How can the ground state electron configuration for a d3 complex correspond to an F term symbol? The ground-state term symbol from a $\mathrm{d^3}$ Tanabe-Sugano diagram is $\mathrm{^4F}$. My question is how the total orbital quantum number $\Lambda=3$, or $\mathrm{F}$ term, arises.

For a $\mathrm{d^3}$ metal, I'd expect the following ground state d-electron configuration:

where notably the $\mathrm{t_{2g}}$ orbitals correspond to $\mathrm d_{xy}$, $\mathrm d_{xz}$, and $\mathrm d_{yz}$ suborbitals.

From the spherical harmonics, $\mathrm d_{xy}$ results from a linear combination of:

$$\begin{align}
Y_2^{-2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-2\mathrm{i}\varphi}\cdot \sin^2\theta \\
Y_2^{+2}(\theta,\varphi) &= \frac{1}{4}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{2\mathrm{i}\varphi}\cdot \sin^2\theta
\end{align}$$

i.e. $m_l=\pm2$ equations.

$\mathrm d_{xz}$ and $\mathrm d_{yz}$ result from:

$$\begin{align}
Y_2^{-1}(\theta,\varphi) &= \frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{-\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta \\
Y_2^{1}(\theta,\varphi) &= -\frac{1}{2}\sqrt{\frac{15}{2\pi}}\cdot \mathrm{e}^{\mathrm{i}\varphi} \cdot \sin\theta \cdot \cos\theta
\end{align}$$

i.e. $m_l=\pm1$ equations.

From this assumption–which I may very well have presented incorrectly – the $\mathrm{t_{2g}}$ orbitals would therefore correspond to $m_l$ values of $\pm1$ and either $+2$ or $-2$. This would mean the greatest $\Lambda$ from one electron in each $\mathrm{t_{2g}}$ orbital (a $\mathrm d^3$ ground-state electron configuration), would be $\mathrm d_{xy} + \mathrm d_{xz} + \mathrm d_{yz} = \pm2+1-1=\pm2$, or a $\mathrm{D}$ term symbol.

Would anyone be able to correct me where my logic went wrong? I have a feeling I'm not allowed to restrict the $\mathrm{t_{2g}}$ orbitals to those $m_l$ values, but why would that not be allowed, when those are the equations that derive the $\mathrm{t_{2g}}$ d-orbitals?

Thank you!

Free ion

The ground-state term symbol is only $\mathrm{^4F}$ in the case of a free ion. If you take a closer look at the Tanabe-Sugano diagram, the $\mathrm{^4F}$ term only appears at the far left-hand side of the diagram, where $\Delta = 0$. $\Delta$ refers to the ligand-field splitting parameter, and $\Delta = 0$ indicates that there is no ligand field, i.e. a free ion.

The quantum number $L$ (total orbital angular momentum) of the ground state can be obtained by coupling the individual orbital angular momenta of the d electrons using a Clebsch-Gordan series. The way to do this is described in most physical chemistry textbooks under the Russell-Saunders coupling scheme. For example in Atkins 10th ed. it is in page 386 under the chapter on "Atomic structure and spectra".

(Note that the symbol $\Lambda$ is used for diatomic molecules, not atoms.)

$L$ is said to be a "good" quantum number, in that the operator $\hat{L}^2$ (almost - this neglects spin-orbit coupling) commutes with the Hamiltonian $\hat{H}$. Quantum mechanically this means that the $\hat{H}$ and $\hat{L}^2$ (almost) share a set of eigenstates, and therefore for every state of the Hamiltonian (which correspond to the electronic configurations we are familiar with) one can (almost) calculate the corresponding value of $L$.

$$\hat{L}^2|\psi\rangle = L(L+1)\hbar^2|\psi\rangle$$

Octahedral complex

The ground state term symbol of a $(\mathrm{t_{2g}})^3$ ion is $\mathrm{^4\!A_2}$, not $\mathrm{^4F}$!

The $\mathrm{t_{2g}}$ set comprises the $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{xy}$ orbitals. These three d orbitals are what we call "real" spherical harmonics, which are linear combinations of the complex spherical harmonics which you have quoted. As such it is not possible to assign $m_l$ values like you have to the $\mathrm{t_{2g}}$ orbitals.

It's not correct to say that $\mathrm{d}_{xy}$ can have "either $m_l = +2$ or $-2$". That would mean that $\mathrm{d}_{xy}$, at any one point in time, is either equal to $Y_2^{+2}$ or equal to $Y_2^{-2}$, which makes no sense. It is not flip-flopping between the two spherical harmonics, it is its own thing: a linear combination of the two spherical harmonics, or a superposition, if you prefer that word. Furthermore, the spherical harmonics only have significance under spherical symmetry, where they act as simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$. In octahedral symmetry, the spherical harmonics aren't of any significance at all and trying to "resolve" the $\mathrm{t_{2g}}$ orbitals into their components as spherical harmonics is physically unmeaningful (it helps with the maths, but that's all).

Under octahedral symmetry, the total orbital angular momentum $L$ is no longer a good quantum number (i.e. $\hat{L}$ no longer commutes with the Hamiltonian) and therefore the term symbol doesn't say anything about it!


Mechanism of styrene dimerization I am struggling to find a mechanism for the above reaction.

So far I have suggested that the proton adds to one of the reacting aromatic species leaving the resonance stabilized secondary cation. This then adds to the ortho position on the second reactant (since alkyl groups are activating). I then get a positively charged intermediate.

For a simpler reaction I would simply remove the proton but I wonder whether the reaction from this step to the product is concerted? I couldn't see an obvious step so I deprotonated and again could find no obvious way to continue apart from a rather elaborate proton transfer from a carbon that does not possess the most acidic proton.

You're right that the first step involves protonation of the terminal carbon in the  styrene double to yield a resonance stabilized benzyl carbocation.  This carbocation then adds to another styrene molecule in the exact same fashion, reacting at the terminus of the styrene double bond to again yield a resonance stabilized benzyl carbocation.  This latter carbocation then alkylates itself (like a Friedel-Crafts alkylation).  Loss of a proton from this final carbocation yields the indicated product.


Ordinary chemical means What is the meaning of ordinary chemical means in the following:

  An element may be defined as a pure substance that can not be
  decomposed by ordinary chemical means.

Essentially it means that you exclude the possibility of splitting an element even further via nuclear means.


How fast does the diamond lose the atoms at sunlight? recently I've read that the diamonds lose their atoms at sunlight on the surface. Would you be able to explain me how fast is the rate of this process? Does the diamond lose the atoms each time when is exposured at sunlgiht? Thank you. :)

Thats true, but even an UV lamp would take 10 bilion years to remove a microgram (one-million part of a gram). It is caused by carbon atoms oxidizing, forming CO2. This process is accelerated by the presence of UV light.


Why is it important that glucose’s third OH group points to the left? My question pertains to the location of the OH group on the 3rd carbon in a glucose molecule. If atoms can rotate freely around single bonds, then why is this important? Why can't glucose be drawn with 'all H on the left'?

This is the concept called stereochemistry.

Each carbon with 4 bonds has an approximately tetrahedral geometry.  We assign a configuration to the situation by numbering the substituents in terms of priority. Hydrogen is lowest.  Oxygen is highest.  The top carbon is second, and the other carbon is third.  We then trace a route from 1 to 2 to 3.  This would be a clockwise circle, and we would call this configuration S.  If we switched sides between the H and the O we would follow this circle widdershins and call the configuration R.  So the substituents can't be exchanged because the configuration would change.

So what?

When there is only one of these tetrahedral centers (called stereocenters) in play for a situation, it does not matter at all which one you have.  But the moment a second stereocenter comes into the picture, chemical properties begin to change.  A fairly tractable analogy is feet and shoes.  When you aren't wearing shoes, your left foot and your right foot are about the same.  But when you try to put shoes on, putting the right shoe on the left foot is slightly more difficult than putting the right shoe on the right foot.  In this loose analogy, the 'reaction' of putting your shoes on is slowed based on how the stereocenters of foot and shoe interact.

Onto glucose ...  Because there are 4 stereocenters in glucose, it has a distinct shape from its isomers.  The other chemicals in your body also have their own shapes/configurations.  Glucose reacts slightly differently than other sugars in the body because of the way these shapes fit together.  Therefore you need to keep the configurations known and well marked.

Most people (about 94% of the rest of you) don't intuitively see stereochemistry concepts and need to make models.  However, pragmatically, if you can take this explanation, and remember that 'it just matters' to keep the notation consistent, you can probably get through most of biochemistry.

If you want to make models, at your next wine and cheese party get some toothpicks, and two pieces each of four different kinds of cheese.  Put four of the cheese chunks on the end of the toothpicks, and then stick the tooth picks in a sausage or something.  Then make an exact copy.  After that is done, exchange two of the cheese chunks on one model and investigate what you have.  There should be no way you can twist or turn or rotate one model to have its kinds of cheeses in the same orientation as the other.  Once you have done this, the following picture will make much more sense:


Electron transfer in reaction between ammonia and hydrochloric acid $$\ce{NH_3 + HCl -> NH_4Cl}$$

The electronic structure of ammonia is such so that the nitrogen atom has an outer shell with a lone pair of electrons.

When ammonia reacts with hydrochloric acid (above), the hydrogen from the $\ce{HCl}$ moves to the ammonia. However, it leaves its electrons behind because there is already a lone pair of 2 electrons on the $\ce{N}$ atom, allowing it to retain a $\mathrm{1s^2}$ shell.

Why doesn't the hydrogen take its electron over and nitrogen doesn't then "donate" an electron to the chlorine?

Chlorine is the more electronegative species. It would thus retain the bonding electrons when the hydrogen is evicted. This hydrogen then forms a coordinate bond with the lone pair on nitrogen.


Which carbons in pyrrole have the highest electron density? In pyrrole, which carbon atoms have the greatest electron density?

In the resonance structures of pyrrole, all carbons can formally bear a negative charge, so it's not immediately obvious which one is most electron-rich:

My book claims that because resonance forms 1c and 1d are greater contributors to the resonance hybrid than 1b and 1e, the α-carbons (which bear the negative formal charge in 1c and 1d) are more electron-rich. Is this really the case?

Interesting question, more subtle than I realized!

Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must also consider the inductive effect of the nitrogen atom, which draws some electron density away from C2 and C5 through the sigma bonds. This would imply that the two proximal carbons (closest to N) are more electron-deficient, and the distal (more distant) pair of carbons are more electron-rich.

