What are the similarities and differences of adiabatic and diabatic states? The idea of an adiabatic potential energy surface or adiabatic state is closely related to the Born-Oppenheimer approximation. This has been discussed on this site before in some detail. People frequently talk about diabatic states as well. It seems based on the reading I have done that a diabatic state is not really counter to an adiabatic state as one might expect from the names, but is sort of a different representation when looking at processes involving multiple potential energy surfaces.

So, my question is, how are adiabatic and diabatic states/surfaces/processes (what word should I be using here??) similar and how are they different? Can both pictures be used to describe the same processes all of the time or only some of the time (or perhaps I'm mistaken and they are never the same)?

Also, I think this is just terminology, but where does a nonadiabatic process fit into all this? I assume this just means there is a crossing of surfaces and the Born-Oppenheimer approximation breaks down, but that isn't quite the same as what I'm asking above I think.

A potential energy surface (PES) is generally the electronic energy written as a function of the nuclear coordinates. This is obtained by solving the Hamiltonian, considering explicit dependence on the electronic coordinates $r$, and parametric dependence on the nuclear coordinates $R$.

$\hat{H}\psi(r;R)=E\psi(r;R)$

Solving this, gives us the electronic energy as the function of the nuclear coordinates, $E(R)$. A plot of this function is the PES. This PES is obtained within the Born-Oppenheimer (BO) approximation, which allows us to assume that the kinetic energy of the nucleus does not matter. The physical interpretation is that the nucleus is very heavy compared to the electrons and, therefore, so much more slower that it can be considered almost stationary. This PES is generally written in the adiabatic basis, where the diagonal terms represent the electronic energy and the off-diagonal terms are the nuclear kinetic energy, $T_n$.

The off-diagonal terms can be neglected, as we already saw that $T_n$ is nearly zero. But in cases where two PESs come very close together (conical intersections and avoided crossings), there is coupling of the nuclear and electronic coordinates. This means that the nuclear kinetic energy can no longer be neglected, and there is a breakdown of the BO approximation. Further, the numbers become very high when the two states (PESs) cross as in a conical intersection and become too hard to handle. In such cases, a unitary transformation is done to change the basis from an adiabatic to a diabatic basis. By doing this, the diagonal term, $T_n$, gets converted to a potential energy term and is easy to handle.

Therefore, adiabatic and diabatic PESs differ by the basis they are written in. They can both be used to describe the same process, but there is no real need to use the diabatic representation when the BO approximations holds.

They can be referred to as adiabatic/diabatic PESs, or states, or surfaces, or representations. The term 'process' here is incorrect as it refers to thermodynamics terminology.


What would this unsaturated cyclic compound be called? I came across this compound recently, and havent been able to find out what it's name could possibly be:

My guess was maybe it's "cyclohexylene cyclohexane", but not quite sure. Can anyone tell me what this compound could be called?

Chemspider tells me that the name should be

  1,1'-Bi(cyclohexylidene)

http://www.chemspider.com/Chemical-Structure.121798.html?rid=e40cd595-d5e5-450e-bdb3-3e558a9da92d

While PubChem mentions:

  Cyclohexylidenecyclohexane

https://pubchem.ncbi.nlm.nih.gov/compound/Cyclohexylidenecyclohexane

  My guess was maybe it's "cyclohexylene cyclohexane"

There's a flaw to this thinking, to name this, I'd suggest you choose one of the cyclohexane rings to act as the word-root. Now you've to just add the name of the substituent.

Recall that (source: http://www.ebi.ac.uk/chebi/searchId.do?chebiId=65236) also click this link for an example (this website particularly has propylidene group as the example):

  An alkylidiene group formed from propane by removal of two hydrogen atoms from the same carbon atom, the free valencies of which are part of a double bond.

So, the substituent in this case would be called "cyclohexyl idene " (without the space). And hence, the name should be cyclohexylidenecyclohexane.


Reducing agent in preparation of CpRuCl(PPh3)2 Synthesis of cyclopentadienyl-bis-triphenylphosphine-ruthenium chloride $\ce{RuCl(PPh3)2(\eta^5-C5H5)})$:

Dissolved $\ce{PPh3}$, $\ce{RuCl3.3H2O}$ and cyclopentadiene in ethanol and heated under reflux to prepare it.

What is the reducing agent and what is the product formed on oxidation of this reducing agent?

The ruthenium is being reduced but i don't know what is being oxidised.

You'd think it's the ethanol solvent that acts as the reducing agent here, and the byproduct is acetaldehyde.


3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile? Which name is the Preferred IUPAC Name of the above molecule? 3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile?

I feel the second one is the PIN, since it's shorter than the first one. But, I haven't been able to navigate the IUPAC Blue Book, regarding such cases.

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. You have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the nitrile group ($\ce{-(C)N}$).

The next rule (P-44.3.2) is looking at the longest chain. Therefore, the parent structure is a butanenitrile.

The following rules (P-44.4.1.1 and P-44.4.1.2) would be looking at multiple bonds, which is not relevant in this case.

Next, Rule P-44.4.1.8 is looking for a lower locant for the attached group that is expressed as a suffix (here: nitrile); however, the nitrile group has the lowest locant ‘1’ in both suggested cases; therefore, this rule doesn't help to make a choice.

The next rule (P-44.4.1.10) again is not relevant in this case.

Finally, Rule P-45.2.1 is looking for the structure that has the maximum number of substituents cited as prefixes to the parent structure. Therefore, the preferred IUPAC name is 3-methyl-4-oxobutanenitrile (two substituents) rather than ‘3-formylbutanenitrile’ (only one substituent).

By way of comparison, if you add more substituents to the methyl group, the principal chain can switch, for example 4,4-dichloro-3-formylbutanenitrile (three substituents) rather than ‘3-(dichloromethyl)-4-oxobutanenitrile’ (two substituents).


Why quaternary nitrogen but not tertiary oxygen? Why do quaternary ammonium ions with a partially positive nitrogen form fairly readily and are often stable but tertiary oxygens, apparently called oxonium ions, are more rare/less stable? The trend seems to continue down the periodic table, too, with phosphonium and sulfonium.

This has to do with the electronegativity of the species and how good of a base (or how good an acid the conjugate acids) are.

$\hspace{9ex}$

$\hspace{7ex}$

[source]

To put this in perspective, nitric acid ($\ce{HNO3}$) has a $\mathrm{p}K_\mathrm{a}$ of $-1.3$. Yes, $\ce{HNO3}$, a strong acid, is more stable than $\ce{H3O+}$! The central oxygen in $\ce{H3O+}$ is too electronegative to effectively share these electrons and form a stable bond. In $\ce{NH4+}$, its lone pairs are far more available due to nitrogen's lower electronegativity.

How common each species is only has to do with the number of acids stronger than $\ce{H3O+}$ or $\ce{NH4+}$. As you can imagine, there are going to be a lot more acids stronger than $\ce{NH4+}$, which starts out around the middle of the $\mathrm{p}K_\mathrm{a}$ spectrum, than compared to $\ce{H3O+}$, which is near the end.


Why for compounds of elements in a group with larger anions have nearly constant lattice enthalpy? Why do compounds of elements in a group with larger anions have nearly constant lattice enthalpy?

In the reasons for the decrease of the solubility of sulphates of alkaline earth metals, it is given that:

  "the size of anions being much larger compared to cations the lattice enthalpy will remain almost constant within a particular group"

I am unable to understand the reasons behind it.

When I was searching I also found that larger cations stablise larger anions and small cations stabilise small anions. I didn't understand this.
Please explain.

If we assume that the lattices have the same structure, the lattice enthalpy is dominated by the charge of the ions and the distance between them. Since we are dealing here with cations from the same group, the charges are the same, so the remaining variable is the distance between them.

With moderately sized or small anions, a smaller cation can get closer, resulting in a shorter overall distance and a larger magnitude lattice enthalpy. In the case of very large anions, however, the anions are already packed as closely together as possible, so if a smaller cation moves closer to one sulfate, it moves farther from another. The equilbrium position of the cation is thus determined solely by the distance between the sulfates and does not vary with cation size.

For the second part of your question, I'm guessing that your statement that "larger cations stablise larger anions and small cations stabilise small anions" might be made with respect to solubility?

Solubility is a function of both the lattice enthalpy and the solvation enthalpy, which are in competition with each other. Generally speaking, small ions have higher solvation enthalpy in water. However, they also can have greater magnitude lattice enthalpy if they are able to pack more closely together.

Empirically, it has been observed that if the cation and anion are both small, the lattice enthalpy "wins" over the solvation enthalpy, and the salts tend to have low solubility in water.

If both ions are large, both the lattice enthalpy and the solvation enthalpy have smaller magnitude, and the empirical observation is again that the lattice enthalpy "wins".

When the sizes are mismatched, however, we find that the solvation enthalpy "wins" and the salts are generally soluble. We can see why in the extreme example from the first part of your question, where a smaller cation does not increase the magnitude of the lattice enthalpy at all, but will definitely increase the magnitude of the solvation enthalpy relative to a similar salt with a larger cation.


What is the major product of the reaction of 2-ethoxypentane with hydrogen iodide? What is the major product formed by heating 2-ethoxypentane with $\ce{HI}$?

The options are:

According to me the correct answer should be option-1 because 2-methylbutyl carbocation is more stable than ethyl carbocation as there are more +I groups attached in 2-methylbutyl carbocation.
But according to my book the second option is correct.
Is my answer correct?
If not please explain

Strictly speaking, none of those answers are correct, provided your stem is right.
The reaction will start with protonation of the ether to give you the following species:

And the iodide nucleophile can attack either side via $S_{N}2$ mechanism. Since the left side is more accessible, it'll probably attack there first giving you:

Then, since we're still in the presence of HI and it is a strong acid, it protonates the alcohol and converts it into a corresponding alkyl iodide via either $S_N2$ like in the example above or $S_N1$:

A general rule of thumb here for the future reference is that in a hydrolysis of an ether with a strong hydrohalogen acid (HI, HBr), you get a mix of two corresponding alkyl halides with halogen where the oxygen used to be.


Oxidation of sodium thiosulfate by iodine When $\ce{Na2S2O3}$ reacts with $\ce{I2}$, it undergoes a redox reaction as

$$\ce{2 Na2S2O3 + I2 -> Na2S4O6 + 2 NaI}$$

Sulfur is being oxidized from a $+2$ state to a $+2.5$ state. Why isn't it getting oxidized any further, say to $3+$, to form a compound like $\ce{Na2HSO3}$?

