Meaning of polarization pulse in microwave (rotational) spectroscopy In the recent rotational spectroscopy literature, "microwave polarization pulse" is often used in conjuction with pulsed Fourier transform microwave spectroscopy (Google Scholar).

Sometimes the pulse angle is also mentioned such as $π/2$. I gather this term is similar to FT-NMR where the net magnetization vector is tipped at a certain angle. When we talk about microwave rotational spectroscopy, what is being polarized?

In other words, why do we call the microwave pulse as a polarizing pulse? I wanted to confirm that this is related to the net electric dipole moment vector of the molecules which is being oriented after excitation?

In NMR the sum of the magnetic dipole vectors is called the magnetisation, and this is induced in the sample by the large fixed external magnetic field acting on the nuclear spins. As you mention it is influenced, i.e. moved about,  by the magnetic field of the rf pulses applied to the sample and so ultimately cause the nmr signal. In optics the similar thing  can occur where an intense optical field can align the electric dipoles and cause a polarisation, i.e. macroscopic dipole in the sample, this is the optical equivalent of magnetisation. The experiment is far more difficult to do in optics, however, as the dephasing time (T2 time in nmr) is several orders of magnitude faster than in nmr as is the lifetime of any level optically excited.

The field is generally called optical resonance spectroscopy and/or coherent transient spectroscopy with effects such as transient nutation, photon echos and self-induced transparency.


Meaning of the term "vertex" in the topic of skeletal formulae From the Wikipedia article on skeletal formula:

  Hydrogen atoms attached to carbon are implied. An unlabeled vertex is understood to represent a carbon attached to the number of hydrogens required to satisfy the octet rule, while a vertex labeled with a formal charge and/or nonbonding electron(s) is understood to have the number of hydrogen atoms required to give the carbon atom these indicated properties. Optionally, acetylenic and formyl hydrogens can be shown explicitly for the sake of clarity.

This is the first use of the word "vertex" in the article. Is a vertex

only the intersection of two line segments, or
the end of each line segment?

I assume the latter is true, but on many places I saw cyclopentyl being represented like this:

which should be methylcyclopentane instead, a completely different thing.

So, it looks like (mostly when describing aryls or cycloalkyls) some people assume that "vertex" means only the intersection of two line segments, otherwise they would have added the $\ce{CH3}$ to the end of the line to indicate methylcyclopentane (in this case). It's certainly inconsistent, but is this justified in any system of structural formulae?

The term vertex (plural vertices) comes from the nomenclature used in graph theory, a branch of mathematics which describes how points in a given space are connected to each other. This turns out to be quite fundamental to chemical nomenclature, as the vertices of a graph can (generally) be chemically interpreted as an atom, and the lines coming out of a vertex can (generally) be interpreted as covalent bonds. Two graphs which contain the same vertices, but where the connectivity between the vertices is different, corresponds in general to different chemical compounds.

To be clear, the term vertex applies to any point in a skeletal formula which is not "in the middle" of a connecting line. This means that every explicitly labelled atom, every "kink" in a chain (which implicitly represents carbon atoms) and every termination (also implicitly a carbon atom) is a vertex. Hydrogen atoms are often not vertices in a skeletal formula, though this can vary.

Regarding to your example, with no other context, chemists would call that a skeletal drawing of methylcyclopentane, which contains six vertices in its skeletal formula, one for each non-hydrogen atom. For it to be understood as a methylcyclopentyl radical/substituent, either the context should clearly state it, or preferably we would use other standardised graphical notation, shown below:


Why formulate drugs as HCl salts when HCl is in stomach acid? I know there are issues of formulation and industrial processing that make it advantageous to produce many amine containing drugs as their salts rather than as freebases. And if giving the drug intravenously, the increased water solubility is clearly an advantage. But if giving orally, is there not sufficient HCl in the stomach to have this occur in vivo? Is it just that some of these drugs are too non polar to even go into solution in the stomach acid? Or do we just need to avoid neutralizing too much of the stomach acid with a drug not intended for that purpose?

Another reason is that many drugs which are conjugated to hydrochloride salts is they are amines, and amines by themselves often stink. So people might be turned off from taking drugs that smell terrible (fishy).

Now, note that even though a label might say something such as "fexofenadine HCl" that doesn't mean there is HCl in the drug. Fexofenadine is an amine, and amines are somewhat basic - and more than basic enough for a strong acid such as HCl to completely protonate.

So if this is what you had in mind for the structure of fexofenadine, it isn't the most correct structure:

This would be a much better structure. The nitrogen in the ring has been protonated and it now bears a positive formal charge. Chloride ion is the counter ion.


pH Difference Our Taste Buds Can Recognise Water at $50~^\circ\mathrm{C}$ has a $\mathrm{pH}$ of $6.63$, while the pH of water at $0~^\circ\mathrm{C}$ is $7.47$.

The temperatures are comparable to a chilled, iced drink (at $0~^\circ\mathrm{C}$) and a hot drink (at $50~^\circ\mathrm{C}$). Would these differences in acidity be noticeable? Would they be noticeable if the drink was more acidic, such as orange juice?

First, the pH of water changes with rising temperature because the total number of $\ce{H+}$ ions (well, solvated hydronium ions, really) increases, not because it has become more alkaline. The water is still neutral, having an equal number of hydroxyl ions as hydronium ions. Heat simply increases the ionization.

Second, it depends on the individual taster, but a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $6.63$ (perhaps acidified with citric acid) would probably taste sour in comparison with a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $7.47$ (perhaps with a bit of baking soda, $\ce{NaHCO3}$).

See Temperature Dependent[sic] of the pH of pure Water for a more detailed explanation.


Stability of beryllium carbonate Beryllium carbonate is reported to be unstable under normal conditions and should only be kept in a atmosphere of $\ce{CO2}$. Why and how this stability is maintained? Why does this compound decompose when kept in an atmosphere of oxygen?

Why and how is this stability maintained?

Beryllium carbonate decomposes into Beryllium oxide and Carbon dioxide.
$$\ce{BeCO3 <=> BeO + CO2}$$

According to Le Chatelier's principle one simply has to store it under a $\ce{CO2}$ atmosphere to push the equilibrium to the left.
$$\ce{BeCO3 <<=> BeO + CO2}~\text{in excess}$$

  Why this compound decomposes when kept in atmosphere of oxygen?

When there is no $\ce{CO2}$ atmosphere the equilibrium is pushed to the right, which would be the "normal" case under standard conditions.
$$\ce{BeCO3 <=>> BeO + CO2}$$


Difference between torsion, out of plane, coplanar and perpendicular bends? In Infrared spectroscopy, stretches are easily understandable. But, how do I visualize (or conceptually understand) the difference between dihedral (torsions), out of plane bends, coplanar bends, and perpendicular bends? Can you give examples?

A molecule with N atoms has 3N degrees of freedom

3 translational modes

if it is linear, it will have 2 rotational modes; if non-linear 3

These leaves 3N-6 (3N-5 if linear) vibrational modes.  Vibrational modes are further subdivided into stretching, bending or torsional motions.

Stretching motions can be symmetric or asymmetric
Bending motions are often referred to as scissoring, twisting, wagging, and rocking
Torsional motions refer to rotations about bonds

This diagram illustrates these various stretching and bending vibrations

Depending on the geometric relationship between the transition moment for a specific vibrational mode and the symmetry axis for the mode, vibrational modes may be further classified as parallel or perpendicular.

Here is a link to a 31 page article (slide show) that has a lot of background and pictures on this subject. Section 4.11 (page 19) might be particularly interesting to you.

One final example that you might be interested in, formaldehyde has 4 atoms and is non-linear so it will have 6 vibrational modes.  These are all pictured on page 2 in this link.


Dangers of corroded mercury battery Disclaimer: Sorry if this is not a pure chemistry question. If it's better suited to another SE, I'm happy to have it migrated there.

I'm in the process of cleaning my father's workshop, and recently discovered an old Duracell mercury battery (picture from Wikipedia):
The battery is corroded, and I would say that the color of the corrosion is brown or dark orange.

The battery was lying in a pile of other things on a workbench. Presumably, the battery has been lying there for many years. As things have been moved around on the workbench over the years, one can expect the corrosion to have been scraped off and been mixed with sawdust, smeared on tools etc.

My concern is that I should basically treat the whole workshop as being contaminated, but still I feel it might be a bit over the top.

I would like some information on just how toxic the corrosion is, and some suggestions for what I should do. Could I swab some sample surfaces to have them tested? How expensive would this be?

Based on the color alone, the corrosion could be mercury (II) oxide, which is nasty. As with other mercury salts, it can end up in your body in a variety of new and exciting ways, such as by inhalation (that is not to say that elemental mercury cannot do so, it is simply more difficult), which make it more dangerous than elemental mercury. For this reason, do not touch it, even if it ends up being another salt.

What I (and I am no hazmat professional) would do is search around on the Internet for my local hazardous materials disposal facility (it may go under a different name) and would call them. I would ask them about the battery whilst giving as much information as possible about it, even if it seems unrelated. If (and only if) they ask you to bring it in, do the following:

Wear at least two pairs of gloves while handling it, and only do so briefly.
Put it in several layers of baggage, something that you will be okay with losing; and make sure it will not break.
Bring it in carefully, being sure not to breach the membrane. Also, be sure not to touch anything with the gloves used.

In addition, be careful with any of the tools that it touched, they could also be highly contaminated. If you want to learn more about mercury (II) oxide, the Wikipedia article seems relatively accurate (the same could be said about mercury batteries): https://en.wikipedia.org/wiki/Mercury(II)_oxide

Disclaimer: As stated, I am no hazmat professional, and my advice does not take authority over the advice of a professional.


Ionization Energy and Electron Affinity This is from my textbook.
But in other reference books I am using and Wikipedia, there is not any mention of absolute zero while defining Ionization Energy and Electron Affinity.
What does it mean? Please Elaborate

This is a practical matter of making sure that there is a common basis for comparing values from all different types atoms and molecules. If it were not the case there would be a range of different values for the same type of atoms or molecules.