Experimental considerations: Orthocresol has a good point that pyrroles are observed to form bonds with electrophiles at the proximal carbons (2/5). However, I think this trend may be due to the superior stability of the intermediate produced by that pathway (3 significant resonance forms) versus a distal (3/4) attack (2 significant forms). The cation from a proximal attack is better stabilized by resonance than the distal-attack cation. I expect that pyrrole forming C2 bonds with electrophiles has more to do with the stability of the intermediate of that pathway than necessarily greater electron density on C2. What I did not expect, but read on Wikipedia, is this:

  Unlike furan and thiophene, [pyrrole] has a dipole in which the positive end lies on the side of the heteroatom, with a dipole moment of $1.58\ \mathrm{D}$.

This is experimental evidence demonstrating that the electron density is lower on the proximal side (C2,5) and greater on the distal side (C3,4). Besides molecular modeling of the electron density clouds, I think the molecular dipole moment is about the most conclusive evidence one can hope for.

It really comes down to the context: in an undergrad first-semester course, you might be expected to say the distal carbons have more density because the nitrogen withdraws from the proximal carbons. In more advanced courses, you might take orthocresol's perspective and base your answer in reactivity, concluding the proximal have more electron density. But I expect a doctorate researcher would probably conduct molecular modeling and compute the electron distributions using quantum mechanics, and perhaps conduct a few experiments to measure the dipole. And at that level, I think C3 and C4 have the honey.


Degree of unsaturation of naphthalene Double Bond Equivalent/Degree of Unsaturation

  25. Degree of unsaturation in

  is

  (a) 6; (b) 7; (c) 8.

There are two rings and six double bonds in structure so degree of unsaturation should be 8, but it is not. Why?

You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct.
You can redraw the structure of naphthalene as such:

and make sure that there are five double bonds and two rings giving degree of unsaturation

$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$

You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there:

$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$

You again arrive at $\text{DU} = 7$ using a simplified formula:

$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$


What happens to the third hydroxyl group in a phosphodiester linkage? A phosphodiester bond occurs when exactly two of the hydroxyl groups in phosphoric acid react with hydroxyl groups on other molecules to form two ester bonds.

But what happens to the third hydroxyl group; i.e. why is it an oxygen and not hydroxyl?

At physiological pH it is simply deprotonated, i.e. it becomes $\ce{-O-}$ instead of $\ce{-OH}$. That's why DNA is said to have a negative charge.

I realise that this image on Wikipedia omits the negative charge, which is strictly wrong (the negative charge is probably implied because writing it out every time would clutter the diagram). A proper biochemistry textbook will show that it is deprotonated. Here's an image from Berg & Stryer's Biochemistry, 7th ed:

Here's one from Alberts et al. Molecular Biology of the Cell, 6th ed:

Whether you draw the negative charge as being delocalised across the two oxygens (as Berg & Stryer have done) or being localised on one oxygen (as it is in MBotC) is not relevant.


what is the physical significance of ∆H=∆U for a chemical reaction? In case of few chemical reactions $\Delta n=0$,so according to the equation $\Delta H= \Delta U+\Delta nRT$ change in enthalpy equals change in internal energy. But what does this in essence actually mean? Like physically what does this indicate and do we get any additional information from that an equation? I got this query seeing the following equation of formation of hydrogen chloride.

$\ce{H2 +Cl2 =2HCl}$ with $\Delta n=0$.

Like can it be associated to why the reaction is zero order?

This equation holds for a reaction involving ideal gases at constant T. Under these conditions, $\Delta (pV) = \Delta n RT$. Assume now that a gas phase reaction occurs at constant pressure and temperature. Then  $\Delta (pV) = p\Delta V$. This is equal to the negative of the pressure-volume work done by the system during the reaction since $w = -p\Delta V$. Therefore, at constant p and T, $\Delta H$ is equal to the change in energy of the system minus the pV work it did. This is consistent with the fact that we equate the enthalpy change for such a process at constant p and T with the heat exchanged, that is, $\Delta H = \Delta U -w = q_p$. Now if $\Delta n =0$ no expansion occurs at constant p and T, and so $w=0$ and $\Delta H = \Delta U = q$.


Do some particles get ionised to 2+ or higher in a mass spectrometer or do they only get ionised to 1+? In a mass spectrometer the sample is vapourised, then ionised, then accelerated.
My textbook says that the high energy electrons from the electron gun knock electrons off the sample molecules and hence ionise the particles by 1+.
But do some particles get 2 or more electrons knocked out or is it always just one ?
I suppose once it has been ionised by 1+ it gets accelerated through the tube so it won't hang around for long but I would think that there is still a small window of time for further ionisations.

In electron impact, you would see mostly +1 charged ions. Electron impact is a "harsh" way of ionization. However, +1 charge is not a universal rule in MS. The state of ionization depends on the ionization source. Later in your courses you might encounter electrospray ionization. There you might have +14 charges, etc.


Is common ion effect applicable on two strong electrolytes Consider the nitration of benzene using mixed conc. $\ce{H2SO4}$
and $\ce{HNO3}$.If a large amount of $\ce{KHSO4}$ is added to the
mixture the rate of nitration will be :

Faster
Slower
Unchanged
doubled

The answer given in my textbook is option 2 , i.e. the reaction will become slow. my teacher told me that it is due to common ion effect.
But in common ion effect, a strong electrolyte decrease the dissociation of weak electrolyte. So,how can common ion effect possibly be the reason here($\ce{KHSO4}$ and $\ce{H2SO4}$ both are strong electrolytes). In my view the answer should be no effect and I'd love to have my concepts about Ionic Equilibrium be corrected if I'm wrong.

Certainly not.

Suppression of dissociation of a strong electrolyte is not a very rational thing to discuss, owing to the unmeasurably large value of $K$.

To see what I mean, try looking up the $K_{a1}$ of $\ce{H2SO4}$.
(At least I couldn't find it!)

Your textbook's given answer is correct.

The equilibrium suppressed by the introduction of $\ce{KHSO4}$ is not the dissociation of $\ce{H2SO4}$, but rather this one:

$$
\ce{HNO3} \ + \  \ce{H2SO4} \rightleftharpoons \ce{NO2+} \ + \ \ce{HSO4-} \ + \ \ce{H2O}
$$

Nothing too auspicious: all that remains is the run-of-the-mill common ion effect we've known from Kindergarten days!

PS.
Conclusion: $\ce{KHSO4}$ sucessfully decreases the concentration of $\ce{NO2+}$, which happens to be the electrophile and a key participant in the rds(rate determining step) of Nitration of benzene.

$$
Rate=k[\ce{NO2+}][\ce{C6H6}]
$$


What is the mechanism of formation of alkali metal superoxide? I was studying about alkali metals. I can get how oxides and peroxides are formed.
In alkali metal oxides, both of the bonds of the dioxygen molecule first undergoes homolytic cleavage, then the nascent oxygen gets two electrons from two metal atoms.
In alkali metal peroxides, one bond of dioxygen molecule undergoes homolytic cleavage and this results in formation of O-O with one unpaired electron on each oxygen then 2 metal atoms donate their electrons to each of the oxygen.
But how superoxides are formed?

Getting to know the superoxide anion radical
Hayyan et al. (2016) have studied and characterized the formation of the superoxide anion radical, $\ce{O2^.-}$. This chemical entity forms as a result of the reduction of diatomic oxygen
$\ce{O2 + e^- -> O2^.-}$
There is some conflicting data on the reactivity of this radical, however it can be isolated:

A few studies have reported that $\ce{O2^.-}$ is relatively reactive. In contrast, a previous study showed that $\ce{O2^.-}$ or hyperoxide, as some scientists name it, is not highly reactive despite being a free radical. The 17-electron $\ce{O2^.-}$ is one of the isolable paramagnetic main-group ions and one of the many intermediate ORR species.

This chemical species can be synthesized in a myriad of ways. It is actually an intermediary in the mitochondrial respiratory chain and it plays a really important role in the activity of respiratory and immune defense systems.

Superoxide can be generated not only enzimatically (as it has been previosuly mentioned), but also electrochemically, photochemically and photocatalitically. Usually, it is either hydrogen peroxide or the peroxy radical used as target-molecules for the synthesis of this species.
Carrying out the synthesis of superoxide photochemically - a mechanism based on solvated electrons
As mentioned earlier, crossing the boundary between diatomic oxygen and superoxide can be done in mutliple ways. Photochemically, this leads to the formation of solvated electrons in the reaction mixture. One electron can attack one oxygen molecule, forming superoxide. Holroyd et al. have concluded in 1978 that photochemical composition of water is the main process leading to the formation of $\ce{e^-(H2O)n}$. The hydroxyl anion $\ce{OH-}$ that forms after the primary dissociation can absorb energy and disproportionate into hydroxyl radical $\ce{OH^.}$ and a solvated electron.
Chemically generating the species
Superoxides of alkali metals have been heavily documented, as well as their synthetic pathways. These alkali cations (such as potassium, sodium and even rubidium and cesium are referred to as superoxide carriers in this context). Zhdanov et al. (2004) have studied the kinetics of synthesis of potassium superoxide. The chemical environment in which this reaction was carried out is alkaline solution of hydrogen peroxide.
$\ce{H2O2 + HO- -> HO2- + H2O}$
Apart from the hydroperoxide species (which involves proton exchange between hydrogen peroxide and the hydroxide species), there has been observed an exchange leading to the formation of an intermediate adduct:

Being a weak acid, hydrogen peroxide enters, under certain conditions, into an exchange reaction with potassium hydroxide to give an intermediate adduct $\ce{K2O2*H2O}$ (potassium peroxide diperoxohydrate).

$\ce{KOH + 3H2O2 -> K2O2*2H2O2 + 2H2O}$
It is this intermediary that disproprotionates upon mild heating into the superoxide species:
$\ce{K2O2*2H2O2 ->[40^o C] 2KO2 + 2H2O}$
This is the most widely used method for obtaining these salts. It is different from the photochemical/electrochemical methods, since the superoxide radical is only generated in situ (if, at all).

Recently, an in situ method was proposed for preparing superoxides by integrating $\ce{H2O2}$ with aqueous $\ce{KOH}$ or sodium hydroxide ($\ce{NaOH}$). $\ce{NaO2}$ was generated by mixing $\ce{NaOH}$ and $\ce{H2O2}$ in aqueous media under ambient conditions (eq 3). $\ce{NaO2}$ is generated by the rapid formation of alkali peroxide, followed by its decomposition in excess $\ce{H2O2}$. However, this method did not provide alkali superoxides in a powder form. By contrast, it was stated that, in this in situ method, $\ce{O2^.-}$ can be generated in solutions for diverse applications.

Conclusions
Two different mechanistic pathways are known for the synthesis of superoxide-based alkali metal salts. One of them involves directly the $\ce{O2^.-}$ species (obtained electrochemically, enzimatically or photochemically) which gets to react with the alkali metal's exposed surface.