I do notice that there's no hydrogen involved in the equation, but maybe the reaction occurs in an acidic medium?

Could someone please tell me whether the above scenario is possible or not and why?

Whether compound $\ce{X}$ is going to be reduced/oxidized and to what degree is mostly dictated by the corresponding redox potential for the given medium. The thiosulfate anion $\ce{S2O3^2-}$ is used in quantitative analysis and in iodometric titration in particular since it is a reasonably strong reducing agent [1 pp. 714-717]:

$$\ce{S4O6^2- + 2 e- <=> 2 S2O3^2-} \qquad E^\circ = \pu{0.169 V}$$
$$\ce{2 S2O3^2- + I2 -> S4O6^2- + 2 I-}$$

However, stronger oxidants such as chlorine are capable of oxidizing thiosulfate further down to sulfur(VI), e.g. forming a sulfate instead:

$$\ce{S2O3^2- + 4 Cl2 + 5 H2O -> 2 HSO4^- + 8 H+ + 8 Cl-}$$
Bromine, being intermediate between iodine and chlorine, can cause $\ce{S2O3^2-}$ to act either as a 1-electron or an 8-electron reducer
according to conditions. For example, in an amusing and instructive experiment, if concentrated aqueous solutions of $\ce{S2O3^2-}$ and $\ce{Br2}$ are titrated, and the titration is then repeated after having diluted both the $\ce{S2O3^2-}$ and $\ce{Br2}$ solutions 100-fold, then the titre will be found to have increased by a factor of exactly 8.

Note, however, that it is not entirely correct to state that sulfur in thiosulfate has oxidation state $+2$. In fact, there are two nonequivalent $\ce{S}$ atoms with oxidation states -1 and +5. Same goes for tetrathionate anion $\ce{S4O6^2−}$ where two central $\ce{S}$ atoms have oxidation state $0$ and the two terminal ones $+5$.
References

Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


What's the convention for the symbols of enthalpies of formation, reaction, etc.? I frequently encounter two ways of writing enthalpies of some process, e.g. for the enthalpy of formation:

$\Delta_{r} G$ (or $\Delta_{R} G$)
$\Delta G_{r}$ (or $\Delta G_{R}$)

On the IUPAC Gold Book webpage they used the first notation (although the use of upper- and lower-case descriptors varies, see here and here). Is this the accepted way of doing it (and the other one is simply a relic) or might this be a matter of preference. And how about the upper- or lower-case descriptor: Is there any convention for that?

I think you have already answered your question… IUPAC's usage should be a pretty good indicator of the proper notation.

There is, however, an even better reference than the Gold Book: IUPAC's Green Book, or Quantities, Units and Symbols in Physical Chemistry (full text here), on section 2.11.1 Other symbols and conventions in chemical thermodynamics (page 59 and following):

  (iii) Examples of the use of the symbol ∆
  The symbol ∆ denotes a change in an extensive thermodynamic quantity for a process. The addition of a subscript to the ∆ denotes a change in the property.

It says later:

  [∆vapH] can also be written ∆Hvap, but this usage is not recommended

The reason is that ∆Hvap could be understand as a change in Hvap, which it is not. It a change of H during a process, so the subscript should be on the ∆.


Largest change in the water level Which would produce the largest change in the $\ce{H2O}$ level when added to water in a $\pu{25 mL}$ graduated cylinder?
(A) $\pu{10.0 g}$ of $\ce{Hg}$ $(d = \pu{13.6 g/mL})$
(B) $\pu{7.42 g}$ of $\ce{Al}$ $(d = \pu{2.70 g/mL})$
(C) $\pu{5.09 g}$ of iron pyrite $(d = \pu{4.9 g/mL})$
(D) $\pu{2.68 g}$ of oak $(d = \pu{0.72 g/mL})$

I approached this question by dividing the mass by density to find the volume change. Thus, I got the answer (D), but the answer is (B).
Can anyone please explain this?

Because it’s density is less than 1, the oak floats.

Therefore it displaced less than its full volume; some of the floating oak is above water.

Since it’s 2.68g of oak, it displaces only the volume equivalent to 2.58g of water.


Standard for Chemistry Supplies In searching for chemistry kits it often leads to stuff like this:

I have seen some nicer "boxed sets" for the ~$500 range, but I don't think this is really ideal. This one looks a little nicer. I am wondering two things:

Where the best "complete" home/DIY chemistry set is that you can get online. Including the chemicals themselves, and the equipment like beakers, pipettes, etc.
Where the standard places are that smaller or bigger labs like University student or research labs get their supplies.

I feel like I read somewhere a while back that it is being outlawed because of bad behavior, but I wanted to ask to see what is possible in terms of doing chemistry experiments/analysis in a DIY/bootstrap fashion to learn about things like way in the past. I don't know much about chemistry but it would be interesting to perhaps learn how to do analysis like you might (I imagine) find in biology, geoscience, ecology, or agriculture.

Answer on question 2.

As a former chemistry student, I saw that most chemicals were ordered at:

Sigma Aldrich
Biosolve
Fisher Scientific
ThermoFisher Scientific
Local pharmacy (at least in Holland)

Equipment was often ordered directly at the manufacturer (like an HPLC) unless the device is pretty general, like a rotary evaporator.


Can carbon monoxide from a floor-heater in a poorly ventilated room eventually be ignited by the fire and explode? I have often wondered about carbon monoxide accidents caused by free-standing heaters burning in a poorly ventilated room causing fatalities. I am aware that CO is explosive between 12 and 75% volume air and wonder why it doesn't eventually ignite from the flames or glow present when the CO level reaches 12%? (This never seems to happen). Or does the decreasing oxygen content in the air due to the combustion cause the fire to die out before the critical level of 12% is reached?

Another possibility:  the "carbon monoxide" formed by combustion is not pure enough CO to get to flammability (12% total CO) in air.  The gas could be mostly carbon dioxide but still enough CO to poison you.


What happens when FeF3 is reacted with benzene? My text book says that halobenzenes can be prepared by aromatic substitution on benzene ring by using Fe or FeX3 as Lewis acid but this method isn't useful for preparation of fluoro benzene because of it's high reactivity.
My question is what will be the product if FeF3 is reacted with benzene, I tried searching a bit on internet but couldn't find anything helpful.

Iron catalyst will not be needed as Fluorine will react directly with benzene explosively,  the probable product formed would be hexaflurocyclohexane , which can predicted by looking at similar reaction of chlorine with benzene in presence of UV rays which is mentioned in NCERT class 11 chemistry  Textbook.

(Replace $Cl$ with $F$  and since fluorine is more reactive UV light is not needed.)


What are the properties of COₓ where x ≥ 3? $\ce{CO_x}$ are part of the oxocarbon molecule class. Commonly known are $\ce{CO}$ and $\ce{CO2}.$ However, there are molecules in which $x$ is equal to or larger than $3.$ Those molecules are barely known, other than the fact that they're unstable.

$\ce{CO3}$ decomposes slowly enough that it may come into contact with other molecules (including biomolecules), but nothing is known of its possible oxidizing potential.

$\ce{CO4},$ $\ce{CO5}$ and $\ce{CO6}$ are very unstable, and could possibly be strong oxidants (though it may never be known since they're too unstable).

Other than that, any information?

At least the trioxide $\ce{CO3}$ exists, and it has even been detected in at least two isomeric forms.  Wikipedia reports that it forms as a transient species in reactions between carbon dioxide and atomic oxygen or ozone, in irradiation of solid carbon dioxide, and even in oxidation of carbon monoxide by $\ce{O2}$.


How does selenium protect against mercury poisoning? I have read that it is not essential to avoid fish with mercury in them if they also have a correspondingly high level of selenium because mercury poisoning is primarily primarily the result of mercury disabling selenium based antioxidants.

How does selenium intake protect against mercury poisoning?  Can enough selenium counteract the effects of high mercury intake and does it matter what kind of exposure took place (breathing vapors etc)?

Selenium has a very high affinity for mercury. It strongly binds mercury to itself, thus neutralizing the reactivity of mercury. The complex formed Hg - Se is not absorbed by body and excreted out from the body.

In this way mercury is unable to lodge in fatty tissues, resulting in no damage to the body.


General expression for entropy and free energy The specific heat capacity of a solid at constant pressure and low temperatures is given by $$c_p=aT+bT^3$$ with $a$ and $b$ are constants.

How can I calculate a general expression for the entropy $S(T,p)$ and Gibbs free energy $G(T,p)$ as a function of $T$, assuming that $S$ and $G$ are zero at $T = 0~\mathrm{K}$?

I started with trying to calculate entropy by
$$\mathrm{d}S = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T = \frac {c_p}{T} \mathrm{d}T$$
and
$$S = \int_0^T \frac{c_p}{T}\mathrm{d}T = (aT + bT^3)\ln T$$

Since G and S are taken to be zero at T = 0, H is also 0 at T = 0.  So you can calculate H by integrating the heat capacity from 0 to T:  $$H=\int_0^TC_pdT=\frac{a}{2}T^2+\frac{b}{4}T^4$$
Then you substitute H into the equation $$\frac{\partial (G/T)}{\partial T}=-\frac{H}{T^2}=-\frac{a}{2}-\frac{b}{4}T^2$$
and integrate between 0 and T.  This will give you G.  You can then solve for S from G = H - TS.  This gives you the same result as integrating directly to get S.


Confusions about inductive effect and acid strength If one of the hydrogens in the alkyl side chain of ethanoic acid is replaced with an atom of Chlorine, the acid has a larger Ka value. My book says this is since the Cl "borrows" electron density from the Carbon it is connected to. And, in this example, this carbon borrows some electron density from the Carbon in the acid group, and thus the oxygens of the acid have less charge density. The bold part is what I don't get.

The C to which this 1 chlorine is attached gets a bit electron deficient but not as much deficient as the C in the acid group (since this has two oxygen atoms attached). So how can it borrow density away from the C in the acid group? In my mind, to borrow some charge density you have to be more electronegative than what you're attached to, and the more deficient carbon should have a greater pull.

What am I missing in this?

The result is correct but the explanation is bad.

The main reason is that when you compare CH3COOH and CH3ClCOOH , the main difference is that the alpha carbon in the former has more electron density than the latter one because of which when the conjugate base gets formed, it is more stabilised than ethanoic acid because of low electron density.

You are correct about this "In my mind, to borrow some charge density you have to be more electronegative than what you're attached to, and the more deficient carbon should have a greater pull".

