Why is SF6 inert towards hydrolysis? Why is $\ce{SF6}$ inert towards hydrolysis? Would not $\ce{H - F}$ bonds be created in water which is much stable and has a high negative enthalpy of formation. This hydrogen bond making must disintegrate the whole compound.

In short its kinetically stabilised, attack at S is close to impossible due to steric hindrance and rigidity of FSF angles. Thermodynamically with water it is very unstable with respect to HF$_{aq}$ and H$_2$SO$_4$. Creepy ...


Decreasing order of basic strength I need to find the decreasing basicity of the following compounds(I, II, III, IV, respectively):

I know that to compare the basic strengths, we need to find the stability of their conjugate acids.
Following are the conjugate acids, I drew (I don't know if they're correct):

According to me, due to resonance of the benzyl groups or lack of resonance, the order of basic strength should be:

$$\mathrm{III > II > I > IV}$$

But the answer given is:

$$\mathrm{IV > I > II > III}$$

Where am I going wrong?

Basicity is how well an atom can donate electrons. More is the electron density more is the basicity of the compound.

In Aniline the lone pair of electrons present on the nitrogen are involved in resonance, so the electron density on N is less than expected.

In compound II (Diphenylamine), since two benzene ring is present so the electron density will be less as now lone pair are involved in resonance of both the ring.
Thus electron density on Nitrogen is less than as compare to Nitrogen atom of Aniline

Similarly, in III compound (Triphenylamine), since three benzene ring is present the lone pair are more localized, and, thus electron density will be much less (least) as compared to the other compounds

In 4th structure, there is no resonance so, lone pair are present with nitrogen itself, and the electron density will be high, also due to inductive effect of the alkyl groups the electron density is more, and thus it can easily donate the electrons as compared to all other compounds. So the basicity is the highest for compound IV.

Therefore, the order of basicity is IV>I>II>III.


How do scientist obtain the value of enthalpy of formation? Do they do it experimentally or do/can they use bond enthalpy to figure it out? For example $\ce{C2H4 + H2 -> C2H6}$. The enthalpy of formation, $\Delta_fH$ of formation for the product is listed as -84, however if I use bond enthalpy to calculate the $\Delta_fH$, it gives be a value of -222, which is considerably different. Shouldn't the difference between bonds formed and bonds broken account for the heat given off my the reaction? With this discrepancy I can only conclude that I mixed up the two concept.

Also if $\ce{C2H4}$  enthalpy of formation $\Delta_fH$ equals to 52,  $\ce{H2}$ = 0 and $\ce{C2H6}$ = -84, the difference between the product and reactant is -136. I am confused about the 3 numbers (-84, -222 and -136) and their significance.

Also, how is $\Delta_fH^\circ$ different from $\Delta_RH^\circ$ of reaction? When do I use each one?

I would really appreciate it if someone can clear this up concisely

There is a consent to start with: the standard enthalpies of formation of elements (like $\ce{H2}$), at standard conditions ($\pu{1 bar}$, $\pu{298.15 K}$) are set to zero.  If there are several forms of one element, than the most stable polymorph is chosen as reference (except for phosphor, ref).

More generally, to determine the $\Delta{}_fH^\circ{}$ of other compounds, you may embark experiments by calorimetry. By means of a calorimeter, you are able to record the energetic changes that occur in the course of a reaction.  For example for the exhaustive combustion of carbon with oxygen:

$$\ce{C + O2 -> CO2}$$

all the heat recorded is due to the formation of carbon dioxide. Since this reaction equation was balanced that it equals only one mol of product ($\ce{CO2}$), the more general standard reaction enthalpy ($\Delta_{\text{R}}H^\circ{}$)  equals to the standard enthalpy of formation ($\Delta_fH^\circ{}$).  And by one experiment, after an other, the body of experimentally determined enthalpies grows.

Based on other experiments than combustion, one was able to determine how much energy is necessary to break a bond between atoms which is called bond dissociation energy. Conversely, this value expresses how much energy over all is released if two atoms establish a bond with each other.  In some instances, these energies are (comparatively) small, for example single bonded $\ce{Cl-Cl}$ ($\pu{242 kJ/mol}$), in other instances (comparatively) large, like triple bonded $\ce{N#N}$ ($\pu{945 kJ/mol}$).

If one knows the structure of a compound, which atoms are connected with each other and how (single, double, triple bond), than these bond dissociation energies may be used to estimate the standard enthalpy of formation of the new compound, for example by an increment system. Alternatively, by virtually setting up for example a combustion reaction of exactly one mole of the new compound into $x$ moles of $\ce{H2O}$ and $y$ moles of $\ce{CO2}$.  Then, the calculation to perform is

$$\Delta_{\text{R}}H^\circ{} = \sum_{\text{all products}}{\Delta_fH^\circ{}} - \sum_{\text{all starting materials}}{\Delta_fH^\circ{}}$$

where $\Delta_{\text{R}}H^\circ{}$ again is the overall reaction enthalpy.  Since these estimations are based on some empiric data, they do not a substitute for an experimentally determined value for your newly prepared compound. Often, increment systems leave some of inter- and intramolecular interactions unconsidered (e.g., ring strain, possible existence of polymorphism) which then yields a difference between the estimated, and the actual value of $\Delta_{\text{R}}H^\circ{}$ or $\Delta_fH^\circ{}$.


On mechanism behind Marcus theory Regarding the mechanism behind Marcus theory, I was wondering if reaction occurs through tunneling when it does not have sufficient activation energy? Wouldn't expecting reactions to occur this way be like expecting random combustion to happen in the arctic?

The reorganisation is just the extension/contraction of of bonds and most importantly solvent motion /reorientation to equalise the energy at which point the electron tunnels and relaxation occurs into the product. Reorganisation happens before the reaction, it is, as it were, how the 'hill' is climbed to reach the transition state thats all. (This is the insight that Marcus had). Consequently it is the rate limiting process, even though it may take less than a picosecond in some reactions.  After reaction, relaxation (reorganisation) also occurs so that the solvent reaches equilibrium with the new charges distribution.

Figure source; R. Marcus, Biochimica et Biophysica Acta., 811, 1985, 265 - 322)


Mechanism of Fischer esterification: Does the carboxylic acid gives off OH- or H+? For the reaction

$$\ce{R-OH + R'-COOH ->[H+] R-COO-R' +H2O}$$

the reaction mechanism is given here.

What seems unusual to me is that the acid appears to be donating $\ce{OH-}$ rather than $\ce{H^+}$.

Can the mechanism be written as follows instead?

$$\ce{R'-COOH <=> R'-COO^- + H^+}$$

$$\ce{HA-> H^+ + A^-}$$

The inorganic acid pushes the equilibrium for the organic acid to the left and it remains undissociated.

Then, $\ce{H^+}$ from the inorganic acid first attacks the $\ce{O}$ atom in $\ce{R-OH}$ resulting in the formation of a water molecule and a $\ce{R^+}$ ion.

Next, as the equilibrium for organic acid is pushed to the right, the bisulfate ion removes the $\ce{H^+}$ from the acid creating a carboxylate ion. This carboxylate ion then reacts with $\ce{R^+}$ ion to form the ester.

In this mechanism, the carboxylic acid donates the $\ce{H^+}$.

I know one can't prove reaction mechanisms to be correct; however, which one the mechanisms given above is more likely to occur?

The accepted mechanism is far more likely than your proposed alternate mechanism for a number of reasons.

First, let’s consider an esterification reaction employing a chiral secondary alcohol such as butan-2-ol. Protonation of the alcoholic oxygen is no difficulty and will happen in equilibrium. To generate the corresponding ester from the conjugate acid of butan-2-ol, we would require some nucleophilic attack mechanism: the acid would either attack according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. If we assume an $\mathrm{S_N2}$-like mechanism, the asymmetric carbon atom would be inverted. If we assume an $\mathrm{S_N1}$ mechanism, it would racemise. Experimental evidence shows, however, that the alcoholic carbon atom strictly retains is stereochemistry — and this principle is used in many total synthesis in which asymmetric alcohols are first generated stereochemically and then transformed into esters of the same stereochemistry. To give you just one of many examples, consider the total synthesis of aetheramide by Gerstmann and Kalesse.[1]

Scheme 1: Example esterification of (R)-butan-2-ol

Scheme 2: Esterification performed during the synthesis of aetheramide by Gerstmann and Kalesse.[1]

Second, and immediately tied to the scheme as referenced above, many alternate esterification methods exist in the literature. In principle, they all work by replacing the acidic particle ($\ce{H+}$) of a Fischer esterification with a less reactive and/or more specific compound. Practically all methods — Steglich or Yamaguchi to name just two — work by adding a compound that is known to interact with the carboxylic acid and not as much with alcohols. However, this is only a weak indication since different methods could follow different mechanisms — even though the quantity of the evidence does seem overwhelming.

Third and finally, the acidic hydrolysis of esters — the reverse reaction of the Fischer esterification — has been studied by $\ce{^18 O}$ marked water. Polanyi and Szabo could show that upon hydrolysis the heavy oxygen atom remains bound to the acid fragment and not to the alcohol.[2] This offers intriguing evidence for the mechanism of nucleophilic attacks on a carbonyl group. I had thought that a corresponding isotope-marking experiment had been performed for the Fischer esterification, too. Unfortunately, I was unable to find a reference. Do note, however, that using isotopically marked oxygen on either the acid or the alcohol would, by careful analysis of the product, rule out or confirm one mechanistic pathway.

Scheme 3: Mechanism of hydrolysis as confirmed by Polanyi and Szabo using $\ce{^18O}$ marked water.[2]

References:

[1]: L. Gerstmann, M. Kalesse, Chem. Eur. J. 2016, 22, 11210–11212. DOI: 10.1002/chem.201602682.

[2]: M. Polanyi, A. L. Szabo, Trans. Faraday Soc. 1934, 30, 508–12. DOI: 10.1039/tf9343000508.


Harnessing phase separation to easily make Li film I want to make a thin, smooth film of pure lithium in a glovebox.

In general, I think that I can use a phase separation approach. I will heat a bath of mineral oil above Li's melting point, then pour liquid Li inside and let it cool slowly. The Li should cool on top and the only curvature should be due to miniscus effect.

I would think that if I used a third, lighter liquid than the Li, I could put a large column of it on top of the Li. However, this requires having a lighter liquid. Li, at melting point, has a density around $\pu{500 kg m-3}$.

Can anyone suggest a suitable lighter liquid?

You are thinking in the wrong direction. Meniscus is not about pressure at all.

You seem to be imagining external hydrostatic pressure as a huge invisible flat lid that we put on top of your molten lithium and kinda force it to stay flat. This is very much not so. The pressure is the same all around, thanks to the Pascal's law. It does not affect the shape. It presses on the curved sides of the meniscus with the same force as it does on top. The very notion of "flat" is simply not there.

If anything, it is gravity force that reduces the meniscus. In effect, adding a liquid on top will partially negate the gravity and thus make things worse. Keep in mind that such liquid, if it is found (I still can't think of any, and I'm not a novice in chemistry, mind you) can't be much less dense than lithium, which almost brings us to the conditions of Plateau's oil drop experiment, with lithium playing the role of oil. Not quite the outcome you wanted, is it?

