What term should I use when I am quantifying the amount of matter, mass or weight? I am writing an academic article but I got confused as I am writing about the weight a specific metal got by using balance scale. Should I use the term mass, which has a unit of kg, or weight, which is a force created by gravity and expressed in N?

Because of the reasoning provided by you, i.e. mass is universal, and weight is the product of mass and magnitude of the local gravitational acceleration, I suggest to stay at mass.  (A purist's view, taking @Ivan's comment into account.)

In addition, referring to the International System of Units, their brochure not only mentions the basic units, but the relevant section 2.1.1.2 starts with:

Unit of mass (kilogram)


Why does rain have that smell? What chemical reactions in the soil cause that smell when it rains? In particular, what reactions in the soil causes the petrichor smell?

Interesting question. I actually had to look up the definition of petrichor.  After getting a little nostalgic about rainfall, I decided to go ahead and forage around for an answer for you.  I found a nice summary article here for those that are interested, and then a really nice bit of history here with a link to a great video about how rainfall helps to produce aerosols, which can contain the petrichor compounds.

Your question regards the reactions in the soil that give the petrichor smell, so let me get to that:

The main constituents of petrichor are volatile plant oils and geosmin. Oxidation of the fatty acids produced by plants can lead to some of the volatile compounds that constitute petrichor. Geosmin is a metabolite of actinobacteria and is produced through a biological pathway.

https://en.wikipedia.org/wiki/Geosmin - a brief intro to Geosmin
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3013058/ - A free paper highlighting the biosynthesis of geosmin


Is lithium is less reactive than sodium? Lithium has most negative electrode potential still its reaction with water is less vigorous than that of sodium

You can't just take the electrode potential as if that's going to correspond to the rate in which something is going to react.  For example, why do you think that aluminum is so stable in atmospheric conditions versus iron?  Aluminum forms a stable oxide layer that serves to protect it against further reaction, while iron oxide is weak, adheres poorly and is and porous.  There are numerous factors which affect a rate of reaction, and they vary in different situations.

Calculating rates of reaction is a very complicated thing, and it's usually easier / more accurate to just determine the Arrhenius parameters experimentally and use those.


What makes carbon special and versatile? My teacher told me that carbon's tetravalency and high catenation ability makes it special and is the reason why there are millions of compounds of carbon.

1) Silicon is tetravalent too but doesn't create as many compounds as carbon.Why?

2) The textbook says that no other element exhibits the property of catenation to the extent of carbon. Why is this so? Does it just happen magically?

There are two key factors that account for the ubiquity of carbon compounds.

Bond strengths: Look at the following table of bond strengths and notice how the strength of both carbon-carbon single and double bonds is much greater than the bond strengths found in other molecules.

\begin{array}\hline
Typical~Single ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C-C & 83    \\ \hline
N-N & 38    \\ \hline
O-O & 35    \\ \hline
Si-Si & 52    \\ \hline
P-P & 50   \\ \hline
S-S & 54    \\ \hline
\end{array}

\begin{array}\hline
Typical~Double ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C=C & 146    \\ \hline
N=N & 109    \\ \hline
O=O & 119    \\ \hline
P=P & 84   \\ \hline
\end{array}
link to reference

These numbers tell us that carbon-carbon bonds are significantly more stable than other bonds and that the driving force for their formation will be more exothermic.  Therefore, large collections of carbon atoms can be bonded together to produce very stable systems.

Hybridization: Saying that carbon is tetravalent doesn't fully convey the versatility carbon can display in forming bonds. Hybridization involves a re-mixing of s- and p-orbitals in response to bond formation and this leads to the formation of more stable bonds. Carbon is capable of extensive variation in the hybridization of the bonds it forms, much more so than most other atoms.  This wide range in hybridization that is possible for carbon allows it to form much more stable bonds in a wide variety of different structural situations.
\begin{array}\hline
Typical~Carbon-Carbon ~Bond~ Strengths~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
alkane & sp^3 & 84    \\ \hline
alkene & sp^2 & 150    \\ \hline
alkyne & sp & 200    \\ \hline
\end{array}


Spectral Peak-Picking in MS Excel I'm working often with spectral data of various gases, and the bulk of the data analysis is assigning the peaks with maximum signal (y-axis) to frequency (x-axis). Often we have several thousand data points, and it can be quite difficult to manually find peaks. I've tried getting around this in Excel with clever IF/AND statements, but the major obstacle is signal noise: I simply cannot devise a clever Excel statement to help me pick peaks.