Edit 1 - The orbitals of CH3COOH can provide more electron density than CH2ClCOOH due to the former having more electron rich orbitals which interferes constructively with the orbitals of other C to satisfy its demands. Think in terms of how Hyper conjugation or resonance works in terms of interference of wave functions.


Find the molarity of the filtrate I can't figure out this stoichiometry question!

10.0 mL of 1.15M copper(II) nitrate reacted with 20.0 mL of .85M ammonium hydroxide.  The solution is filtered. Determine the molarity of the filtrate.

I figured out that the net ionic equation is Cu + 2OH -> Cu(OH)2, so the 2NH4NO3 dissolves, and the Cu(OH)2 probably doesn't pass through the filter, but I don't know where to go from there. Thanks in advance!

$$\ce{Cu(NO3)2}+2\ce{NH4OH} =>\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$
Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)

First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$
$$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$

Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry.
So actually, $\ce{NH4OH}$ our limiting reagent.
Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became  $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%.

0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$  Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.


Identifying the number of bonding and lone pairs without a dot and cross diagram I've just learned how to predict the shapes of molecules in class today using VSEPR theory.
I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?

There is a simple, four-step calculation that you can perform:

count the valence electrons the atoms you are bonding have.
count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)
substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.
substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

Using a simple example such as sulphur dioxide:

$6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$
$8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$
$24 - 18 = 6$, i.e. three bonding electron pairs.
$18 - 6 = 12$ i.e. six lone pairs.

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:

$$\ce{O=S+-O-}$$

With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.


Is there any theory more advanced than MOT? I am currently going through molecular orbital theory(MOT), I should appreciate it's a very nice theory and it's helping me to fill the gaps of normal hybridization concepts and valance bond theory(VBT). But I am unaware about it's limitations seemingly it's the most advanced theory and have the capability of explaining everything. Although we know everything has it's limitations so if exist then what are they?
It will be also interesting to know about some modern viewpoints, accepted or proposed theory, and concepts regarding this stuffs.

MO theory is more of a descriptive guide than a rigorous mathematical construct. The main idea of molecular orbital theory is that the electrons in atoms can be described by linear combinations of atomic orbitals (LCAOs), which themselves are the exact solutions to the Schrodinger equation for single electron atomic systems.
This idea can be applied in a more mathematically quantitative way by use of the Hartree-Fock (HF) method. HF is essentially MO theory put into practice. HF takes as input the geometry of a system (i.e. the spatial coordinates of all atoms in your molecule) and a set of "basis functions" which are usually sets of 3D gaussian functions centered on the atomic nuclei of the system. These basis functions are your "atomic orbitals" in MO theory. The HF method takes linear combinations of the atomic orbitals to generate a set of molecular orbitals which are usually more delocalized. These are the molecular orbitals of MO theory, and we can calculate them using the HF method running on a computer.
There are several approximations involved in HF theory, which the descriptive MO theory inherits.

The Schrodinger equation (SE) itself is an approximation. The SE is not exact as it fails to account for so-called relativistic effects. These are effects originating from Einstein's theories of special and general relativity. Some examples of relativistic effects are spin-orbit coupling and relativistic contraction of orbitals, especially in heavy atoms. The straightforward HF method doesn't take relativistic effects into account. Starting instead from the Klien-Gordon or Dirac equations will recover (some of) the missing relativistic effects.

The HF method assumes a complete separation between the movement of electrons and nuclei, known as the Born-Oppemheimer approximation. Basically, electrons are assumed to equilibrate instantaneously to the positions of atomic nuclei. This usually isn't a problem because electrons are much lighter than atomic nuclei so they do move much faster. Not a bad approximation in most cases.

Electron correlation. Even if we ignore the previous two approximations, the HF answer (and therefore the MO theory prescription) misses out on a very chemically important effect known as electron correlation. Electron correlation arises because a single set of molecular orbitals combined into a Slater determinant (in order to satisfy the Pauli exclusion principle) is not enough to fully describe the true exact solution to the Schrodinger equation for a molecular system. One must take linear combinations of many Slater determinants to improve on the HF answer. Electron correlation is responsible for effects such as the Van der Waals dispersion force. Another way of thinking about Electron correlation is that it is the energy that arises due to the dynamic Coulomb repulsion between moving electrons (although this picture is incomplete).

In summary, an exact molecular theory would need to account for Einstein's theory of relativity, combine the movement of atomic nuclei with electrons, and account for the dynamic correlations between different electrons in a system, in addition to being a proper quantum mechanical theory.
There are some quantum chemistry theories out there that incorporate relativistic effects into them. For example, the Douglas-Kroll-Hess approach is similar to HF theory, but begins from the Dirac equation and so incorporates relativistic effects explicitly. I believe there are also approaches that don't use the Born-Oppemheimer approximation, though I'm unfamiliar with them.
There is a massive amount of research and development into the so-called post Hartree-Fock methods. All of them attempt to account for electron correlation in various different ways. The main issue with the post HF methods is that the results (energies and wavefunctions) are often incredibly computationally demanding to actually compute, except on very small molecules. The most exact of the post-HF methods is the Full Configuration Interaction (FCI). With a sufficiently large input basis set, FCI produces an exact answer to the Schrodinger equation under the Born Oppenheimer approximation. Other post-HF methods include: Møller-Plesset perturbation theory, Coupled Cluster theory, and Multi-Configurational Self Consistent Field theory. All of these post HF methods are more accurate than the MO theory picture, but they are also way more difficult to visualize and understand for new students.


Why can't I dry/concentrate ethanol using calcium chloride ... or can I? My (very unreliable) school textbook tells me:

  Lower alcohols form a solid derivative with certain metal salts.

and,

  It is for this reason that ethanol cannot be dried/concentrated using anhydrous calcium chloride.

Nowhere else in the book, is there any reference to this supposed reaction between a lower alcohol (ethanol) and a metal salt (calcium chloride). So I couldn't read more about this from there...

Googling wasn't much help; though I did find this paper that seems to concur with my textbook (albeit, with more detail) ... the paper is dated all the way back to 1923, so I suspect the modern Chemist's view of the matter would be different (the paper makes use of the term "alcoholate", which no longer carries the same meaning).

Question:

Does ethanol really react with calcium chloride to form some "solid derivative"?

Additionally,

Do lower alcohols (methanol and ethanol to name a few) really react with "metal salts" (textbook didn't specify, so i would use the following examples: copper sulfate and magnesium chloride)?

The apparent answer is yes, to quote from a source (composed from extracts of old chemistry journal):

  Calcium chloride forms addition compounds with the alcohols. On evaporating a solution in ethyl alcohol at a low temperature rectangular plates of $\ce{2CaCl2.7C2H5OH}$ are deposited. The compounds $\ce{CaCl2.3C2H5OH}$ and $\ce{CaCl2.CH3OH}$ have also been separated, as well as a mono- and a di-acetone compound, and compounds with isobutyl and amyl alcohols.

So, if you are working with a cold solution of $\ce{C2H5OH}$ with $\ce{CaCl2}$ exposed to air undergoing evaporation, it is possible to form an addition compound (taking the form of rectangular plates) with the formula, $\ce{2CaCl2.7C2H5OH}$ or $\ce{CaCl2.3C2H5OH}$.


Why does ink form tendrils on wet paper When a drop of ink or watercolour paint is placed on wet paper, rather than uniformly diffusing in all directions it forms tendrils which rapidly expand out from the spot.
Why is this and what is the chemistry or physics occuring?
This video may further show the type of thing I am referring to
https://motionarray.com/stock-video/black-silver-tendrils-pack-216423
It appears that the pigment solvent forms discrete channels outwards through the water. I wondered if this might be a surface tension interaction or something else.

If you’re talking about a gentle placement of the drop on the paper, I’d guess it’s likely due to the fibrous structure of the paper providing “easier” paths for the ink to flow. If you’re talking about an impacting drop, the drop will vibrate when it impacts the paper until the forces of surface tension are exceeded and the drop “splatters”.


What is the structure of the hypothetical xenon octafluoride? According to Frlec et al.,[1] theoretical studies indicated that there is a possibility (albeit slim) that neutral xenon octafluoride ($\ce{XeF8}$) might exist.

I am curious as to what structure this molecule might possibly have. In terms of hybridisation theory, I have only learnt cases up to seven-coordination ($\mathrm{d^3sp^3}$ or $\mathrm{sp^3d^3}$), where the molecular geometry is pentagonal bipyramidal. However, answers based on other frameworks (e.g. molecular orbital theory) would also be welcome.

My conjecture: could its structure possibly be hexagonal bipyramidal?

Frlec, B.; Holloway, J. H.; Slivnik, J.; Šmalc, A.; Volavšek, B.; Zemljič, A. An examination of the possibility of the existence of xenon octafluoride. J. Inorg. Nucl. Chem. 1970, 32 (8), 2521–2527. DOI: 10.1016/0022-1902(70)80296-2.

There does seem to be a good body of evidence, including your reference, for the existence of xenon octafluoride.

Regarding it's structure, according to VSEPR theory as tabulated here, the structure is square anti prism:


Will ground glass joints relieve pressure non-catastrophically? I need to heat a fluid for seven days at a time. The fluid goes in round bottom flask that is suspended in a water bath. The test fluid will decompose and generate a small amount of gas over time.

Gas generation is variable and dependent on what material I am testing in this material compatibility experiment.

I have experience in plumbing and fluid systems but not chemistry. In a plumbing system I would add a relief valve to relieve the pressure generated over time, without allowing potentially dirty air to contaminate the experiment.

I have a ground glass joint. I found a source that says the ground glass plug will lift to relieve pressure, even if I put a plastic clip on it, and the plastic clip would ensure the plug would not go flying.

I still have concern since this is an environment I am not familiar with, and I am not keen on exploding glassware in the facility. Past experimenters have placed small beakers upside down on top of the flask, and have also used aluminum foil as a cover (I will not do this so the vapor contact with foil does not affect the experiment results). I tried it with a beaker but a moderate amount of fluid condensed on the surface, which was not ideal.

Will I be fine securing a ground glass joint with a plastic clip, should I use an upside down beaker, or is there a more obvious solution?

Ground-glass joints are notorious for seizing at the wrong time, particularly if reactants are alkaline.

If pressure might change, I'd add some relief mechanism, such as a water-trap or rupture disk. Without knowing your setup, something as simple as a balloon might work, if not attacked by reactants.


