica-based TLC plates stable? I am trying to understand why the experiment asks for raising the pH of a fairly acidic solution. The analytes should not be affected by it, so I thought the silica plate might be. Are silica-plates vulnerable in acidic conditions and, if yes, what happens?

Assuming "normal" TLC with silica as stationary phase, and a mix of ethyl acetate / hexane as eluent, charged compounds (i.e., salts) do not elute well; their adsorption is too tight to move forward.

Hence, if your reaction is performed under acidic conditions (sulfonation reaction, for example), drop a small sample of your reaction into a test tube or small vial, gently quench it against an aqeuous solution of $\ce{Na2CO3}$, extract it with some ether / MTBE / methylene chloride (well...) and spot the extract on the TLC plate.  With some training, this works easily around o.5 mL of total volume.

Conversely, if your reaction proceeds under basic conditions (like a Grignard reaction, for example), bring the sample back to neutral, for example with dilute HCl (or, with more fragile products like tertiary alcohols, with an aqueous solution of $\ce{NH4Cl}$), and then spot the organic extract on the plate.

I referred to "normal" TLC as there are other stationary phases than silica, too.  Alumina (basic, neutral, acidic) works with the same principle as silica.  These two contrast to applications of RP-TLC, were the stationary phase is coated with alkyl chains, and much more polar solvents are deployed as eluents, hence with much lesser issues of free salts / free acids. But the price tag for these plates is higher than for normal phase TLC.


EDTA sedimentation problem? I tried to create a suspension of EDTA powder in Glycerin with 1:1 ratio. I keep facing problem of EDTA sedimentation. I've tried other ratios also but this was the only one which formed least amount of sediment.
When I used coarse powder, the sediment formed within 24 hours but when finely ground powder was used, the same occurred but after 3 days. I even tried heating the powder & glycerin together on low & medium heat but EDTA sediment still forms but with varied time.
I'd like that the suspension remain consistent irrespective of how much time goes by.
I read on the internet that anti-settling agents (like xanthan gum & microfibrillated cellulose) can be used but I haven't tried that yet.
Can anyone please help & guide me?

I see 4 possibilties to avoid EDTA sedimentation:

Mechanical ways like shaking, mixing or container turning - but you say it is not applicable.

Increasing viscosity by thickening glycerol to  decrease the rate of sedimentation. Gel stages could probably even stop the sedimentation, but may not be applicable. You may want to try

Carboxymethyl cellulose (CMC)
Polyvinyl alcohol(PVA)
Polyvinylpyrrolidone/polyvidone/povidone (PVP)
Polyethylene glycol(PEG)

Other way would have been adjusting the glycerol density to match EDTA density if EDTA had not had too low density $\pu{0.86 g/cm^3}$ ( while glycerol  $\pu{1.26 g/cm^3}$). OTOH, if there was a chance to replace glycerol by ethanol 73 w/w %, having the same density as EDTA has ... but I guess  there is not.

It may be worthy to try some emulsifiers, but EDTA particles may be too large and emulsifiers may not be applicable for the suspension purpose.

Soy lecithin is another emulsifier and thickener
Pickering stabilization – uses particles under certain circumstances
Sodium phosphates - not directly an emulsifier, but modifies behavior of other molecules, e.g. caesin
Mono- and diglycerides - a common emulsifier found in many food products (coffee creamers, ice-creams, spreads, breads, cakes)
Sodium stearoyl lactylate
DATEM (diacetyl tartaric acid esters of mono- and diglycerides) – an emulsifier used primarily in baking
Simple cellulose – a particulate emulsifier derived from plant material using only water


Does pressure affect non-gaseous reactions? Pressure holds a great significance when it comes to reactions involving gaseous species. Changing pressure can affect the kinetics and even mechanisms of such reactions.
However, does pressure hold any significance in reactions involving solid and liquid reactants only?

Examples, if any, would be appreciated.

Yes. Pressure, as a concept, is simply the force exerted by random molecular motion per unit area, and since all matter is composed of molecules, and all matter above absolute zero has some random motion, everything has molecules exerting some force over some area. (This pressure is the vapor pressure of the solid, which is typically extremely low. If the external pressure on the solid is higher than the vapor pressure of the solid, the solid molecules will not fly off and escape. Things with high vapor pressure exert a high "pressure" on their surroundings and thus escape easily - jet fuel or nail polish remover have very high vapor pressures.)