Sodium metal, coated in a mixture of superoxide, peroxide, oxide, hydroxide and possibly carbonates
The other mechanism involves the disproportionation of the alkali peroxide-hydrogen peroxide adduct, freeing the superoxide species in the reaction mixture (apologies, but I have found no concrete mechanistic details for this rather bizarre disproportionation).
References

https://en.wikipedia.org/wiki/Sodium_superoxide
Hayyan m., Hashim M.A., Al Nashef I.M., Superoxide Ion: Generation and Chemical Implications, Chem. Rev., 2016, 116 (5), pp 3029–3085.
Sawyer, D. T. Superoxide Chemistry, McGraw-Hill,
D. V. Zhdanov, M. A. Ul’yanova & Yu. A. Ferapontov, A Study of the Kinetics of Synthesis of Potassium Superoxide from an Alkaline Solution of Hydrogen Peroxide, Russian Journal of Applied Chemistry
Richard A. Holroyd and Benon H. J. Bielski, Photochemical Generation of Superoxide Radicals in Aqueous Solutions, Contribution from the Chemistry Department, Brookhaven National Laboratory, Upton, New York 11 973, Received March 9, I978


What's the relationship between supersaturation and temperature? In gravimetric analysis, what does supersaturation mean? And, how is it achieved by lowering the temperature, even though increasing the temperature makes a higher number of solute particles more soluble?

Supersaturation means simply that more of the substance is dissolved than should be. Supersaturation is an unstable condition. Given some "trigger" precipitation will occur and equilibrium conditions will then be restored.

The trigger could be a seed crystal, scratching a glass rod on a side wall or a piece of dust which acts as a nucleation point.


Question about the role of d-orbitals in polarisation I know that, although many elementary books say that d-orbitals are involved in the bonding of hypervalent main group compounds period 3 or higher such as the sulfate ion but in reality they are only barely, if at all, involved in bonding.

The question is- is it true that "the d-orbitals are not directly involved in bonding but rather in polarisation" and what does that exactly mean? I saw that statement on the comment sections of a post of mine but don't know the proof.

Well, please be more specific about the molecule(s) in question for that quotation.  d-orbitals are certainly involved in some bonding, especially for the transition metals, but you are right that the participation of d-orbitals in the bonding of main group covalent molecules and polyatomic ions has been overstated in the past and is in fact fairly minimal.

As you mentioned, there is very little, if any, d-orbital participation in the bonding of sulfate.  As for d-orbitals being involved in polarization but not bonding, that literally does not make sense for something like sulfate.  Sulfur's d-orbitals are empty, so it is impossible for them to polarize without bonding.  Only when there are occupied d-orbitals (as in transition metals) can non-bonding d-orbitals affect things like the shape and polarity of the compound (as in crystal field theory/LFT).

So, long story short, I think that comment you read is simply wrong, and your original understanding is correct.  See the below references if you want to read more about it.

References

Reed, A. E.; Schleyer, P. v. R. Chemical Bonding in Hypervalent Molecules. The Dominance of Ionic Bonding and Negative Hyperconjugation over d-Orbital Participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. https://doi.org/10.1021/ja00160a022.
Cunningham, T. P.; Cooper, D. L.; Gerratt, J.; Karadakov, P. B.; Raimondi, M. Chemical Bonding in Oxofluorides of Hypercoordinate Sulfur. Faraday Trans. 1997, 93 (13), 2247–2254. https://doi.org/10.1039/a700708f.


Reaction between 1-methylcyclopentanol and potassium permanganate In my basic organic chemistry class, I learned that only primary and secondary alcohols undergo oxidation (using $\ce{KMnO4}$).

However, what is the outcome of reacting tertiary alcohol (1-methylcyclopentanol) with a strong oxidant like $\ce{KMnO4}$ with no heat?

P.S. I am specifically interested to know what happens with no heat, because according to my minimal knowledge I know if heat was added then it is dehydration.

There is no reaction.
t-BuOH may be used as a solvent for permanganate reactions. For example, see Abiko et al. [1].

References

Abiko, A.; Roberts, J. C.; Takemasa, T.; Masamune, S. $\ce{KMnO4}$ Revisited: Oxidation of Aldehydes to Carboxylic Acids in the Tert-Butyl Alcohol - Aqueous $\ce{NaH2PO4}$ System. Tetrahedron Letters 1986, 27 (38), 4537–4540. https://doi.org/10/c3t9xj.


Phenol preparation from Grignard reagent Reaction: Phenyl magnesium bromide in presence of excess oxygen, on acidification yields phenol, as below.

My instructor told me that the above reaction occurs, and if the reagent is in excess then phenol is the end product. However, I'm not able to figure out the mechanism.

I've tried and got stuck at this point:

I wish to know the complete mechanism of this reaction.

As requested by the OP - the route that lab chemists use to form phenols from aryl halides (not aryl fluorides) is via formation of the aryl boronic acid or boronate. These may be formed by either metallation of the aryl halides to form the aryl lithium or Grignard and reaction with a trialkyl borate $\ce{B(OR)3}$, or by Pd catalysed reaction with pinacol borane (Suzuki-Miyaura reaction). These boronates or boronic acids may be isolated, but in practice an oxidative workup, most often with basic peroxide, yields the phenol in good to excellent yield.

(read more links: 1, 2, 3)

image from US7678158B2.


What type of bond(s) are present in the borohydride ion? In the $\ce{BH4-}$ ion, what bonds are present between boron and hydrogen?

Are there three covalent bonds between boron and three hydrogen atoms (using the boron 2s, 2px, and 2py orbitals), with the fourth hydrogen being a hydride ion $\ce{H-}$ which forms a coordinate or dative bond to the boron 2pz orbital?

No, this is simply a bond between four hydrogen atoms ($\ce{H}$) and a boron anion ($\ce{B-}$). These bonds are identical to and indistinguishable from the other $\ce{B-H}$ bonds. They are all covalent bonds of length $1.21\:\mathrm{Å}$.

Your idea of $\ce{H-}$ completing $\ce{BH3}$'s $2\mathrm{p}_z$ orbital isn't quite right either. Atom's individually have atomic orbitals, but when they form bonds these atomic orbitals combine to form molecular orbitals. This is picture more accurately demonstrates the bonding between $\ce{B-}$ and the $\ce{H}$ atoms.

[Photo Source]


Fixing incongruities in PES Scan I tried to parameterize a particular x-y-z-t dihedral angle of novel molecule using a Quantum Mechanical scan, but I got the strange result below.

The energy changes sharply around 130° and $E_\text{dihedral}(0°) \neq E_{dihedral}(360°)$.
I performed PES scan in gaussian09 and psi4 and I got the same results in different softwares.

I'm uncertain what approach to take to fix the scan. I saw some parameterization articles which didn't perform PES scan 0-360 degree. Other authors performed scan in the range of 0-60 degree or 0-120 degree and I didn't predict I will face a result like this. What is going wrong with my scan and how can I fix the issue? Would performing single point energy for each angle instead of PES scan help?

Even without seeing your molecule, I can say the referee is very likely correct. If you think of doing a scan of the dihedral of hydrogen peroxide, by symmetry the path from $\theta=180\to360$ should just be the reverse of $\theta=0\to180$. In your case, I don't know the symmetry around the dihedral, but at a bare minimum the energy should be the same after a $360$ degree rotation. The fact that it isn't suggests that some scan step (probably that point around $130$) irreparably distorted the molecule geometry so that additional steps didn't bring it back towards the initial geometry.

Generating the scan point by point may help, but you still have to ensure the rotations you are making are physical and aren't distorting the molecule too much. Sharp changes in the energy between steps are a sign that something is going wrong.

You don't mention this detail in your question, but its also important to consider whether what you want is a rigid scan (where you just change this angle) or a relaxed scan (where you change this angle and then allow the rest of the structure to relax).


How can I recognize aromatic rings in a connection table? Given a connection table (that is, a list of atoms, their connectivities/bonds, and the type of bonds), how can I detect the presence of an aromatic ring?

For reference, the aromatic rings in the structure will always be depicted in Kekule form (alternating single and double bonds).

It is simple enough for me to use a cycle finding algorithm on the graph representing the bonds to find candidates. The problem is now recognizing which of those rings can be aromatic.

A naive approach for at least six-membered carbocyclic rings would be to check if the ring bonds alternate between single and double bonds. But, the structures below would be problematic:

where the center ring in phenanthrene and the six-membered ring in 2H-indene are decidedly aromatic, even if there is no complete alternation of single and double bonds. If you try to check exocyclic double bonds, the leftmost structure could be a false positive.

Pointers to algorithms would be very much appreciated. I am ignoring heterocyclics like thiophene or imidazole, or cations like tropylium for now, but if you know how to handle them too, it would be a nice bonus.

I haven't gotten an adequate explanation of why people thought the question was "too broad"; thus, to restrict the question, here is a drastically short reformulation:

  Given a connection table for a molecule that contains any number of six-membered rings (possibly fused, spiro, etc.) and is represented in Kekule form, can one write an algorithm to determine which rings are aromatic and which are not? Atomic coordinates are available if necessary.

I will reiterate what I have written in a comment earlier, as of why I find this a bit problematic (especially for a cheminformatics approach).

There still is no clear cut definition of aromaticity, there even is an (at the time of writing) upcoming conference Aromaticity 2018 to get more clarity on the matter.

There are a couple of questions on the site, where I have written an answer for which demonstrate the issue. I don't want to be shamelessly promoting my own posts, but I also don't want to repeat myself. I think one of the crucial examples is How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures?

From examples like this, it seems clear that from structure and Kekule form alone it appears impossible to determine whether a ring has aromatic character or not. Mithoron stated in the comments something quite categorical, which I find in its extend debatable; but I wouldn't want to dismiss it either:

  I wouldn't say 2H-indene is aromatic, much like 1,2-benzoquinone also isn't.

Take the first as an example, it has a high C-H acidity (higher than indene), because the resulting structure would have a fully conjugated system very closely related to naphthalene. This structure is undoubtedly more aromatic (handwavy, but if you could quantify it, it probably would), so any kind of structural diagram, yes/no-parameter would misrepresent that.

  Given a connection table for a molecule that contains any number of six-membered rings (possibly fused, spiro, etc.) and is represented in Kekule form, can one write an algorithm to determine which rings are aromatic and which are not? Atomic coordinates are available if necessary.

I'm afraid the answer to this question is no.

What you could do, and where you'll probably be right many times is using the bond lengths as an indicator. The bonds in such rings should all be very, very similar (maybe ±10%), an given Pekka Pyykkö's data (pdf from his homepage) fall somewhere around 140 pm. Couple that with the ring-finder and I would assume it's going to be close many times.


On Electrochemical Cells and Salt Bridges We're studying the topic Electrochemistry at school now, and I've got a few doubts regarding electrochemical cells and salt bridges.