Solvents in substitution reactions My book says that to favour a reaction by SN1 mechanism we need to use a polar protic solvent and a polar aprotic solvent in a reaction by SN2.
* My Prior Knowledge:*

  There are two types of polar sovents : Polar Protic Solvents contain a hydrogen bonded to a highly electronegative atom. They can solvate both positively and negatively charged species. On the other hand Polar Aprotic Solvents can only solvate positively charged species to an appreciable extent as there is no hydrogen bonding.

A question which arises now is that why should we use a Polar Aprotic Solvent in an SN2 reaction. The leaving group (which usually is negatively charged) needs to be stabilized as well. Wouldn't it be perfectly fine if I used a Polar Protic Solvent? If this is fine then why does presence of Polar Protic Solvent favor SN1 over SN2?
Also I asked this question on SE a while back. I am looking for a more satisfactory answer apart from the ones given. here

Take the example of methyl bromide reacting with chloride ions:

In protic solvents the chloride ion is strongly solvated by hydrogen bonding and so the energy of the reactants is lowered, compared to in aprotic solvents. For the SN2 mechanism, the transition state is less polar than the reactants because the negative charge is spread out across the chlorine and the bromine and so the transition state is less stabilised by protic solvents than the reactants, increasing the activation energy of the reaction.

In aprotic solvents, the reactants and transition state are less stabilised because of the lack of hydrogen bonding but crucially the difference between the stabilisation of the reactants and the transition state is less and so the activation energy is lower, meaning that the SN2 reaction proceeds more quickly in aprotic solvents than in protic solvents.

The reverse argument applies to SN1 reactions where the first transition state is more polar than the reactants and so is more stabilised by protic solvents.


Determine polarity of covalent bond with formal charge In a covalent molecule, can we conclude that if the formal charge between two atoms are +1 and -1 respectively, then the chemical bond acting between them is a polar covalent bond (such as the bond between N and single-bonded O in Nitrate Ion)? On the other hand, if both of the atoms has a formal charge of 0, then the chemical bond acting between them is a nonpolar covalent bond (bonds in sulfur dioxide and sulfur trioxide)? How about bond between atoms with formal charge of +1 and 0 (P with single-bonded O in phosphate ion)?

All bonds between two atoms of different elements are polar to some extent. To determine how polar, you need to check the elements’ electronegativities and only them.

Formal charges are exactly what their name implies: a formality. When drawing Lewis structures you should attempt to generate all-octet structures but often this isn’t possible without moving electrons entirely between atoms. In that case, formal charges are assigned to rationalise e.g. a nitrogen with only four electrons.

Check out these structures of nitric acid, sulphur trioxide and carbon monoxide. (Resonance structures applicable to $\ce{HNO3}$ and $\ce{SO3}$ not displayed.)  All bonds are polarised towards the oxygen, since oxygen’s electronegativity is $3.5$; the second highest amoung the non-noble-gas elements. Note especially that the bonds in $\ce{CO}$ are polarised towards oxygen despite oxygen bearing a formal positive charge. (There is a lot more to $\ce{CO}$ which is shown well in Martin’s answer including the molecular orbitals)


How to balance the redox reaction between bismuth(III) sulfide and nitrate anion? I have to solve the following redox reaction which occurs in an acidic environment, using the half-equation method.

$$\ce {Bi2S3_{(s)} + NO3-_{(aq)} -> BiO3-_{(aq)} + SO4^2-_{(aq)} + NO_{(g)}}$$

There seems to be a lot elements that change their oxidation numbers, except for oxygen:

$\ce{Bi}$ from $\ce{+III}$ to $\ce{+V}$
$\ce{S}$ from $\ce{-II}$ to $\ce{+VI}$
$\ce{N}$ from $\ce{+V}$ to $\ce{+II}$

What should I begin with? How would one go about grouping three redox pairs?

While in general a reaction involving three independent redox pairs may not have unique coefficients, this particular one contains one coupled oxidation half-reaction and one regular reduction half-reaction.

The key is to note that both elements in $\ce{Bi2S3}$ are oxidized, and hence we have a stoichiometric relationship between the two oxidation half-reactions. This suggests we write the oxidation half-reaction as follows: $$\ce{18H2O + Bi2S3 -> 2BiO3- + 3SO4^{2-} + 36H+ + 28e-,}$$ and we can proceed with the usual algorithm for balancing half-reactions.


What function is used to graph the potential curve of two hydrogens with parallel spins? TL; DR
I am asking for

an analytic potential curve function for the triplet state of $\ce{H2}$ (preferred) with experimental parameters, or
experimental data (enough points to realise a smooth-looking curve).

Also see solid red curve pictured below.

To characterise the energy of a two-atom system, Morse potential is applied as a second approximation after the harmonic oscillator model. This is an analytic function of the form $$E=D_\mathrm{e}\left\{1-\exp[-a(r-r_\mathrm{c})]\right\}$$
where the parameters $D_e$, $a$ and $r_c$ depend on the atoms in play. Surely enough, better results are achievable. For examples on simple molecules, I found the paper by Hulburt and Hirschfelder, 'Potential Energy Functions for Diatomic Molecules', J. Phys. Chem., ($1940$) to be rather useful. Have a look at equations $(14)$ and ($15$) specifically (available more easily here). The one I used was

$$E_{\mathrm{P}}=\mathrm{D_e}[(1-\mathrm{e}^{-z})^2+\mathrm{c}x^3\mathrm{e}^{-2z}(1+\mathrm{b}z)],$$
where
$$z=\frac{\omega_\mathrm{e}}{{2}\sqrt{\mathrm{B_eD_e}}}\left(\frac{x-\mathrm{r_0}}{\mathrm{r_0}}\right)$$

This is for the case where spins are antiparallel.
If atoms of parallel spin approach one another, the qualitative picture is quite different. A smaller distance will equal higher energy regardless of separation, i.e., no stable equilibrium state exists. Unfortunately, finding a nice function for the antibonding situation proved less fruitful. An example of what I'm after is the solid red curve found in this Wolfram demonstration.

I believe that you wont be able to find a closed analytic expression for $\mathrm{H_2}$, since no closed analytic expresion is possible for a many-body problem (more than 2 bodies). That said you can calculate as many points on the triplet curve as you like by using a QM program able of doing unrestricted Hartree-Fock (UHF). Here you should be aware that UHF does not give very good potential curves, because the Hartree-Fock approximation lacks the electron correleation energy. $\mathrm{H_2}$ is a quite small system, so you will be able to make a potential curve using unrestricted MP2 or unrectriced CCSD. I do believe that PySCF is capable of doing such calculations. Regarding the experimental data, I am not aware of any databases for such kind of data. I would suggest finding a paper where a curve is presented and then use engauge digitizer to extract the points you need.


Halogens and the free energy of activation in a resonant benzene system If we look at the right-hand-side table here: http://en.wikipedia.org/wiki/Hammett_equation, we notice something particular.  It appears that the free energy of activation (for example, measured in terms of the acidicity of a carboxyl group on a benzene ring), and with respect to substituents in the para- position on the benzene ring, increases as one moves down column 17 on the periodic table.

Why is this happening?  Shouldn't we expect the opposite to be true?  Consider that, as we travel down column 17 corresponding to the halogens, electronegativity decreases and the probability of resonance contributions via pi-bonding should increase.

Resonance between a carbon and a halogen atom decreases as we move down the series $\ce{F, Cl, Br, I}$.  This is because overlap is most effective between orbitals of the same size.  Therefor, carbon-fluorine overlap, both involving $\ce{2p-2p}$ orbitals should be much stronger than carbon $\ce{2p}$ orbital overlap with the $\ce{3p, 4p, etc_{.}}$ orbitals in the larger halogens.  Consequently carbon-halogen resonance becomes less significant as we move down the halogen series.  If we now consider that 1) the resonance and inductive effects for the para-halogens are in opposition to each other (tend to counteract each other) and 2) the resonance effect decreases as we move down the halogen series, then the order of halogens in the Table you mentioned makes sense.


Why is carbon monoxide a neutral oxide? Carbon monoxide ($\ce{CO}$) is considered to be a neutral oxide, not acidic or basic. But why? It reacts with hot concentrated sodium hydroxide solution to give a solution of sodium methanoate:

$$\ce{NaOH + CO -> HCOONa}$$

Doesn't the reaction of carbon monoxide with the basic hydroxide ion show that it must be acidic?

Well, carbon monoxide can be created from formic acid by adding sulphuric acid which will dehydrate said formic acid:

$$\ce{HCOOH ->[\ce{H2SO4}] CO + H2O}$$

Therefore, we can imagine the reverse reaction theoretically, which would make carbon monoxide an acidic oxide. However, the forward reaction does not proceed easily and it needs both the high acidity of sulphuric acid and its strong dehydrative properties to actually work. And your question mentions using hot, concentrated sodium hydroxide to make the reverse one work.

Most oxides that are classified as acidic or basic either have a very electrophilic central atom (e.g. $\ce{CO2}$) which can be attacked by the weak nucleophile water (which in turn can then release an acidic proton), or they have a high charge density on the oxygen which allows it to abstract a proton from water directly. Carbon monoxide is neither. If you check out its molecular orbitals, you will notice that even though carbon is partially positive it has the largest HOMO contribution, meaning a proton would be more likely to attatch to the carbon side — which doesn’t want one at all. The LUMO is, luckily, also more carbon-centred, meaning nucleophilic attacks on carbon are possible. However, it is also degenerate due to the double bond so that an attack is not favoured.

Thus, the carbon monoxide molecule is one that won’t react with water at all and totally defies the concept of acidic/basic oxides.

Abbreviations:

HOMO is a widely used abbreviation for the Highest Occupied Molecular Orbital, i.e. the one with the highest energy that still contains electrons. It is usually the orbital that will attack nucleophilicly or that will be attacked electrophilicly.
LUMO is a widely used abbreviation for the Lowest Unoccupied Molecular Orbital, i.e. the virtual (unoccupied) orbital that has the lowest energy. When considering a nucleophilic attack, the attacking electrons will usually interact with the LUMO. Electrophiles attack with other molecules’ HOMO with their LUMO.


Decomposition products of magnesium ethanoate What does magnesium ethanoate decompose to? The Wikipedia page states that one of the products is magnesium oxide. What's the other product? I can come up with some guesses:

Ethanoic anhydride ($\ce{MgC4H6O4(s) -> C4H6O3(g) + MgO(s)}$)
Propanone and carbon dioxide ($\ce{MgC4H6O4(s) -> C3H6O(g) + CO2(g) + MgO(s)}$)

EDIT: Ok, here's your answer (Thermochimica Acta, 78 (1984) 17-27): It decomposes into the oxide, carbon dioxide and propanone:

$Mg(C_2H_3O_2)_{2(s)} \rightarrow MgO_{(s)} + CO_{2(g)} + CH_3COCH_{3(g)}$

PRE-EDIT:

The Russian Wikipedia favors the second reaction. This seems to support the idea.