For example, using an =IF(AND(greater signal than adjacent points),"max","") statement gives tons of extraneous hits where the noise causes a slight fluctuation in the data. Similarly, creating a column of slopes and doing an =IF(slope goes from + to -),"max","") statement also gives bad data.

The most elegant solution I have is to pick an arbitrary value # that most of my peaks are above and use an =IF(AND(slope goes from + to -,B>#),"max","") statement. That seems to work for stronger peaks (maybe 70% of the peaks on a good day), but I still have to go in and manually look at the weaker peaks (and often when doing these spectra many energetic transitions that need labeling are statistically unlikely-->low signal). I'm wondering if anyone here has a better solution able to detect local maxima not resulting from noise with a better success rate.

One thing you can do before searching for peaks is to smooth the data with a 9 point (or 15 point) moving average. I'm not sure how the ends are handled, but the average of 9 points replaces the data point so noise fluctuations are reduced.


Break down hemoglobin into heme Is there chemical that can break down the red blood cells/hemoglobin into smaller protein sub units/globins?

I would like to extract the heme molecules from the blood but not necessarily purify it,or separate it.I would like to get solution containing liberated heme units,its not problem if other stuff,like proteins and globins are floating in that solution,I just want to detach the heme part from the protein.Heme is iron containing coordination complex that binds to oxygen and give blood its red color.

I believe the chemical you are talking about that can break down the red blood cells/hemoglobin into heme and globins is carboxymethyl cellulose (CMC; see Ref). There is a US patent claiming a method for separating blood hemoglobin into heme and globin from blood cells. The abstract states that:

  The invention relates to a method of dividing blood hemoglobin into heme and globin by suspending blood cells, separated from blood, in an acid aqueous solution, by precipitating the heme by means of carboxymethyl cellulose, and by separating the precipitated heme-CMC complex out from the aqueous solution of globin.

Title of the patent: Method for dividing blood hemoglobin into heme and globin

Patent: Karin Autio, Martti Kiesvaara, and Yrjö Malkki, US Patent 4,518,525, 1985 (11 pages)


Do you include the "10 H2O" in the formula for borax? When you calculate the mass of borax needed to make up a standard solution, do you include $\ce{\cdot 10 H2O}$ in the formula for borax? Why or why not?

You are talking about the waters of hydration in crystal structure $\ce{Na2B4O7 \cdot 10H2O}$, and the answer is invariably—yes.

These water molecules account for 47.24% of the crystals weight! If you wanted to make $1\: \mathrm{L}$ of $1\: \mathrm{M}$ of $\ce{Na2B4O7}$, and you did not account for these waters of hydration, you would end up making a $0.4818\: \mathrm{M}$ solution. Less than half the desired concentration!

This is the result of a double effect. Not only are you not adding as much salt as you thought you were, but because the salt contains water it is diluting the solution even further. In the case above $95.05\: \mathrm{mL}$ of water are actually added to the solution just in the salt!


What kind of mechanism is adopted here? The relative reactivities of acyl compounds towards nucleophilic substitution is acyl chloride> anhydride>ester>amide.
I would like to know how  this kinetic property is decided. Is it on the basis of better leaving grp or is there another explanation for it?,

It is likely to be due to the different amounts of partial positive charge which exist on the carbonyl carbon.

We can consider for example the rate of reaction of a given amine (such as aniline) with the acid chloride, acid anhydride and an ester such as the methyl ester. The partial positive charge argument would explain this reactivity series.


Does the change in Gibbs free energy depend on initial and final concentrations? Suppose there is a decomposition reaction, and by varying different factors, the equilibrium can be disturbed to alter the final concentration of compounds. So, does the value of ∆G here depend upon the initial and final concentration? The basic rule of ∆G = G(products) - G(reactants) seems unsatisfactory, because, it seems pretty absurd to me, that the value of ∆G is the same when 0.01% of a gas decomposes, and when 85% of it decomposes.