What I am going to do is listing out (point-wise) what I understood from class, and then pose my questions (in bold). The logic behind listing out what I understood about the topic beforehand? I've a very strong feeling I'm harboring some serious misconceptions regarding the topic, so if I list out what I do know, then hopefully someone can alert me to an incorrect point, and also correct it.

LHC - Left half cell (contains the anode, which is negatively charged in electrochemical cells)
RHC - Right half cell (Contains the cathode, which is positively charged in electrochemical cells)

During the course of the reaction,

In the LHC, metal atoms of the electrode release their electrons to the anode and enter the solution as positive ions.
The positive metal ions, under the influence of the negatively charged anode, accumulate around the anode, resulting in a Solid-Liquid Junction. The Solid-Liquid Junction is also known as the Helmholtz Double Layer.
Similarly in the RHC, negatively charged ions (formed from the electrolyte solution) accumulate around the cathode, resulting in another Solid-Liquid Junction.
It is the potential difference between the LHC and RHC Solid-Liquid Junctions potentials that gives rise to the emf which results in the flow of electrons (electric current) from the LHC to the RHC via the metal wire that connects their corresponding electrodes.
However, as the reaction proceeds, the quantity of positive and negative ions builds up (in the LHC and RHC respectively). So much, that these ions under their mutual attraction, override the force of attraction that binds them to the electrodes and accumulate at the porous asbestos partition that separates the two half cells. This is called the Liquid-Liquid Junction. The formation of the Liquid-Liquid Junction comes at the cost of the Solid-Liquid Junctions, which disintegrate.

As the Liquid-Liquid Junction potential doesn't involve the electrodes, it cannot set up a potential difference across them, therefore no current flows along the wire. This is why an electrochemical cell stops generating current after some time has elapsed.
This problem can be overcome by use of a salt bridge, which in effect, neutralizes the net charges on either side of the asbestos partition, thus releasing the ions back into the solution. These ions are once again attracted by the electrodes, and the two Solid-Liquid Junctions are re-established. So current once again begins flowing through the wire.

I'd appreciate it if anyone could proof-read those points. Also I have doubts whether the Solid-Liquid Junction can also be called the Helmholtz Double Layer.
I made these points in class, and I let my teacher go through them, and he's in agreement with them, so I don't think this is a case of talking down the points wrong, at least not with respect to what was mentioned in class.

Now I asked my teacher, Q1: Screw the partition. Why not simply have the two half cells in separate compartments kept at a distance from each other, instead of using a 'partial' barrier like the porous asbestos partition? Wouldn't that eliminate the possibility of a Liquid-Liquid Junction being formed?

His answer: Well they're ions buddy, so they'll attract each other anyway and accumulate at the sides of the compartment, after all....they're charged. So even though that wouldn't result in a Liquid-Liquid Junction, there still wouldn't be any potential difference across the electrodes, so nil current. Plus, if you separate the compartments like that, it'll just take up space;  compact electrochemical cells are more convenient to use.

Q2: Is my teacher entirely right here?

And yet again, Q3: Are all the points I've mentioned correct?

You have it all backwards. Liquid-liquid junction does not exist at all in your supposed meaning. Solid-liquid junction does, but it is very much not what you think it is, to the point that you might just as well assume it does not exist either.

The complete story would hardly fit in an answer (and, moreover, it is explained in any decent textbook on the subject), so I'll focus on the details that are especially wrong. First of all, the double electric layer, being electrically charged, creates an electric field of its own, and thus shields the rest of the solution from the electrodes and their field. Second, the electric forces are really, really strong (like a thousand times stronger than you seem to think, and even stronger than that), and therefore the said layer at any point contains but a very tiny portion of metal ions. That portion suffices for creating an electric field which would completely negate the field of electrodes. The vast majority of metal ions float freely inside the solution, having no idea at all that they are being electrolyzed.

As the ions build up, the shielding becomes ever more perfect. Dissolve some $\ce{NaCl}$ in a glass of water without any electrodes or anything; the ions would float freely in the solution. Do they feel mutual attraction? Well, in a way, they do. Do they assemble into strange walls across the glass? I've never seen anything like that. Now, the ions in you cell (the majority that does not participate in the double layer on either side) feel exactly the same way as ions in that glass.

Electrochemical cell, if set up properly, will give current until one of the electrodes is completely eaten up. The need for a salt bridge has absolutely nothing to do with electric effects. Rather, it serves to prevent the mixing of solutions, which are presumably different and better left that way, for fear that otherwise they might react with the electrodes directly.

As for the cell with separate spaces and no bridge, it will stop working right away, much like the light bulb connected with a single wire to one pole of a battery.


What is GMEC (Global Minimum Energy Conformation)? In a paper I have read the term GMEC, which was not further explained.
GMEC stands for Global Minimum Energy Conformation. But what is it?
Is it just the conformation with the lowest energy?

Yes. It is the conformer with the lowest energy out of all possible conformers.

There are also local minima which have a lower energy than conformations either side of the them on the energy diagram but do not have the lowest possible energy.

In this diagram showing the conformers of butane we can see that all three stable conformers; the two gauche conformers and the anti conformer are at local energy minima but the anti conformer is the global minimum.


Does CO2 form carbonic acid in the atmosphere? Since when $\ce{CO2}$ goes into water it makes carbonic acid, then what happens when it rains? Some of the $\ce{CO2}$ in the air joins the rain drop as it falls and while the rain drop falls and changes shape, it releases then regains $\ce{CO2}$? Is there a critical point or point of saturation within the raindrop for $\ce{CO2}$? It’s an interesting dynamic considering the complexity of storms and weather systems.

Short answer

It does form carbonic acid, as rain drops are water and carbon dioxide is soluble in water.
But the content of carbonic acid is minimal as you can guess from $\mathrm{pH=5.6}$

Longer answer

Majority of dissolved $\ce{CO2}$ does not form carbonic acid, 99.83% stays as hydrated oxide.

The equilibrium $\ce{CO2}$ concentration in water follows the Henry's law and is proportional to the $\ce{CO2(g)}$ partial pressure.

Therefore, as a drop falls, the partial pressure of $\ce{CO2(g)}$ grows together with the atmospheric pressure and  $\ce{CO2}$ solubility in water would increase as well, if temperature did not raise with decreasing altitude, typically $\mathrm{ 0.0065 K/m}$.
The total effect is general decreasing $\ce{CO2}$ solubility when a drop is falling.

Sure, some drops may evaporate during their fall ( seen as precipitation patterns without raining ) and the dissolved $\ce{CO2}$ becomes gas again.

Natural rain has $\mathrm{pH=5.6}$  because of dissolved $\ce{CO2}$ and equilibrium concentration of $\ce{H2CO3}$.

If all  CO2 converted to H2CO3, the natural rain and soda water would be much more acidic, because carbonic acid is quite strong acid with pKa=3.6, instead of formal 6.3 if 100% conversion is supposed.

In carst areas, there is ongoing reversible reaction of dissolving calcium carbonate by surface rain water going down the soil and precipitating it again in caves :

$$\ce{CaCO3(s) + H2O + CO2(aq) <=> Ca(HCO3)2}(aq)$$

The calcium bicarbonate comes also to surface water and makes frequently the majority of the calcium content.( and similarly for magnesium l


Di-alpha-halogenation of ketones in acidic medium I was studying alpha-halogenation of ketones from LibreTexts Chemistry,

  Under acidic conditions the reaction occurs through the formation of an enol which then reacts with the halogen${^1}$.

1) Protonation of the carbonyl

2) Enol formation

3) SN2 attack

4) Deprotonation

It also says (look at the highlighted text):

Overreaction during base promoted α halogenation

  The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used${^2}$.

Why can't we just reprotonate the monohalogen in the same way as the initial ketone to form a dihalo compound?

Referance

${^1}$:https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones

${^2}$:https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones

Acid Catalysed Halogenation of ketone

In the halogenation of ketones, both in acidic and basic medium, enols are reactive intermediates. After addition of first halogen to double bond, deprotanation occurs immediately (Scheme 1). The second halogen does not add$\ce{^1}$.

Scheme 1

The introduction of second Bromine is slower then the first ,since cabocation is destabilized by the electron-attracting polar effect of two Bromines (**Scheme 2).

Scheme 2

Consequently, reprotonation of the monobrominated ketone does not occur.

Base Catalysed Halogenation of ketone

The mechanism of this reaction involves the formation of an enolate ion as a reactive intermediate. After mono bromination, the enolate ion of the alpha-bromo ketone is even more stable the enolate ion of the starting ketone.(Scheme 3). Consequently, a second bromination occurs$\ce{^2}$.

Scheme 3

As discussed above, in acid catalysed halogenation  the intermediate is unstable ,hence reprotonation to give dihalo products is less (If one equivalent of Bromine is taken).

Note$\ce{^3}$ :

Comparatively, in base calalysis the intermediate (Bromo enolate anion) is quite stable leading to di and tri bromination (provided acidic H is present).

Reference

Organic Chemistry, 5th Edition  by Prof. Marc Loudon
Organic Chemistry, 5th Edition by Prof. Marc Loudon
Organic Chemistry, 2nd Edition ,Jonathan Clayden , Nick Greeves ,Stuart Warren.


Does acetamide respond positively to iodoform test? This answer explains why acetic acid does not give a positive iodoform test.

Does acetamide respond positively to the test?

From wikipedia article of haloform reaction:

  Acetyl chloride and acetamide don't give this (iodoform or in general haloform) test.

Why?

  Acetamide is not a ketone which is substantially more acidic than the
  amide.  It's about a million times too weak an acid at the alpha
  protons. Hydroxide is too weak a base. (Quora)


Can freezing temperatures cause chalk to fragment? I am a physics student, with little experience in chemistry. I am researching for a project which involves freezing and its capabilities on humans and different minerals like chalk.
I have been researching chalk for the last couple of months as part of my project. Can the cracks generated cause the mineral to break into fragments under freezing conditions?
It would be appreciated if anybody can explain this to me or even send/link me some papers that I can read further about this effect.

Chalk cannot at given conditions explode chemically, spontaneously turning into other compounds.
It could locally "explode" mechanically as consequence of mechanical stress caused by thermal dilation due changes to temperature. But calling it explosion would be very much exaggerated. It would be rather just cracking.
There is a way such cracking would be more probable, if the chalk is porous, soaked by water. The internally forming expanding ice could crack the chalk easily  But it would not happen at the first freezing event. It would need probably repeated freezing and melting, similarly like long term freeze weathering of rocks. The speed of such process depends on porosity..
Note that such cracking would be rather progressive than sudden, with cracks or crevices developing progressively, with occasional small pieces or blocks  falling away. Sudden break of the whole block into small pieces is highly non-probable


Why exactly does precipitation occur? In a solution, we have ions floating around but when we have a precipitate, they're arranged as they would be in a solid. This conversion should cause a decrement in the entropy of the system and cause the reaction to be non-spontaneous, but that is not the case. Data of several $K_\mathrm{sp}$'s suggests that the reactions are very spontaneous, with $\log{K_\mathrm{eq}}$ around $30-40$ for some salts. Why does this happen?