Depending on temperature the ionisation energy would be less due to a molecule's rotational and vibrational energy, and vary with temperature because the number of levels populated depends on the temperature. For some types of atoms there may be low lying electronic levels that could be populated at higher temperatures, say at 1000 K in a high temperature gas. Again the measured ionisation energy would depend on temperature.

Also, if this was not done the $\Delta H$ values would in effect have double counted the effect of temperature, once in the ionisation energy and again in the heat capacity term.


How to identify the most stable Lewis adduct of boron trichloride? Which of the following is the most stable Lewis acid–base adduct?
  (a) $\ce{H2O\bond{->}BCl3}$
  (b) $\ce{H2S\bond{->}BCl3}$
  (c) $\ce{H3N\bond{->}BCl3}$
  (d) $\ce{H3P\bond{->}BCl3}$

This is a question from KVPY 2016 examination. In my first attempt, I focused on the positive charge that is going to appear on the donating atom. So the less electronegative the donor, the better it is at donating its lone pair to the vacant orbital on boron. However, the answer key states that the answer is option (c).

I am unable to rationalize why (c) is the most stable adduct.

You have to consider also the inherent strength of the covalent bond.  A strong covalent bond can entice an electronegative atom to accept some positive charge or give up some negative charge.

Roughly speaking, hydrogen and period 2 nonmetals form much stronger bonds among one another than with other elements, and this tends to override the electronegativity difference.  For instance, fluorine is much more electronegative than chlorine, but fluorine bonds so much more strongly with hydrogen that hydrofluoric acid is only a weak acid in water whereas hydrochloric acid is strong.  Note that the dissociation of hydrogen chloride in water creates hydrogen-oxygen covalent bonds.

Here, boron binds especially strongly with the period 2 nonmetals nitrogen and oxygen.  So (a) and (c) have this advantage, and after that (c) wins between these two by electronegativity difference.


What causes the smell in rotten meat-grease and how can I get rid of it thoroughly? We've just spilled the can of used meat-grease we collect so that it doesn't go down the drain and it turns out it was rotten. Now the granite counter-top and the wood-laminate floor of the kitchen smell very bad even after repeated cleaning with (1) dish-soap, (2) vinegar and (3) Murphy's Oil Soap as well as an enzymatic product called anti-icky-poo.

Extensive googling suggests that the smelly chemical is either putrescine or cadaverine. Am I right about that? How can I clean my kitchen such that it does not smell so bad?

If the stench is caused by putrescine or cadaverine, then you are in luck! Both putrescine and cadaverine are amines ($\ce{RNH2}$), which react with acids to form water-soluble ammonium salts ($\ce{RNH3+}$). Most soaps, surfactants, and other cleaning products are basic to help remove fats and oils (by hydrolysis). Try white vinegar, which contains acetic acid. If that fails, you can buy muriatic acid (a dilute solution of hydrochloric acid), which is used to clean pools. Find it wherever pool supplies are found.

Caution: While both of these solutions are fine for the granite, muriatic acid may not be fine for your floor - test in a small hidden area first. Vinegar should be okay for the floor as long as you don't let it soak.

Putrescine:

Cadaverine:


Estimating BDE Values I'm having trouble understanding how to "estimate" the value of the BDE for the compound shown below (I believe it is a secondary benzylic C-Br value) shown below (the 58 kcal/mol). We are supposed to be able to estimate the value from the table values shown below. My teacher said to use the C-H values and C-Br values from the table below to estimate it, but I don't understand how they are able to indicate the value. Any helpful drawings would be nice since I'm new to organic chemistry and have trouble just understanding the names! Thank you.

What you have shown in red is the second propagation step in the benzylic free radical bromination of ethylbenzene. The bond dissociation energy  (BDE) of bromine is +46 kcal/mol while the overall heat of reaction of the step is -12 kcal/mol. Thus, the formation of the C-Br bond must be -58 kcal/mol. The BDE for the C-Br bond in (1-bromoethyl)benzene is +58 kcal/mol.
However, the BDE for bromine is not listed in your tables. So how can one estimate the BDE of the C-Br bond in (1-bromoethyl)benzene using the data in the tables? A simple ratio and proportion will suffice. In equation A compare the primary C-H bond of ethane to that of toluene. This ratio is compared to the ratio of the BDE of the secondary C-H bond of propane and its counterpart in ethylbenzene, which is the unknown "x". An estimated value of
82 kcal/mol is obtained. In equation B, this ratio compares secondary C-H bonds with secondary C-Br bonds. The estimate for the BDE of the C-Br bond in (1-bromoethyl)benzene is the value of "y" which is equal to 59 kcal/mol. Pretty close, wouldn't you say?


What does the volume of a gas depend on? In my chemistry textbook, it states that the volume of a gas is directly proportional with the amount(number of moles) of that gas. However, it also mentioned before that the volume of a gas depends on its container. So if the volume of a gas depends on its container then wouldn't the volume not change as you add or take away gas molecules? As long as the container doesn't change wouldn't the volume of the gas stay the same?

I must clear one thing up. gases always fill their container. If you take away gas particles, it will always fill its container. The volume of gas is proportional to many different factors, though. Based on the ideal gas equation, volume is proportional to moles and temperature. It is inversely proportional to pressure. when the text say that volume is proportional to amount, it is assuming all other variables are constant, and not changed. So yes, in general, as there is an increase in volume, there is an increase in number of moles of gas.


Why does Azo coupling of β-naphthol takes place at alpha position and not at gamma position I have searched for coupling reaction of $\beta$-naphthol with benzene diazonium salt at quite a few places. But everywhere the coupling has been shown at alpha position. Why doesn't it take place at gamma position to minimise steric hindrance.

Chemistry LibreTexts

Chemguide

My book

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.

If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 , with aromatic rings are possible.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.

References:

https://pubs.rsc.org/en/content/articlehtml/2015/gc/c4gc02381a

https://en.wikipedia.org/wiki/Azo_coupling

Mechanism of formation of 2-naphthol red dye (aka Sudan 1)


Is there any experimental evidence for OClF5? On page 88 and 89 of this document, the author laments that $\ce{OClF5}$ has still not been discovered. He says it would be "just about the ultimate possible storable oxidizer" (for rockets).

(Note: it's pages 88 and 89 as labeled by the writing on the page, not the pdf metadata.)

The text seems a little dated. Has anyone ever found/produced any $\ce{OClF5}$? A related compound, $\ce{OClF3}$, was made in 1970 (footnote on page 89). So this seems to indicate that one of those weird loopy pairs of electrons can be jammed/bonded with an atom of oxygen.

Yes there has been experimental evidence published.  Reaction of CIF5,  with OF2: Preparation and Properties of Chlorine Oxide Trifluoride and the Existence of Chlorine Oxide Pentafluoride, Angewandte Chemie, Internation Edition, vol. 11, pages 1094-5, purports to show a series on F-19 NMR spectra of the compound, at temperatures ranging from 34 to -76 degrees C.

The series shows a relatively sharp peak at 34 degrees C, which progressively broadens and shifts upon cooling to until about -40, upon which further cooling sharpens to a peak with much different chemical shift than the room temperature peak.  The authors conclude that at room temperature there is fast exchange among the 5 F atoms, and at low temperature 4 equivalent equatorial F atoms are observed, while the peak for the axial F atom is too broad to be observed.

However, the above claim is considered refuted by "Chlorine Oxyfluorides" Advances in Inorganic Chemistry and Radiochemistry vol. 18, pages 319-98.

See On the XeF+/H2O System: Synthesis and Characterization of the Xenon(II) Oxide Fluoride Cation, FXeOXeFXeF+ J. Am. Chem. Soc., vol. 131, pages 13474–89 for further discussion and references:

  Chlorine oxide pentafluoride, ClOF5, would be the highest performing earth-storable, liquid rocket-propellant oxidizer. Although the synthesis of ClOF5 had been claimed in 1972,(1) this claim was subsequently refuted.(2) Continued efforts to prepare this compound by oxidative fluorination of substrates such as ClOF3(3-5) or ClOF4−(6, 7) have failed because the oxygen ligand is more easily oxidized than the central chlorine atom. This failure prompted a search for alternate synthetic approaches, such as the reaction of ClF5 with an oxidative oxygenator. Potential candidates for oxidative oxygenators include hypofluorous acid, HOF, a well-known unstable compound,(8, 9) and its protonated form, the H2OF+ cation,(10, 11) if indeed it should exist.


Spin interaction term for a two-particle system of fermions with a potential that contains spin-spin interaction Consider a two-particle system consisting of two identical fermions in a potential $$V(\vec{r})\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$$ where $V(\vec{r})$ is the spatial part of the potential and the indices $1$ and $2$ correspond to each particle respectively and $\vec{\sigma}=\sigma_1i+\sigma_2j+\sigma_3k$ with $\sigma_{i}$ the $i$-th Pauli matrix.

I want to find the contribution of the spin interaction term to the energy of the system. Now, working with singlet and triplet states, I know that the energy will be analogous to the eigenvalue of each state respectively.
Working things out, I have found that the eigenvalue of the singlet state for $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$ is $-3$ times the eigenvalue of the triplet state.
By also trying out as $V(\vec{r})$ a potential used in physics(Yukawa potential), I also found that the triplet state has the same energy as the case in which we have two bosons in that potential.

So, denoting the ground state energy of that potential for two identical bosons as $E$, the triplet state gives $E_{triplet}=E$ and the singlet state gives $E_{singlet}=-3E$.

So, the question is:
What is the physical explanation of this?.
And, why is the spin-spin interaction more "intense" for the singlet state than for the triplet state(by taking the absolute value of each state's eigenvalue)?

EDIT: Since I don't consider spin-orbit interaction, I think that the answer might have to do with spin-spin coupling because of the dependence on $\vec{\sigma_{1}}\cdot\vec{\sigma_{2}}$.