It is the surface tension that matters, and adding liquid might change that in an unpredictable way. Then again, you can achieve the same result without any extra liquid. Just use different material for the container. I guess using metal instead of glass would make the meniscus bend in the opposite direction. With some tweaking, it might be possible to find the material with wetting angle of $90^\circ$ that would make lithium surface perfectly flat, but I don't think it is worth the effort.

So it goes.


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

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

Figure 1: Molecular orbital scheme of an octahedral complex including σ and π interactions from the ligands to the metal. Based on the schemes given by Professor Klüfers in his coordination chemistry course at the LMU Munich.

Complexing a metal is especially favourable for the ligands since they are able to lower their electrons’ energy levels significantly by forming dative σ bonds — and in the cases of halides or other full shell ions, even the electrons in π-symmetric orbitals benefit a little by π forward-bonding.

Therefore, the question should better be rephrased as:

  Why are noble gases not known to form coordinate bonds?

Adding an electron to an atom means that all its orbitals are raised in energy: There is additional negative charge repulsing the electrons. This brings the energy levels of the valence electrons closer to those of the metal they are bonding with, and only for similar energy levels is the energy gain from interacting maximised. Note that electronegativity (that inversely correlates rather well with the energy level of the highest occupied orbital) rises from left to right across a period, so the noble gases have highest electronegativity and therefore the lowest valence energy levels. (The correct physical observable would be ionisation energy, but the same holds generally true for that.)

Therefore, the orbitals of noble gases should be substantially lower than those of halides, which are again notably lower than those of oxido/sulphido ligands. This gives a large energy difference between metal and potential ligand orbitals meaning that noble gases would easily be displaced by better ligands such as water.

However, I could imagine without having checked the literature that metals with high oxidation states may be able to be complexed by noble gases as their orbitals are again substantially lowered.


Does Finkelstein's reaction proceed through an SN2 mechanism? I was dealing with a question whose answer was justified only when it was assumed that the Finkelstein reaction proceeds through an $\mathrm{S_N2}$ mechanism. So, does it really follow an $\mathrm{S_N2}$ mechanism or is there some other mechanism at play?

The Finkelstein reaction is a nucleophilic substitution reaction. As such, it generally has the choice of proceeding via the $\mathrm{S_N1}$ or the $\mathrm{S_N2}$ mechanism.

The reaction is typically conducted in acetone; a polar but not very protic solvent. The solvent is used because all the sodium halides except for sodium iodide are not very soluble in it and precipitate out. It does not show any real preference for either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

Therefore, we should study the nucleophile and leaving group. Iodide is both a good nucleophile and a good leaving group; the former is important as it allows us to assume a viable $\mathrm{S_N2}$ mechanism. The other halides that you are substituting with iodide are not as good leaving groups but still rather okay. This again does nothing really to tilt the scales towards either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

The final choice we have is that it depends on the substrate. Probably, if you are replacing a primary halide chances are that the reaction will proceed according to the $\mathrm{S_N2}$ mechanism while an $\mathrm{S_N1}$ mechanism should predominate for tertiary halides.


Mechanism for formation of Quinol from a Phenolic Ether What's the mechanism for formation of Quinol from the given phenolic substrate? The reaction seems to be an acid catalysed hydrolysis, but I'm not able to reach the final product. I guess the first step should be addition of water to the nucleophilic π bond, but what after that?

Here is a mechanism for this enol ether hydrolysis.


How to calculate ∆rGº from entropy and ∆fGº in different temperatures? "The total oxidation of glucose occurs according to the following chemical equation:
C6H12O6 (s) + 6O2 (g) -> 6CO2 (g) + 6H2O (l)
The following table gives us the free energies of standard formation and the standard molar entropies of
compounds involved in the previous reaction.

Compound
∆fGº(298 K) kJ/mol
Smº (298 K) J/K/mol

C6H12O6
-917,2
212,10

O2
0
205,14

CO2
-394,36
213,14

H20
-273,13
69,91

Based on the previous data, determine the ∆rGº
for glucose oxidation at 308 K"
It's a question of a doctoral selection process and I want to figure out if it is formulated wrong or if there is a way to an answer.

For a small temperature change like that, you can make use of dG=-SdT.  Determine $\Delta rG^0$ and $\Delta r S^0$ at 298, multiply the latter by 10 , and subtract it from the former to get $\Delta rG^0$ at 308K.


The concept of normal coordinates of a molecule Sometime ago I had posted the What is the origin of "normal" in normal coordinates and normal modes? in math & sci history problems. Nobody was sure for the reason for using the word normal there - it was an interesting discussion though. The concept of normal coordinates is transiently brought in vibrational spectroscopy in undergraduate textbooks but rarely explained in detail. Certainly, the concept is borrowed from mechanics.

How can we explain the concept of normal coordinates? What is the need of a new set of coordinates if x, y, z can show also the displacements of an atom from its mean position? Let us take the example of a non-linear molecule like water.

Edits
More specifically, how would we mathematically define a normal coordinate for the oxygen(s) and hydrogen during a normal mode vibration of water molecule? Or a "normal coordinate" is a global displacement coordinate for the whole molecule for a given normal mode vibration. Water has 3 IR active modes. Do we have only 3 normal coordinates $q's$ for water? Wikipedia shows a simplistic picture of a normal coordinate $q$ as a measure of displacement but from little bit of reading Herzberg's other specific monographs by Wilson, the concept of normal coordinate does not seem that trivial.

Since I was reviewing this in some detail recently, I will try to supplement Prof. Hutchison's already good answer with some more detail as to why one would choose to work with normal coordinates, rather than the already quite simple Cartesian coordinates.

The short answer is that, under the assumption of small displacements, normal coordinates decouple the vibrational degrees of freedom from the rotational degrees of freedom and make all vibrations mutually orthogonal, such that the total energy of the nuclear motion of the molecule can be expressed as a sum of translational kinetic energy (3 dimensions), rigid rotation at the fixed equilibrium geometry (3 dimensions), and vibrations about the equilibrium geometry ($3N-6$ dimensions).

Now, it is not so difficult to show that if we allow a molecule to move in all $3N$ degrees of freedom, that the total energy of the system will have a term which couples the rotation of the molecule to the vibrations of the molecule, the so-called Coriolis energy. This is because if we work in strictly Cartesian coordinates, the axes are fixed, which means that as a molecule rotates, a simple displacement of two atoms (imagine a diatomic vibration) will not have the expected periodic behavior in Cartesian coordinates, but will appear to shift amplitude among the Cartesian coordinates.

So, we would like to work with a set of coordinates which circumvents this problem as we can see that it is an artifact of the coordinates we chose from the outset and does not have any useful physical meaning.

As it turns out, it is rather non-trivial to thoroughly show that there is a particular choice of coordinates which transforms the Cartesian coordinates such that the frame of molecular vibrations is fixed and hence eliminates the coupling between vibrational and rotational degrees of freedom. This is where I should mention that, if you like, the Bible of molecular vibrations is the book Molecular Vibrations by Wilson, Decius, and Cross[1]. Chapters 2 and 11 fully motivate and derive the transformation which completes this procedure. The coordinates turn out to be what we now call normal coordinates, and I will only point out that the transformation takes the following form,
$$
q_i=\sqrt{m_i}\Delta \alpha_i
$$
where $i$ is atom index which runs from 1 to $N$ and $\alpha=x,y,z$, so that we rescale the $3N$ Cartesian coordinates into $3N$ mass-weighted coordinates.

It shouldn't be too hard to see that the kinetic energy in these new coordinates takes the simple form,
$$
2T=\sum_{i=1}^{3N}\dot{q}_i^2
$$

Then, using a quadratic approximation to the potential energy, the potential energy takes the form,

$$
2V=\sum_{i,j=1}^{3N}f_{ij}q_iq_j
$$

where $f_{ij}=\left(\frac{\partial^2V}{\partial q_i\partial q_j}\right)$ is the harmonic force constant.

One can show that if you solve Newton's equations of motion using these expressions for the kinetic and potential energy, that one will have to solve an eigenvalue equation where the eigenvalues, $\lambda_k$, correspond to the frequencies of vibration and the eigenvectors correspond to the normal modes, i.e. the actual motion of atoms which vibrate at the the frequency $\lambda_k$.

Now, these $q$'s are not quite normal coordinates, which I will call $Q_i$, but they are very closely related to normal coordinates. Namely, what we call normal coordinates are the ones which have one additional nice property of having a potential energy of the form,

$$
2V=\sum_{k=1}^{3N}\lambda_k'\dot{Q}_i^2
$$

That is, we want the potential energy to not contain any terms involving more than one coordinate. Thus, we have to determine the coefficient $\lambda_k'$ which makes this transformation.

In Ref. [1], sec. 2-4, this derivation is performed and one finds the marvelous result that $\lambda_k'=\lambda_k$. Hence, one can make this representation of the potential energy diagonal by multiplying by the characteristic frequency which we got from each $q_k$.

Lastly, lest you convince yourself that vibration and rotation are not accounted for anywhere in this, when you solve the eigenvalue I mentioned above, you will find six eigenvalues which equal zero. This is only the case because we made this transformation to mass-weighted coordinates, and these six zeros tell us that six of our degrees of freedom appear to be completely stationary. These are translations and rotations.

So, people work with normal coordinates because they have these nice properties of providing a diagonal represent of the kinetic and potential energies when vibrations are harmonic, and giving a mutually orthogonal set of coordinates with which to work. This is nice because even when we wish to include anharmonicities and other corrections, we have a set of coordinates which have desirable properties and which are a usually a good approximation to the true "vibrational motions" of the system.

I put "vibrational motions" in quotes because it isn't clear that talking about the motions of atoms in a vibrational state is all that meaningful in quantum mechanics, and one can easily find a unitary transformation which causes the motion to look different in Cartesian space, but which leads to the same observable properties.

[1]: Wilson, E. B., Decius, J. C., & Cross, P. C. (1980). Molecular vibrations: the theory of infrared and Raman vibrational spectra. Courier Corporation.


Do polarizability and electron delocalization have analogous effects on leaving groups? With respect to bimolecular substitution and bimolecular elimination reactions, both electron delocalization and polarizability have an analogous effect on a species, making them good leaving groups.

For example I- is a good leaving group since its negative charge is spread out over a large area.

Also, the tosylate ion is a good leaving group because of electron delocalization where its electrons are spread out over a large area due to resonance stabilization.

However, the analogy seems to break down when considering their nucleophilicity in aprotic media, I-  is a good nucleophile and tosylate is not.

Is there a fundamental problem of the analogy between polarizability and electron delocalization?

TL;DR: Nucleophilicity is primarily a kinetic phenomena.  Leaving group ability is a thermodynamic phenomena.

The stabilization of an anion by delocalization of charge means that the resulting products will have lower energies than alternate situations. In accordance with the Hammond postulate (a few logical steps skipped), transition state barriers with low energy products will be smaller than those with higher energy products.