This site mentions magnesium carbonate and propane-2-one as products, but this seems to be questionable, given the other sources.


Converting SMILES to .sdf files I have around 1000 SMILES entries in Excel and I have to generate .sdf file for all. Is there any way to automate this whole process, preferably with an offline tool?

Datawarrior lets you do it.
Export a csv sheet with two columns: The name of the compound and its SMILES code.
Be sure to use a comma as a delimiter.
Your output file, opened with a text editor, should look like this:

Name,SMILES
Hexazine,n1nnnnn1
Furazan,c1nonc1
Benzisoxazole,c1noc2c1cccc2

From DataWarrior, File>Open and open your file.
Structures are automatically generated from SMILES codes.

Then, File>Save Special>SD-File, and you are done.


how to calculate the initial concentration of NaOH in a mixture? I have doubts about how to find the equation for the calculation of the initial concentration of $\ce{NaOH}$. I have the following information:
 in the kinetic study of a reaction,

We measure $\ce{20.00 ml}$ of $\ce{CH_3COOC_2H_5}$ solution of 0.0400 $\ce{mol/L}$ and also $\ce{20.00 ml}$ of $\ce{NaOH}$ solution of 0.0400 $\ce{mol/L}$

We mix both substances in one Erlenmeyer.

The question is:  find the equation to calculate the concentration of $\ce{NaOH}$ at $\ce{t=0}$.

It is a bit weird to me because a concentration of the $\ce{NaOH}$ solution is already given.

The best (and the only) that I can think is this:
if the concentrations of the reactives are equal, then:
$$\ce{C_iV_i = C_fV_f}$$

then we have,
$$\ce{C_i=C_fV_f/V_i}$$

$\ce{C_i}$ is the initial concentration of $\ce{NaOH}$, $\ce{V_i}$ is the initial volume of $\ce{NaOH}$, $\ce{C_f}$ is the final concentration of the whole mixture ( which I don't know how to find), and $\ce{V_f}$ is the final volume of the whole mixture

Probably this is wrong but I don't have any other ideas.
 any clues?

Thanks.

(Note: apologies for my poor English)

As @MaxW pointed out, if you assume that the volumes are additive implies that your total volume is 40 mL. Now, remember the definition of molar concentration:
$$C=\frac{\text{moles solute}}{\text{total volume}}=\frac{n}{V}$$

That means that the concentration and the volume are inversely proportional, thus if initially, you have a $\text{0.0400 } \frac{\text{mol}}{\text{L}}$ in 20 mL and then you increase the volume twice, then your concentration decrease to the half of initial value. In numbers:
$$\frac{\overset{\text{this are your moles of solute}}{\overbrace{\frac{\text{0.0400 mol}}{\text{1 L}}\times{\text{20 mL}}}}}{\underset{\text{this is your final volume}}{\underbrace{\text{40 mL}}}}=\text{0.0400 }\frac{\text{mol}}{\text{L}}\times\frac{1}{2}=\text{0.0200 }\frac{\text{mol}}{\text{L}} $$


Why isn't viscosity considered a property of solids? Viscosity is defined as a measure of resistance to flow which arises dues to internal friction between layers of fluid as they slip past one another while liquid flows.

Why can't we say that viscosity is a property of solids as well? Except, it would then be said that solids have infinite viscosity.

The concept of deformations of solids under loading has a long history. It just isn't called 'viscosity' - instead this is just part of the mechanical properties of solids.

Relevant places to start include:

Wikipedia on Plasticity

Wikipedia on Deformation

The Non-Destructive Testing org on Elastic/Plastic Deformation

In the purely elastic regime, if you pull on a steel bar it will extend under the load, and will return to its original dimensions when the load is removed. If you apply too large of a load, it will plastically deform, so that the dimensions will not be the same when the load is removed. Under some circumstances even fairly small loads over long times can result in permanent deformation.

Physically, these can be a result of:

Dislocation formation and movement in crystals (e.g. slip bands)
Grain boundary motion
Creep (point defect formation and diffusion)
Crazing in plastics (formation of open volumes)
And a number of other, less common, mechanisms.

These topics are typically covered in metallurgy and materials science courses/textbooks. Again, even in amorphous solids, one would not tend to call it viscosity - that is reserved for liquids.


Why are acid catalysed reactions of carbonyl compounds reversible? I am currently studying aldehydes and ketones and their reactions. I've noticed that acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. Why is this so? Is this because the addition of the nucleophilic base stabilises the electrophilic carbon? Or is there some other reason?

Update: Sorry, "all" would not be appropriate here. In my text-book, it is mentioned only once. Nucleophilic substitution by weak nucleophiles is carried out in a weakly acidic medium, while a basic medium is used for strong nucleophiles. The corresponding mechanism use the convention I've spoken about. I can't see the use of reversible arrows elsewhere. I wanted to know why is the nucleophilic addition reversible (as per my book) in a weakly acidic medium.

My text book (not a standard one): Nootan ISC Chemistry XII

The pictures of the reactions (I'm not able to rotate them for some reason)

This is a picture from wiki (Simple aldol reaction). Both Acid and base catalyzed reactions are reversible up to aldol intermediate. Dehydration is the only irreversible step. So this statement " I've noticed that all acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. " is incorrect.

Single arrow means that the step is irreversible. Reversible arrow means that reaction can go both forward and backward. Usually reversible steps are happening before the irreversible step and a happening faster than irreversible step.

This is easy to demonstrate with a pool table that nave a small obstacle near the hole. Reversible reactions are balls that move over the the surface of the table. They can get from any point to any other point of the table. Sometimes they get on that small obstacle near the hole and roll back to the table (they were not able to overcome activation barrier). Irreversible reactions are represented by balls that were able to overcome the barrier and now ended up in the hole. Now they have no chance to go back to the table on their own.

What makes reaction irreversible vs reversible. From thermodynamics perspective irreversible reactions are reactions with a large change in $\Delta$G. From organic chemists perspective reversible reactions are typically those that include protonation/deprotonation steps. Irreversible reactions are those where a strong bond like C-C or O-P forms.

Reversibility is a subject to conditions. 2H -> H2 (recombination of radicals) is effectively irreversible at room temperature, but is reversible at 3000K.

Why do we need to account for reversible reactions? (1) To understand the mechanism. (2) To know how to increase concentration of the reagents right before the first irreversible step to accelerate the overall reaction.

tl;dr: in reactions you are asking about formation of C-C bond is effectively irreversible step.


Unequal ionization energies of methane Why does methane have two different ionization potentials? How does this work? I understand that MO theory predicts C-H bonds of differing strength, while hybridization predicts C-H bonds of varying strength.

http://books.google.com/books?id=gY-Sxijk_tMC&pg=PA827&dq=methane+ionization&hl=en&sa=X&ei=VrMVVN3JLI-vogTv8IGABw&ved=0CCgQ6AEwAA#v=onepage&q=methane%20ionization&f=false

Is methane ionized successively and is are there two different ionization energies found? Doesn't this make intuitive sense; by ionizing methane we end up with a fundamentally different molecule which would more than likely have different ionization energies.

Also, how are these ionizations performed? The reference given above makes no mention of whether the bonds are broken homolytically or heterolytically (although ionization implies heterolytic bond cleavage). Even then, there are two ways bonds can be cleaved heterolytically.

In the photoelectron spectroscopy (PES) experiment a sample is irradiated with photons of differing energy and the kinetic energy of any ejected electrons is measured.  By knowing the energy of the incoming photon and the kinetic energy of any ejected electrons, we can calculate the energy required to ionize the electron as the difference between these two recorded values. An electron will only be ejected if the photon energy is equal to or greater than the energy required to ionize an electron in the molecule.  See the section entitled "Photoelectron Spectroscopy" in this link if you would like to read more.

In the case of methane, all of the bonds are equivalent as required by symmetry.  There is only one $\ce{C-H}$ bond strength for methane.

When we run the PES experiment with methane we see 3 signals (as pictured in your link; two of the signals are due to photoionization of electrons from the $\ce{C-H}$ bond, and the third signal, at much higher photon energy, is due to photoionization of the carbon 1s electrons).  We know that the 4 bonds in methane can be described by 8 orbitals, 4 bonding, 4 antibonding.  We can transform these 8 localized orbitals into 8 molecular orbitals; the 4 bonding MO's are pictured in the link you provided.  Notice from the picture in your link that 3 of the 4 MO's are degenerate, but there is also a different 4th MO.  All of these bonding MO's are filled and ionization can occur from either one of the 2 types - hence 2 signals from the occupied $\ce{C-H}$ MO's (in the PES experiment, these are not sequential ionizations, e.g. these are not first and second ionizations; rather they are independent ionizations that will occur only when our incoming photon energy matches the energy of the orbital containing electrons).  During the PES experiment with methane, no bonds are broken, rather an electron from an orbital is ejected and the methyl radical cation is formed.


How do I arrive at the mass of solvent from a molar fraction concentration? So the question says:

  If the mole fraction for $\ce{H2SO4}$ in an aqueous solution is $0.325$, what is the mass of water (in grams) in $100\ \mathrm{mL}$ of solution?

Me, and my other classmates tried to solve this one but we didn't reach a fully correct answer and our teacher arrives at a different answer.

My answer is:

Since X(H2SO4) + X(H2O) = 1

So: 1-0.325=0.675

X(H2O)=0.675

The mole percent is (0.675*100)=67.5%

1 mole percent = 0.01 mol

H2O moles = 0.675mol

H2O mass (g) = (0.675) * (18) = 12.15 g

Teacher is not convinced with this answer and say its wrong because the total must be 100g, and if the H2O mass is 12.15g then H2SO4 mass will be 31.85g, and 12.25+31.85 dont equal 100.

His answer is simple as follows:

1-0.325=0.675

0.675=67.5% of H2O

H2O mass= 67.5g

Both your teacher and your answer are wrong!

There are two steps to this problem.  First, we need to convert between mass fractions and mole fractions.  Second, we need to convert from a per-amount basis to a per-volume basis.  "Per-amount" is my made up word for all units such as mass fraction, mole fraction, molality, etc that are expressed per amount (whether moles or mass) of substance (whether total or solvent).  "Per-volume" units include things like molarity, grams per liter, etc., where the basis is the volume of solution.  Converting from an amount basis to a volume basis requires knowing the density of the solution!