The answer to ' why is the spin-spin interaction more intense for a singlet than triplet'  is due to the fact that any internal interaction cannot change the overall energy  of the system. Put informally its a sort of 'centre of gravity' thing and applies in all cases. The total energy change is $-E_S + gE_T$, where the $g=3$ is the multiplicity of the triplet. The energy change is zero as $E_S = 3E_T$.

The spin-spin interaction has a standard calculation for example, for the hyperfine structure splitting of hydrogen atom ground $1$s $^2\mathrm S_{1/2}$ ground state, which produces the 21 cm line of radio astronomy and the hydrogen maser.  The potential in which the spins exist (a proton and electron in this case) only comes into the calculation in determining the overall magnitude of the splitting and then as $|\psi(0)|^2$ sometimes called the Fermi contact interaction.

(By standard methods I mean the interaction is of the form $H=A\vec I\cdot \vec J$ where the magnetic moment is generated by nuclear spin angular momentum $\vec I$ and $\vec J$ that due to electron spin. The total angular momentum of the atom is $\vec F= \vec I + \vec J$ is constant and the expectation value is the energy $E=A<\vec I\cdot \vec J>= \frac{A}{2}(F(F+1)-J(J+1)-I(I+1))$. In the case of two spin $1/2$ particles $J=1/2$ and $I=1/2$ and the energies are $A/4$ for $F=1$ and $-3A/4$ for $F=0$. The scaling term A contains constants and  $|\psi(0)|^2$ )


Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane? E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane.

I can't think of a reason that they would need to be in the same plane, so what is the reason for this strange phenomenon?

When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond.  For these two p orbitals to form a pi bond they must be in the same plane (or very close).

This is illustrated in the following figure.  This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.

image source

Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.

This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.


Do I use the Nernst equation when the concentrations of electrolyte in both half cells are equal? I'm doing a chemistry report, and for my experiment I investigated the amount of voltage produced by different half cells in a voltaic cell.

However, I tested them under nonstandard conditions as the temperature at the time was $\pu{18^\circ C}$, and the concentration of the electrolytic solution in the half cells was $\pu{0.1 M}$.

I tried the Nernst equation, but since the concentration of both my electrolytes were $\pu{0.1 M}$, calculating the reaction quotient gave me an answer of $1$, and $\log 1 = 0$, therefore I'm left with the same cell potential I started with (cell potential at standard conditions) and I'm stuck there.

Here is an example for a magnesium-zinc voltaic cell at $\pu{291 K}$ ($\pu{18^\circ C}$):

$$E_\text{cell} = 1.593 - \frac{0.05778}{2} \log_{10} \frac{0.1}{0.1} = \pu{1.593 V}$$

Could you please explain the solution to me?

Yes, you still use the Nernst equation, but one of your implicit assumptions is not correct.

A point about the Nernst equation that often confuses people is that, at first glance, it doesn't predict any change in $E$ as temperature changes, as long as $Q$ remains equal to $1$, and $\ln Q = 0$.

$$E = E^\circ - \frac{RT}{\nu F}\ln Q \tag{1}$$

However, a slightly more careful analysis reveals that if you set $\ln Q = 0$, it only implies that $E = E^\circ$. The conclusion that $E$ doesn't change is based on an assumption that $E^\circ$ doesn't change, which is not true.

"Standard conditions" do not stipulate any temperature. So $E^\circ$ itself is a function of temperature. The standard cell potential $E^\circ$ at $\pu{298 K}$ is going to be different from the standard cell potential $E^\circ$ at $\pu{250 K}$, and hence $E$ will vary with temperature, even if $\ln Q = 0$.

So, how does one determine the temperature dependence of $E^\circ$? The best way is to measure it, but if that is not possible, then you have to somehow find an expression for it. The Nernst equation alone has no answer for this. Instead you would have to turn to thermodynamics. We know that

$$E^\circ = -\frac{\Delta_\mathrm rG^\circ}{\nu F} \tag{2}$$

Note that $\Delta_\mathrm rG^\circ$ is a function of temperature, and hence so is $E^\circ$. So, the problem becomes one of determining the variation of $\Delta_\mathrm rG^\circ$ with temperature. The most primitive way is probably to look up the data for $\Delta H_\mathrm{f}$ and $S_\mathrm{mol}$ of each of the compounds, calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ for the reaction (using Hess's law), and then find

$$\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ \tag{3}$$

Of course, when you look up data, the chances are that the data you find are specified for $T = \pu{298 K}$. So, when you use these data to calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$, you are finding the values of these quantities at $T = \pu{298 K}$. If you plug these values into equation $(3)$, then you also assume that the values of $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ at your desired temperature are equal to their values at $\pu{298 K}$.

Depending on how much accuracy you need, this may or may not be tolerable, and there are certainly more thorough ways of calculating it. However, that discussion is best left for another question.


Ethanol and water hygroscopic equilibrium concentration I am working with a company to help improve their reaction process where they use SDA 2B Alcohol (Karl Fischer titration has it at 99.66% EtOH, 0.3375% toluene). Specifically, I am attempting to determine what theoretical %wt or %vol of water is pulled from atmospheric air into pure ethanol given an infinite amount of time? Does it ever reach an equilibrium state and, if so, what is it?

My gut guess is ~95.5%wt EtOH/4.5%wt H2O, the azeotropic concentration.

Assume constant relative humidity and room temperature. The ethanol is stored in a closed tote (~220 gallons). Initially, the tote is full of as-shipped ethanol (negligible air volume). However, half the ethanol is removed as part of a batch run. So, really, what I'm trying to get at is what happens to the leftover ethanol that is now sitting in a closed tote with ambient conditions.

Please let me know if additional details will help answer the question. Thanks!

You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air.  There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below).

If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote.  Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water.  That's 0.01%.  Whether this much water ruins your alcohol or not, you will know better than anyone.

Every time you open the tote, you're potentially letting in 40g of water.
I say "potentially" because, since the gas content is not instanty exchanged, much less water than that might go in.  But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.


Does iron(VI) fluoride exist? Why does $\ce{FeF6}$ not exist? There are hexavalent iron compounds, so that is not the problem. There is the $\ce{[Fe(CN)6]^3-}$ ion, so there doesn't seem to be a steric problem.

Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer.  Iron is fairly early in the transition metal series, so when pushed beyond the $+3$ oxidation state it has few $d$ electrons in the central core.  As explained in the referenced answer, this makes the iron strongly pi-electron accepting, and therefore more favorable for combining with a stronger pi-donor ligand such as oxide instead of fluoride.


Why will CuSO4 solution be acidic? An end-chapter problem posed in an 11-grade texbook asks to analyse whether a water solution of CuSO4 will be acidic.

The standard answer is yes, because "$\ce{H2SO4}$ is a strong acid, hence $\ce{SO_4^2-}$ is a weak base. As a weak base, it will not be too hot about grabbing $\ce{H^+}$ from the environment. On its part, $\ce{Cu^2+}$ will strongly attract $\ce{OH^-}$ groups, and overall protons will prevail over hydroxides."

But I was told in the chat that the real answer is more interesting and complicated, so I post this question in the hope that someone will explain the process in-depth.

Because the second pKa of sulfuric acid is weak, sulfate withdraws protons from water.  Therefore, the contribution of sulfate alone is to make the solution slightly more basic than pH 7.

However, as quantified in CE4501 Environmental Engineering Chemical Processes, Cu will form species such as CuOH+, which is the reason the solution will be acidic overall.  This could be approximated by thinking of $\ce{Cu(H2O)6^{2+}}$ as having a pKa of ~6.3.

Because  $\ce{HSO4-}$ is a stronger acid than CuOH+ is a base, the solution is acidic.


Would Z and E isomers show different optical activity? Consider this compound, which is in E-configuration and that red asterisk denotes the chiral carbon. If I were to somehow cause the E-configuration of the alkene part to Z-configuration keeping the rest of the molecule same, would it have the same optical rotation?

Not necessarily. Although the change you suggest would not alter the absolute stereochemistry (S in both cases) of the asymmetric center starred, the two molecules are diastereomers and therefore will have different physical and optical properties. Even if they did both rotate light the same way, they probably won’t have the same degree of rotation for the same concentration (specific rotation).

The problem boils down to whether we can predict the direction (or degree) of rotation from a drawn absolute stereochemical configuration. The answer to this is still “no”. We don’t currently have a theory that will allow us to say that, for a certain molecule, R gives us d or l, etc.

Some studies have been able to come up with some rules of thumb for small molecules, but there is no comprehensive theory to make this prediction in general. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2704476/


Reaction between potassium permanganate, alum, and sodium hydrogen sulfite When I was a teenager, I played with a chemistry set. I tried adding potassium permanganate, alum, and sodium hydrogen sulphite (not sodium hydrogen sulphate) to water.

Upon adding the third ingredient, the water would rapidly turn clear, and a brown precipitate would form. After filtering this out, I allowed the water to evaporate, which left a white crystalline deposit. All three ingredients were definitely needed in order for the reaction to happen, which is why I remember it, since none of the other reactions I knew about needed three reactants. (This was a long time ago, so there's a very slight possibility I've mis-remembered the ingredients, but I'm 99% sure they're right.)

At the time, neither my teachers nor my parents could help me work out what kind of reaction it might be, so out of sheer curiosity, I thought I'd ask here. I'm no chemist, so simple explanations would be appreciated.