However, for a bimolecular reaction to occur, the molecule must have 'proper collision speed and orientation'.  Diffuse charges (over multiple atoms) generally make this more difficult.  This extra facet of the of the geometry of the transition state is why the equivalent stability argument does not apply to both leaving group ability and nucleophilicity.


Will 2-methylbutanal undergo aldol condensation Will 2-methylbutanal undergo aldol condensation?

We are taught that it will not because the +I effect makes the carbanion unstable, but if we make an enolate ion, isn't it resonance stabilized by the C=O group?

There is no reason why the aldehyde cannot undergo an aldol reaction with itself. Enolates are not rendered unstable by the presence of additional alkyl substituents; in fact they are considered to be more stable. (See: formation of silyl enol ethers with $\ce{Et3N}$/$\ce{Me3SiCl}$.)

However, there is no way it can eliminate $\ce{H2O}$ to form the α,β-unsaturated ketone, since the aldol product does not have any α-hydrogens. Strictly you cannot call it an aldol condensation, although often no distinction is made in the literature between an aldol addition and a condensation.

The formation of the β-hydroxyketone above was reported as long ago as 1924, the reagent being potassium carbonate. More recently, in an attempt to make a 1,3-dioxolane from 2-methylbutanal with potassium t-butoxide and 2-chloroethanol, it was found that the enolisable aldehyde underwent "self-condensation" instead: Org. Lett., 2006, 8, 3745.


What are the practices to naming organic chemicals by abbreviations? Is there a preferred method to writing organic chemicals by abbreviation?

I am finding it hard to decipher the methods I see in writing.

For example $\ce{EtOH}$ is ethanol, which makes sense, but $\ce{PhAc}$ is acetophenone, where the constituents appear to be backwards.

What about more complicated chemicals like 1,1-diphenylethanol, would $\ce{EtPh2OH}$ be considered appropriate, or maybe $\ce{Ph2EtOH}$?

The only accepted practice without any further constraints is to only use element symbols and sum or constitution formulae. Taking your and a few other examples, that would result in:

$\ce{C2H5OH}$ or $\ce{H3CCH2OH}$
$\ce{C6H5COCH3}$
$\ce{H3C(C6H5)2COH}$
$\ce{H3CCOOC2H5}$
$\ce{(CH3)3COCH3}$

The more complicated these get the less clear everything is. Thus, many people make use of a set of common abbreviations. However, these must be defined in a list of abbreviations, footnotes or wherever.

$\ce{EtOH}$
$\ce{PhAc}$[1]
$\ce{MePh2COH}$
$\ce{EtOAc}$
$\ce{$t$BuOMe}$

Where the symbols $\ce{Et, Ph, Ac, Me, $t$Bu}$ would need to be defined to mean ethyl, phenyl, acetyl, methyl and tert-butyl, respectively.

Finally, of course, you can suggest other common or your own abbreviations and use them (consistently!). For example, in my group it is rather common to abbreviate $\ce{EtOAc}$ as $\ce{EE}$ (because of its retained German name Essigester). Or a common way to denote the last compound in that list is $\ce{MTBE}$ or $\ce{TBME}$ (methyl-tert-butylether or tert-butylmethylether). Again, these abbreviations must be defined in footnotes or a list of abbreviations in best practice.

Finally note that an abbrevation typically includes all the hydrogens. Thus, methyl is always $\ce{CH3}$; $\ce{MeCl}$ is permissable for chloromethane but $\ce{MeCl2}$ does not mean dichloromethane. Consequently, neither $\ce{EtPh2OH}$ nor $\ce{Ph2EtOH}$ are permissable for 1,1-diphenylethanol, since carbon $\ce{C{1}}$ has no remaining hydrogens (not 2 as $\ce{CH3CH2}$ — ethyl — would imply).

[1]: Note that $\ce{Ac}$ can also mean actinium. If your paper uses actinium anywhere, do not use $\ce{Ac}$ as an abbreviation for acetyl or it shall cause confusion. The same is true for $\ce{Pr}$ and praseodymium and a few even less common others.


Converting bromobenzene to 1‐bromo‐3‐methylbenzene I started by taking $\ce{FeCl3}$ to reduce electron-giving capacity of $\ce{Br}.$ Then we need a methyl at meta, so I thought about Friedel–Crafts acylation, but then also methyl will come at para.

This is one of those pointless paper exercises that you would never do in the real world. I would not start from bromobenzene as $\ce{Br}$ directs ortho/para, so this is going to take several steps.
You could make the phenyl Grignard or phenyllithium from bromobenzene quenching with DMF followed by aqueous acid workup to give benzaldehyde. Brominate this with $\ce{BrCl}/\ce{AlCl3}$ described here to give 3-bromobenzaldehyde then do a Wolff–Kishner reduction $(\ce{H2NNH2}/\ce{KOH})$ on the aldehyde to give 3-bromotoluene. I would not risk a Clemmensen $(\ce{Zn}/\ce{Hg}$ & conc. $\ce{HCl})$ here as I think there is a danger of reducing the $\ce{Ar-Br}$ to $\ce{Ar-H}.$


Why second ionisation energy greater than the first? Why is the second ionisation energy for atoms with more than 1 electron in the outermost sublevel greater than that of the first, if you are removing an electron from the same sublevel ? Why is there a greater attraction for an electron in the same sublevel?

The ionisation energy basically measures how tightly an electron is held. That in turn is determined by a couple of factors, namely:

Nucleus-electron attraction. If this increases, the ionisation energy increases.
Electron-electron repulsion. If this increases, the ionisation energy decreases.

More formally these can be quantified by $Z$, the nuclear charge, and $\sigma$, the shielding parameter. The effective nuclear charge, which is hopefully a term you have come across, is defined as

$$Z_\mathrm{eff} = Z - \sigma$$

and the larger $Z_\mathrm{eff}$ is, the larger the ionisation energy. You don't necessarily need a formula to prove this, as it should be pretty intuitive. Larger effective nuclear charge $\implies$ electron is held more tightly $\implies$ more energy needed to ionise. But here is the mathematical relationship:

$$\text{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$

As long as you are staying within the same subshell, i.e. the principal quantum number $n$ is not being changed. So, the ionisation energy is determined by $Z_\mathrm{eff}$.

When you remove an electron, you decrease the electron-electron repulsions; this means that $\sigma$ decreases. As a result $Z_\mathrm{eff}$ decreases and the ionisation energy decreases. (Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus. However, within this inorganic chemistry framework, the nuclear charge $Z$ is simply the number of protons in the nucleus and therefore remains constant.)


Calculating boiling point of a solution What is the boiling point of an aqueous solution that has a vapor pressure of $23.0~\text{Torr}$ at $25~^\circ\text{C}$?
$(P^\circ_{\ce{H2O}}=23.78~\text{Torr};~K_\mathrm b= 0.512~^\circ\mathrm{C\ kg/mol})$

To solve this I used the equation $\Delta T_\mathrm b =m\times K_\mathrm b$. I calculated a molality of $1.88247\ \mathrm{mol/kg}$ which I then used to calculate a boiling point of $-0.9638\ \mathrm{^\circ C}$. This answer is apparently wrong and I don't understand why.

If molality is 1.88, then the change in boiling point is -0.96 degrees C, not the boiling point itself.


What are good books about MO theory that don't require a lot of math background? Are there any books about MO theory that explain the theory to a certain detail but don't rely on a lot of math background or explain the math within the book?

I'd suggest the seminal Orbital Interactions in Chemistry by Albright, Burdett and Myung-Hwan Whangbo.

The second edition came out in 2013 and I'm sure there are many copies of the first edition around as well.

They focus a lot on pictoral MO theory - indicating how orbitals of different shapes and symmetry combine.

To truly understand MO theory at a fundamental level, it really helps to study some quantum mechanics (and thus mathematics) but you can grasp a lot from the approach used in this book.


How can I determine the highest boiling point given a list of molecules? I know that the highest boiling point has to do with which has the strongest intermolecular force. I also know that the strongest would be ionic, then hydrogen bonding, then dipole-dipole, then london dispersion.

Can someone show me how I am supposed to solve these questions step by step though?

Am I supposed to create the Lewis dot structure for all of them and then see which is nonpolar vs. polar?

Also, once I come up with the Lewis structure, is that all I need to determine polarity? I just look to see if it's symmetric or not?

I know the 3D VSEPR shapes, so am I supposed to come up with that and then see which is nonpolar or polar?

Can someone provide me with a step by step solution as to how I can solve any of these problems from start to finish so that I have a good methodology?

Thank you!

Q1 and Q2 have the same approach. One notices that they are homologous alkanes and picks the biggest and the smallest, having the most London dispersion/van der Waals force and least, respectively. Other forces are known not to be present between alkanes.$^1$ (There is a small difference between branched and linear alkanes, but that is negligible compared to another $\ce{CH2}$ group.)

For Q3, one sees that they are combinations of hydrogen and halogen. While van der Waals forces are also present here, they are likely to be dominated by dipole-dipole interactions and hydrogen bonding (where applicable).$^2$ Which of the two is actually determining here, does not matter, since both point to a high boiling point in $\ce{HF}$.

In Q4, one notices that the question deals in homo-atomic molecules, so there can be neither hydrogen bonding nor dipole-dipole. Thus, the more electrons, the easier it is to polarize the molecule, and the heaviest molecule wins.

To summarize: To me, the questions are more about test-taking than actual chemistry. I cannot say whether the boiling point of $\ce{C4H10}$ or of $\ce{HF}$ is higher without looking up the values or knowing them and I do not expect most chemists to. Look for which forces may apply and reason about the trends. To do so, one may have to derive the molecular structure, but I would expect that in a test, one either does not need to or has the time to do so.

$^1$ This is where practice and knowledge are indispensable, which is unfortunate for the learner.

$^2$ In this case, this is trivial, but there are cases where exact quantitative calculations would be necessary. One can construct systems in which van der Waals forces outweigh ionic repulsion.


How much time liquid mercury takes to evaporate? I asked before about a thermometer that I broke yesterday. To summarize, I broke a mercury thermometer in my bedroom, which is a place where is difficult to have sure that I get rid of all mercury, but I cleaned the large amount without touching it.

However, I have more specific questions now. I discovered that I, by my lack of attention, have spread mercury along the house. I found a very tiny drop (really difficult to see) on the floor near to my desktop, which is in the living room, not in my bedroom. I also found a few same little drops in the bedroom, but it was really tiny. So, since I can assume that I will not be able to get rid of all mercury now, I have the following question:

How much time mercury takes to evaporate?

I think that wait the mercury to evaporate it is the only way that I can get rid of them now. Also, I have to say that I live in a country which doesn't have a service to check the amount of mercury in the air of the house neither get rid of the mercury.

Edit:

I found what maybe is more mercury tiny drops under a table in living room, but this time I tried to estimate the size of the possible drops.

Three really tiny mercury beads (zoom in if you could not clearly see the white points)

I know is it difficult to see, but I placed a piece of adhesive tape on the floor, which has the same size as my thumb's nail (about 1.4cm).

I don't know if is a good way to measure something, but I cut one of the three beads in a imagem editor and placed copies of it from on side until another. It takes about 30-40 copies to made the path. So, 1.4cm/30 results in each bead being at least about 0.47mm.