Step by step:

Converting to mass fraction

$$ 0.325 \frac{\mathrm{mol\;\ce{H2SO4}}}{\mathrm{mol\;total}} \Rightarrow  \frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}}$$

$$\frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}} \times \frac{98.1\mathrm{\frac{g\;\ce{H2SO4}}{mol\;\ce{H2SO4}}}}{18\mathrm{\frac{g\;\ce{H2O}}{mol\;\ce{H2O}}}}=2.62\mathrm{\frac{g\;\ce{H2SO4}}{g\;\ce{H2O}}}\Rightarrow \frac{2.62}{2.62+1}\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}=0.724\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}\Rightarrow (1-0.724)\mathrm{\frac{g\;\ce{H2O}}{g\;total}}=0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}$$

This step is pretty easy to do using the information in the problem.  There are 72.4 grams of sulfuric acid present in 100 g of the solution.  So far, so good.

Converting to a per volume basis

But at this point the problem gets very tricky:

  ...what is the mass of water (in grams) in 100 mL of solution?

It says 100 mL, not 100 g.  This makes the problem much harder.  This also makes your teacher's answer wrong:

  Teacher is not convinced with this answer and says its wrong because the total must be 100g

The total doesn't need to be 100 g because we are apparently dealing with 100 mL of solution.  Depending on the density of the solution, the total mass will be more or less than 100 g.

According to Wikipedia, sulfuric acid at a mass fraction of 0.7 has a density of 1.60 kg/L and at a mass fraction of 0.78 has a density of 1.70 kg/L.  Let's suppose a mass fraction of 0.724 has a density of ~1.65 kg/L.

$$0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}\times\frac{1650\mathrm{\;g\;total}}{\mathrm{L}}= 46\mathrm{\frac{g\;\ce{H2O}}{L}}$$

With the final number, now it is easy to see that if there are 430 grams of water per liter, then in 100 mL there are 46 grams of water.


Work done in Irreversible Isothermal Expansion This is the graph for the Isothermal Expansion of an Ideal Gas. Sorry for the extremely basic doubt, but I just could not understand how the expansion was Isothermal if the graph was a straight line. The pressure is changed once suddenly and then remains constant.
Why don't we consider this as an Isobaric expansion instead of Isothermal if the graph is not even an isotherm? If a gas is expanding but the pressure has to be constant then why doesn't the temperature increase?

There is some confusion about terminology. The figure shows two processes. The curved path represents a reversible expansion in which there is pressure balance between system and surroundings throughout, so that the applied pressure can be described by the ideal gas law (or Boyle's law $p=\frac{p_1V_1}{V}$). This process is clearly isothermal, since $pV=nRT$.
The second process shown as two straight paths consists of a pressure drop and and an expansion against constant external pressure and is also isothermal (in thermodynamic lingo). The fact that the initial and final points are at equal T makes it isothermal. The temperature (and other properties except presumably composition) of the system during the process is on the other hand not well-defined. This process is irreversible.


If you put two blocks of an element together, why don't they bond? Say you have two lumps or blocks of an element, like lithium for example, say in the form of two bars.

Why, when you bring the two bars together so that they touch each other, do they not instantly bond with each other forming one larger bar or block? We can weld elements together so they 'stick' to each other, but what is the process that actually causes two like elements to bond together? Why do we need to 'weld' two bars together - why don't they just bond on their own?

Why, when you bring the two bars together so that they touch each other, do they not instantly bond with each other forming one larger bar or block? ... Why do we need to 'weld' two bars together - why don't they just bond on their own?

The problem is generally one of two things: gases (air) or metal oxides get in the way.

You can actually bond two pieces of metal this way: it's called cold welding. But, in order to get it to work with large pieces of metal, you have to (1) get all the air out of the way, and (2) you have to clean both surfaces very thoroughly in order to remove all traces of surface oxides, and (3) you have to make sure both surfaces are perfectly matched, either precisely flat or with precisely the same curvature.

Once you've taken care of those preconditions, my expectation is that all metals should cold weld. No guarantees as to how hard it might be to actually satisfy these conditions, though.

  We can weld elements together so they 'stick' to each other, but what is the process that actually causes two like elements to bond together?

In the most common welding methods, both of the surface to be welded are actually melted in the area right around the weld, which takes care of (1) and (3) above.  Requirement (2) is dealt with in a variety of ways, such as by surrounding the weld area with an inert gas blanket or by using a 'flux' material that either chemically reacts with any metal oxides to turn them back into metal, and/or that "floats" on top of the weld puddle to protect it from oxidation.

ADDENDUM: After reading DarioOO's answer, I realized that I should note that cold welding does not satisfy the "forming one larger bar or block" aspect of your question. An assembly of cold-welded parts is not attached as strongly as it would be if it were machined from a single piece.


Does iron(III) sulfate react with copper? As my understanding, there should be an oxidation-reduction reaction:

$$
    \ce{2 Fe^3+ + Cu → 2Fe^2+ + Cu^2+}
$$

However, I always see the process using $\ce{FeCl3}$ to etch copper, but I never see people using $\ce{Fe2(SO4)3}$ to do this. Is it because such reaction does not occur, or is it because $\ce{FeCl3}$ is acidic in aqueous solution?

In $\ce{FeCl3}$ etching of copper, the chloride ion is extremely important.

Chloride ions coordinate to $\ce{Fe^3+}$, $\ce{Cu+}$ and $\ce{Cu^2+}$.

See Copper Etching in Ferric Chloride Ind. Eng. Chem., 1959, 51 (3), pp 288–290 for the relative concentrations of the particular chloride complexes.

In contrast, if $\ce{Fe2(SO4)3}$ were used, there would only be aqueous ions.

At the copper surface, there is only a one-electron oxidation of $\ce{Cu}$ to $\ce{Cu+}$.  As the above article explains, without coordination of chloride, the $\ce{Cu+}$ would be essentially insoluble.

The rate limiting step of the etching reaction is diffusion of ions from the surface of the copper.  Coordination of chloride provides solubility and enables diffusion.  This is the main reason that $\ce{FeCl3}$ is needed.


Differences between phenols and alcohols I know phenols are more acidic as compared to alcohols, but are they considered different from alcohols?

Sure, you can study something as a subset, but are phenols considered a subset of alcohols, or are they considered as completely different from alcohols? My confusion is due to the fact that both of them contain a hydroxyl group.

Nope. Alcohols consist of an -$\ce{OH}$ group bonded to a saturated carbon ($\mathrm{sp^3}$ hybridized, no multiple bonds).
IUPAC says:

alcohols
Compounds in which a hydroxy group, -$\ce{OH}$, is attached to a saturated carbon atom $\ce{R3COH}$. The term 'hydroxyl' refers to the radical species, $\ce{HO^.}$.

and

phenols
Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g., 2-naphthol:

(source: iupac.org)

A phenol consists of an -$\ce{OH}$ bonded to an unsaturated $\mathrm{sp^2}$ carbon. Thus, it does not qualify as an alcohol. One can classify it as an enol, though.

Really, to me, the classification doesn't matter. Classifications are artificial, what is important is, how well the properties fit in the classification.
Many of the alcohol properties depend upon:

Its unsaturated nature: oxidation to ketone/aldehyde/acid
The weaker $\ce{R-O}$ bond and its ability to easily break and form an $\ce{R+}$ cation (this makes it a good participant in $\mathrm{S_N1}$ reactions)

Phenol can obviously not be oxidised at the $\ce{OH}$ to a ketone/acid (though one can do stuff to make it into a quinone). Phenylic carbocations are unstable, thus we don't get any $\mathrm{S_N1}$ reactions, and the $\ce{Ph-O}$ bond stays put.
On the other hand, most of the reactions of phenol depend upon its

Aromatic phenyl ring: All the EAS reactions
Weaker $\ce{O-H}$ bond (i.e., acidic nature): Reimer-Tiemann reaction, etc.

Thus phenols and alcohols don't have too many reactions in common. So, in this case, they have been classified in a sensible manner--if phenols were classified as alcohols, we would basically be clubbing two radically different classes of compounds under one umbrella.


Seeking word equations is there a general word equation so I can learn how to write out the following:

With chlorine, it seems to be: Hydrocarbon + Chlorine ---> ____ + HCL

With Hexane, it seems to be: Hydrocarbon + Br ---> ____ + HBr

I am not sure how to derive the first part of the final reaction (____)

It seems that methane has lost one hydrogen atom.

It seems that hexane has lost one hydrogen atom too.

Does that mean, if it was say Pentane with Chlorine, the first part of the final reaction would be CH4CL (not ch5cl) ? Likewise, would Pentane with Bromine be C6H12BR (and not c6h13br)

It feels like i'm rote-learning a skill, wonder where my knowledge is missing hmm..

I don't understand your question completely, but If you want to understand what's happening in an organic reaction you need to know which type of reaction is in question and you have to look at the mechanism. This is a radical reaction with the following mechanism (excluding the termination step):


Identify KNO3 and KH2PO4 at home I am not expert in chemistry. I need to find a way to identify a $\ce{KNO3}$ and $\ce{KH2PO4}$ powder at home. I suspect the seller gave me the wrong powder. What is a simple way to test between these compounds at home?

In the absence of other chemicals you can utilize the fact that nitrates(V) are much stronger oxidizing agents than phosphates(V).
You can soak a piece of newspaper in a saturated solution of the salt, let it dry well in a ventilated place and try to ignite it.

If the salt was $\ce{KNO3}$, it will burn vigorously and fast (that's why kids are often using paper soaked in sodium nitrate as primitive fuse for home-made firecrackers and rockets).

$\ce{KH2PO4}$ doesn't promote combustion of paper, so it will burn normally or even slower.


Formation of coordination complexes? I'm having trouble understanding how exactly coordination complexes are formed.

Compounds such as $\ce{[Cu(H2O)6]++}$ are the ligands filled into the empty 3 of 4p orbitals+ 1 of 5s orbital + 2 of 4d orbitals or how?

If you want to analyse a compound, it is best to start with the $d$-electron count of the metal. For $\ce{Cu^{2+}}$ we get $d^9$, since $\ce{Cu^0}$ would be $d^{11}$.

Now, consider the following (simplified) molecular orbital diagram for a $d^1$ compound:

As you can clearly see, the ligand electron pairs mix with some of the $d$, the $s$ and $p$ orbitals of the metal.