The fact that the solution turned clear is a pretty good guarantee that this was a redox reaction: the permanganate oxidised some other species, and in turn was reduced, probably to manganese dioxide, the brown precipitate you noticed. Unfortunately, neither of the other reactants you mention are reducing agents!
Do you think it's possible that the third reagent was actually sodium hydrogen sulphite, NaHSO$_3$? If so, this would make a lot more sense: the (hydrogen)sulfite ion would have been oxidised to (hydrogen)sulfate. This reaction requires an acidic medium, and the reason the alum (potassium aluminium sulfate) was necessary was to lower the pH.
These individual steps have the following equations:

Al$^{3+}$ + 4 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 4 H$^+$
MnO$_4^-$ + 4 H$^+$ + 3 $e^-$ $\rightarrow$ MnO$_2\downarrow$ + 2 H$_2$O
HSO$_3^-$ + H$_2$O $\rightarrow$ HSO$_4^-$ + 2 H$^+$ + 2 $e^-$

(note that the last two are half-equations -- they can’t occur on their own, but I’ve written them separately for clarity) giving an overall reaction of

Al$^{3+}$  + 2 MnO$_4^-$ + 3 HSO$_3^-$ + 3 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 2 MnO$_2\downarrow$ + 3 HSO$_4^-$

The white crystalline deposit in this case would just be alum again, more or less. You have some sodium in your solution now, from the sodium hydrogensulfite, so you'd get some sodium aluminium sulfate as well, but the only anion in your solution will still be (hydrogen)sulfate, and my guess is that with aluminium this will crystallise out as the sulfate.


Can an organic compound have an ionic bond? Is there any organic compound that has another type of bond other than the covalent bond?

There's a large number of salts with organic anions (e.g. potassium hydrogen tartrate) or organic cations (e.g. tetrabutylammonium hexafluorophosphate)

There are even salts, in which both the anion and the cation are organic compounds with covalent bonds. This is often used in pharmaceuticals. Here the active component appears in protonated form and is served as a salt with the conjugated bases of carboxyclic acids (e.g. citrates, succinates, etc.)


What is the signal at around 2090 wavenumbers in the nitroguanidine FTIR spectrum? There is a signal at around 2090 apparent in the FTIR spectrum of nitroguanidine  from Sigma-Aldrich (first image). It's also apparent in FTIR spectra we've taken in the lab (second image), but when assigning peaks to bonds, we are unclear to what it is.

We are unsure of how Sigma-Aldrich prepared their pellets, but we used KBr. We are thinking perhaps KCN, but then again - how could that possible form?

Its common Nujol, a hydrocarbon oil used to prepare paste-samples for IR from powdered materials. Hence the C-H valence around 2090 cm-1


Can green vehicle antifreeze (Ethylene Glycol) be diluted in a glass container? I picked up a container of undiluted green antifreeze, and oops, I don't have anything to mix it in! I do have a glass container though at the top of my recycling bin that is about a half gallon. I've heard some plastic containers (soda bottles, plastic milk jugs) can't be used for anti-freeze, but what about glass? Two questions arise from this

As Long as I rinse the previous contents (milk) out, can I safely use this glass container for diluting antifreeze and putting it into my vehicle?
Once I'm done can this go back into the recycle bin, or will trace amounts of antifreeze spoil it for recycling?

As Long as I rinse the previous contents (milk) out, can I safely use this glass container for diluting antifreeze and putting it into my vehicle?

Not sure if you are talking a bout antifreeze for the radiator system or windshield deicers, but both systems are rugged enough that the do not require very pure water to be used. A washed milk container would suffice. The antifreeze wont disrupt the bottle and the bottle wont adulter the antifreeze

  Once I'm done can this go back into the recycle bin, or will trace amounts of antifreeze spoil it for recycling?

This somewhat depends on how it will be recycled. If you are putting it in general glass recycling then the bottle will become cullet that will heated to temperatures high enough to burn off any residue and it wont matter.

If you are returning the milk bottle to the producer for reuse then they have to clean them well enough to remove milk reside which is harder to remove than either your antifreeze or the bitterants in antifreeze. Any reputable bottler would expect some amount of non-milk substances in the bottles and clean them sufficiently well to make sure every bottle is clean and safe. Though I would recommended that you do rinse it before furnishing it to them out of courtesy.


Mathematically balancing chemical equation yields weird results I found this equation while testing my equation balancer:
$$
\ce{a C7H6O3 + b C4H6O3 -> c C9H8O4 + d C2H4O2}
$$

Obviously, $a = b = c = d = 1$ is a solution. However, if we were to bring in some maths:

$
\begin{cases}
7 a + 4b &= 9c + 2d\\
6a + 6b &= 8c + 4d\\
3a + 3b &= 4c + 2d\\
\end{cases}\\
\begin{cases}
a &= \frac{11}{9}\alpha - \frac{2}{9}\beta\\
b &= \frac{1}{9}\alpha + \frac{8}{9}\beta\\
c &= \alpha\\
d &= \beta\\
\end{cases}
$

This leaves us with many ways of balancing the equation:

when $\alpha = 1, \beta = 1$, $a = b = c = d = 1$; $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$
when $\alpha = 1, \beta = 2$, $\begin{cases}
a &= 7\\
b &= 17\\
c &= 9\\
d &= 18\\
\end{cases}$; $\ce{7 C7H6O3 + 17 C4H6O3 -> 9 C9H8O4 + 18 C2H4O2}$
when $\alpha = 2, \beta = 1$...

From what I know, this happens when the equation consists of 2 or more independent equations (as is the case here: More than one way of balancing a chemical equation). However, here the two equations are:

$$
\ce{11 C7H6O3 + C4H6O3 -> 9 C9H8O4}
\\
\ce{-2 C7H6O3 + 8 C4H6O3 -> 9 C2H4O2}
$$

What is concerning here is that the second equation has a negative stoichiometric coefficient. My question is, is a negative coefficient acceptable in an equation, and in this case which is the correct way to balance the equation?

As @Ivan Neretin and @Tyberius pointed out, the negative coefficient of a reactant can be flipped to be positive by moving the reactant to the products' side.

So, in my case, the equations become
$$\begin{align}
\ce{11C7H6O3 + C4H6O3 &-> 9C9H8O4}\\
\ce{8C4H6O3 &-> 2C7H6O3 + 9C2H4O2}
\end{align}$$

Now, since the original equation $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$ consists of 2 independent equations which can be separately balanced and combined in an arbitrary proportion. See  More than one way of balancing a chemical equation for a more detailed explanation.


Do I really need 4 different water treatment systems with this water analysis? We installed an artesian well and we had some surprises. We asked a few places around and they all told me that we needed 4 different systems to solve our water problem. I am ready to accept this fact but before having our basement transformed to a factory I just wanted to confirm that this is the only solution.

Here is the result of my analysis.

The problems are:

Barium  $6.9 \:\mathrm{mg/L}$    (2nd analysis revealed same results)
Apparent Color 17.5 UCA
Iron  $0.47 \:\mathrm{mg/L}$
Total sulfur $0.07 \:\mathrm{mg/L}$

Also the water is hard ($210 \:\mathrm{mg} \ce{CaCO3}/\rm L$) which is not recommended

My question is, do I really need 4 different systems to treat my water?

You shouldn't need 4. The only dangerous thing is the barium, and permeakra is right that you should keep an eye on the equipment so that it doesn't silently fail on you. You should check perhaps once a month. It's only ~3.5 times the EPA limit, so a failure won't be immediately be a hazard (unless the sequestered barium is released all at once, but that seems unlikely).

The other 3 problems and the hardness are, at those levels in the water, are simply nuisances. So they will discolor your water, produce a weak sulfurous odor, and eventually clog the pipes, but they will not harm anyone. If you fixed the iron and sulfur, that would probably improve the color, so only 3 systems are needed. And you don't have to worry about the system failing silently since you will see it.

Check these statements before making a decision, this is too important to trust only one source!


Comparison of basic strengths of hydroxides of alkali metal group I have read that $\ce{LiOH}$ is less basic than $\ce{KOH}$. But I don't know its reason. I have tried to frame a possible explanation.

My explaination:

I think that as $\ce{Li}$ undergoes more hydration than $\ce{K}$, $\ce{OH-}$ group will stick stronger to $\ce{Li+}$ ion. Due to this $\ce{Li+}$ will be reluctant to loose $\ce{OH-}$ and thus will be less basic. Is my explanation correct?
And, is this explanation correct for any other pair of hydroxides, eg, $\ce{Sr(OH)2}$ & $\ce{Ba(OH)2}$?

Yes that is it.

To paraphrase the wikipedia article HSAB theory

"Hard and soft (Lewis) acids and bases" (HSAB), also known as the Pearson acid-base concept, HSAB is widely used in chemistry. It assigns the terms 'hard' or 'soft', and 'acid' or 'base' to chemical species.

'Hard' applies to species which are small, have high charge states (the charge criterion applies mainly to acids, to a lesser extent to bases), and are weakly polarizable.
'Soft' applies to species which are big, have low charge states and are strongly polarizable. The concept is a way of applying the notion of orbital overlap to specific chemical cases.

Pearson evidently had revised values in the paper (which is behind a firewall):

Pearson, Ralph G. "Absolute Electronegativity and Hardness: Application to Inorganic Chemistry." Inorganic Chemistry, volume 27, number 4, 1988, pp 734–740. https://doi.org/10.1021/ic00277a030

The data from that paper is on this webpage: http://www.knowledgedoor.com/2/elements_handbook/chemical_hardness.html

\begin{array}{|c|c|} \hline
Ion & Chemical\ Hardness\ (eV) \\ \hline
\ce{Li+} & 2.39\\ \hline
\ce{Na+} &   2.30   \\ \hline
\ce{K+} &  1.92    \\ \hline
\ce{Rb+} & 1.85     \\ \hline
\ce{Cs+} &   1.71   \\ \hline
\end{array}


How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride? How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride?
I can't seem to single out how many atoms of Lithium are in lithium fluoride.

I know that the first step is to convert 14.3 grams to moles, so 14.3 multiplied by the molar mass of LiF divided by 1 is 0.55 moles. But where do I go after this? Do I just multiply 0.55 by Avogadro's number?

Convert 14.3 grams of Lithium Fluoride to moles which is the same for lithium and lithium fluoride since there is one atom of lithium per molecule of lithium fluoride (carry an extra significant digit to round later to 3 significant figures).

  $\mathrm{moles(Li) = \dfrac{g(LiF)}{mw(LiF)} = \dfrac{14.3}{25.939} = 0.5513\text{ }moles(Li)}$

Convert moles to atoms by multiplying with Avogadro's number, and round to 3 significant figures.