So, with this information, how many of that beads are needed to make the air dangerous?

I found this statement in a paper about $\ce{Hg}$ contamination from dental work:

  The evaporation rate of elemental mercury at room temperature ($\pu{20 ^\circ C}$) is approximately $\pu{50 \mu g\:cm-2h-1}$ (range of $\pu{40-60 \mu g\:cm-2h-1}$).

They cited the following work as a source for that number:

  Gary N. Bigham, Wanyu R. Chan, Manuel Dekermenjian, Ali Reza, "Indoor Concentrations of $\ce{Hg}$ Vapor Following Various Spill Scenarios," Environmental Forensics 2008, 9(2-3), 187-196 (https://doi.org/10.1080/15275920802121975)

German regulations (as an example) say that 0.035 µg/l of metallic mercury vapour in air pose no risk under lifelong exposition, and 0.35µg/l can be harmful under long (months?) exposition for some people. The math is full of imponderables, esp. the actual amount that was spilled and size of drops.

Swiss officials https://www.toxinfo.ch/quecksilberhaltige-fieberthermometer say that a single broken fever thermometer can lead to intoxications, but the risk can usually be averted by collecting as much as possible and ensuring proper, regular ventilation.

The following publication suggests that a properly cleaned up spill from a fever thermometer should become undetectable within several weeks, but the airborne Hg concentration can be significant in that time, and regular ventilation is highly recommendable.

  E. Martin Caravati et al. Elemental mercury exposure: An evidence-based consensus guideline for out-of-hospital management, Clinical Toxicology 2008, 46:1, 1-21, https://doi.org/10.1080/15563650701664731

But one thing is quite clear: A single, large blob that you miss (under furniture, wooden floorboard etc.) can stay for a very long time, and lead to potentially harmful concentrations in the air. There are a lot of reports about little mercury puddles that were found under floorboards during laboratory renovations, even decades after the last mercury instrument had been thrown out.


Why don't covalent compounds dissociate in water? I understand that ionic compounds dissociate when dissolved in water. Why don't covalent compounds do the same?

Many 'covalent' molecules do dissociate in water, HCl (as pointed out in the comments), phenol, acetic acid, for example, whereas some 'ionic' compounds do not to any appreciable extent, e.g. silver chloride, lead sulfate. The general reason is that the 'system', that is molecule plus solvent, will move into its lowest energy state. This means that if breaking the weakest bond and solvating the ions produced, is of lowest energy then this will occur. In the case of HCL, the polar water molecules can solvate making $\ce{H3O^+_{aq} and Cl^-_{aq}}$ where both ions are surrounded by a few water molecules. Water has a very high dielectric constant (relative permittivity) which means that electric fields are 'damped', i.e they are attenuated to short range so that oppositely charged ions can easily separate and hence exist independently in solution. With carboxylic acids the reasoning is similar, the weakest (O-H) bond dissociates, and the ions are solvated. However, dissociation may not be 100% indicating that the energy gain on dissociating is not that great. In insoluble salts, the lattice binding energy and also bond energy is too great compared to the energy gain expected upon solvation and they remain as molecules in a solid. Organic molecules with strong bonds e.g. cyclohexane, benzene, with only C-C and C-H, hardy dissociate in water. The bonds are too strong compared to energy gain on solvation after dissociation. (If dissociation produced radicals, there will be no effective solvation in a polar solvent so would remain at high energy and hence unlikely to occur)


Polarity of Methyl Radical Is methyl free radical a polar molecule?
According to Advanced Organic Chemistry by Jerry March, it is given as non-polar.
But then wat about the moment of the single electron?

I think you are getting confused see firstly methyl free radical is a planar with two hydrogens facing each other . The dipole moment vector of the single electron is away from the central carbon
Carbon has Electronegativity value 2.55 while for H it is 2.2 so all the hydrogens have a small positive charge and the carbon has a small negative charge . the direction of the dipole is from negative to positive so the dipole moment vector of the single electron egts cancelled by the resultant of that of the C-H bonds hence it is a non polar structure


Definition of nucleophile I came across this definition.

When discussing nucleophilicity we’re specifically talking about donating a pair of electrons to an atom other than hydrogen (usually carbon).  When a species is donating a pair of electrons to a hydrogen (more specifically, a proton, H+) we call it a base.

Website:

http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/

Does this make this drawing of mine wrong? I labeled the base here a nucleophile even though the nucleophile is lending n electron pair to the hydrogen. Are all bases nucleophiles?  Or should I stick with the definition above? (Forgive the forgotten arrow from the bond between oxygen and hydrogen which should be pointing at oxygen's valence environment.

All nucleophiles are Lewis bases. They are just electron pair donor.

A molecule or ion is called a "base" (or, “Brønsted base”) when it interact with proton, it is called a "nucleophile" when it interact with other atom centers, which can be carbon, nitrogen, sulfur and so on. So the row "base" and "nucleophile" may overlap. For example, $\ce{OH-}$ is a base when it take proton off an acid, and it can also be a nucleophile when it hydrolyzes ester (nucleophilic attack to carbonyl group).

The broad nucleophilicity definition is how good the nucleophile interact the positive or partial positive atom center. When the atom center is proton, nucleophilicity is then basicity. Because proton center is hard and usually less steric hindred, a bulky and hard (for hardness $O$ > $N$ > $S$ center) electron donor will be likely to prefer proton over the softer and more hindred non-proton center. These electron donors are called Non-nucleophilic bases. Potassium tert-Butoxide is a good example.


How do you recognize a carbohydrate molecule? I am studying carbohydrates in organic chemistry and I am confused a bit on what they are and how you recognize whether a molecule is a carbohydrate or not. For example, will a carbohydrate always have an $\ce{OH}$ group with either an aldehyde or a ketone group? Can it also contain other atoms like nitrogen or bromine? Can it have ether, ester?

The name "carbohydrate", which literally means“carbon hydrate,” arises from their chemical composition, which is roughly $\ce{(C.H2O)_n}$, where $n\ge 3$. The basic units of carbohydrates are monosaccharides.

Monosaccharides or simple sugars are aldehyde or ketone derivatives of straight-chain polyhydroxy alcohols containing at least three carbon atoms. Such substances, for example, D-glucose and D-ribulose, cannot be hydrolyzed to
form simpler saccharides.

Examples of monosaccharides:

Classification

Monosaccharides are classified according to the chemical
nature of their carbonyl group and the number of their C atoms.

If the carbonyl group is an aldehyde, as in glucose, the sugar is an aldose. If the carbonyl group is a ketone, as in ribulose, the sugar is a ketose. The smallest monosaccharides, those with three carbon atoms, are trioses.

Those with four, five, six, seven, etc., C atoms are, respectively, tetroses,
pentoses, hexoses, heptoses, etc. These terms may be combined so that, for example, glucose is an aldohexose, whereas ribulose is a ketopentose.

However, there are similar molecules with different atoms such as nitrogen which are also considered carbohydrates. A class of polysaccharides known as glycosaminoglycans (unbranched polysaccharides of alternating uronic acid and hexosamine residues).Practical examples are Hyaluronic acid(linked disaccharide
units that consist of D-glucuronic acid and N-acetyl-Dglucosamine) and Heparin (sulfated disaccharide units):

Nevertheless, when carbohydrates bond with other molecules to form other newer bonds, they are given different names and are not necessarily carbohydrates but are given new names - a good example are glycoproteins (proteins covalently associated with carbohydrates.)

Reference

Biochemistry Voet and Voet, Grisham


What are the uses of ethyl phthalate The last time I went to my junior college I remember seeing huge blue colored drums labeled ethyl phthalate. So just out of curiosity I decided to check upon this compound. Wikipedia entry containes some information about it's toxicity and usage, but I couldn't understand that part clearly. Could someone help me by listing the possible uses and properties of ethyl phthalate?

German Wiki article contains more use-cases for ethyl phthalate:

Formulation auxiliaries for plant protection products
Fixer for fragrances;
In the process of ethanol denaturing (in cosmetics this product is denoted as Alcohol denat.);
In manufacturing of cellulose esters as a very well-gelling and light-resistant plasticizer;
A plasticizer for cellulose-based varnish binders and as enteric coating for medicaments;
A part of cosmetic products due to its film-forming, softening and hair-conditioning properties;
Explosives phlegmatizer;
A carrier in the dyeing process of cellulose acetate or triacetate fibers.


Does increasing the entropy of a system cause it to absorb heat? This is my first time posting here, so bear with me. It was stated (in this video: https://youtu.be/7jT5rbE69ho?t=325) (Watch about the next fifteen seconds, until 5:40) that an increase in entropy leads to a decrease in heat (or temperature or absorbs energy - unsure of the correct terminology). (If I'm understanding correctly.) Now, I've never taken a thermodynamics class or anything past high school chemistry, but this is a little confusing to me and I'm hoping to get an explanation. What follows below is my best attempt at understanding why. It is most likely mis-guided, so feel free to correct at will.

I understand that entropy can be partially understood as disorder. When the Helium-3 enters the Helium-4, that constitutes an increase in entropy.
The change in entropy is defined as Heat/Temperature (see below)

  Rudolf Clausius invented the idea of entropy in such a way that the change in entropy is the ratio of the heat exchanged in any process and the absolute temperature at which that heat is exchanged. That is, he defined the change in entropy DS of an object which either absorbs or gives off heat Q at some temperature T as simply the ratio Q/T.

I then figured that maybe the Q-T ratio (entropy) increased by the total energy (of the helium system) increasing, requiring it to pull heat from somewhere

Miscellaneous thought: My above reasoning confused me because I thought it required energy to decrease entropy locally (add order to a system), so it seems strange that a system increasing in entropy
(like my helium example) would absorb energy.

I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.)

Yes, entropy is a measure of disorder (sort of).
No, $dS={\delta Q\over T}$ only for reversible processes.
And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals with infinitesimal increments and not with finite quantities. Second, because like I just said, that formula is not universally applicable.

All in all, the increase of entropy alone is not necessarily accompanied by the heat flow to the system. It may be so, but it may be otherwise. You may imagine pretty much any kind of possible outcome:

You dissolve some $\ce{NH4NO3}$ in water (which surely increases entropy), and the beaker gets palpably cold. The system starts sucking heat from the surroundings, just like you anticipated.
You mix your $\rm^3He$ with $\rm^4He$, and... nothing happens (except the increase in entropy, of course, but that's a thing you can't see or touch). There are no thermal effects to speak about, and hence no heat flow.
You burn a candle in a closed vessel (supposedly, containing enough air), and guess what? Entropy increased, but heat is flowing outside.

As for your last remark: true, it takes some energy to decrease entropy locally (say, in your fridge). But in doing so, we inevitably increase entropy elsewhere, so the overall balance is positive.

So it goes.


How is the aromaticity in graphene different from the aromaticity in benzene? The web page Aromaticity in Graphene and other 2-D Systems begins:

I. Graphene
While the σ-bonding in graphene is assumed to be a rigid honeycomb framework built out of two-center two-electron (2c-2e) C-C σ-bonds, the π-bonding is supposed to be delocalized. However, according to the adaptive natural density partitioning (AdNDP) analysis and the electron sharing indices, graphene is aromatic, but its aromaticity is different from the aromaticity in benzene, coronene, or circumcoronene. Aromaticity in graphene is local with two π-electrons located over every hexagon ring. (emphasis added)
II. BC3 Honeycomb Epitaxial Sheet
[...]