Now, instead of $d^1$ we have $d^9$, so we just start filling in 8 more electrons into the MOs. This gives us a full $t_\text{2g}$ MO and 3 electrons in the $e_\text{g}$ MO. ($t_\text{2g}$ and $e_\text{g}$ are ways of describing the symmetry of the orbitals. This comes from applied group theory.)

So you see, the $5s$ orbital never comes into play here, only the $3d$, $4s$ and $4p$ orbitals of the metal take part in bonding.

You might want to look at Ligand Field Theory and Molecular Orbital Theory to further your knowledge in inorganic chemistry. Not everything is as simple as the "fill in the boxes" method.


QM9 dataset with chirality I am trying to find chiral centers of molecules in QM9 dataset. Browsing their SMILES representation, I noticed SMILES yield using

datasets.get_qm9(GGNNPreprocessor(), return_smiles=True)

don't have @ or @@, but it seems that some of these molecules have alternative representations. For instance, both 'Cc1ccoc1C(C)O' and 'CC@Hc1c(C)cco1' represent the same molecule, but of course the latter one outputs the chiral center:

[(1, 'S')]

while only the former is available in the dataset. I found this at https://github.com/chemspacelab/qm9-C7O2H10. Does anyone know if this can be generalized to all the molecules in the QM9 dataset? Thanks!

I don't know about that particular GitHub archive, although I'll point out that the smiles.csv contains several kinds of SMILES - the final column containing the correct SMILES.

IMHO, this is an unfortunate result of people mis-using SMILES. As you note, it's perfectly possible to take a chiral molecular structure and output a SMILES with no stereochemical information.

Take alanine - to a chemist, if you simply say 'alanine,' they immediately think 'amino acid' and translate to L-alanine, or C[C@@H](C(=O)O)N as the so-called "isomeric SMILES":

Unfortunately, if you go to PubChem for L-Alanine, you'll see a "canonical SMILES" which sounds official and is CC(C(=O)O)N - no stereochemistry!

The official version of QM9 can be found on FigShare based on the corresponding publication.

While the official files are in a somewhat non-standard XYZ format, they do have the stereochemically-correct SMILES at the end of the file, e.g. C1N2[C@H]3[C@@H]4[C@@H]5O[C@]13[C@H]2[C@H]45 Just watch that some numbers are formatted like -8.0386*^-6 (i.e., $-8 \times 10^{-6}$) and can sometimes be misinterpreted by programs like Open Babel or RDKit .. practically everything I've seen.


What makes the cyclic structure of glucose more stable than the open chain structure? Is it tautomerism? Or some other factors involved?

You get to form a C-O $\mathrm{\sigma}$ bond at the expense of a C-O $\mathrm{\pi}$ bond. The single bond has a higher bond energy, even though it is somewhat de-stabilized by the anomeric effect. It's still an overall win, despite taking the entropic hit to confine your molecule into a ring.


Is Tollen's test safe? Tollen's test is an important test to distinguish between aldehyde and ketone.

$$\ce{2 [Ag(NH3)2]+ + RCHO + H2O → 2 Ag(s) + 4 NH3 + RCO2H + 2 H+}$$

But is this reaction safe? Is there any threat in the formation of silver nitride (fulminating silver)? It is formed when the products of most of the reaction of silver($\ce{I}$) salts are kept aside undisturbed. It act as a potential explosive or a  contact explosive. If there is any threat in the formation of silver nitride, what precaution must be taken to prevent it?

The Tollen's test is safe, but silver amine solutions should be prepared as needed just before use.  The formation of explosive silver nitride is a slow process and can be prevented by treating the solution with mineral acid after use.


How to control current and voltage in electrolysis I am doing an experiment that aims to determine if the rate of electroplating changes with the temperature. I am using a copper rod, a solution of copper(II) ions and a neutral rod to be coated. I also want to keep every other variable the same, which means the voltage across the rods, the concentration of the solution and the current that goes through the circuit. But the times my professor tried to demonstrate electrolysis, either the current or the voltage was always dropping. How do I prevent this from happening? How can I ensure that the voltage and current remain unchanged for the entire duration of a trial?

You can't control both voltage and current electrically:  Your power supply will hold one or the other constant, but the properties of the electrolytic cell will then determine the other one.

So you need to hold all of the chemical parts of the experiment constant:  Have a large ion bath so that the concentration doesn't change much as the plating takes place.  Stir well to keep things mixed. Have large electrodes (i.e. small current density) so that small surface changes don't dominate.  Etc.


Why are steric effects repulsive? Normally, I think about orbitals interacting in the context of bonding. When two atomic orbitals overlap, they can do so in-phase or out-of-phase. The in-phase overlap results in a bonding molecular orbital with lower energy than the original AOs; conversely, the out-of-phase overlap results in an antibonding MO with higher energy. Valence electrons fill the bonding MOs first, resulting in a bond.
The explanation of steric effects that I often hear is that they are caused by the overlap of two electron clouds, resulting in repulsion. Since electron clouds are just composed of occupied orbitals, how is this "overlap of electron clouds" different from the picture of bonding I described above? Does the steric repulsion just come from occupied antibonding MOs? I imagine that inner shell electrons and Coulomb forces must play a role as well, right?
@matt_black answered a related question about steric vs eletronic effects by saying:

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds

I guess I'm essentially asking a follow-up question on how effects are transmitted through space.

It's all about the Pauli exclusion principle. Assume the available MOs in each of two approaching molecules are fixed, and that each molecule is in a closed shell configuration (all electrons are paired). The electrons in closed shell systems cannot accept additional electrons that share principal, azimuthal and magnetic quantum numbers, since all the spin QMs are taken up. In fact, other than filling higher energy (LUMO) orbitals there is no way to accept additional electrons while retaining the current structure. The solution is to deform the structure of the molecule so as to generate new orbitals that allow formation of novel MOs. The deformation is associated with a potential energy barrier, since it is assumed that the colliding molecules are in low energy electron configurations. Therefore the instantaneous response to the collision is repulsive.


Why are lactones more acidic than esters? Lactones ($\mathrm pK_\mathrm a \sim 25$) tend to be more acidic than esters ($\mathrm pK_\mathrm a \sim 30$). I know that the lactone is fixed in a (E)-conformation, due to the ring:

and that in normal esters, the (Z)-conformer is favoured based on an anomeric effect, namely the donation of the $\mathrm{sp^2}$ lone pair on the ester oxygen into the C–O σ* orbital (image taken from Clayden, Greeves & Warren, Organic Chemistry, 2nd ed. (OUP), p 805):

Is this effect responsible for the increased acidity? If so, how?

Overview

The only interesting thing about lactones is that if the ring size is relatively small, they necessarily adopt the (E) conformation. Therefore, the question essentially boils down to: for a generic ester, why is the (E)-conformer 1 more acidic than the (Z)-conformer 2? This is a question that is easier to investigate, since we remove all other variables e.g. molecular structure.

Before diving into the details, I'll state the answer: 2 is more stable than 1. Likewise, if one compares the enolate anions, 2a is more stable than 1a, but the difference in stability between the anions is not as large as in the esters. Hence, $\Delta G^\circ$ for the deprotonation of 2 is larger than that for the deprotonation of 1, and 2 is less acidic. Schematically:

In physical organic chemistry the difference between the two values of $\Delta G$ is referred to as $\Delta \Delta G$. In this case, we can define $\Delta G_2 - \Delta G_1 \equiv \Delta \Delta G$, which we expect to be positive.

Neutral molecules

The two conformations of esters have been studied for quite a long time; as far as I can tell there seem to be two major reasons put forth as for why 2 is more stable than 1.

In the literature, the accepted argument seems to be based on a minimisation of dipole-dipole repulsions in 2.1–4 In particular, Roos et al.2 investigated the three possible contributing factors of (1) dipole-dipole repulsions, (2) repulsions between lone pairs on oxygens, and (3) steric repulsions between R and R'. Their conclusion was that the dipole-dipole repulsions were the most significant, and this analysis seems to have been accepted by later authors.

In textbooks, however, the n-to-σ* anomeric donation seems to prevail.5–9 However, no literature references are given, and Deslongchamps6 admits that "there is no direct experimental evidence to show the importance of [these] secondary electronic effects" (p 57). This was written in 1983, and even computational studies since then have not mentioned the existence of any such effect.

What everybody can agree on, though, is that 2 is more stable than 1. The energy difference was determined by IR spectroscopy to be $\pu{4.75 kcal mol-1}$ for methyl formate, and $\pu{8.5 kcal mol-1}$ for methyl acetate (the latter being larger because of steric repulsions between methyl groups).10 These values are corroborated by multiple computational studies.2–4

Enolates

I actually get the feeling that some of the textbooks get sloppy, or gloss over this aspect. Reading Clayden5 for example, the impression I get is that because 1 is more unstable than 2, it reacts more readily (with bases or nucleophiles). However, in comparing $\Delta G$ for two different reactions, one must be careful to look at the relative stabilities of not just the reactants, but also the products. If 2a is also more stable than 1a, and to the same extent ($\Delta\Delta G = 0$), then there is no reason to expect any difference in acidity.

The question that we now turn to is therefore: why is the relative stabilisation of 2 partially lost upon deprotonation to form 2a? The answer will, of course, depend on which of the two above explanations you subscribe to.

First, the dipole minimisation argument. Upon deprotonation, negative charge develops on the α carbon; hence as a whole, the enolate moiety has a smaller dipole moment than an undeprotonated carbonyl moiety. (Since the carbonyl oxygen bears more negative charge than the α carbon, the direction of the dipole moment is roughly the same.) Houk3 has reported this in the form of Mulliken partial charges on atoms, and Wiberg4 in the form of electron populations, but the key point is the same: when the dipole of the carbonyl group is smaller, the dipole-dipole interactions are weaker. Consequently, while 2a is still more stable than 1a, the energy difference is not as large as it is in 2 versus 1. Houk's ab initio calculations (6-31+G*//3-21G) on methyl acetate3 gives the energy difference between 2a and 1a as $\pu{3.8 kcal mol-1}$, and that between 2 and 1 as $\pu{9.2 kcal mol-1}$. Therefore, the difference $\Delta\Delta G$ is roughly $\pu{5 kcal mol-1}$.

Extra: in a later article, Houk and Jorgensen11 revisited the topic to calculate the potential effects of solvation. A polar solvent would be expected to preferentially stabilise the more polar (E)-conformers 1 and 1a (which are more polar since the dipoles reinforce each other). It turns out that solvation in either water or acetonitrile only reduces $\Delta \Delta G$ by $\pu{1 kcal mol-1}$.