  $\ce{atoms(Li) = moles(Li)\text{ }(N_{a})} \mathrm{ = 0.5513(6.022E23) =  3.32E23}$


Titration of magnesium sulfate with potassium permangante One common method for measuring the amount of $\ce{Fe^2+}$ in $\ce{FeSO4}$ is by dissolving it with $\ce{H2SO4}$ and then titrating with $\ce{KMnO4}$.

However, is it possible to measure the amount of $\ce{Mg^2+}$ ion in magnesium sulfate by dissolving it in $\ce{H2SO4}$ and then titrating it with $\ce{KMnO4}$? Why or why not?

The $\ce{Fe^2+}$ concentration can be measured in this fashion since the iron(II) cation can be oxidized by a suitable oxidant, like the permanganate ion $\ce{MnO4-}$.

$$\ce{MnO4- +4H+ +3Fe^2+ -> MnO2 + 2H2O + 3Fe^3+}$$

Since permanganate is purple, and manganese dioxide is an insoluble brown solid, we have a built in indicator for redox titrations like these.

However, magnesium ion content cannot be determined this way since $\ce{Mg^2+}$ is the highest usual oxidation state for magnesium. The electron configuration for magnesium is $1s^2 2s^2 2p^6 3s^2$ or $[\ce{Ne}]3s^2$. The $\ce{Mg^2+}$ ion has the electron configuration of $\ce{[Ne]}$, which is relatively stable. Remove an additional electron from $\ce{Mg^2+}$ to get $\ce{Mg^3+}$, you get an ion with the electron configuration of $1s^2 2s^2 2p^5$, which is the same electron configuration as electron-hungry fluorine atoms, but with a nucleus containing three more protons. $\ce{Mg^2+}$ cannot be chemically oxidized further.

Now, one could use this method to determine the concentration of many other metal ions that can be oxidized, including even $\ce{Mn^2+}$.

$$\ce{3Mn^2+ + 2MnO4- + 2H2O -> 5MnO2 + 4H+}$$


Dehalogenation of vicinal dihalides Why, among different metals, only zinc is favourable for the dehalogenation of vicinal dihalides?

Zinc is used effectively in the dehalogention of vicinal dihalides because a carbon zinc complex is formed in the mechanism which is probably more stable. This reaction is also known as Simmon' Smith reaction.

The stability and usefulness of the Simmons-Smith reagent (zinc metal)
may be attributed in part to the higher covalency of the carbon-zinc
bond together with solvation and internal coordination of the zinc (which makes the zinc complex stable). (Source)
Reduction of vicinal dihalides by zinc metal proceeds by way of
oxidative insertion of Zn(0) into a C-X bond, producing an organozinc
species (stable) which undergoes dehalogenation:

(source)


Why is MnO2 not a peroxide? As I know, $\ce{MnO2}$ is called manganese(IV) oxide.

Why can't we name it manganese peroxide since manganese(II) exists too?

In other words, how do we know that the oxidation number of manganese in $\ce{MnO2}$ is $4$ and not $2$?

Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.

Peroxide is well known as a ligand to transition metals (though whether it is better described as peroxide or superoxide is sometimes not always clear, e.g. in some cobalt species). An example which you might come across in the lab is chromium(VI) oxide peroxide, $\ce{CrO5}$.


Derivation of formula for the degree of unsaturation Degree of Unsaturation (DU) of an organic compound tells us about the total number of $\pi$ bonds (both double bonds and triple bonds) and number of rings/cyclic chains in the compound.

Degree of unsaturation forms an integral part of the four tools used to identify the structure of an organic compound (some of the others being structure after hydration and ozonolysis (there's one more I don't remember)).

Wikipedia states the formula of degree of unsaturation as:

  $$\text{Rings + $\pi$-bonds} = C - \frac{H}2 - \frac{X}2 + \frac{N}2 + 1$$
  where $C$ = number of carbons, $H$ = number of hydrogens, $X$ = number of halogens and $N$ = number of nitrogens

A more general formula is stated as

  $$DU = 1 + \frac{1}{2}\sum{n_i(v_i - 2)}$$
  where $n_i$ is the number of atoms with valence $v_i$.

What I'm interested is the underlying concept from which this formula is derived.

Well, that is simple.

A Carbon (C) makes 4 bonds, a Hydrogen (H) or a halogen (X) makes 1 bond and Nitrogen (N) makes three bonds. So in a molecule with $c$ Carbons, $h$ Hydrogens, $x$ halogens and $n$ Nitrogens, we will have $4c+h+x+3n$ "bonding capacity".

Assuming there are no rings, no double/triple bonds, we must have $c+h+x+n-1$ bonds in between these atoms [think like how many dashes you can draw between $m$ $\ce{O}$'s: $\ce{O-O-O}$ two dashes for three $\ce{O}$'s apparently].

Since each bond uses one "bonding capacity" from each of the two atoms its binding, $c+h+x+n-1$ bonds are using $2(c+h+x+n-1)$ bonding capacity. So the remaining capacity is:

$$(4c+h+x+3n)-2(c+h+x+n-1)=2c-h-x+n+2$$

If this is larger than zero, it means our molecule is unsaturated, i.e. still has a capacity to make bonds. And our assumption that "there are no rings, no double/triple bonds" is wrong. This bonding capacity is used in forming new bonds either in rings or in pi-bonds.  Since each bond uses two capacity one from each side we should have $c-h/2-x/2+n/2+1$ "unsaturated" bonds.


How to draw a reaction coordinate diagram for SN1 mechanism? How would you draw a reaction coordinate diagram for an SN1 mechanism with the following conditions?

each step in the mechanism (there are 2 steps) is favorable
the second step of the mechanism is the rate-limiting step

I believe the RC diagram should look some thing like this, but I am not completely sure.

Yes, that diagram would be correct for the generic reaction. However in the specific case of an SN1 reaction the intermediate is almost always higher in energy than the reactants or the products, as they usually involve a carbocation. It is also typical of SN1 reactions for the first step to be rate-limiting, as carbocation formation from a neutral reactant is thermodynamically unfavorable unless specific reaction conditions are present. The three general rules for reaction coordinate diagrams are as follows:

The number of “peaks” should be equal to the number of steps in the reaction.

The height of each peak relative to the valley to its left is representative of the activation energy of that step. Larger vertical distance between peaks = greater activation energy = slower step. The rate-limiting step of any reaction will have the greatest distance between its peak and the valley preceding it.

The height difference between your reactant(s) and product(s) is equal to your total change in enthalpy (delta H) for the reaction. If the products are below the reactants then enthalpy change will be negative, and the reaction is exothermic. The inverse is true of endothermic reactions.

It may be worth clarifying that this diagram is supposed to represent an SN1 mechanism and the second step is rate-limiting. The diagram of these reactions should look like this:

Hopefully this is helpful!


Dissolution of CO2 in water The following question has troubled me for a while.

For gases that are slightly soluble in water, there is a proportional relationship between the partial
pressure, $P$, and the mole fraction, $x$, of the gas molecules dissolved in water (Henry's law):
$\ce{P = k_H*x}$
A container ($V$ = $570$ $mL$) is filled with water ($V_l$ = $500$ $mL$) and pressurized with $CO_2$ gas ($P_0$ = $50$ $atm$), before it is allowed to stand at $10^\circ$C until the vapor–liquid equilibrium is established.
Calculate the pressure of $CO_2$ $[atm]$ in the container and the amount of $CO_2$ $[mol]$ dissolved in the water. The Henry coefficient of $CO_2$ for water at $10^\circ$C is $k_H$ = $0.104 × 10^4$ $atm$, and we will consider that the reaction of $CO_2$ in water can be ignored.

I did the following calculations but I am unsure with the outcome.
$\ce{n_{total}\approx n(H_2O) = \frac{\rho V}{M(H_2O)} = 27.753 mol}$
$\ce{x = \frac{P_0}{k_H} = 0.048077}$
$\ce{n(CO_2)_{dissolved} = x*n_{total} \approx 1.3343 mol}$
$\ce{P_1 = \frac{nRT}{V} = \frac {1.3343*0.0831451*1.01325*283}{0.570} = 55.8 atm}$
Shouldn't the final pressure be smaller than the initial pressure ($50$ $atm$)?

Assumptions:

Reaction of carbon dioxide with water is neglected.
Vapour pressure of water is negligible.
Volume of solution does not change on dissolution of carbon dioxide.
Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2_{(aq.)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$

Initially no $\ce{CO2}$ is dissolved in water and so initial moles of $\ce{H2O_{(l)}}$ and $\ce{CO2_{(g)}}$ can be calculated as follows:
$$
n_{{\ce{H2O_{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\
n_{{\ce{CO2_{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\
$$
Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.
Final pressure in the container can be written as:
$$
P_f= \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
And by Henry's law we can say,
$$
P_f= K_H \times X_{\ce{CO2_{(aq.)}}}
$$
Thus from both these equations we can conclude:
$$
K_H \times  \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
We also known that,
$$
n_{{\ce{CO2_{(g)}}}_f} = n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f} \\
$$
So,
$$
K_H \times  \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f})RT}{V}\\
$$
Solving this we get,
$$
n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\
$$
This correspondingly means:
$$
n_{{\ce{CO2_{(g)}}}_f}= \pu{0.015 mol}\\
P_f= \pu{5 atm}\\
$$
The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $P_f=P_i$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.


What is the molecular formula of 2-ethynyloctahydro-1H-indene? I thought the formula for this $\ce{C11H15}$, but it's not. I need assistance in finding the proper formula:

I suspect you forgot one hydrogen at the terminal carbon in alkyne group $\ce{R-C#C-H}$. With this in mind the formula would be $\ce{C11H16}$:


-ic -ous nomenclature I recently came across a practice problem in my textbook asking me to name a few compounds using -ic and -ous endings. The exact wording is:

Write the name of each of the following ionic substances, using -ous and -ic endings to indicate the charge of the cation.

The first one was $\ce{CoCl2}$. However, cobalt has more than just two oxidization states (3, 2, 0, and -1).