Update: The wording is similar to what is written in Is Graphene Aromatic? Popov, BozhenkoEmail &  BoldyrevEmail (2012), Nano Research, 5, (2), pp 117–123 (open access here):

ABSTRACT
We analyze the chemical bonding in graphene using a fragmental approach, the adaptive natural density partitioning method, electron sharing indices, and nucleus-independent chemical shift indices. We prove that graphene is aromatic, but its aromaticity is different from the aromaticity in benzene, coronene, or circumcoronene. Aromaticity in graphene is local with two π-electrons delocalized over every hexagon ring. We believe that the chemical bonding picture developed for graphene will be helpful for understanding chemical bonding in defects such as point defects, single-, double-, and multiple vacancies, carbon adatoms, foreign adatoms, substitutional impurities, and new materials that are derivatives of graphene.

Question: How is the aromaticity in graphene different from the aromaticity in benzene? It is likely that those familliar with the topic will already understand the explanation in the block quote, but is it possible to find a simpler if less accurate way to explain the difference, in a similarly compassionate way to how this answer was extremely helpful to my question about double bonds and chromophores?

Simpler explanations, can by their nature be less accurate than a detailed description, but are helpful in beginning to understand things.  Getting to the heart of your question:
“How is the aromaticity in graphene different from the aromaticity in benzene? . . .  but is it possible to find a simpler if less accurate way to explain the difference,”
Perhaps a productive way to approach your question [portion given above] is to consider the molecular structure of coronene and circumcoronene while contrasting them to the interpretation of aromaticity of graphene given in your cited article.
 circumcoronene
 CORONENE
Graphene from article cited in question.
According to the computational method employed “adaptive natural density partitioning method” [we can choose to agree or not agree with their interpretation as indicated by a comment to your post by @A.K.], the aromaticity of graphene is centralized within the set of hexagonal rings.  To quote the authors “Aromaticity in graphene is local with two π-electrons delocalized over every hexagon ring.”  The salient point here being the “delocalization” is within and isolated to or restricted to [my words] a six-membered ring.
As discrete molecules, the aromaticity in coronene and circumcoronene, as depicted the PubChem structures, is delocalized throughout the entire discrete molecule and not restricted to a discrete set of hexagonally bonded carbon atoms.
In other words, it comes down to electron density distribution.  In graphene, according to the authors of the article cited, the areas of highest electrons density are within the center of the six-membered ring while in coronene and circumcoronene the electron density is more uniformly distributed over the entire molecule.
Ref:
https://pubchem.ncbi.nlm.nih.gov/compound/coronene
https://pubchem.ncbi.nlm.nih.gov/compound/Circumcoronene


NMR: Magnetic and chemical equivalence in benzene According to my textbook, all the protons in $\ce{^{12}C_6{}^{1}H_6}$ are magnetically equivalent, but those in $\ce{^{13}C_6{}^{1}H_6}$ are not.

I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule. I do not understand this with respect to benzene as I expect Hc to couple differently to Ha than Hb to Ha and thus for them to be magnetically inequivalent irrespective of the isotope.

All of the protons in $\ce{^{12}C_6{}^{1}H_6}$ benzene are equivalent by symmetry, therefore they are chemical shift equivalent.

You wrote, "I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule."

I think the correct statement would be, for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.  If there are no other non-chemical shift equivalent nuclei in the molecule, then the chemical shift equivalent nuclei are also magnetically equivalent.

In the case of $\ce{^{12}C_6{}^{1}H_6}$ benzene, all of the protons are equivalent by symmetry, therefore they are chemically equivalent and since there are no "different" nuclei (non-chemical shift equivalent nuclei) for them to couple with, they are also magnetically equivalent.

In the case of $\ce{^{13}C_6{}^{1}H_6}$ benzene all of the protons are equivalent by symmetry so they are all chemical shift equivalent.  Since all of the carbons are $\ce{^{13}C}$ (spin=1/2), there are now "different" (the $\ce{^{13}C}$ nuclei are non-chemical shift equivalent) nuclei for the chemical shift equivalent protons to couple with; therefore the protons will be magnetically non-equivalent.  The spectrum is now an AA'A'A"A"A"'XX'X'X"X"X"' (A=proton, X=carbon-13) spectrum.  Since the protons are not magnetically equivalent (at least some of) the couplings will be observable.


How can you simulate the effect of a solvent on a polymer in an MD simulation? I'm doing a molecular dynamics simulation of polymer melts and I want to include the effect of a good/bad solvent on the melt. How can this be done?

My simulation models polymers using a bead-spring model and the interactions forces are Lennard-Jones for pair interactions and FENE for bonding forces.

Edit:

FENE stands for Finite Extensible Nonlinear Elastic. It's a log based elastic force.
The potential function used, Lennard Jones + FENE is described here (http://lammps.sandia.gov/doc/bond_fene.html)

You have generally two choices a) implicit and b) explicit solvent. Presumably, you don't have good interaction parameters for explicit solvent, since you are likely doing a generic polymer melt.

There are a variety of solvation models; LAMMPS, for instance, implements a dielectric pair style for which you can attenuate the dielectric constant of your solvent (and a few other things). Other popular approximations are generalized Born models.

This paper uses GROMACS to simulate poly(styrene sulfonate) and poly(allylamine hydrochloride) in explicit and implicit solvent, as an example from the literature.


Why is it necessary to heat an esterification reaction? Why do we heat the mixture acid and alcohol?

The reaction of esterification is exothermic so it is not affected by increasing temperature, then what is the use of heat ?

Basically the use of heat is to liberate the water molecule from the mixture of acid and alcohol that is $\ce{OH-}$ ion from alcohol and $\ce{H+}$ from acid that results in formation of ester.Moreover it helps in evaporation of water since esterification reaction forms an equilibrium.The removal of water from mixture forces the reaction to move forward.It helps us to get a large amount of product.Alternative to heating some dehydrating agents like conc. $\ce{H2SO4}$ can also be added.


The mathematical condition for a complete complexation reaction I'm being asked to find the volume of a $\ce{NH3}$ solution ($0.1\ \mathrm{M}$) to be added to $10\ \mathrm{mL}$ of a $\ce{AgNO3}$ solution ($0.1\ \mathrm{M}$) to assure the 'complete formation of $\ce{[Ag(NH3)2]+}$'. Both constants of formation are given:

\begin{align*}
\beta_1 &= \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}} =10^{3.3} \\
\beta_2 &= \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}} =10^{7.2}
\end{align*}

As far as I know, a complete complexation would mean that $[\ce{Ag+}]<10^{-5}$. I tried working it out with the common method of solving those kind of complex equilibria, but the equations I get seem pretty much impossible to solve because of the volume that I have to find, which gets incorporated in them.

\begin{align*}
\dfrac{\beta_2}{\beta_1} &= \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}} \\
\ce{[Ag+]_0} &= \ce{[Ag+]} + \ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]} \\
\ce{[NH3]_0} &= \ce{[NH3]} +2\ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]}
\end{align*}

$\ce{[Ag+]_0}$ and $\ce{[NH3]_0}$ can be expressed with respect to the volume of the $\ce{NH3}$ solution, but the equations I get are really complicated...

What approximations should I go for besides $\ce{[Ag+]} << (\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]})$?

Their solution is really weird, but much simpler; they invoke this mathematical condition, which makes the problem easily solvable:

$$\dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+]}}>10^2$$

Starting from this, I can find $[\ce{NH3}]$ present in the solution, therefore the volume.

Where did it come from?

Ok, to start let's just check the given answer of 24.3 mL.

The total solution is 34.3 ml (10 mL Ag + 24.3 ml NH3).

We'll assume that all 0.001 moles of silver are as species $\ce{[Ag(NH3)2^+]}$ so:

$\ce{[Ag(NH3)2^+]} \approx (\dfrac{10}{34.3})0.100 = 0.0291 $

$\ce{[NH3]_{initial} =}(\dfrac{24.3}{34.3})0.100 = 0.0708$

We'll assume $\ce{NH4^+}$ and $\ce{[Ag(NH3)^+]}$ are both negligible so:

$\ce{[NH3] =} 0.0708 - 2(0.0291) = 0.0126$

Now from the equation

$\dfrac{\beta_2}{\beta_1}= 10^{3.9} = 7.94 \times 10^3 = \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$

$\ce{[Ag(NH3)^+] =} \dfrac{[\ce{Ag(NH3)2^+}]}{(7.94 \times 10^3)\ce{[NH3]}} = \dfrac{0.0291}{(7.94 \times 10^3)(0.0126)} = 2.91 \times 10^{-4}$

$\ce{[Ag^+] =} \dfrac{[\ce{Ag(NH3)^+}]}{(2.00E3)\ce{[NH3]}} = \dfrac{2.91 \times 10^{-4}}{(2.00E3)(0.0126)} = 1.15 \times 10^{-5}$

fraction of total Ag as $\ce{Ag(NH3)^+}$

$\dfrac{2.91 \times 10^{-4}}{0.0291} = 0.01 $

fraction of total Ag as $\ce{Ag^+}$

$\dfrac{1.15 \times 10^{-5}}{0.0291} = 3.97  \times 10^{-4}  $

Limits on ratios

Given the two equations:

$\mathrm{K}_1= 10^{3.3} = \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}}$

$\mathrm{K}_2 = 10^{7.2} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}$

then we can derive a third

$\mathrm{K}_3 = 10^{7.2 - 3.3} = 10^{3.9} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$

It is possible to solve for some lower limits on the ratios assuming that the $\ce{[NH3]}$ can at most be 0.1 M. So:

$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)+]}} \geqslant \frac{1}{\mathrm{K}_1\ce{[NH3]}} = 5.01 \times 10^{-3}$

$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_2\ce{[NH3]^2}} = 6.31 \times 10^{-6}$

$\frac{\ce{[Ag(NH3)+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_3\ce{[NH3]}} = 1.26 \times 10^{-3}$

So the absolute upper limit on the % of the silver as $\ce{[Ag(NH3)2^+]}$ is 99.874%.

Quick and dirty calculations...

To check this I brute forced some calcs in a spreadsheet.