$$\mathrm dG = -S\mathrm dT+ V\mathrm dP + \sum \mu_i \mathrm dN_i$$
If (products) and (reactants) are at the same temperature and pressure then $\mathrm dT$ and $\mathrm dP$ are both $0$, and $\mathrm dG = \sum \mu_i\mathrm dN_i$. Integration gives $\Delta G = \sum \mu_i\Delta N_i$. $\mu$ is the chemical potential which is just the specific molar Gibbs free energy. For the decomposition of $\ce{AB -> B + A}$ this is $$(\mu_\ce{A}\Delta N_\ce{A} + \mu_\ce{B}\Delta N_\ce{B}) - \mu_\ce{AB}\Delta N_\ce{AB}$$


Importance of halogens and nitrogen in degree of unsaturation Why is oxygen (or other elements) isn't present in the formulae for calculation of the value of degree of unsaturation $(\mathrm{DU})$ as it usually more common than nitrogen and halogens.
And whats so special about nitrogen and halogens to be in the formulae
$$\mathrm{DU} = C + 1 - \frac{H + X - N}{2},$$
where $C,$ $H,$ $N,$ $X$ are the numbers of carbon, hydrogen, nitrogen and halogen atoms present in the compound, respectively?

So here is the method I learned years ago without memorizing an awkward formula. The concept is to reduce the molecular formula to a hydrocarbon and compare it to the most saturated hydrocarbon bearing the same number of carbon atoms.  Consider the infamous free-base, hydroxychloroquine, $\ce{C18H26ClN3O}$.

The following atom replacements may be done in any order. Ignore divalent atoms ($\ce{O,S}$). The formula is now $\ce{C18H26ClN3}$. Replace all halogens with an equal number of hydrogens $\Rightarrow \ce{C18H27N3}$. Substitute $\ce{CH}$ for each nitrogen or trivalent atom $\Rightarrow \ce{C21H30}$. The most saturated $\ce{C21}$ hydrocarbon is $\ce{C21H44}$. Now do the math: $\ce{C21H44}$ - $\ce{C21H30}$ = $\ce{H14}$. Now divide 14 by 2. The Degree of Unsaturation is 7.
As to the question as to why oxygen is dropped from the equation, if the oxygen in hydroxychloroquine weren't there, the degree of unsaturation would still be the same. For example, the Degrees of Unsaturation of tetrahydrofuran and cyclobutane are the same, D.U. = 1.


Does broader peak mean stronger absorber in UV/VIS? I want to ask a question about peaks in a UV/VIS spectrum.

My lecturer in our introductory course mentioned the following:

  Two molecules may have the same molar extinction coefficient, but if I have two molecules, A and B, and B has a peak that is broader than molecule A, it will be a stronger absorber.

He failed to qualify this further or provide any reasoning.

My thought was that if it is a broader peak, then molecule B will absorb a wider range of wavelengths in a UV/Vis spectrum, but I couldn't see what he meant by a "stronger" absorber.

Does he mean that molecule B absorbs more wavelengths, so it is "stronger" or is there another qualification for using the word "stronger" to describe B?

The classical concept of  oscillator strength $f$ is useful here, but it should be used only in a qualitative way. This is defined as $f=a\int\epsilon_\nu d\nu$ where $a$ is set of units with value $4.3 10^{-9}$ if the extinction coefficient is in units  dm$^3$/mol/cm and $\nu$ is in wavenumbers. The maximum value of $f$ is unity and is close to this for an allowed transition in a dye molecule, such as rhodamine, but can be as small as $10^{-9}$ for a forbidden transition.

In your case if two molecules have similar $\epsilon$ at some wavelength (or frequency) but one has a narrow transition compared to the other, then the one with the narrow transition will have a smaller integral and hence the smaller oscillator strength. In this sense only is the transition 'weaker', i.e. it is not the probability of  absorbing at a given wavelength that is being considered but that over all wavelengths, so it really depends on how you want to define a 'strong' vs a 'weak' transition.  In QM the transition moment integral is used instead of oscillator strength and gives a proper description of absorption.


Identify the structure of the compound (C3H5NO) given its NMR chemical shifts I am trying to figure out the structure of the compound $\ce{C3H5NO}$ given just the following data from the 1H NMR spectrum:

$\delta 5.8 ~ \text{(singlet, 1H)}; \delta 6.2 ~ \text{(triplet, 2H)}; \delta 7.4~ \text{(triplet, 2H)}$

From 2-2-1 combination I have the following candidates:

Only the first and second one satisfy the multiplet data but I can't understand how their chemical shifts are so high. For a $\ce{-CH2}$ adjacent to $\ce{-C#N}$, the chemical shift reaches the value around $1.3 + 1 = 2.3$ but the given values are almost thrice of it. The $\ce{-OH}$'s proton's value seems fine nevertheless. Can anyone explain the reasoning behind the two other values? Or is the compound I have predicted wrong?