So if it is one of the states in the middle, how do I decide whether to use -ous or -ic

Cobalt may possibly have all those other oxidation states (and others too) for this nomenclature system, we only care about the two most common oxidation states in ionic compounds: $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$. Cobalt(II) compounds would thus be named  cobaltous and cobalt(III) compounds would be cobaltic. Which is yours?


How are standard bond enthalpy values experimentally produced? The question's essentially it. I've always wondered about how these values are produced, since we refer to them heaps in the lab to assess our experimental results, but I don't actually know how they're produced.

As defined by the IUPAC Gold Book, mean bond energy (standard bond enthalpy) is defined as:

The average value of the gas-phase bond dissociation energies (usually at a temperature of $298\ \mathrm{K})$ for all bonds of the same type within the same chemical species.
Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.

These energies are calculated by studying the homolytic cleavage of a bond into its respective radicals, or the colligation of two radicals to form a particular bond. Though measuring the heat from the colligation of radicals might be the most direct method to determine these values, it is also the most difficult to perform with any accuracy, and is not the preferred method.
There are a variety of ways to measure bond dissociation energies through bond cleavage—by irradiating the molecule with electromagnetic waves, bombarding it with a beam of electrons, measuring the equilibrium constant of the reaction $\ce{A-B <=>A. + B \cdot}$, or by measuring the rate by which a radical reaction proceeds$^{[1]}$.

$[1]$ Szwarc, M. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 1951, 207 (1088), 5–13.


C-H Bond Dissociation Energy varies with substitution What is the reason behind it?

"Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons."

Nature always try to acheive least potential energy for stability . Any organic reaction which makes the product less stable is not favourable .

If the C-H bonds are cleaved homolytically in all cases , the resulting stability of the free radicals will tell us about the feasibility of the reaction .

The primary free radical is the most unstable and to bring about its C-H bond breaking would mean increasing its energy comparitively . Hence its bond energy is the most and so its the stronger bond of the three .

Tertiary free radical is the most stable while secondary lies in between them . The same explanation can be used to draw out their comparitive strength of the C-H bonds .


What are the oxidation states in Borohydride (BH4-) The CRC Elecrochemical Series and the solution to our exercise suggest that the oxidation states of borohydride are -V for Boron and +I for Hydrogen.

Since H (2.2) is more electronegativ than B (2.0), I would have expected the oxidation states to be +III for B and -I for H.

What am I missing here?

The solution to our exercise lists this (WARNING: it turned out that the solution was wrong):

And the CRC Elecrochemical Series lists this (also suggesting -V for B):

I would have expected the oxidation states to be +III for B and -I for
  H.

You're right, boron is in the (+3) oxidation state in these equations.

In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral

In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)


Functional variation in derivation of the Hartree-Fock equations In Szabo and Ostlund's Modern Quantum Chemistry, the procedure of single determinant energy minimization is presented. Omitting the whole procedure, I have a question about functional variation during the derivation of the Hartree-Fock equations.

Given the single determinant $| \Psi_0 \rangle = | \chi_1 \chi_2 \ldots \chi_a \chi_b \ldots \chi_N \rangle$, the energy $E_0 = \langle \Psi_0 | \mathscr{H} | \Psi_0 \rangle$ is a functional of the spin orbitals $\{ \chi_a \}$. $E_0$ is the expectation value of the single determinant $\vert \Psi_0 \rangle$,

$$
E_0[\{ \chi_a \} ] = \sum\limits_{a=1}^N [a\vert h \vert a] + \frac{1}{2} \sum\limits_{a=1}^N \sum\limits_{b=1}^N [aa \vert bb] - [ab \vert ba].
$$

This equation can be varied:

$$
\begin{align*}
\delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] + [\chi_a | h | \delta \chi_a] \\
&+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \delta \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] + [\chi_a \chi_a | \chi_b \delta \chi_b] \\
&-\frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + [ \chi_a \chi_b | \delta \chi_b \chi_a] + [\chi_a \chi_b | \chi_b \delta \chi_a] \label{1}\tag{1}
\end{align*}
$$

Authors suggest to the reader as an exercise to manipulate this equation for $\delta E_0$ to show that

$$
\delta E_0 = \sum \limits_{a=1}^N [\delta \chi_a | h |\chi_a] + \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] - [\delta \chi_a \chi_b | \chi_b \chi_a] + \text{c.c.} \label{2}\tag{2}
$$

It is clear that the first sum in $\eqref{1}$ can be easily converted to that in $\eqref{2}$ because

$$
[\delta \chi_a | h | \chi_a]^* = [\chi_a | h | \delta \chi_a]
$$

Analogously, for the second sum in $\eqref{1}$ one can show that

$$
[\delta \chi_a \chi_a | \chi_b \chi_b]^* = [\chi_a \delta \chi_a | \chi_b \chi_b]
$$

and

$$
[\chi_a \chi_a | \delta \chi_b \chi_b]^* = [\chi_a \chi_a | \chi_b \delta\chi_b].
$$

I did the same manipulations with the third sum in $\eqref{1}$ and obtained

$$
\begin{align*}
\delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] \\
&+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] \\
&- \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + \text{c.c.} \label{3}\tag{3}
\end{align*}
$$

So, comparing $\eqref{3}$ and $\eqref{2}$ I suppose that one should show that both terms in each sum are equal, i.e.

$$
[\delta \chi_a \chi_a | \chi_b \chi_b] = [ \chi_a \chi_a | \delta \chi_b \chi_b]
$$

and

$$
[\delta \chi_a \chi_b | \chi_b \chi_a] = [ \chi_a \delta \chi_b | \chi_b \chi_a].
$$

So, how to do that?

Is it not just because of the fact that $[aa|bb]$ is in chemists notation, meaning that $[aa|bb] = [a(1)a(1)|b(2)b(2)]$ and so you can just switch them around?


The carbon-13 NMR for 1,1,1-trichloroethane I was working the following spectroscopy problem from Clayden 2e that asks us to identify the structure of a compound from information about its carbon-13 NMR peaks.

  Problem 3.5: The solvent formerly used in some correcting fluids is a single compound $\ce{C2H3Cl3}$, having 13C NMR peaks at 45.1 and 95.0 ppm. What is its structure? How would you confirm it spectroscopically?

I initially thought that the structure would be 1,1,1-trichloroethane (shown below) since there will be two different carbon chemical shifts: one bonded to three $\ce{H}$ atoms and one $\ce{C}$ atom in the 0-50ppm region (corresponding to the 45.1 ppm peak) and another peak for a carbon bonded to three $\ce{Cl}$ atoms in the 50-100ppm region (corresponding to the 95.0ppm peak).

Although it considers 1,1,1-trichloroethane, the solutions manual to the book concludes that the correct structure is 1,1,2-trichloroethane (shown below), the other possible structure with formula $\ce{C2H3Cl3}$.

However, this compound would have both peaks close to eachother in the 50-100ppm range.

Wouldn't the correct molecule be 1,1,1-trichloroethane, and wouldn't the solutions manual be incorrect? The NMR specs below that I found online corroborate my thinking.

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

https://www.chemicalbook.com/SpectrumEN_71-55-6_13CNMR.htm

https://spectrabase.com/spectrum/5WUteTf4M4u

I fully agree with your assessment and assume that this is a mistake in the book. I wonder what exactly the authors suggested as spectroscopic verification. I personally would suggest using 1H-NMR as a verification method. In 1,1,1-trichloroethane, only one hydrogen peak should be observed as the three protons of a methyl group are always homotopic. In 1,1,2-trichloroethane, two signals in a $1:2$ ratio should be observed as a triplet and dublet, respectively. Depending on what the book suggests as verification, it should be easy to confirm the nomenclature error.


Reaction rate unit conversion: how to convert from molarity to ppm How can I convert a reaction rate $k$ with units of $\mathrm{1\over M~~s}$ to $\mathrm{1\over ppm~~s}$?

For context: I want to make a numerical simulation of a system of reactions that happens in air (and in water droplets suspended in the air), I have already built a part of the model for which I used $\mathrm{ppm}$ for my concentrations and thus, e.g., $\mathrm{1\over ppm~~s}$ as the units for the $k$ values of second-order reactions in the system. (Reactions of different orders have different units, obviously, but those details are irrelevant to the question.) The article I have on hand lists a $k$ value for the reaction I need in $\mathrm{1\over M~~s}$ and the reaction takes place in the liquid phase.I assume that the PPM is weight-based although the article in question does not actually specify this.

I am asking the above question because I am not sure if this conversion is necessarily as straightforward as a simple factor of $10^{-6}$.

We can do it in two ways:

1) For dilute solutions, ppm and mg/L is equivalent. So, for dilute solutions, the straightforward conversion scheme should look like::
$$ppm=\frac{mol}{L}\times MW\times 1000$$

2) with no assumption::

first, remember that $ppm=y_i \times 10^{-6}$

$$ppm=\frac{mol}{L} \times \frac{x gm}{mol} \times \frac{1 L}{10^{-3} m^3} \times \frac{m^3}{y~gm} \times 10^{-6}$$

here, x is the MW of solute and y is the density of the solvent (for dilute solutions) or solution in $\frac{gm}{m^3}$


What is the precise definition of standard molar entropy? I understand what the standard molar entropy is, and how to use it in calculations, but I'm interested in understanding exactly how it's defined and measured.

My recollection from a long time ago is that it goes roughly like this:

for each element, assume that its entropy goes to zero at 0 kelvin when it's in some standard state
then, for the compound of interest, calculate and/or measure the entropy difference between the compound and its constituent elements at 0 K.

However, I'm not even sure that's exactly correct, and there are a lot of details that need to be filled in if it is right. Most of the following issues will be covered by the exact definition, but to be more elaborate, these immediately come to my mind: what exactly is the standard state for each element? Do all elements form a perfect crystal at 0 K, and how can we prove this? Should I think of the third law (zero entropy at zero kelvin) as being a definition that sets an otherwise arbitrary baseline, or can it be proven that the quantum mechanical von Neumann entropy really is zero for pure elements at 0 K?