One possible explanation I could think of was that all precipitation processes are enthalpy driven instead of entropy driven but in those cases enthalpy should either be really high and all solubilities must increase upon increase in temperature. I haven't studied these enough to know if that is the case or not, so help from someone more experienced would be appreciated.

Short story

Maybe it will help to think first about a ridiculous case. Say you have a large hydrophilic solid, and a small drop of water is added on top. We would  not expect the entire solid to disappear into the liquid. Yes, there will be some adsorption and solubility here, yet most of the solid will remain undisturbed. In some sense, the solid 'does not fit' into the liquid droplet over a certain limit.

Similar things are true,  again in a certain interpretation, for more realistic situations. What does this 'fitting' mean exactly? It definitely does not mean there is not enough physical space per se. Instead as you increase the concentration of your solute, the distances between solute entities starts to decrease on average. Hence, at a certain radii the intermolecular forces become strong enough to pull the solutes together. There is often some energy barrier to this approaching, as colloid chemists will tell you, but if the end result is some energy minimum, the process will occur at some rate.

If the ratio of the amount of solvent to the amount of solute increases, that is we add more solvent, we should expect the solubility to increase. This is indeed what we observe. In summary, the amount of 'space' with respect to solubility is a function of the strength of intermolecular forces and the amount of solvent, among other things. Here you might also discover yourself that smaller particles dissolve more (not only faster!) than bigger particles.

You may compare precipitation to condensation of water in air. There is plenty of actual physical space to go around but once a certain limit is crossed (the dew point) the water will become liquid. Nitrogen and oxygen will not condense at these temperatures since their intermolecular forces are much weaker.[a]

An equally short story

Note first that for $\Delta_\pu{r} G \leq 0$ ($\Delta G$ was poor notation on my part) to be the criterion (necessary and sufficient) of spontaneity, some five conditions must be met: constant temperature, no pressure change, only expansion work is done, and the application of the thermodynamic limit. Fifthly, only one chemical reaction is assumed to take place.[b] Provided these prerequisites stand, we may look at the terms in

$$\Delta_\pu{r} H-T\Delta_\pu{r} S\le0.$$

(We should not really compare entropy directly to enthalpy because they have different dimensions.) Jumping in quickly, we might say immediately that if the solute can wriggle around more in a solution than in a solid, then certainly $\Delta_\pu{r} S>0$ for the transition $\text{solid}\to\text{solute}$. The only thing standing in our way is that damned enthalpy! Sure enough, that may be the case for some solutions but it is not the whole picture.

Since the solute entities have a solvent layer in solution, there is an entropy effect for the solvent as well. If the solvent is tightly bound to the solute, entropy decreases for this solvent entity. There is also an enthalpy contribution from solvation, not only from the act of breaking or forming bonds in a solid. I would not really dare to generalise which is the bigger contributor 'for most cases'. In general,

\begin{align}
\Delta_\pu{r} H_{\pu{solid to ideal gas}} & + \Delta_\pu{r} H_{\pu{breaking bonds in solvent}} + \Delta_\pu{r} H_{\pu{solute forming bonds with solvent}}\ + \\ &+ \Delta_\pu{r} H_{\pu{residual}} - T\cdot(\text{corresponding ‘entropy’ terms}) \le 0
\end{align}

where 'residual' signals the  solute-solute interactions in the solution.

A sentence or two on $K_{\pu{sp}}$ (with concentrations)

From the IUPAC Gold Book,

  solubility

  The analytical composition of a saturated solution, expressed in terms of the proportion of a designated solute in a designated solvent, is the solubility of that solute. The solubility may be expressed as a concentration, molality, mole fraction, mole ratio, etc.

As pointed out by Nicolau Saker Neto in a comment, we should not overemphasise the role of $K_{\pu{sp}}$ in determining solubility. Again there are cases where it is a good indicator, and situations where it fails poorly. There are assumptions which must be satisfied:

there actully exists an equilibrium $$\ce{AB(sln) <=> A+(sln) + B+(sln),}$$ that is not all of the solute "molecules" are dissociated, not really the case for strong electrolytes in a dilute solution,
that the simpification of $a \approx c$ is good enough ($a$ is activity, $c$ is some type of concentration),
provided that the above holds, the concentration of $\ce{AB(aq)}$ should be constant wrt additon of another electrolyte (in some calculations).

If these do not hold, $K_{\pu{sp}}$ (with concentrations alone) is a poor approximation for true solubility. See Denbigh, K. G. 'The Theory of Solubility Product'. J. Chem. Educ., 1941, 18 (3), 126. DOI: 10.1021/ed018p126.

[a] I have largely ignored the discussion of surface energies and surface tension, and how the interaction is often more favourable in the bulk than on the surface, and the fact that $r^3$ grows faster than $r^2$, except for small $r$ when it does not, and hence we need some seeds or nuclei to start the process, and supersaturation, and how particle sizes change etc. The example of smaller particles dissolving more is, of course, a surface tension effect.

[b] In the thermodynamic limit, the general condition of spontaneity comes from Clausius inequality. This is written as

$$\small{T\pu{d}S \ge \delta q.}$$

With the first law of thermodynamics, this is equivalent to

$$\small{T\pu{d}S\ge\pu{d}U-\delta w}$$

or

$$\small{-(\pu{d}U +p\pu{d}V - T\pu{d}S)\ge-\delta w(\text {other})}.$$

With constant temperature and pressure,

$$\small{\pu{d}(U +pV - TS)\le\delta w(\text {other})}.$$

Finally, if only expansion work is done,

$$\small{\Delta_\pu{r} G\le 0}$$

where $\small{\Delta_\pu{r} \equiv \partial / \partial \xi}$; $\small{\xi}$ is the extent of a reaction. The last result also assumes only one reaction takes place, otherwise it is no longer a necessary–sufficient condition for spontaneity.


NMR - coupling of chemically equivalent protons In Klein's Organic Chemistry 3rd Edition page 671, it states

  This observation, called the n + 1 rule, only applies when all of the neighboring protons are chemically equivalent to each other.

However, in the following page, it states

  Equivalent protons do not split each other

The first statement infers that for the n + 1 rule to be valid, the protons must be chemically equivalent - inferring that equivalent protons can couple, while the second statement directly contradicts this.

Could my misunderstanding be clarified?

The "n+1" rule refers to a situation where you have a proton of type A with $n$ protons of type B next to it. Proton A's signal will be split $n+1$ times by the B's. However, none of the B's will split each other because they are equivalent.


What determines the speed of charging of a battery? I was told on physics.stackexchange that this question belongs to Chemistry, what do you think?

Recently we have seen the surge of superfast mobile phone charging. Just an example: http://www.bbc.com/news/technology-30708945

What underlying physical principle determines the speed of charging of a battery?

There are many factors limiting speed of charging. Here are just a few:

Run-away overheating -- some battery chemistries, such Li-ion, may even self-ignite.
Spalling of electrodes -- the physical stress of rapid charging can buckle electrodes and cause a composite electrode to break apart.
Side reactions -- to increase the charging current, higher voltage is required. This over-voltage causes undesired side reactions, such as lithium plating (and out-gassing in Ni-MH and lead-acid batteries).

To make cells charge more rapidly, each of these potential issues must be analyzed and both the cell and the charging system must be designed to overcome the problem. In some cases, there have been spectacular failures, even when millions have been spent on design.


How to tell how far equilibrium is? I am trying to solve a question that asks to compare Q vs. K. K = 4.2, and Q = 1.0.
Problem Details:

2.0 mol of Samples of A, B, and C
$\ce{A + B \rightleftharpoons 2C}$

I know that the reaction equilibrium is to the right, but, from this information, how can I tell if it is far to the right or slightly to the right?
Thanks.

Try writing out the K expression. Remember that in general,

${K = \frac{[product]}{[reactant]}}$

In your case, ${K = \frac{[C]^2}{[A][B]}}$

Or in other words,

$K = \frac{[product]^2}{[reacant][reactant]} = \frac{[product]^2}{[reacant]^2}$

I combined the $[reactant]$ terms because the equation you have, there is 1:1 ratio of both reactants.

We can take the square root of both sides:

$\sqrt{K} = \sqrt\frac{[product]^2}{[reactant]^2} = \frac{[product]}{[reactant]}$

So now just plug in your numbers to determine how far to the right the equilibrium lies.

$\sqrt{K} = \sqrt{4.2} = \frac{[product]}{[reactant]} = 2.0$

Therefore (upon consideration of reaction stoichiometry), there is a 2:1:1 ratio of products to reactants at equilibrium.

We can also plug in the Q (reaction quotient) value; remember that K is just a special Q value (K is the Q value when the system is at equilibrium):

$\sqrt{Q} = \sqrt{1.0} = \frac{[product]}{[reactant]} = 1.0$

The Q value indicates that the system currently lies away from the equilibrium mix of a 2:1:1 ratio between products and reactant gases. At Q = 1.0 we have a 1:1:1 mix of A, B, and C.


Conditions for the reduction of carboxylic acids by LiAlH4 From my textbook, the reduction of carboxylic acids by LiAlH4 "has to be in anhydrous conditions, such as dry ether, followed by aqueous acid." I read from the internet that the reduction reaction has to be in anhydrous conditions since LiAlH4 reacts violently with H2O. But why does the reduction reaction need to be 'followed by aqueous acid'? Isn't that contradictory to the fact that LiAlH4 needs to be in anhydrous conditions?

When the reaction is done, you need to quench it to (1) destroy the remaining LiAlH4 (2) protonate the lithium alkoxide to give you the alcohol. The aluminium salts formed can be separated from the alcohol. If you don't quench the reaction it makes separation essentially impossible.

Of course, if you add the water before the reaction is done, then that defeats the purpose of using an anhydrous environment.


Electrical conductivity of graphite On this Wikipedia page, the electrical conductivity of various materials are given in the third column ($\sigma \text{ (S/m) at 20}^\circ \text{C}$).  I am interested in the entry for Carbon (graphite):

$2$ to $3 \times 10^5 \text{ S/m} \perp \text{basal plane}$
$3.3 \times 10^2 \text{ S/m} \parallel \text{basal plane}$

Why is conductivity lower parallel to the plane than in the direction perpendicular to the plane?

I am surprised by the relative magnitudes of these $\sigma$ values because I thought that one of the amazing things about graphene -- an atomic layer of graphite -- was that its conductivity in the plane (i.e., parallel to the basal plane) is very high.  I would expect a similar trend parallel to the basal plane in graphite.