The calculations below ignores the reactions below (alkali solutions do not produce appreciable amounts of silver hydroxide due to the favorable energetics for the reaction to the oxide):
$$ \ce{NH3 + H2O <-> NH4^+ + OH^{-}}$$
$$\ce{Ag^+ + OH^{-} <-> AgOH(s)}$$
$$\ce{AgOH(s) -> Ag2O(s) + H2O} \text{ (pK = 2.875)}$$

 NH3    [Ag(NH3)2+] [NH3]tot    [NH3]       [Ag(NH3)+]  [Ag+]       mol(Ag)     F*
 20.1   3.32E-02    6.68E-02    3.32E-04    1.26E-02    1.90E-02    1.950E-03   4.87E-01
 20.2   3.31E-02    6.69E-02    6.62E-04    6.30E-03    4.75E-03    1.334E-03   2.50E-01
 20.3   3.30E-02    6.70E-02    9.90E-04    4.20E-03    2.12E-03    1.191E-03   1.61E-01
 20.4   3.29E-02    6.71E-02    1.32E-03    3.15E-03    1.20E-03    1.132E-03   1.17E-01
 20.5   3.28E-02    6.72E-02    1.64E-03    2.52E-03    7.68E-04    1.100E-03   9.11E-02
 20.6   3.27E-02    6.73E-02    1.96E-03    2.10E-03    5.35E-04    1.081E-03   7.46E-02
 20.7   3.26E-02    6.74E-02    2.28E-03    1.80E-03    3.95E-04    1.067E-03   6.31E-02
 20.8   3.25E-02    6.75E-02    2.60E-03    1.57E-03    3.03E-04    1.058E-03   5.47E-02
 20.9   3.24E-02    6.76E-02    2.91E-03    1.40E-03    2.40E-04    1.051E-03   4.82E-02
 21.0   3.23E-02    6.77E-02    3.23E-03    1.26E-03    1.95E-04    1.045E-03   4.31E-02
 21.1   3.22E-02    6.78E-02    3.54E-03    1.14E-03    1.62E-04    1.041E-03   3.91E-02
 21.2   3.21E-02    6.79E-02    3.85E-03    1.05E-03    1.36E-04    1.037E-03   3.57E-02
 21.3   3.19E-02    6.81E-02    4.15E-03    9.69E-04    1.17E-04    1.034E-03   3.29E-02
 21.4   3.18E-02    6.82E-02    4.46E-03    9.00E-04    1.01E-04    1.031E-03   3.05E-02
 21.5   3.17E-02    6.83E-02    4.76E-03    8.40E-04    8.82E-05    1.029E-03   2.84E-02
 21.6   3.16E-02    6.84E-02    5.06E-03    7.87E-04    7.77E-05    1.027E-03   2.66E-02
 21.7   3.15E-02    6.85E-02    5.36E-03    7.41E-04    6.91E-05    1.026E-03   2.50E-02
 21.8   3.14E-02    6.86E-02    5.66E-03    7.00E-04    6.18E-05    1.024E-03   2.36E-02
 21.9   3.13E-02    6.87E-02    5.96E-03    6.63E-04    5.56E-05    1.023E-03   2.24E-02
 22.0   3.13E-02    6.88E-02    6.25E-03    6.30E-04    5.04E-05    1.022E-03   2.13E-02
 22.1   3.12E-02    6.88E-02    6.54E-03    6.00E-04    4.58E-05    1.021E-03   2.03E-02
 22.2   3.11E-02    6.89E-02    6.83E-03    5.72E-04    4.19E-05    1.020E-03   1.94E-02
 22.3   3.10E-02    6.90E-02    7.12E-03    5.48E-04    3.85E-05    1.019E-03   1.86E-02
 22.4   3.09E-02    6.91E-02    7.41E-03    5.25E-04    3.54E-05    1.018E-03   1.78E-02
 22.5   3.08E-02    6.92E-02    7.69E-03    5.04E-04    3.27E-05    1.017E-03   1.71E-02
 22.6   3.07E-02    6.93E-02    7.98E-03    4.84E-04    3.04E-05    1.017E-03   1.65E-02
 22.7   3.06E-02    6.94E-02    8.26E-03    4.66E-04    2.82E-05    1.016E-03   1.59E-02
 22.8   3.05E-02    6.95E-02    8.54E-03    4.50E-04    2.63E-05    1.016E-03   1.54E-02
 22.9   3.04E-02    6.96E-02    8.81E-03    4.34E-04    2.46E-05    1.015E-03   1.49E-02
 23.0   3.03E-02    6.97E-02    9.09E-03    4.20E-04    2.31E-05    1.015E-03   1.44E-02
 23.1   3.02E-02    6.98E-02    9.37E-03    4.06E-04    2.17E-05    1.014E-03   1.40E-02
 23.2   3.01E-02    6.99E-02    9.64E-03    3.94E-04    2.04E-05    1.014E-03   1.36E-02
 23.3   3.00E-02    7.00E-02    9.91E-03    3.82E-04    1.93E-05    1.013E-03   1.32E-02
 23.4   2.99E-02    7.01E-02    1.02E-02    3.70E-04    1.82E-05    1.013E-03   1.28E-02
 23.5   2.99E-02    7.01E-02    1.04E-02    3.60E-04    1.72E-05    1.013E-03   1.25E-02
 23.6   2.98E-02    7.02E-02    1.07E-02    3.50E-04    1.63E-05    1.012E-03   1.22E-02
 23.7   2.97E-02    7.03E-02    1.10E-02    3.40E-04    1.55E-05    1.012E-03   1.19E-02
 23.8   2.96E-02    7.04E-02    1.12E-02    3.31E-04    1.47E-05    1.012E-03   1.16E-02
 23.9   2.95E-02    7.05E-02    1.15E-02    3.23E-04    1.40E-05    1.011E-03   1.13E-02
 24.0   2.94E-02    7.06E-02    1.18E-02    3.15E-04    1.34E-05    1.011E-03   1.10E-02
 24.1   2.93E-02    7.07E-02    1.20E-02    3.07E-04    1.28E-05    1.011E-03   1.08E-02
 24.2   2.92E-02    7.08E-02    1.23E-02    3.00E-04    1.22E-05    1.011E-03   1.06E-02
 24.3   2.92E-02    7.08E-02    1.25E-02    2.93E-04    1.17E-05    1.010E-03   1.03E-02
 24.4   2.91E-02    7.09E-02    1.28E-02    2.86E-04    1.12E-05    1.010E-03   1.01E-02
 24.5   2.90E-02    7.10E-02    1.30E-02    2.80E-04    1.07E-05    1.010E-03   9.93E-03
 24.6   2.89E-02    7.11E-02    1.33E-02    2.74E-04    1.03E-05    1.010E-03   9.73E-03
 24.7   2.88E-02    7.12E-02    1.35E-02    2.68E-04    9.89E-06    1.010E-03   9.55E-03
 24.8   2.87E-02    7.13E-02    1.38E-02    2.62E-04    9.51E-06    1.009E-03   9.37E-03
 24.9   2.87E-02    7.13E-02    1.40E-02    2.57E-04    9.15E-06    1.009E-03   9.20E-03
 25.0   2.86E-02    7.14E-02    1.43E-02    2.52E-04    8.82E-06    1.009E-03   9.04E-03

(1) In the table the column [Ag(NH3)2+] assumes that the 1 millimole of silver is all $\ce{Ag(NH3)_2^+}$. The value is expressed as a concentration so it falls due to the dilution of the extra $\ce{NH3}$. In the column mol(Ag) all the silver species are totaled. The value should be 1.00E-03 moles or 1 mM Ag. Values significantly above that indicate that the calculations are in error. (Note that the way the calculations are done the mM Ag will always be too high.)

(2) The column [NH3]tot is the concentration of all NH3 (mMol NH3)/(total volume)

(3) The column [NH3] is the "free" ammonia. It assumes ammonia is either free $\ce{NH3}$ or $\ce{Ag(NH3)_2^+}$ and thus it ignores the single amine complex $\ce{Ag(NH3)^+}$. It calculates free ammonia as [NH3]tot - 2*$[\ce{Ag(NH3)_2^+}]$

(4) The column labeled F* is the fraction of silver as $\ce{Ag(NH3)^+}$ and $\ce{Ag^+}$. Thus the real decision is what fraction is low enough since the fraction continues to fall as excess ammonia is added.

Conclusion

The choice of the "end point" is arbitrary...

mL        end point
----      -----------------------------------
21.4      97% [Ag(NH3)2^+]
21.6      [Ag^+] 10% of [Ag(NH3)^+]
22.1      98% [Ag(NH3)2^+]
24.3      [Ag(NH3)^+] 1% of [Ag(NH3)2^+]
24.4      99% [Ag(NH3)2^+]
24.6      [Ag^+] = 1*10^-5


Metal carbonate + carbonic acid? As metal carbonates are basic would there be a reaction if a sample of, say potassium carbonate, was added to carbonic acid? If so what would be the products? A balanced equation would be useful.
Will this apply to other basic substances if they are added to an acid from which they already have an ion of?

$\ce{Na2CO3 + H2CO3 -> 2NaHCO3}$, yielding sodium "bicarbonate", as an example, and in similar fashion, $\ce{Na2SO4 + H2SO4 -> 2NaHSO4}$, or sodium "bisulfate". The "bi" in this nomenclature meant two different anions, and is confusing, as it is the opposite of "bi" in mercury bichloride, two of the same cations. For that reason, sodium hydrogen sulfate or sodium acid sulfate might be preferred.


Is this buffer possible? Question
Can a buffer solution be obtained by mixing-
A) $\ce{NH4Cl}$ + $\ce{CH3COONa}$
B) $\ce{CH3COOH}$ + $\ce{NH4OH}$
My Thoughts
I know that a buffer solution can be obtained if we mix a acid and its salt of conjugate base or a base and its salt of conjugate acid.
But I was wondering whether a buffer solution can be prepared if we mix a weak acid and weak base or if we mix an acidic salt and basic salt?

Both cases are equivalent in the case of    molar ratio 1:1. It would lead to a double buffer of $\ce{HAc/Ac-}$ and $\ce{NH4+/NH3}$. But the buffer capacity would be inferior, as the resulting $\mathrm{pH}$ is too far from the both respective $\mathrm{p}K_\mathrm{a}$.


Does DCl have a higher rotational partition function (qrot) than HCl? And if so why? I know that DCl has a higher moment of inertia than HCl, but I can't really explain how this actually affects the rotational partition function, I know larger molecules have a higher moment of inertia and this means the energy levels are closer, but I don't know how to explain how this has an effect on the partition function of rotation, can someone help?

The partition function is the sum over states, which means over all the energy levels, thus
$$Z=\sum _i g_i \exp(-E_i/(k_BT))$$
where $g_i$ is the degeneracy of each level with energy $E_i$. The rigid rotor rotational energy levels of a diatomic molecules are $$E_J= \hbar^2J(J+1)/(2I)$$ where I is the moment of inertia $I=\mu r^2$ for reduced mass $\mu=m_1m_2/(m_1+m_2)u $ (in kg) and bond length r. The degeneracy of rotational levels is $g_J=2J+1$.

As the moment of inertial of DCl is greater than that of HCl (due to reduced mass effect) the energy levels are lower in energy in DCl at the same J than in HCl and so at the same temperature more levels are populated and the partition function is greater. This means that the exponential term in the partition function is greater in DCl at a given J and so this increases the partition function. (recall that $e^0 = 1$ and $e^{-\text{big number}} \rightarrow 0$). The energy levels of DCl and HCl are shown in the left figure and the terms in the partition function on the right as $g_J \exp(-E_J/(k_BT))$. (The curves are plotted for clarity assuming that J is  continuous not discrete). The respective partition function is the sum of the values at each J in the right-hand figure.