If you have a liquid reaction between molecules A and B, bringing the liquid to a higher pressure will increase the frequency of collisions between molecules of A and B (by the definition of pressure). Since there is a finite probability of the reaction proceeding for any given collision, increasing the collisions increases the rates.

Practically, for you to see that kind of effect in a laboratory would require a very pressure-sensitive reaction, large molecules whose shape changes with pressure, or a reaction with a large change in the number of moles. Biological chemistry is rich with examples pressure-sensitive non-gaseous reactions, mainly due to pressure changing the shape and behavior of large complex molecules.

There's also the effect of osmotic pressure, which is a pressure induced not by random molecular motion but by the tendency of the substance to seek a lower free energy state. Osmotic pressure will affect concentrations (the system will seek to reduce the concentrations of ions), which will affect reaction rates.

Increasing the pressure of solids and liquids can also change their physical properties (heat capacity and thermal conductivity are two excellent examples). These physical properties are innately tied to the reaction mechanism. Since most reactions occur in sequence, if the physical properties are different (and if they change over the course of the reaction), pressure will exert an effect there.

Finally, there is an effect technically excluded by your question, but important nonetheless, and that is the effect of pressure on solid-gas reactions (sublimation or deposition). The manufacture of integrated circuits involves depositing materials onto chips in layers, and pressure is an important way to control that rate.

In general, the more nonlinear the reaction mechanism, the more likely it is that the small effects that pressure creates will cascade into larger changes that are what you would call "significant."


Removal of oxygen from pyridine N-oxide Under what condition do the reaction proceed with sole removal of oxygen in contrast with the product also ending up with chlorine substituent?

I'm familiar with both reaction mechanism, but can't seem to figure out when will a certain pyridine end up with or without chlorine. I guess one factor would be electron richness of the ring system, chlorine being deactivating subsituent ends up only on the electron rich pyridines.

I'm not sure whether $\ce{PCl3}$ can lead to the installation of a chlorine atom at C-2. I performed a cursory Reaxys search and in the first page saw several examples of deoxygenation without chlorination, but no examples of concomitant chlorination.

As far as I know, treatment of pyridine N-oxides with $\ce{PR3}$ simply leads to deoxygenation, with $\ce{P(III)}$ being oxidised to $\ce{P(V)}$.

However, if you were to treat pyridine N-oxide with phosphorus oxychloride $\ce{POCl3}$, then you would see chlorination at C-2.

If you have a source that says that $\ce{PCl3}$ can deoxygenate and chlorinate, please do inform us. Otherwise as far as I can tell, you need $\ce{POCl3}$ to effect a chlorination.

References

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010; pp 121, 155.
Weickgenannt, A. Pyridine N-Oxides, Baran Group Meeting, June 9, 2012; p 3.


Why does diazo coupling preferentially occur at the para position? When the diazonium ion couples with dimethylaniline, why is the para position favored?

Is this only due to steric factors or do electronic factors (such as ortho-para directivity) play a role here too? In particular, I suspect ortho-para electronic factors might be significant because some of the resonance structures have a cation directly next to the electron-donating group.

Both steric and electronic factors play a role in this reaction.  Resonance structures such as those shown below favor electrophilic attack (the diazonium ion is an electrophile) at the ortho and para positions. Since there are twice as many ortho positions as para positions in N,N-dimethylaniline, on a purely statistical basis we might expect a 67:33 ortho/para product distribution if only electronic effects controlled the reaction.  Since diazonium coupling usually produces a majority of the para coupled product, steric factors must be playing a role as well.  The somewhat bulky dimethylamino group does make it more difficult for an electrophile to approach the ortho position and the diazonium ion is relatively large.  These two steric factors combine to reduce the amount of the ortho isomer and the para isomer predominates.


How can I make a black precipitate without a sulfide? I'm supposed to make a black precipitate using only the following chemicals: copper sulfate(II), lead nitrate, acetic acid, barium chloride, sodium hydroxide, copper(II) carbonate, methanol, potassium permanganate, hydrochloric acid, hydrogen peroxide, magnesium sulfate, and potassium dichromate.

I know that solid sulfides (used in qualitative analysis) are typically black in colour, such as copper(II) sulfide, but I don't have any source of the sulfide ion here. Is it actually possible to make a black precipitate?

Hint:  Manganese dioxide is pretty black.

Look up potassium permanganate and note its reactions especially the reduction in neutral solutions.