Do you have any suggestions for reading that I can do to understand this?

Good catch - you are absolutely correct. This is a mistake in the Wikipedia page, as a quick check with the reference listed there confirms. The values are the wrong way around.

Changing it now...

Edited to add: this of course raises the question of why it should be this way around. A somewhat hand-wavy answer is that, just like in benzene, the $p_z$ orbitals overlap within each graphene sheet so that the electron contributed by each carbon is delocalised in this plane (the other three are bound up in the three $\sigma$ bonds). So in a sense this is conjugation, but over a whole 2D plane rather than around a ring or along a single chain of atoms. On the other hand, the graphene sheets are rather further apart from one another than the covalently bonded atoms within a sheet (as you might expect since the sheets are only weakly bound through van der Waals interactions), and so the overlap is greatly reduced and conduction is much more difficult.

A more complete answer would rely on a study of graphite's electronic band structure, which is an active research area and beyond the scope of this response!


Reduction using sodium borohydride When we react a ketone with NaBH4, it reduces to alcohol. I coudn't understand the mechanism of this reaction. The first step in my text book states that the hydrogen from NaBH4 attacts the carbonyl carbon. I didnt't understand why this happens. Specifically, I coudn't understand why the Boron didnt attack oxygen instead.

Sodium borohydride is a complex unit made from the let's call it molecule $\ce{BH3}$ and the salt $\ce{NaH}$. Group 3 compounds like boron only have what we call an electron sextett instead of an octett. That makes them good Lewis acids, hence they tend to react with Lewis bases to form adducts. A famous example is the $\ce{H3B-NH3}$ adduct. In your example the hydride anion $\ce{H^-}$ is the Lewis base that forms an adduct with $\ce{BH3}$ to give $\ce{Na[BH4]}$. This means that at least one $\ce{H}$ in $\ce{[BH4]^-}$ can be somewhat easily removed again. The others tend to react as well but let's just call the $\ce{BH3}$ unit a stable molecule and assume it's a combination of $\ce{BH3}$ and $\ce{NaH}$. It's just a good way to store this hydride-anion so that it can be easily removed again. So once it sees your carbonyl group that has a positive partial charge on the carbonyl carbon (as the more electronegative oxygen pulls the electrons closer to itself) it can attack there as nucleophile. As you drew it already the double bond will turn to a single bond giving a negative charge on the oxygen.

Next you have to assume that $\ce{B}$ and $\ce{O}$ form quite stable bonds, hence the formation of a $\ce{B-O}$ bond is somewhat favored. And of course once the hydride ion is at the carbon we end up with an electron sextett on the boron again. Therefore it's the same thing, the carbonyl oxygen with an extra negative charge behaves like a strong Lewis-base again forming an adduct with the $\ce{BH3}$ unit.

This is later treated with water or acidified water to get the alcohol. To the question on the final product. I once talked to a boron chemist about that. He doubts that all four hydrogens will act as reducing agent under normal conditions. It's more likely that they will get hydrolyzed during workup.

Edit: Thinking about it. If you aren't sure about what attacks first you could probably also consider a transition state. I drew in the arrows here but you shouldn't do that (although you see it too often). I guess this would just be an alternative view on the reaction.


Solution of very small particles with neutral buoyancy and distinguishable to the naked eye I'm trying to study the Tea leaf paradox, and asked a question about it on physics.stackexchange.com. To get further on the problem, I want to make a solution of very small objects in water that are distinguishable to the naked eye, and have (very close to) neutral buoyancy. I teach at a high school and have access to the chemistry lab here. Suggestions?

I think that the cheapest and fastest way is that of working on natural products, like actual tea leaves, or ground coffee (I just tried it with coffee, and it can be done):

If you want only particles with neutral buoyancy, or, if I understand well, floating in the solvent, you can use said natural products, which contain particles of different shapes and densities, and discard by fractionation the unwanted ones.

What I did:

Place the ground coffee in a glass of water and stir a lot.
Let the suspension sit. You will notice that part of the parcticles are on a top layer, some are floating in water, and some are at the bottom.
Put the glass in a sink, and gently add water to it, so that the top layer eventually falls out of the glass, together with some water
Let the suspension sit: if needed repeat the last step, otherwise go on
To discard the bottom layer, gently pour the solution in another glass, by leaving some water in the first glass: the first glass will contain the heavier bottom layer, and the second one will contain only floating parcticles
If needed, repeat the last step


Why unsaturated fats are usually cis? I've been studying organic chemistry and I think trans fats are actually more stable than cis fats. This is because the steric hindrance (for trans) would be lower and the molecule energy would be lower too (more stable).

Why are unsaturated fats usually cis at room temperature if the steric hindrance is higher?

Is that because the room temperature provides enough energy to break the pi bond in the double covalent bond (for example) making the molecule rotate and have a isomer of higher energy?

Nature produces cis fats exclusively (someone may prove me wrong with some weird example.) This is why unsaturated fats such as vegetable oil have lower melting point than saturated fats of the same MW (the cis double bonds lead to a more compact molecule which gets less tangled with its neighbours.)

The reason nature produces cis bonds is due to the enzymatic mechanism in which these bonds are produced. Therefore this is down to reaction kinetics, and as often happens the product which is not the most thermodynamically stable of all possible products is produced. It just has to be more stable than the reactants.

When fat is artificially hydrogenated with a metal catalyst (for example nickel) to increase its viscosity/raise its melting point (to turn vegetable oil into margarine) some of the cis double bonds that escape being hydrogenated are able to change from cis to the more thermodynamically stable trans. These trans fats are unnatural and are therefore widely claimed to be unhealthy.


Working with stainless steel in 3 M HCl solution We are trying to mix a material using 316 stainless steel blades in a $3~\mathrm{M}\ \ce{HCl}$ solution for 4 hours in a PE container? Will the blades stand for 4 hours or do they start to corrode in 4 hours?

The austenitic, chromium-nickel based stainless steel SAE 316 approximately corresponds to the material number 1.4401 or 1.4436. Especially due to the addition of molybdenum in its composition it is more resistant to corrosion overall than the usual type 304 (1.4301). However, it is generally not intended to be used in hydrochloric acid.

Hydrochloric acid lacks the oxidizing properties that are needed to maintain the passive corrosion-resistant surface layer on stainless steel.

The typical corrosion rate in $1\ \%$ hydrochloric acid at room temperature is about $0.1\ \mathrm{mm\ a^{-1}}$, which is approximately $0.1\ \mathrm{g\ m^{-2}\ h^{-1}}$. At higher concentrations or higher temperature, the corrosion rate is strongly increased. Note that your concentration of $c=3\ \mathrm{mol\ l^{-1}}$ $\ce{HCl}$ corresponds to about $10\ \%$ hydrochloric acid.

In addition to this uniform corrosion, stainless steels in media containing chloride ions are particularly susceptible to pitting corrosion, i.e. highly localised corrosion with discrete pits.

Furthermore, stainless steel SAE 316 is prone to chloride-induced stress corrosion cracking at elevated temperatures (over about $50\ \mathrm{^\circ C}$).


Finding the frequency factor when raising temperature The question is:

  A reaction rate doubles when the temperature increases from $\pu{25^\circ C}$ to $\pu{40^\circ C}$. Calculate $E_\mathrm a$ and the frequency factor.

I found the activation energy to be $\pu{35.8 kJ}$ using the two point form of the Arrhenius equation. What I'm having trouble with is finding the frequency factor. I have two unknowns, $k$ and $A$, and to me it seems like this is impossible to solve without knowing what the rate constant $k$ is. All the examples in the book solves this problem graphically, but apparently you can solve this another way according to my teacher.

The answer given for $A$ is $1.9 \times 10^6$ but what method do you use to solve for this?

This question has no answer.

The Arrhenius equation is:

$$ k = A e^{-\frac{E_a}{RT}}$$

A linearized form of the Arrhenius equation is

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T^{-1}$$

This equation linearly relates $\ln{k}$ to $T^{-1}$: the intercept is $\ln{A}$ and the slope is $-\frac{E_a}{R}$.

To fully define a line, we need two parameters.  This can be two completely specified points that lie on the line, or any single point on the line plus a slope for the line.  For this problem that would mean either (a) two temperatures and two rates, or (b) one temperature, one rate, and one slope.

Using the information we are given:

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T_1^{-1}$$
$$ \ln{2k} = \ln{2} + \ln{k} = \ln{A} - \frac{E_a}{R}T_2^{-1}$$

Any way we combine those two equations will only yield an equation equivalent to

$$\ln{2} = -\frac{E_a}{R}\left(T_2^{-1} - T_1^{-1} \right)$$

in which $\ln{k}$ and $\ln{A}$ have both cancelled out.  That is because the starting two linear equations have the same coefficients for $\ln{k}$ and $\ln{A}$ in each equation.  Similarly, the two equations $2x=y$ and $2x+2=y+2$  can't be solved for $x$ and $y$.

The problem as stated gives us only a slope, but not even a single point that lies on the line.  The rate could double by going from 1,000,000 $\text{s}^{-1}$ to 2,000,000 $\text{s}^{-1}$ (a very fast reaction!) or by going from 0.1 $\text{yr}^{-1}$ to 0.2 $\text{yr}^{-1}$ (pretty slow).  There is no way to find the intercept of a line when we are given only the slope.  Thus, there is no way to solve for $A$ using the information given.


Is the term "heteroatom" reserved for exclusive elements? From my experience, typically in organic chemistry, the implication of "hetero-atom" in some molecule is understood as B, N, O, S, P or a halogen -- or any nonmetallic element, perhaps.

If one describes a nanostructure that contains inorganic and organic ligands (like metal-organic frameworks), is it safe to describe the metals as heteroatoms? Or should the term be reserved for covalent-type atoms only?

For example, is this an acceptable statement?

  Snurr and co-workers showed results for simulated H$_2$ sorption in a
  variety of rht-MOFs containing only carbon and hydrogen in the
  linkers, as originally these MOFs did not contain heteroatoms
  other than oxygen and the metal.

The IUPAC Goldbook provides no 'official' definition of the term heteroatom,and as such you're somewhat free to use (and abuse) the term as you see fit.

Since no IUPAC definition exists, it may be helpful for you to define what you plan on calling a heteroatom at the beginning of whatever you're writing, to inform the reader (common practice in scientific literature for poorly defined terms). Clearly an organic chemist would have a very different definition to someone involved in purely metallic synthesis.

I would however, point to several other IUPAC documents in which, whilst no 'definition' of a heteroatom is given, they do suggest lists of possible heteroatoms. For example, the following statement from the IUPAC nomenclature guidelines which includes many elements (including metallic ones, possibly going against your suggested definition).  This priority order of heteroatoms is repeated multiple times.