Dhruba Charan Dash published the problem in his analytical chemistry textbook (ISBN 9788120353008). This is the picture that goes with the question:

Notice that the scale of the chemical shift increases with the strength of the magnetic field $B$. This means they are using the $\tau$ scale instead of the $\delta$ scale. The modern description of the signals would be $\delta = 10 - \tau$, where $\tau$ is the value given in the figure.
So we have a singlet at 4.30 ppm (B = broad?) and triplets at 3.83 ppm (typo in the figure, should be 6.17 instead of 5.17) and 2.63 ppm.
$$\delta\ 2.63 \text{ (triplet, 2 H)}$$
$$\delta\ 3.83 \text{ (triplet, 2 H)}$$
$$\delta\ 4.30 \text{ (broad, 1 H)}$$
For completeness, I will post the answer given by the author Dhruba Charan Dash:

PROBLEM 13.17 PMR spectrum of the compound $\ce{C3H5ON}$ in an acidic medium is shown below. Suggest its structure. Given that the peak 1 disappears in the presence of $\ce{D2O}$.
Solution Analysis of the spectrum:
(i) Since the peak 1 disappears in the presence of $\ce{D2O}$, it may represent a group containing active hydrogen. This group may be a hydroxyl group.
(ii) The peaks 2 and 3 must represent the hydrogen atoms attached to the methylene group.
(iii) The analysis of integral trace indicates the presence of 1, 2, and 2 protons within the peaks 1, 2 and 3 respectively.
The point (iii) suggests the presence of two methylene groups. These must be attached each other (point (ii)). The only structure consistent with the above facts is $$\ce{HOCH2CH2CN}$$

It is clear why the question posed by the OP (and by the teacher of the OP, as I learned in chat) was hard to solve. A single problem transcribing the question (replacing $\tau$ with $\delta$) obfuscated the intended chemical shift values.
For reference, here is the problem as presented to the OP:


How can multiplets in para-disubstituted benzene rings be described? How should I label the multiplicity of the aromatic signals in an 1H-NMR spectrum with the following substitution pattern?

I have seen lots of people labeling both signals in this kind of system as doublets, probably because the resolution of the spectrum was not high enough, or because we are often told, these look like doublets. However, what looks like a doublet doesn't necessarily have to be one and thus, should not be called doublet, because it doesn't reflect the actual couplings in this system:

The ortho-coupling (3J) for both signals should be about 6.0 and 9.0 Hz.
The meta-coupling (4J) for both signals should be between 0.9 and 3.0 Hz.
The para-coupling (5J) for both signals should be between 0.0 and 1.0 Hz.

(These numbers are from the Hesse-Meier-Zeeh, a German textbook about spectroscopy in organic chemistry).

I am inclined to call these signals multiplets rather than doublets, because I can only guess the coupling constants, especially for the right signal at about 7.36 ppm.

Or is it good practice to call them pseudo-doublets? And what would be the symbol (like s for singlet, d for doublet, etc.)?

Many people report these as doublets. However, you are correct that these are not doublets - there are second order effects in AA’XX’ spin systems (note that a first-order analysis will fail, the pattern is not a ddd - see https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-aabb.htm for more info).

Using the phrase “apparent doublet” (abbreviate as app d) is probably a good compromise between being technically correct and retaining useful information. You can still report the “coupling constant” between the two largest peaks, but with the caveat that it is not a first-order doublet.

If you want to be very precise, then multiplet is the way to go. However a typical organic chemist would also be interested in the splitting between the two largest peaks in your multiplet (which if I am not mistaken, corresponds to the sum of $^3J_\ce{AX}$ and $^5J_\ce{AX’}$). Since the para coupling is tiny, the splitting is essentially representative of the ortho coupling. In my opinion it is helpful to leave that information in.


Direction of gas molecules in kinetic model of gases? So when learning about the kinetic model of gases, my teacher said that in a volume $A*v_{x}*\Delta t$, exactly half the molecules will have a velocity component of $v_x$ and the other half will have a velocity component of $-v_x$. But I think they are wrong!