An internet search turned up plenty of information about the values of standard molar entropy for various compounds and how to use them, but I wasn't able to find a resource with a detailed description of its definition.

What is the precise definition (preferably with reference) of standard molar entropy?

Not quite an answer, but too long for a comment:

Here's a link to the official CODATA site showing the values:
codata.info (via the Internet Archive).
It references the following source:
Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989.
Unfortunately, that source isn't readily viewable online (all I found was what appears to be a German version, at Berichte der Bunsengesellschaft für physikalische Chemie 1990, 94 (1), 93–93 (It's behind a paywall.)  But if you could get your hands on it (interlibrary loan, say), it should contain a detailed description of how the values were actually calculated.

FWIW, here's a link showing how standard molar entropies might be calculated (but I don't know if the procedure described here is what is used to determine the official CODATA values): Standard Molar Entropy of Solid Aluminum Oxide, Comments to Tandy Grubbs, Stetson University (via the Internet Archive).
Essentially, it mentions using the Debye extrapolation I mentioned in an earlier comment up to $\approx\pu{15 K}$, and then using a differential scanning calorimeter to determine the reversible heat flow between $\pu{15 K}$ and $\pu{298.15 K}$

And here's a purely theoretical approach (unfortunately, everything except the abstract is behind a paywall):
Pascal, T. A.; Lin, S.; Goddard Iii, W. A. Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics. Phys. Chem. Chem. Phys. 2011, 13 (1), 169–181. DOI: 10.1039/C0CP01549K.

Also, some general comments on the issue of the entropy of a substance at absolute zero:

It is possible the question of whether substances need to be perfect crystals to have zero entropy at $\pu{0 K}$ does not enter into the calculation of standard molar entropies.  Consider the following:

Any calculation of standard molar entropy requires determining the initial reversible heat flow starting at $\pu{0 K}$, at standard pressure:
$$
S^\circ(\pu{298.15 K}) =
\int_{0}^{\pu{298.15 K}}\frac{\mathrm{đ}q_\mathrm{rev}}{T}
$$
The initial part of this integral (starting at $\pu{0 K}$) can't be done experimentally, and thus has to be based upon a model.
We may not know how to model whether a substance being a perfect crystal at $\pu{0 K}$ will affect the amount of reversible heat flow needed to increase its temperature.  In addition, we may not know how to predict the actual state of substances at $\pu{0 K}$ (i.e., the extent to which they are not perfect crystals–and to the extent that they are not, what the distribution of their forms actually is).
Further, even if there were an effect on heat flow from not being a perfect crystal, and we did know how to model it, the effect may be small compared to the experimental errors that are present in determining standard molar entropies.
It's still an open question whether a substance needs to be a perfect crystal to have zero entropy at $\pu{0 K}$.  Some in the field argue it does, some argue it doesn't.  Those in the latter camp argue that, given that no other states are accessible at $\pu{0 K}$, the fact that the system is locked into only one microstate (even if it's not a perfect crystal) gives it zero entropy.


Does heat loosen "corrosive bonds" in iron compounds... permanently? One [obvious] way to remove a seized nut is to heat it [thus enlarge it]... more than the bolt. But that's not the only way to use heat in this application. Apparently you can also let the stuff cool down and the nut will still be significantly easier to remove than before applying theat. Why does this happen? The description of this latter video says something about heat "separating the corrosive bonds". Is there real science behind this claim?

If by corrosive bond they mean a physical bond not a chemical bond then yes, there is a very strong force of dry friction between the two parts.

Bolts and nuts are made so that the two fit together tightly so that they do not wiggle loose but not so tight that they are impossible to fasten. The problem is is that when iron corrodes the oxide has more volume than the iron. In fact when iron oxidizes, its size increases 70% (See: Pillings-Bedworth ration). This expansion puts a lot of force between the nut and the bolt. Since dry friction increases linearly with force, friction between the two increases also making the bolt harder to remove.

When you expand the nut and/or shrink the bolt some of this force is relieved and friction decreases making the nut easier to turn.


Does salt hydrolysis takes place in buffer solution? I am a bit confused whether hydrolysis of salt takes place in buffer solutions because the books which I read doesnt consider them.But I think since weak base or weak acid salt is present hydrolysis should take place.

Water autohydrolyzes, or autodissociates, or autoionizes, to produce a very small amount of H+ and OH- ions. The product of their concentrations is 10e-14.
Most salts that dissolve in water dissociate into ions. Often, one of these ions will have a strong attraction for H+ or OH- and tends to reduce the concentration of that ion in the solution. But since the autodissociation of water is an equilibrium process, if an ion (like acetate ion) combines with the H+, more water will dissociate to produce H+ and also OH-. When sodium acetate is added to water, it dissociates into Na+ and CH3CO2- ions; the Na+ ion attracts several water molecules around it without a preference for OH-. The CH3CO2- ion exhibits a strong preference for H+ over plain H2O, and so would reduce the H+ concentration in the solution. Water compensates by producing more H+ and OH- to keep their product equal to 10e-14.

Acetate ions continue to absorb H+ (it also has an equilibrium to maintain), and finally the acetate ion may be largely present as acetic acid (HCH3CO2). The extra OH- ions that came from the water hang around increasing the pH.

So, the answer to your question is yes, the salt hydrolyzes; in fact, the hydrolysis of the salt is what allows a buffer to be made. You establish a ratio of concentrations of the salt and its acid to give a desired pH. Instead of adding tiny amounts of acid to affect a tiny equilibrium product like 10e-14, you use the ratio of larger amounts of buffer ingredients, governed by the acid dissociation constant.

If the salt doesn't hydrolyze, like NaCl, which ionizes without changing the pH of water, it isn't useful as a buffer material - you can't use it to affect the concentration of H+ or OH- in water.


Predicting the product in a series of reactions I am stuck in finding the final product $\ce{C}$ formed in the following series of reactions:

  $$\ce{Bu-C#CH ->[NaNH2] A ->[Ph-CHO][H2O] B ->[MnO2] C}$$
  where $\ce{Bu}$ denotes the butyl substituent and $\ce{Ph}$ the phenyl substituent.

I could easily predict product $\ce{A}$ and $\ce{B}$. $\ce{A}$ is simply formed by replacement of hydrogen with sodium and $\ce{B}$ by the meta substitution in benzene ring.

However, I am having trouble predicting the final product. I know that $\ce{MnO2}$ is a mild oxidising agent.

You seem very sure about your intermediate B; we’ll come back to that in a second.

For now, let’s concentrate on the reagents for your final step.

$$\ce{B ->[MnO2] C}$$

Manganese dioxide is the product you get from heating any manganese salts in the bunsen burner until red-hot; thus, one might call it the most stable or thermodynamic oxide. It seems very unreactive and in fact I only know of one type of reaction that it is able to perform on organic molecules: allylic and benzylic alcohols are oxidised to the corresponding carbonyl compounds while fully saturated (i.e. not allylic) alcohols are untouched.

Since you have that reagent, it might be helpful to go backwards; assuming that the final step is an oxidation of a benzylic or allylic alcohol to a corresponding carbonyl. Thus, $\ce{C}$ would either be a ketone or an aldehyde and $\ce{B}$ would have to be an alcohol. But your proposed structure $\ce{B}$ does not include an alcohol! Therefore, your proposed reaction of the second step is very likely wrong.

$$\ce{A ->[Ph-CHO][H2O] B}$$

Indeed, here you assumed that the anion $\ce{A}$ you proposed reacts with the phenyl ring of benzaldehyde in an electrophilic aromatic substitution. Take a good look at the two emboldened words and consider whether an anion can really be electrophilic. Then ask yourself what anions are in philicity terms. Once you have done that, attempt to locate the position of benzaldehyde with the corresponding electronic properties.

The solution is given in the spoiler below; please try to find it yourself, though; it is a much better exercise than just looking things up.

   Anions are negatively charged as are electrons. Thus, anions are practically never drawn towards negative charges — remember that like and like repulse each other. Instead, anions are drawn to positive charges: they are nucleophilic.

 Nucleophiles are drawn to the most positively charged bit of a molecule they are attacking. In the case of benzaldehyde, the carbon side of the $\ce{C=O}$ bond is most positively charged due to the resonance depiction we can draw below:

 $$\ce{C=O <-> \overset{+}{C}-\overset{-}{O}}$$

 Therefore, the acetylide anion will attack the carbonyl carbon of benzaldehyde nucleophilicly, creating a benzylic alcohol. It is now no secret that this is cleanly oxidised to the corresponding alkynophenone.

   $$\ce{Bu-C#C^- (A) ->[Ph-CHO][H2O] Ph-CHOH-C#C-Bu (B) ->[MnO2] Ph-CO-C#C-Bu (C)}$$


Why do Electrons leave the Zinc in a Galvanic Cell What induces  electrons to leave the atoms around the part of the zinc metal in a solution to travel up along that zinc metal and above the solution, through the wire and to the copper metal and reduce the copper ions in that solution.

Is it a result of the positive metal ions in a 'sea of electrons' being attracted to the negative ions of the solution making the electrons surrounding them leave, are the electrons repelled by the electrons in the solution or do these electrons 'sense' the copper ions on the other side waiting to be reduced and flows towards them?

I'm just not sure what prompts them to leave in the first place if the positive copper ions are not directly in contact with the zinc.

First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}<=>Zn^2+_{(aq)} +2e-}$$

At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode. This excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.

But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- <=> H_{2(g)} +2H2O}$$
The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.

To go back to the core of your question,

I'm just not sure what prompts them to leave in the first place if the positive copper
ions are not directly in contact with the zinc.

If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a salt bridge.

The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.

This paper is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).


Why is there no neutron in protium? My chemistry teacher once told me, that every atom should have at least as many neutrons as protons, to keep the electrons from crashing into the nucleus.

Protium, however, doesn't even have neutrons to begin with. How does this work? Or was my chemistry teacher wrong about this?