  P-22.2.2.1.3 If two or more kinds of heteroatoms occur in the same name, their order of citation follows the sequence: F, Cl, Br, I, O, S, Se, Te, N, P, As, Sb, Bi, Si, Ge, Sn, Pb, B, Al, Ga, In, Tl, Hg.


Removing titanium from tantalum crucible I'm trying to remove titanium filling from a tantalum crucible for a high-temperature effusion cell.  I've managed to get a majority of the material out with tweezers, but some remains and is difficult to remove.  I understand that sulfuric acid will corrode titanium, but not tantalum, and was considering this as a method for extraction.

If anyone has some advice on a method to (safely) remove $\ce{Ti}$ from the $\ce{Ta}$ crucible without destroying it, I would very much appreciate it.  I am certainly not a chemist and could use some help.

Straight hydrochloric acid- Say 20% is virtually  inert to Tantalum- but significantly corrosive to titanium. If you heat it to boiling point of hydrochloric- it will eat through the titanium.

http://www.parrinst.com/wp-content/uploads/downloads/2011/07/Parr_Titanium-Corrosion-Info.pdf
http://www.tantaline.com/Tantalum-Corrosion-71.aspx


Mercury brain half-life If have read conflicting reports on the half-life of mercury in the brain. I would like to hear some estimates from this community.
I am also wondering how long chelation therapy would take to remove mercury from the brain.

Since you're concerned about mercury from fillings, you may have already taken a look at the wikipedia article about dental amalgam toxicity. It would be worth reading through some of the sources listed, too.

I assume you're an adult, I don't know if you're male or female. Children are more sensitive to mercury than adults because their brains are actively developing. If you're female, you may become pregnant at some point and there would be some concern about fetal mercury exposure.

However, according to a 2003 WHO report, the amount of mercury ingested from dental amalgam is low, with exposures estimated from 1ug per day to 12.5ug per day, but the vast majority of people are exposed to less than 5ug per day. The mercury from fillings is elemental mercury, which is poorly absorbed, less than 0.01% is taken up, with smaller amounts reaching the brain.

Removing your existing fillings probably isn't worth it, as the removal process would expose you to more mercury than just letting the filling sit there.

I don't know if this will make you feel better or worse, but coal fired power plants are probably a larger source of mercury exposure than your fillings. The mercury from the coal plants is methylmercury, which is much more efficiently absorbed than elemental mercury. This bioaccumulates in the environment, so that fish have relatively high levels of mercury in them.

As far as your perceived concentration problems, we really can't say if it's caused by the mercury or not. If it concerns you, you should see a doctor. And it may be more efficient to treat the symptoms through ADD medication than trying to remove mercury.


Does aminosugars give positive response in alkaloid tests? Does aminosugars give positive response to test for alkaloids such as tests with Dragendorff's reagent, mayer's reagent, Hager's reagent, Wagner's reagent etc?

Most alkaloids are tertiary amines. Almost all tests for alkaloids, reagents of which are  are solutions of the salts of heavy metals, are precipitation tests. For example, all of tests with Dragendorff's reagent, Mayer's reagent, Hager's reagent, and Wagner's reagent give color precipitates: The heavy metal atom in the reagent with the nitrogen in the alkaloid would form ion pairs. These ion pairs form an insoluble precipitate. The color of the precipitate depends on reagent, plant species, etc. (Ref.1). Most amino sugars are either primary amines or acetamides, thus, I doubt they'd give precipitates with sought reagents. However, some reagents give false-positive tests for non-nitrogenous extracts indicating the presence of alkaloids. The best example is Dragendorff's spray reagent (Ref.2).

Mean time, some reagents are giving negative results even with alkaloid extracts, e.g., Mayer's reagent (Ref.3). Therefore, my answer is purely opinion based.

References:

Felix G. Coe, Gregory J. Anderson, "Screening of medicinal plants used by the Garífuna of Eastern Nicaragua for bioactive compounds," Journal of Ethnopharmacology 1996, 53(1), 29-50 (https://doi.org/10.1016/0378-8741(96)01424-9).
Abdel‐Azim M. Habib, "False‐positive alkaloid reactions," Journal of Pharmaceutical Sciences 1980, 69(1), 37-43 (https://doi.org/10.1002/jps.2600690111).
J. W. Field, M. Kandiah, "A note on the use of mayer's reagent for the detection of quinine in alkaline urine," Transactions of the Royal Society of Tropical Medicine and Hygiene 1935, 28(4), 385-390 (https://doi.org/10.1016/S0035-9203(35)90133-X).


What is the electrical conductivity of pool water? What is the electrical conductivity of pool water? I tried to find some information, unfortunately without success. I need to know it to make an underwater electronic project.

The pH should be around 8.4 and the conductivity in the range of 2.5 - 3 mS/cm. The lower value is appropriate for freshly prepared water. The conductivity is used for the determination of TDS (total dissolved particles).


How would using different salts in a salt bridge effect the voltage generated by a voltaic cell? If I use a range of different nitrate solutions (e.g. calcium nitrate, iron nitrate) for the salt bridge in a voltaic cell, what effect would it have on the flow of ions in the voltaic cells?

Will the different properties of the salts such as ionic radius effect the voltage generated by the voltaic cell?

If a certain property would most likely have an effect, what would it be and why?

(Both the cathode and the anode are kept constant with copper in copper nitrate solution and zinc in zinc nitrate solution)

The salt bridges used by mine (so far) were filled with aqueous electrolytes with solutions of either potassium chloride, potassium nitrate, or sodium chloride.  For an electrochemical cell of $\ce{Zn | Zn^2+||Cu^2+|Cu}$ described by you, i.e. based on

$\ce{Cu^2+ + Zn \rightleftharpoons Cu + Zn^2+}$

these appear to me a resonable choice, because

Their ions (of KCl, KNO3, NaCl) would not interfer with the redox reactions, so the potential between the two electrodes (the "voltage" you mention) were not affected.  Referring to the standard potentials, $\ce{Fe^3+}$ of iron nitrate in contrast, is more likely to interfer with your system than the $\ce{Ca^2+}$ of calcium nitrate, esp. in aqueous solution; but the design of the salt brigde (diaphragm) may reduce this electrochemical interference.
Beside electrochemical inertia in the window of investigation, an additional criterion to select electrolytes for salt bridges is their cations and their anions should possess a similar mobility in the electrolyte of tehe salt bridge (the flow you mention).  The rate of the ions to move not only depends on the viscosity of the medium and the charge of the specis, but on the efficient radius of the ion (the ion plus the hydration sphere), too; yet this should not influence the potential between the electrodes, rather than the current between the electrodes if the set up is intended to transform chemical into electrical energy.

In short: within reason, keep the KCl in the salt bridge.


How to calculate the pH of a propanoate buffer after addition of sodium hydroxide? I've calculated the pH of the buffer solution, as seen.
The question I'm referring to is part d), which asks: 'Calculate the pH of the solution after 0.01 mols of NaOH are added to  500$cm^3$ of the solution'.

The addition of the NaOH will "use up" 0.01 mol of propanoic acid and "create" 0.01 mol of propionate. To make things easier, just work in molarity. Thus 0.02 M NaOH.

The pH of the final solution can be found by adjusting the equation you used in part b,

$pH = -log \left( 1.26 \times 10^{-5}\frac{0.1-0.02}{0.05+0.02} \right)=4.84$

The result agrees with our expectations. Since the solution is a buffer, the pH is not expected to drastically change with the addition of a strong base. It is buffered.


Classification of oxides,transition elements I understand  what  basic,acidic and amphoteric oxides are.
But how do we classify oxides on the basis of metallic nature?Explain how we can deduce if the oxide is metallic or not.
For example,what does it mean when it is stated that "CrO2 is a metallic oxide."

This was an answer to one of the questions in an exam that I wrote recently.

'Metallic oxide' has nothing to do with basicity/acidity of the oxide and is  about solid phisics.

Basically it means, that the solid is electrically conductive like a true metal. This involves a half-filled conductivity zone (that can be roughly imagined as a number of half-filled overlapping orbitals of same energy), so electrons can move freely through the crystal lattice.

There is no simple rule to guess metallicity of multi-element compounds, unfortunately; because this property is strongly tied to electronic structure of the solid and crystall lattice. As an example: of $\ce{TiO2}$, $\ce{VO2}$, $\ce{CrO2}$ and $\ce{MnO2}$ only $\ce{CrO2}$ is truly metallic, while $\ce{VO2}$ looses metallic conductivity at lower temperatures (it has two allotropes, one metallic and one insulating). For TiO2 it is, at the very least, expected, as it is a $d^0$ compound, but for other metallic oxides it is unclear, or, at least, aboive average chemist's level.

=================

As for acidic/basic classification of oxides, there are few solid rules. Generally, any oxide with oxidation state 5+ is a strong suspect for acidic nature, with 3-4 being anyware from acidic to basic and +1 usually basic except for halogens.

For an oxide to be basic, corresponding metallic cation must be reasonably stable. This implies relatively large radius, and moderate to low charge density. Consequently, +1 metal oxides are practically exclusively basic, +2 metal oxides are usually basic except for very small cations (Be and few late d-elements like copper and zinc) and +3 are usually amphoteric except for very large cations (like La) Anything of higher oxidation state is usually acidic or at most aphoteric, as corresponding cation has too high charge density to be stable in water solutions. The only true salts of +4 cation I can come with are $\ce{Zr(SO4)2}$ and $\ce{Hf(SO4)2}$ and even for them the corresponding hydroxide is classified as amphoteric.

Still, in some cases such oxides still may act as bases, but corresponding cation has lower charge. Probably, the simpliest examples are $\ce{VO2}$ and $\ce{UO3}$, with corresponding salts containing $\ce{VO^{2+}}$ and $\ce{UO2^{2+}}$ respectively. This, however, is not usually condidered as an excuse to classify these oxides as basic/amphoteric.

Some oxides are not acidic/basic/amphoteric as there is no 'simple' salt corresponding ot the oxide in question. This claim is usually debatable, as such a salt usually may be found (for example, NO, typically classified as not producing salt, may be tied to salts like $\ce{Na2N2O3}$ aka  Angeli's salt), but for simplicity are not considered. There are reasons for it, as basic/acidic oxide classification is very rough, and should not be relied upon in extreme, borderline and simply too complicated cases.


How to write down the composition of eluent mixtures for TLC I obtained some Rf values via TLC by adding two or three drops methanolic triethylamine to an eluent mixture of let's say DCM/MeOH = 9:1. I couldn't find an example of how a mixture like that is written down in short in publications. I think that calculating the absolute content of dissolved ammonia wouldn't be suitable.

Describe what you did. The key is make sure your communication allows your experiment to be reproduced.

  ... Rf 0.34 (10 mL of 9:1 DCM/MeOH + 3 drops of 1.0 M Et3N in MeOH).