(To plot the curves I used a mass of Cl = $35$ and H and D, $1$ & $2$ respectively. The unified atomic mass unit $u = 1.660\cdot 10^{-27}$ kg, and assume a temperature of $300$ K. The bond length is $74.15$ pm.)

(Note that the often used integral form of the partition function has a systematic error that does not imporove as J increases.)


Is there a reaction happening between Coconut Oil & Hydrogen Peroxide? On a whim, I mixed together some coconut oil and drug store (3%) hydrogen peroxide and washed my hands with it. It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)

Did any chemical reaction occur between the coconut oil and $\ce{H2O2}$ to make a new compound? My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?

I only had high school chemistry so I could be wrong. Just my guess. Anyone know for sure?

Did any chemical reaction occur between the coconut oil and H2O2 to make a new compound?

Most likely, no chemical reaction occurred.

Whether any reaction occurred depends on the conditions you used when mixing the oil and hydrogen peroxide. How long did they mix for? What temperature was the mixture at? Did you add anything else to the mixture?

Hydrogen peroxide is used in the production of epoxidized vegetable oils. However, the reaction is performed for prolonged times at elevated temperatures with concentrated hydrogen peroxide in the presence of a catalyst.

I am guessing that you just poured the two together at room temp, then soon after used the mixture to wash your hands. Compared to the commercial reaction, you are missing time, heat, and a catalyst, so it is very unlikely that any reaction occurred.

  My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?

Oils are composed of triglyceride esters of fatty acids, not fatty acids themselves. So you would not see any peroxylauric acid.

  It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)

It felt less greasy because it was less greasy: you added water to it (3% hydrogen peroxide is $\ce{H2O2}$ in water). You probably made an emulsion of oil and water and used that to wash your hands (kind of like using soap).


What exactly does a "system" include in chemistry? If we have reactants $\ce A$ and $\ce B$ and the reaction(s) between them haven't started yet, are those reactants our "system"?

After the reaction takes place, are products then also included in the "system"?

Basically, is it the case that the term "system" refers to all the chemicals we have in a reaction?

Note: This was not originally written by me. This was first written by @airhuff. I just thought it was relevant in this question as well.
According to this Wikipedia article:

...a physical system is a portion of the physical universe chosen for analysis. Everything outside the system is known as the environment. The environment is ignored except for its effects on the system.
The split between system and environment is the analyst's choice, generally made to simplify the analysis. For example, the water in a lake, the water in half of a lake, or an individual molecule of water in the lake can each be considered a physical system.

Every system has some relationship to it’s surroundings. As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.

Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.

Open system: Both energy and matter can pass into and out of the system.


Why does the US not manufacture sodium thiopental themselves? A shortage of a given chemical (sodium thiopendal) has prompted Utah to allow firing squads for executions. This shortage is due to mostly European chemical factories refusing to export the drug to the US.

I am against the death penalty, but can't help wondering why USA, arguably the most powerful nation in the world, can not manufacture this chemical themselves. Is it difficult to produce ?

I […] can't help wondering why USA […] can not manufacture this chemical themselves.

They can and they did!

Back in the nineteen thirties, Ernest H. Volwiler and Donalee L. Tabern filed a couple of patents on sedative thiobarbituric acid derivatives for the Illinois-based Abbott Laboratories. Among these was US 2,153,731 for sodium thiopental.

Abbott Laboratories produced it for a long time and when the company was restructured, the still Illinois-based split-off Hospira continued the production.

Note that it is not a "bad compound" as such and actually, it is still listed as an alternative to propofol on the WHO Model List of Essential Medicines.

According to a 2010 article in the German ÄrzteZeitung (online, 29.09.2010), the FDA was aware of the thiopental shortage and a speaker of Hospira told the NYT that his company is rather unlucky about the fact that this anaestetic found it's way into death row.

Around this time, the production of thiopental at the facilities in Illinois was seemingly stopped.

In a 2011 article in the ÄrzteZeitung (online, 24.01.2011) it is stated that Hospira had initially announced to continue the production at a facility in North Carolina and in Liscate, Italy.

However, it seems that the production in North Carolina never started. Moreover, the Italian authorities demanded a declaration that the thiopental produced in Liscate would not be used for lethal injections. Obviously, Hospira could not comply.

Personally, I think that Hospira wasn't overly unlucky about the legal situation in Europe. They are a health care company and I suppose that they prefer to make profit by improving the medical situation of their customers, rather than by killing them!


How do I calculate the density of propane assuming an ideal gas? Assuming an ideal gas, calculate the density in $\mathrm{g/cm^3}$ of propane, $\ce{C3H8}$ at $0.000~^\circ\mathrm{C}$ and $1.000~\mathrm{atm}$. (3 significant figures)

I have figured out that I need this formula:
$$\frac{n}{V} = \frac{p}{RT}$$

I have a total of $44.09562~\mathrm{g/mol}$

I am stuck at this point.

As you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\frac{n}{V}= \frac{P}{RT}$$

To find the density, we can further modify the equation as,
\begin{align}
\frac{m}{MV} &= \frac{P}{RT}\\
\frac{\rho}{M}&= \frac{P}{RT}\\
\rho &= \frac{PM}{RT}\\
\end{align}

Now we just have to replace the parameters with their respective values.

The value of temperature should be converted to Kelvin as $(0 + 273)~\mathrm{K} = 273~\mathrm{K}$.

Likewise, the value of gas constant $R$ should be chosen as  $0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}}$.
\begin{align}
\rho &= \frac{1~\mathrm{atm} \cdot 44.09562~\mathrm{g~mol^{-1}}}{{0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}} \cdot 273~\mathrm{K}}}\\
&=1.9683~\mathrm{g/L}\\
&=0.0019683~\mathrm{g/cm^3}\\
\end{align}


Why does silicon tetrafluoride have a higher melting point than sulfur tetrafluoride? So looking at the Wikipedia pages of sulfur tetrafluoride and silicon tetrafluoride, the melting points are −121 °C and −90 °C respectively, and so $\ce{SiF4}$ has the higher melting point. However, their boiling points are −38 °C and −86 °C, respectively, giving $\ce{SF4}$ the higher boiling point.

I can justify that $\ce{SF4}$ has the higher boiling point because it is more polar since its Lewis structure has a lone pair on the sulfur atom, and so it experiences greater dipole-dipole forces. However, why does $\ce{SF4}$ have a lower melting point?

High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).

$$\begin{array}{lrr}
\hline
\text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\
\hline
\text{pentane} & −130 & 36.1 \\
\text{isopentane} & −160 & 27.2 \\
\text{neopentane} & −18 & 9.5 \\
\hline
\text{octane} & −57 & 125 \\
\text{2-methylheptane} & −110 & 117 \\
\text{2,2,4-trimethylpentane} & −107 & 99 \\
\text{2,2,3,3-tetramethylbutane} & 101 & 107 \\
\text{cubane} & 131 & 133 \\
  & & \text{(sub. at r.t.)*} \\
\hline
\text{decane} & −30  & 174 \\
\text{isodecane} & −75 & 167 \\
\text{adamantane} & 270 & \text{sub.} \\
\hline
\end{array}
$$

* sub. — sublimes; r.t. — room temperature.


How long the block starting with element 121 will be? I remember from my chemistry classes that (after the initial irregularities) a new block of elements starts every two periods. After the initial s-block and p-block following it shortly, we have d-block starting at period IV, and f-block starting at period VI.

Now that Element 118 has been discovered, we're about to open period VIII and we're two elements short of a new block.

What block will it be? How many groups, what name etc?

As you noted, this is a very appropriate question in light of the IUPAC announcement that we have just finished filling Period 7!

The names of the subshells s, p, d, and f are named after the old spectroscopic terms sharp, principal, diffuse, and fundamental. We ran out of fancy names after that, so the subsequent subshells are named in alphabetical order - g, h, and so on - which means that after the 8s block is filled, we would theoretically have a 5g block.

The orbitals in the g subshell would be labelled with the quantum number $l = 4$, so $m_l$ would take integer values between $-4$ and $4$ (inclusive) giving a total of nine g-orbitals. Each g-orbital could hold two electrons with opposite spins, so the g-block would have $18$ electrons.

However, it is worth noting that the electronic configurations may or may not obey the aufbau principle fully. Whether the 5g orbitals will actually be filled or not will certainly not be easy to determine, considering how short the half-lives of those elements are likely to be.

Wikipedia has an article which talks about it.


Converting w/v to mole fraction If I have 10 mL 0.9% w/v $\ce{NaCl}$ solution, I can deduce that the solution contains 0.09 g of $\ce{NaCl}$.
Imagine that this is standard laboratory conditions and 1 ml = 1 g is acceptable.
Is the mass of $\ce{H2O}$ simply $\pu{10 g} - \pu{0.09g}$?
This would lead to mole fractions of $\pu{76.43}$% $\ce{H2O}$ and $\pu{23.56}$% $\ce{NaCl}$.
Update
I realised I had made some errors in my original question. Do the above percentages look right and should I express them as unitless fractions?

You say that you have

10 mL 0.9% w/v NaCl solution

This is an unusual specification and as user M. Farooq pointed out it is ambiguous. So let's assume that this is a reagent that I am to add for some reaction. Is this:
$\quad$(1) 10 ml water (10 grams) and 0.09 grams of NaCl
$\quad$(2) 9.91 ml water (9.91 grams) and 0.09 grams NaCl
$\quad$(3) 10 ml aliquot of 99.1 ml water and 0.9 grams NaCl
Each of these would be slightly different. The gist is that (1) and (2) are different because the densities of the solutions will be slightly different as well as the volumes. (2) and (3) are different because the volumes will vary even though the densities are the same.
Pragmatically such reagent solutions are generally used in a large excess for whatever their use is, so the slight variations don't make any difference in the reaction. However it is just sloppy to give an ambiguous specification when a specific specification could and should be given.
Significant Figures
The other inconsistency with the question posed is with significant figures. The OP has specified 0.9% NaCl, but calculated "mole fractions of 76.43% H2O and 23.56% NaCl". You can't get four significant figures for the mole fraction when the NaCl concentration only has one significant figure.


You can reorder the Z-matrix with a free program like Molden. For example open the molecule in Molden and press the ZMat editor. From there you should see a button that says reorder Z-matrix. Click the atoms one by one, making sure to select the HCH atoms sequentially so they have an explicit angle. Dummy atoms are another good option.


Preparation of acetate buffer from sodium acetate and hydrochloric acid A $\pu{1.0 L}$ buffer solution of $\mathrm{pH}~4.43$ is made up of $\pu{0.30 M}$ sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions. Calculate the volume of sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions used to prepare this buffer solution.

The answer given for volume of $\ce{HCl}$ is $\pu{750 mL}$ and for the volume of volume $\ce{CH3COONa}$ is $\pu{250 mL}.$

My approach is to assume $\ce{HCl}$ as the limiting reactant and use ICE table to determine the equilibrium concentration of the ethanoic acid and sodium ethanoate at the end of the reaction. Then I use the Henderson–Hasselbalch equation.

The answer I got was $\pu{500 mL}$ for both the conjugate base and weak acid. Is the answer wrong?

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is given ($\mathrm{p}K_\mathrm{a} = 4.76$).