Your chemistry teacher is making a few simplifications there that make the statement false on a black-and-white true-and-false scale. Protons would repel each other electrostaticly due to their same charges. Neutrons interact with protons by the so-termed strong interaction (because it is stronger than the weak interaction; props to physicists for inventing names) which lead to stability of the nucleus, preventing it from falling apart due to the repulsion.

Hydrogen, specifically the $\ce{^1H}$ isotope, does not need neutrons to stabilise the nucleus because one proton cannot repel itself.

Note that ‘electrons crashing into the nucleus’ is a very strong simplification that you should eliminate from your language. Electrons are waves for all that matters so they are all around the nucleus at the same time. In fact, the highest probability of detecting them is right there, in the nucleus. (On a simpler scale, Bohr once invented ‘orbits’ that you can rationalise much like the orbit of a satellite, which also does not crash onto Earth. But don’t tell a quantum chemist or nuclear physicist that I said that!)

Finally, it is wrong to say ‘at least as many neutrons as protons’. $\ce{^3He}$ is a stable isotope with less neutrons than protons in the nucleus.


Can ionic compounds can ever "dissolve" as molecular compounds do? Here's an example. Say you take a 25˚C solution of NaCl, and lower the temperature suddenly. The equilibrium constant (solubility product) is immediately decreased, so ions start to join together into ionic compounds that float freely in the solvent before settling to the bottom. (In some cases they may be visible as precipitate.)

Can we refer to these free-floating ionic compounds as "dissolved" solute?
Does "XY(aq)" (the aqueous solution of an ionic compound) seem like a good description of this state?

Can we refer to these free-floating ionic compounds as "dissolved" solute?

No. Dissolved implies that the ions are solvated. If the ions are in a solid ionic lattice then they are not solvated by the solvent and so would not be described as part of the dissolved solute.

  Does "XY(aq)" (the aqueous solution of an ionic compound) seem like a good description of this state?

Again no. $\ce{XY(aq)}$, in the context of ionic compounds, refers to $\ce{X^{n+}}$ and $\ce{Y^{n-}}$ ions solvated by water molecules. The dissociation is generally regarded as implicit. For example, $\ce{NaCl(aq)}$ implies an aqueous solution containing $\ce{Na+}$ and $\ce{Cl-}$ ions. A better description of the state you are thinking of would be $\ce{XY(s)}$, using state symbols, or even better you could describe it as a suspension.

From the IUPAC Gold Book:

  Suspension:
  A liquid in which solid particles are dispersed.

This accurately describes your system.


What happens if the electronegativity difference is exactly 2.1? I'm learning about electronegativity in atoms and how their difference affects the type of bond they make. I know that no difference makes a non-polar covalent bond, a difference less than 2.1 is a polar covalent bond, and a difference above 2.1 is an ionic bond. But what kind of bond forms if the difference is exactly 2.1?

As mentioned above there is no hard line between an ionic and a covalent bond, they are just two ends of a spectrum where a bond can have a certain degree of ionic or covalent character. At the lower end of this spectrum, bonds with small electronegativity differences between the atoms have very little ionic character because the charge distribution in the bond is almost symmetrical between the two atoms. As the EN difference increases the charge distribution becomes asymmetrical and this results in an increasingly polarised bond, which therefore has an increasing degree of ionic character. Eventually at the other end of the scale you have a bond where almost all of the charge is concentrated on one atom and this is generally termed to be an ionic bond.

So in answer to your question; what happens if the EN difference is exacly 2.1; nothing happens, you have a relatively strongly polarised bond which under some definitions might be classed as ionic and under others as strongly polar covalent but in reality it has a large degree of both ionic and covalent character.


How is the charge of covalently bonded atoms determined? My question is about what it is - at the subatomic level - that determines the charge on atoms which participate in covalent bonding in reality. I understand that formal charge is just that - it's a formality, which gives human beings a very useful tool for modelling reality and making predictions. I, though, would like to know what is responsible for the charge we find in covalently bonded molecules.

Let me explain. If we picture an ionic bond as a highly polar covalent bond, we can see that in an atom like NaCl we have the state of affairs where the highly electronegative chlorine atom has "pulled away" an electron from the sodium atom.

(Orbitals omitted for ease of drawing)

Ultimately this means that Na, with its 11 protons has just 10 electrons, which gives it a positive charge, while Cl, with its 17 protons, now has 18 electrons, giving it a negative charge. Electrostatic forces between the two nuclei result in the bond.

Of course, it might be more accurate to say that the Na nucleus simply ceases to be significantly affected by its former 3s1 electron. As a result the 11 protons in the Na nucleus, which result in a +11 charge, are countered by only 10 electrons, giving a +1 charge overall. Meanwhile, the 17 protons of the Cl nucleus (+17 charge) experience the negative charge of 18 electrons (-18) giving a -1 charge overall.

Let's compare this with the situation we find in the carbonate ion. (Link) For the purposes of my question I won't worry too much about resonance, I don't mind too much whether we have the accurate -2/3 charge, but let's keep it simple and just look at one of the contributing structures.

(Orbitals again omitted for ease of drawing. Orange indicates an additional electron from some other source. Oxygen has only eight electrons ordinarily, but one will be gained when forming this ion)

In this case we have covalent bonding. In the case of the singly bonded oxygens, we have 8 protons and 10 electrons. This is two more negatively charged electrons than positively charged protons. This, though, is not ionic bonding, it is covalent bonding and the polarisation is not such that the electron contributed by carbon dissociates completely from the Carbon. Instead it is "shared" between the two. Nevertheless, the "sharing" is not equal and oxygen attracts the electrons more strongly than the carbon, so we have a -1 charge.

My question, though, is - why is it that the charge works out like this? Why is it that when you have oxygen (8 protons) (charge of +8) and a filled outer shell (8 electrons plus buried inner core gives -10 (charge of -1 on each electron)) you have a charge of -1 not -2, is it simply the covalent bond, is it simply the fact that the two electrons in the sigma bond are sufficiently far enough away in reality?

But then why should it be that the doubly bonded oxygen, which also has ten electrons surrounding it, has no charge at all? Meanwhile, the carbon atom in the  middle is surrounded by 10 electrons also, whereas it has only 6 protons and yet carries no charge at all.

Naively the answer seems obvious. The bonds are covalent and the electrons are "shared." So, in the double bond (which, of course, has a different nature to the single bond) one must assume that the electrons are positioned in such away that they are balanced by the protons of the oxygen nuclei. There is therefore no charge on this atom.

In the case of the Carbon atom we surely again understand that the right balance of charges and distance from the nucleus is arrived at.

Overall we have 32 electrons and 30 protons so the overall -2 charge on the whole atom makes sense.

The ultimate question I've been trying to lead up to is what is it that dictates the distribution of charge. What is the difference - at the subatomic level - between two ionically bonded atoms, a covalently bonded molecule like the CO32- ion which carries a charge, and a covalently bonded molecule like H2O - which is also polar (hence we can have hydrogen bonding) but isn't usually assigned a charge. On what basis do we assign charge?

As you point out, formal charge is a "bookkeeping" method of assigning electric charges to atoms in a way that lets us easily approximate where the electron density is.

Chemistry is made up of layers of successively less detailed approximations. The concept of "formal charge" is near the top of the stack - very little detail is involved. It is only slightly more detailed than Lewis structures, in that we are at least trying to acknowledge that covalent bonds don't really share electrons equally.

With that in mind, we will look at the next higher and next lower level of detail, to see why formal charges are useful.

Whenever we draw a Lewis structure, we imply that each line is made up of two electrons that share a molecular orbital that exists somewhere in the space between the two atoms that it connects. Without knowing any better, we would assume that each electron can be wherever it "needs" to be to fill the valence shell of an atom. So, in the case of $\ce{CO3^2-}$, the carbon has four bonds and no lone pairs, giving it a full octet. The oxygen atoms have either three lone pairs and a single bond, or two lone pairs and a double bond, again giving each a full octet.

Note that we counted each electron twice for the purpose of checking the valence configuration. It's as if we are saying "each atom can borrow electrons from the other whenever it needs them." We are effectively ignoring any tendency of atoms to pull electrons towards each other - they float between atoms as needed, as if maintaining a full octet were the most important job in the world.

Next we look at the formal charges. The rules for determining formal charge say that we count the number of valence electrons an atom normally has (without bonds), then subtract the electrons we assign to it. We assign to an atom all of the electrons that are in lone pairs, plus one half of the bonding electrons. In other words, instead of saying that each atom gets the electrons whenever it wants them, we are saying "each atom in the bond divides the electrons equally between themselves." When you do this, you find that some atoms have "extra" electrons (more than they would normally have in the valence shell) and some have too few - thus, a formal charge.

This starts to give us a slightly more correct picture of how electrons are distributed inside of a molecule, but it doesn't really tell the whole story. Notice that we didn't care about what type of atom it was when we split up the electrons. We only looked at that to find the "natural" number of electrons it would have.

The next layer of detail involves electronegativity. In any covalent bond between two different elements, one element "likes" electrons more. The reasons for this are complex, and you have to go down another couple of layers to get to the reason. For now, just know that it is related to quantum mechanics.

If we know the relative electronegativities of elements, we can start to account for this tendency as well. We do this when assigning oxidation numbers. With oxidation numbers, you assign to each atom all of its loan pair electrons. Then, for each bond, you give all electrons in the bond to the more electronegative atom. Then we subtract the total from the number of valence electrons the atom normally has.

Notice what we are doing. First we said "all electrons are shared equally all the time." Then we said "Divide the electrons equally." Finally we said "the most electronegative atom gets all of the electrons."

None of these is actually true, but each gives a quick approximation that is useful for a particular type of problem. To get a real picture of what is happening to the electron density, you have to do detailed quantum mechanical calculations. They are really hard to do! We typically use computers to solve them. But, if we know what the limitations are, we can use quick approximations to get an idea of where the electrons are at that is good enough to, for example, make reasonable predictions in organic chemistry.

To summarize:

Valence count - assume all electrons are shared equally
Formal charge - assign one half of bonding electrons to each atom
Oxidation number - assign all bonding electrons to the most electronegative atom