As for the stereoelectronic argument, only Alabugin7 offers any possible explanation:

  A number of factors can contribute to this difference but, at least partially, this effect can be explained by the reduced acceptor ability of the C–O bond of the enolate where the oxygen atom has significant negative charge and by the subsequent decrease in the energy of the nO to σ*C–O stabilisation.

Further reading

Much of the research carried out has been targeted at understanding the anomalously high acidity of Meldrum's acid, and a literature search for this will yield many of the references cited here. The $\mathrm pK_\mathrm a$ values below (DMSO) are taken from Arnett and Harrelson12:

The dramatic increase in acidity upon "cyclisation" of a diester - to the point where the cyclic diester (Meldrum's acid) was more acidic than a cyclic diketone (dimedone) - was traditionally explained with the above information (i.e. it is more acidic because it has two (E)-esters).

However, newer research suggests that there is also an extra stereoelectronic factor that accounts for the unusually high acidity, namely donation from the ester oxygen lone pair into the σ* orbital of the adjacent $\ce{C(sp^3)–O}$ single bond. This stabilising effect is more pronounced in the conjugate base because the oxygen has greater electron density and the donor orbital has a higher energy.13

Notes and references

Huisgen, R. Neuere Beiträge zur Chemie mittlerer Ringe. Angew. Chem. 1957, 69 (11), 341–359. DOI: 10.1002/ange.19570691102. Disclaimer: I can't quite read German, but none of the figures indicate any sort of stereoelectronic effect.
Wennerstrom, H.; Forsen, S.; Roos, B. Ester group. I. Ab initio calculations on methyl formate. J. Phys. Chem. 1972, 76 (17), 2430–2436. DOI: 10.1021/j100661a015.
Wang, X.; Houk, K. N. Theoretical elucidation of the origin of the anomalously high acidity of Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1870–1872. DOI: 10.1021/ja00214a032.
Wiberg, K. B.; Laidig, K. E. Acidity of (Z)- and (E)-methyl acetates: relationship to Meldrum's acid. J. Am. Chem. Soc. 1988, 110 (6), 1872–1874. DOI: 10.1021/ja00214a033.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012.
Deslongchamps, P. Stereoelectronic Effects in Organic Chemistry; Pergamon Press: Oxford, U.K., 1983.
Alabugin, I. V. Stereoelectronic Effects; Wiley: Chichester, U.K., 2016.
Kirby, A. J. Stereoelectronic Effects; Oxford UP: Oxford, U.K., 1996.
D. A. Evans' famous Chem 206 notes also only mention this stereoelectronic effect. The set of notes is downloadable here (the relevant part is in Lecture 2).
Blom, C. E.; Günthard, H. H. Rotational isomerism in methyl formate and methyl acetate; a low-temperature matrix infrared study using thermal molecular beams. Chem. Phys. Lett. 1981, 84 (2), 267–271. DOI: 10.1016/0009-2614(81)80342-9.
Evanseck, J. D.; Houk, K. N.; Briggs, J. M.; Jorgensen, W. L. Quantification of Solvent Effects on the Acidities of Z and E Esters from Fluid Simulations. J. Am. Chem. Soc. 1994, 116 (23), 10630–10638. DOI: 10.1021/ja00102a032.
Arnett, E. M.; Harrelson, J. A., Jr. Ion pairing and reactivity of enolate anions. 7. A spectacular example of the importance of rotational barriers: the ionization of Meldrum's acid. J. Am. Chem. Soc. 1987, 109 (3), 809–812. DOI: 10.1021/ja00237a028.
Byun, K.; Mo, Y.; Gao, J. New Insight on the Origin of the Unusual Acidity of Meldrum's Acid from ab Initio and Combined QM/MM Simulation Study. J. Am. Chem. Soc. 2001, 123 (17), 3974–3979. DOI: 10.1021/ja001369r.


Why do shape and size have the greatest effect on the optical properties of nanoparticles? The LSPR properties of nanoparticles depend on a number of factors, including shape, size, material, crystal structure, dielectric environment, etc. However, it has been consistently shown that shape and size have the greatest effect on the optical properties of nanoparticles. Exactly WHY, however, is this true? Does anyone have an explanation at the molecular level?

As Surface Plasmon Resonance phenomenon is mainly seen in metal nanoparticles, I don't know whether this answer will help you? The size dependent optical phenomenon of semiconductor nanoparticles (e.g., $\ce{CdS}$) can be explained using the effective mass approximation model.

where r is radius of the particle (the rest details of the above equation can be find in the link).

so the energy gap is inversely proportional to the radius of the particle.

When the radius of the particle  decreases, the band gap increases so it will emit high energy wavelength (i.e., wavelength in blue region) and when the particle size is large the energy gap will be less so it will emit light in red region (lesser energy).


Why is polonium tetrafluoride non-existent? I learnt that the group 16 elements can form halides, but with the exception that polonium cannot form a tetrahalide with fluorine.
However, I thought that it was possible. This is because the electronegativity difference between polonium and fluorine is very high since polonium is a metal and fluorine is a highly electronegative atom and so should be able to form a salt.
Why is this wrong?

Before going into details, let me state Wikipedia:

$\ce{^{208}PoF6}$ was probably successfully synthesised via the same
reaction (passing fluorine gas over $\ce{^{208}Po}$) in 1960 where a
volatile polonium fluoride was produced, but it was not fully
characterized before it underwent radiolysis and decomposed to
polonium tetrafluoride.

Even attempts were made to dissolve polonium metal in hydrofluoric acid but no fluoride salts were isolated as a free salt as they radiolysed and decomposed. According to this 1961 paper:

Although no fluoride salts of $\ce{Po}$ have been isolated, $\ce{Po}$
is quite soluble in $\ce{HF}$. Solutions containing 2  curies of
$\ce{Po}$ per ml of $\ce{1 N HF}$ are reported to be unsaturated.
Attempts to prepare a volatile fluoride by reaction with $\ce{F2}$ gas or
with liquid $\ce{BF3}$ have been unsuccessful. However recent work with $\pu{1
 mg}$ of $\ce{^{208}Po}$ indicated the formation of a volatile $\ce{Po}$ compound,
which decomposed before it could be isolated.

Polonium tetrafluoride is very difficult to isolate as it is seen to form either a complex or oxidize/decompose radiolytically. Also, bringing analytical data on such compounds is hard as they are produced in minute amounts possible in the range of micrograms. Below is a text from this e-book 1

A white solid possibly polonium tetrafluoride is obtained by treating
polonium hydroxide/tetrachloride with aqueous dilute hydrofluoric
acid. Treating this solid suspension in sulfur dioxide yields a bluish
grey product (possibly $\ce{PoF2}$) which quickly reverts back to the
solid on standing owing to radiolytic oxidation. The solubility of
$\ce{Po(IV)}$ increases with the concentration of hydrofluoric acid
indicating a complex formation
There has been unsuccessful attempts to make volatile hexafluoride by
passing fluorine to $\ce{^{210}Po}$ but recently a fluoride has been
made with $\ce{^{208}Po}$ plated on platinum. This product appers to
be quite stable in vapor phase but on cooling, a nonvolatile compound
is formed possibly polonium tetrafluoride resulting from radioactive
decomposition. Analytical data are not recorded for any polonium
fluorides owing to difficulty of determining fluoride ion at microgram
level.

Reference

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1962
https://www.osti.gov/servlets/purl/4367751/4367751.PDF


Is a beta pleated parallel sheet always formed from more than one peptide? I can conceptually grasp the beta pleated anti-parallel sheet by imagining a curved length of string i.e. one polypeptide, starting with the N- terminus on the left and ending with the C-terminus at the end of the string, opposite the N terminus.
I can't do the same for the beta pleated parallel sheet if it is only formed of one polypeptide. How could a fold in a single strand cause the N and C termini to be parallel with each other?
Or is it that parallel sheets are always composed of more than one polypeptide?

Would single chain parallel sheet look more like this:

While it is true that some antiparallel beta sheets have neighboring strands that are adjacent in primary sequence, other cases do exist. As for parallel beta sheets, the polypeptide chain between strands could be connected via a loop or a helix if the two strands are close in primary sequence.
The mapping of strands on the primary structure has been studied thoroughly, and there are two types of diagrams to quickly show what is going on. For the "TOPS" diagram, we pretend to look down the strands, and up and down triangles represent strands in one or the other direction:

Source: https://onlinelibrary.wiley.com/doi/pdf/10.1002/pro.3285
Circles represent helices. If you look closely at the connecting lines, you can see whether the connection is above or below the plane of the paper.
The other representation has the sheets in the plane of the paper, e.g.:

Source: https://image.slideserve.com/258316/beta-sheet-topology-diagrams-l.jpg
Take-home messages

Most sheets are mixed (combo of parallel and antiparallel).
Most neighboring strands are not adjacent in sequence.
When two neighboring parallel strands are adjacent in sequence, there is a loop or a helix between them (or a strand from a different sheet) to connect them in the proper orientation.


What is the highest the boiling point of water can be raised? As I understand it, the boiling point elevation in a particular solvent (in this case water) is a function of a particular constant of that solvent, and the molality of a dissolved substance. In turn, the maximum molality is a function of the solubility of the compound, as well as the van 't' Hoff factor. Thus it would follow that the highest possible elevation of water's boiling point would be the compound with the greatest ratio of solubility to van 't' Hoff factor. So the question remains: which compound is that?

The "particular constant" $k$ you refer to in the question isn't an arbitrary constant.  This "law" is usually written as

$\Delta T = km = \frac{M_1 RT_b^2}{1000 \Delta H_1} m$

where k is the "constant you refer to; $M_1$ is the molecular weight of the solvent, $R$ is the gas constant, $T_b$ is the boiling point of the pure solvent, the constant 1000 makes the formula applicable only to water, and $\Delta H_1$ is the heat of vaporization of the solvent, here assumed to be water.  $m$ is the molality of the solute.

This equation is derived in courses in physical chemistry and is itself a simplification to allow slide rule simplification (this is true for a lot of formulas in chemistry).

So what you want to do to make $\Delta T$ as large as possible is to find a solvent with a large molecular weight and a high boiling point and a low heat of vaporization.  The constant 1000, which is the number of grams of solvent in a liter of solvent will also have to be adjusted.  And of course $m$ can be played with.

I don't think anyone has ever worked out what works best for a large $\Delta T$ since boiling point elevation, once used for molecular weight approximation, isn't used for that purpose any more -- and thermometers are a lot more sensitive and accurate than they used to be.