When added together, $\ce{HCl}$ (a strong acid) and $\ce{NaOAc}$ (weekly basic salt) react completely to give $\ce{HOAc}$ and $\ce{NaCl}$, according to the following equation:
$$\ce{NaOAc + HCl -> HOAc + NaCl} \tag{1}$$

Suppose you added $x$ volume of $\ce{NaOAc}$ and $y$ volume of $\ce{HCl}$ to make the $\pu{1.0L}$ of buffer ($\mathrm{pH} = 4.43$). Thus if units of $x$ and $y$ are $\pu{L}$, then:
$$x + y = 1 \tag {2}$$

And, after acid-base reaction, according to the chemical equation given above and equation $(2)$, remaining amounts of $\ce{NaOAc}$ in $\pu{mol}$ is $(0.3x-0.2y)=\left(0.3x-0.2(1-x)\right)= (0.5x-0.2)$ and amounts of $\ce{HOAc}$ formed in $\pu{mol}$ is $0.2y= 0.2(1-x)=(0.2-0.2x)$. Thus, since total volume of buffer is $\pu{1.0 L}$, respective concentrations of $\ce{NaOAc}$ and $\ce{HOAc}$ in the solution are as follows:
$$\ce{[NaOAc]} = (0.5x-0.2)/1.0=(0.5x-0.2) \text{ and } \ce{[HOAc]} = (0.2-0.2x)/1.0 = 0.2-0.2x$$

Now you can substitute these values to Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[NaOAc]}}{\ce{[HOAc]}}$$

$$4.43 = 4.76 + \log \frac{(0.5x-0.2)}{(0.2-0.2x)}$$
Thus,
$$ \log \frac{(0.5x-0.2)}{(0.2-0.2x)}= 4.43 - 4.76= -0.33$$
$$ \frac{(0.5x-0.2)}{(0.2-0.2x)}= 10^{-0.33} = 0.468$$
Once you solved for $x$, you would get, $x=0.495$ and $y=1.0-0.495=0.505$ in $\pu{L}$, respectively (or $495$ and $\pu{505 mL}$). Thus, I'd say the given answers for your question is incorrect.


What substance is used for staining brass? While working at CNC engraving, I had at my disposal a small bottle of liquid used to stain brass for creating signs etc - engrave a writing, stain the engraving, then grind/polish the flat topmost layer to remove the colored layer and leave it only in engraved depressions.

The bottle didn't reveal much about composition of the liquid, beyond some warnings about toxicity. It was a brass staining liquid, period.

The coloration it created was dark brown, that seemed black by comparison with the surrounding metal. The reaction took only a few seconds until full pitch of coloration was achieved and the liquid itself was transparent with very mild cyan-blue hue. The coloration was quite persistent; even roughly 10 years afterwards it's still good as new on the pieces I kept and required rather deep brushing to remove from the topmost surfaces.

Can you tell me what substance it was?

(below a sample with dark parts achieved through that liquid)

Edit: I found a very similar liquid on sale, but the page is equally devoid of composition details. The instructions say: [the liquid] is to be used in room temperature. It allows the finish in tones between black and light brown. The time of reaction is almost immediate. The color and speed can be controlled through dilution in water up to 50%.

This site regarding patina forming solutions might help:  http://www.sciencecompany.com/-W160.aspx#1


Is it possible predict the oxidizing/reducing nature of a compound? I have recently begun with Inorganic Chemistry and am finding it totally difficult to understand things (I am trying to minimize memorization and approach logically).

One of the things I am having problems with is understanding oxidizing and reducing agents.
Given a compound(and a few other necessary conditions in some cases), can I predict whether the compound will behave as a reducing or an oxidizing agent?
Well obviously if there is any such rule, there will be exceptions. But is there anything as such to reason out a few statements and take me partly towards the answer?

For example I have begun with Group 1 and the book says at some point that peroxides and superoxides of alkali metals act as strong oxidizing agents, group 1 hydrides are powerful reducing agents etc. Well I find it difficult to understand these statements and my book offers no explanation.

You may or may not be familiar with the concepts of electronegativity and oxidation states, but ultimately here is the rule of thumb you seek. If your compound contains highly electronegative elements not in their most negative oxidation state, expect it to be an oxidizer. If it contains elements with low electronegativity which are not in their most positive oxidation state, expect it to manifest at least some reducing properties.

On top of that, just like you anticipated, there is a whole lot of exceptions and subtleties which together constitute the magnificent field of elementary inorganic chemistry.


Why is BDE low in Iodine, but not behaves like Fluorine? When halogens have a low Bond dissociation energy, BDE, their free radicals are stable.

$\ce{Cl-Cl}$ (BDE: $\pu{58 kcal/mol}$) and $\ce{F-F}$ (BDE: $\pu{38 kcal/mol}$). $\ce{Cl}$ free radical is less stable than $\ce{F}$ free radical. But why that doesn't happen with iodine ($\ce{I-I}$, BDE: $\pu{36 kcal/mol}$). Iodine free radicals are unstable.

Where have you found the information that the iodine free radicals are unstable? It does not sound right for me.

BDEs of chlorine, bromine and iodine decrease down the group as the size of the atom increases: 243, 192, and 151 kJ/mol. The BDE of fluorine, is however, lower (158 kJ/mol) than those of chlorine and bromine. Here you were right! But the free radical stability depends also on other factors. It increases with increasing the atom size and with decreasing the electronegativity. Additionally the stability of the free radical decreases as the orbital is held closer to the nucleus. Thus, as we can expect, the iodide radical will be the most stable, and the fluoride - the least./

Btw correlation does not mean causality! (:


Effective nuclear charge on an inner electron Is the shielding effect in a multi-electronic species caused due to mutual repulsion of electrons?

If it were so, while calculation the $Z_\text{eff}$ on an electron in say the $\mathrm{2p}$ orbital of a $\ce{Na}$ $\mathrm{(1s^2,2s^2,2p^6,3s^1)}$ atom, why don't we also consider the repulsive force of outer electrons in $\mathrm{3s}$ and add it to $Z_\text{eff}$ instead of $Z_\text{eff} = Z - \text{repulsion due to inner electrons}$?

Why doesn't the equation read

$$Z_\text{eff} = Z - \text{repulsion due to inner electrons} + \text{attraction due to outer electrons}?$$

Firstly, effective nuclear charge is just a model for predicting trends in atomic properties. The actual solution of the quantum mechanical equations does use the full nuclear charge and explicit calculation of the all electron-electron repulsions (and electron-nuclear attraction).

The point of $Z_\text{eff}$ is to approximate the value of the nuclear charge felt by electrons at a given distance. The answer to your question then is Gauss's Law,  which states that, to determine the field at a given point, you consider all and only the charge inside a certain surface (on which your point sits). Moreover, it states that the field there is equivalent to if all of the charge inside that surface was a point charge at the center.

So if we apply Gauss's Law to our atom, we add up all the charge inside a sphere centered on the nucleus ($Z-\text{effective number of core electrons}$), and regard the system as an effective nucleus with that charge, no core electrons and the electron of interest being the "first" electron and so we ignore the outer electrons.

Here are some physics discussions for why, intuitively, Gauss's Law allows you to ignore outside electrons from the physics stackexchange.


Voltaic cell and Nernst equation It is given voltaic cell
  $\ce{Cu}|\ce{CuSO4}(\pu{10^-3 M})||\ce{CH3COONa}(10^{-2})|\ce{Pt}(\ce{H2})(\pu{1 atm})$.  Find the potential at $\pu{50^\circ C}$.
  Given $K_\mathrm{a}$ for $H= 1.8 \times 10^{-3}$.
  $E_{\ce{Cu^2+/Cu}} = \pu{0.336 V}$

When voltaic cell starts working: $\ce{H2}$ becomes $\ce{2H+}$. In the other hand $\ce{CH3COO-}$ hydrolizes forming $\ce{CH3COOH}$ and $\ce{OH-}$. We know $K_\mathrm{a}$ and we can find $K_\mathrm{b}$. And from that we can find the concentration of $\ce{OH-}$ and after that using the formula $\mathrm{pH}=14-\mathrm{pOH}$  we can find the concentration of $\ce{H+}$.

Now how we can find the concentration of $\ce{Cu^2+}$ in order to replace it into Nernst equation?
\begin{align}
  E &=E^\circ - \frac{RT}{nF} \cdot \ln\{\text{Qc}\}
  \text{, where}&
  \text{Qc} &= \frac{[\ce{H+}]}{[\ce{Cu^2+}]}.
\end{align}

A correct standard notation for the galvanic cell would also include the physical state of each species involved. As such, the correct notation for the given cell is:

$$\ce{Cu(s)}~|~\ce{CuSO4(aq)}~(10^{-3}~\mathrm{M})~||~\ce{CH3COONa~(aq)}~(10^{-2}~\mathrm{M})~|~\ce{Pt(s)}~(\ce{H2(g)})~(1~ \mathrm{atm})$$

Recall that aqueous solutions of salts are generally accompanied by their molarity in solution. Also, metals are usually solid, while gaseous species are accompanied by a reference electrode (platinum in this case) and their pressure value.

In your case, notice that the ionic salt $\ce{CuSO4}$ is assumed to be a $100\%$ dissociated ionic salt, hence, its aqueous solution will produce $[\ce{Cu^{2+}}]=[\ce{SO4^{2-}}]=10^{-3}\mathrm{M}$. That should answer your final query.

Apart from that, your expression for $\mathrm{Q_c}$ is wrong. It should be $\displaystyle\mathrm{Q}=\frac{p_{\ce{H2}}}{[\ce{Cu^{+2}}]}$. Since $\ce{H2}$ is present in gaseous form, use its pressure value in the expression for $\mathrm{Q}$  (and not molarity which doesn't exist)

PS: The concentration ($\pu{10^−3M}$) that's given to you is the concentration of copper ions after equilibrium has been achieved for the interconversion of $\ce{Cu+2(aq)}$ and $\ce{Cu(s)}$.


How a plane of symmetry makes compounds lose their optical activity I understand that if a compound has a plane of symmetry it loses its optical activity, it is very easy to understand this when we have four different substituents on a single carbon atom but when we have larger more complicated structures for example rings with many substituents which are themselves stereogenic how can we be so sure it wont have any enantiomers if we only find one fortunate plane which produces a perfect mirror image?

That's really simple. Imagine two different vessels which are precise mirror images of each other; so are all the molecules inside. Imagine that you shine polarized light on them. Being the mirror images of each other, they have to rotate light in opposite directions. But as the compound itself is symmetric, they are in fact identical, so they have to rotate light in the same direction. How can these conditions be reconciled? Just like that: don't rotate light at all.

So it goes.


For what processes is Mayer's formula applicable? I know that

$$C_p - C_V = R, \label{eqn:1}\tag{1}$$

but I have also seen

$$C_p - C_V = R/M \label{eqn:2}\tag{2}$$

being used in case of ideal gas.
In what cases \eqref{eqn:1} is used and in what cases \eqref{eqn:2} is used?

The former difference is for molar heat capacity in J/K/mol, the latter for specific heat capacity J/K/kg, or J/K/g, depending on M unit.