If you do all of your TLC with the additional of triethylamine, then you can include a sentence in the general methods about this procedure.

  All TLC plates were developed in a 9:1 mixture of DCM and MeOH that was made alkaline by the addition of three drops of 1.0 M Et3N in MeOH for every 10 mL original solution.

The procedure you describe is common, especially in the elution of basic molecules (since silica is slightly acidic). You might find examples of how to state this sort of procedure in the synthesis of amines and alkaloid natural products. Ultimately, in a publication, you should err on the side of thoroughness for your experimental methods. If it's too long, a reviewer of the editor will tell you.


What happens when deodorants or perfumes expire I know that deodorants and perfumes have alcohol as their main ingredient so I was thinking what happens when their expiry date passes?

I tried to search on google but got mixed results. Some say that they do not ever expire but some say that an expired deodorant can burn the skin.

What i think is  that as they contain mainly alcohol so maybe a major part of the alcohol gets oxidised to ketones or acids which may in turn harm the skin. Does anyone of you know what happens when deodorants expire?

The meaning of expiry dates

The date of expiry is widely interpreted by consumers as the date after which a product is known to detiorate.  However, the meaning of expiry dates varies widely from product to product, and from country to country.  Often times, the date of expiry is not a "first known bad" date, but a "last known good" date.  In that latter case, the dates indicates that manufacturers know their product will last up until the expiration date, but after that, they don't know what happen.  Their product could last 1000s of years, completely stable and undeteriorated.  Or it could degrade instantaneously the day after expiry.

I don't know for sure what perfume manufacturers mean by their expiry dates.  You could try asking the manufacturer for clarification.  Ask for their data from their product stability studies.

Likely routes of perfume degradation

All of the above notwithstanding, let's suppose that perfumes do degrade over a time span of several years.  Subject to that assumption, what could be happening happening?  First, I agree with the commenters that ethanol oxidation to acetaldehyde or acetic acid is unlikely to be a major contributor to perfume degradation.

However, the same is not true of other perfume components.

Oxidation

Aldehydes such as octanal, benzaldehyde, citral, are an important ingredients in many perfurmes.  Citral degradation has been studied and is accelerated by acid and oxygen, with the main ultimate product being p-cymene.  Benzaldehyde is oxidized in air to benzoic acid, and octanal is oxidized to octanoic acid.
Terpenes are also important pefume components.   Limonene is a component of many perfume blends and is a representative terpene.  It is also oxidatively unstable, with products such as limonene oxide, carveols, and carvones being formed.  Some of these oxidation products can trigger allergic reactions, which while not precisely "burning the skin", could lead to such a perception.

Other possible degradations

Perfume, especially if opened frequently, could also change by absorbing water from the atmosphere, leading to dilution and an alteration of its olfactory properties.  Frequent opening or exposure to the atmosphere could also result in differential evaporation of the more volatile components, which would also change its olfactory properties.

The bottom line

Expired perfume or deodorant might be just as good as regular perfume.  Or it might be degraded.  The degradation products might possibly include compounds that have been reported to form rashes.  It is unlikely, however, that the chemical mechanism for this is the formation of large amounts of acid.


gas chamber for calibrating gas sensors I have $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ sensors and need to calibrate them. Can anybody help me to produce the $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ in laboratory for calibrating the sensors (the gas should be produced in known in $\mathrm{ppm}$)

To be accurate you will need standard gas cylinder and pneumatic connections (pressure regulator, needle valves or better mass flow controllers, pipes). You will need a measuring chamber with an overflow. As a rule of thumb, if the chamber geometry is not exotic (lets say a cube with no dramatic dead volume), you need to flush it, at least  10 times, with each concentration you want to measure.

You can do a 2 points calibration for O2 by measuring atmosphere where O2 concentration can be assumed constant (span) and then flush with pure nitrogen (zero). For CO2 and CO you will need standard cylinders.

Anyway a 2 points calibration is a little bit risky because you will not be able to assess linearity and then you will have no idea in which extent your sensor is linear (dynamic range). Properly calibrating sensor requires multiple points calibration and therefore a dilution device and standards.

Caution: When dealing with gas you must take care of exhaust and prevents risk such as intoxication (especially CO and in a fewer extent CO2) or asphyxiation (due to a lack of oxygen) and explosion (CO is explosive).

You should find those materials, resources and skills in a University (Chemistry or Physics Department) or an Air Quality Laboratory.

Note: If your sensors are electrochemical ones, you need also to assess their correlations/deviations with temperature and humidity.


Acidity of aldehydes Which is more acidic between methanal ($\ce{HCHO}$) and ethanal ($\ce{CH3CHO}$).
Please explain using General organic chemistry basic concepts.

My Effort:

I saw the stability of the conjugate base.

It would be $\ce{HCO-}$ and $\ce{CH2^{(-)}CHO}$.
Due to resonance I can say that second one is more stabilized and thus a better acid.

But answer says that methanal is more acidic than ethanal.

Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data.  If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively.  A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below.

Now if someone asked, "would pulling the same proton off from ethanal be easier or harder?", you could answer "harder" because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon.  But it turns out that when ethanal is treated with base a different proton is removed!  The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed.  Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?

Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.


Why do these two calculations give me different answers for the same acid-base titration? In class, I performed an weak acid-strong base titration using commercial white vinegar and sodium hydroxide with the aim of finding the concentration of of ethanoic acid in the vinegar, and I used stoichiometric calculations as well as the $\mathrm{p}K_\mathrm{a}$ equation to find the concentration from the data I collected. However, these two methods are giving me different answers and I have no idea if I did something wrong in my calculations, or used the wrong method outright.

As a bit of background info, the label on the vinegar says it has a $\ce{CH3COOH}$ concentration of $5\%$, which according to my calculations (I measured the density of the vinegar to figure this out) is around $0.9\mathrm{~mol~dm^{-3}}$.

Here's the data from the titration:

Volume of $\ce{CH3COOH}$: $3.1\mathrm{~cm^3}$ (diluted with water to the $50\mathrm{~cm^3}$ mark for experimental purposes)

Concentration of $\ce{NaOH}$: $0.1 \mathrm{~M}$

Volume of $\ce{NaOH}$ added: $33.1\mathrm{~cm^3}$

Initial $\mathrm{pH}$ of $\ce{CH3COOH}$: $3.0$

Half-equivalence $\mathrm{pH}$: $4.6$

So, using the stoichiometric method, we have the formula

$$M_\mathrm{acid} V_\mathrm{acid} = M_\mathrm{base} V_\mathrm{base}$$

so,

$$M_\mathrm{acid}(0.0031) = (0.1)(0.0331)$$

$$M_\mathrm{acid} = \frac{(0.1)(0.0331)}{0.0031} = 1.067 \mathrm{~mol~dm^{-3}}$$

This answer is fairly close to the literature value, so I'm assuming it's accurate. (The percentage concentration works out to be about $6.1\%$.)

On the other hand, the $\mathrm{p}K_\mathrm{a}$ equation is $K_\mathrm{a} = \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$

Since $\mathrm{p}K_\mathrm{a} = \mathrm{pH}$ at the half-equivalence point, the $K_\mathrm{a}$ is $10^{-4.6} = 2.5 \times 10^{-5}$; similarly the initial $\mathrm{pH}$ is $3$, so the initial $[\ce{H+}]$ is $10^{-3}$.

So,

$$2.5 \times 10^{-5} = \frac{(10^{-3})^2}{[\ce{HA}]-0.001}$$

Entering this into a calculator returns the value $[\ce{HA}] = 0.041 \mathrm{~mol~dm^{-3}}$.

So, why is the second answer so much smaller than the first? Have I used one of the equations incorrectly or made an error in calculation? Does one of the methods not work for the acid-base titration in question (and why would that be?)

I sincerely apologize for the messy formatting; I literally had to learn LaTeX from scratch just to post this question. Thank you so much in advance.

Part of the reason why there is such a large discrepancy is because in your second method of calculating $[\ce{HA}]$, you failed to take into account the fact that your vinegar was diluted from $3.1\mathrm{~cm^3}$ to $50\mathrm{~cm^3}$.

Effectively, if the original concentration of $\ce{HA}$ was $x$, then your dilution would have brought it to a concentration of $\frac{3.1}{50}x$. And you plugged the $\mathrm{pH}$ of this diluted solution into the the $K_\mathrm{a}$ equation. Let's adjust for that dilution:

$$\begin{align}
\frac{3.1}{50}x &= 0.041\mathrm{~mol~dm^{-3}} \\
x &= 0.661\mathrm{~mol~dm^{-3}}
\end{align}$$

As you can see, there is still some difference between the two values. However, you can calculate the theoretical $\mathrm{pH}$ of a $1.067\mathrm{~mol~dm^{-3}}$ solution of $\ce{CH3COOH}$, using the fact that the $\mathrm{p}K_\mathrm{a}$ is $4.6$. Here, I will make the simplifying assumptions that the final and initial concentrations of $\ce{HA}$ are the same, i.e. $[\ce{HA}] = c_{\ce{HA}}$, and also that $[\ce{H+}] = [\ce{A-}]$:

$$\begin{align}
\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} &= 10^{-4.6} \\
&= 2.511\times 10^{-5}\mathrm{~mol~dm^{-3}} \\
[\ce{H+}] &= \sqrt{(2.511\times 10^{-5})(1.067)}\mathrm{~mol~dm^{-3}} \\
&= 0.00518\mathrm{~mol~dm^{-3}} \\
\mathrm{pH} &= 2.28
\end{align}$$

You could go all the way and do mass-balance and charge-balance equations in order to avoid making that assumption, but I doubt it alters the result significantly. So, if you really had a concentration of $1.067\mathrm{~mol~dm^{-3}}$, your observed initial $\mathrm{pH}$ would not have been $3.0$. There is likely some experimental error present, or perhaps the $\mathrm{pH}$ meter that you used to measure the initial $\mathrm{pH}$ is not calibrated well.


What would happen if the reaction of reduction of carboxylic acids by LiAlH4 is carried out in alcohol? Which is the intermediate compound in the reduction of carboxylic acids by $\ce{LiAlH4}$ and what would happen if the reduction of carboxylic acids is carried out in alcohol?

From my knowledge I know that a geminal diol is formed and then elimination takes place to form $\ce{R-CH=O}$. If this is reacted with alcohol instead of $\ce{LiAlH4}$ again what are the expected products?

Alcohols are not an ideal solvent for LiAlH4, but the main issue is that there is no carboxylic center to where the hydride transfer from LiAlH4 can take place.

So in this case, the side-reaction between alcohol and LiAlH4 can be regarded to be minimal and the major product will be the alcohol produced from the carboxylic acid.

P.S. : A diagrammatic representation of the reactions for LiAlH4