Because there should be some small portion of gas molecules that will have a velocity component of: $v_x = 0$

Am I correct? So is the term for the average number of collisions with the wall during the time interval $\Delta t$, just an approximation?:

$(0.5)*A*v_{x}*\Delta t*n*N_a/V$

This is clear with an example. If there are three molecules in a container, one moving left on the x axis, one moving right on the x axis, and one moving directly up the y axis,  then on average, only 1/3 of the molecules is moving towards the right wall.

You have pointed out the problem of how to handle the infinitesimal in probability theory. So yes a molecule could traveling in the $yz$ plane and have 0 $x$ velocity. However the probability of the motion being in exactly the $yz$ plane can be made as small as you like.

Let's make the problem one dimensional and choose a random angle between 0 and 359.999... degrees. So the question is what is the probability that the chosen angle will be for a vector on the x-axis?

To a degree if I chose 0 or 180 then the vector will be on the x-axis. Odds 2/360
To a tenth of a degree if I chose 0.0 or 180.0 then the vector will be on the x-axis. Odds 2/3600
To a hundredth of a degree if I chose 0.00 or 180.00 then the vector will be on the x-axis. Odds 2/36000

and so on...

So what probability should be assigned to a randomly chosen angle being exactly along the x-axis?


Finding percent weight I have a mixture of $\ce{NO2(g)}$ and $\ce{N2O4(g)}$ at $63\ \mathrm{^\circ C}$ and $750\ \mathrm{mmHg}$. I need to find the weight percent of the first compound. Density $d=1.98\ \mathrm{g/L}$.

I assumed that we have $M=aM_1+bM_2$ where $M_1$ and $M_2$ are the molar masses of the first and second compound. I know that $m=nM$ so the weight percent would be $$\frac{m_1}{m}\cdot100\ \%=\frac{aM_1}{aM_1+bM_2}\cdot100\ \%$$
My problem lies in finding the coefficients. Can someone give a hint please?

We can find the average molar mass with the density formula involving pressure and temperature $\overline M=\frac{\rho TR}{p}$.

We also know that, in this case, $\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$. Since we only "care" about $M_1$ we can write the average molar mass as $\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$ and then find $\frac{n_1}{n_1+n_2}$.
$$\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2}^{(*)}=\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+\frac{n_2}{n_1+n_2}M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}=\frac{m_1}{m_{total}}$$
Compute $(*)\cdot100\%$. The development of the fraction was to show why it works.

Another possible solution is this :
$$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%$$
$$\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%$$
$$n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$


Can imines be catalytically hydrogenated? Can imines be catalytically hydrogenated? According to the DEA this appears to be possible. However I cannot find any other references to catalytic hydrogenation of imines anywhere else.

http://thedea.org/synthesis.html

Several methods on the reduction of imines are outlined on Wikipedia. Another method using catalytic hydrogenation is given here.


Types of argon gas regulator? I have a high pressure bottle of argon gas (99.98 %) that I need to use in an experiment. The problem is that the regulator I have is a generic type that is seemingly optimized for high flow rates (welding). I need a fairly low flow rate of under 200 cm³/min. However, I do not know what kind of regulator would be suitable. Any suggestions for search description?

No cylinder regulator that I know of will provide fine control of flow rates on the order you require.  The purpose of a regulator is to control the line pressure, not the line flow rate.  In theory you could stick a flowmeter in-line and try to finesse the regulator pressure and shutoff valve (if there even is one) to reach the flow you want, but it would probably be a maddening exercise.

What you need is a mass flow controller (MFC) of some kind.  Depending on your budget and your needs for adjustability and precision of the flow rate, there is a variety of options:

If you know you will only require a specific, set flow rate (or, a small number of set flow rates) you might do okay with a flow orifice: set the upstream pressure appropriately and the flow should match the spec.
If you need to be able to vary the flow rate and an approximate value is acceptable, something like these dial-knob flow control valves would probably work okay.
If you need precise control of the flow rate, you're probably up against something like these Porter mass flow controllers. They're quite pricey (a few thousand USD) but they work extremely well.  You'd need both the in-line MFC unit and the separate control box to make it work.

IMPORTANT: One thing to be aware of is that all of these technologies are gas-specific to a greater or lesser extent.  The orifices and dial-knob controllers depend at least on the viscosity and density of the gas, and the Porter MFCs depend also on the specific heat of the gas.  The differences will be minor, so I wouldn't worry about it for the first two, but if you need highly precise flow rates, you will have to compensate for the differences if you should need to switch gases.  Porter actually calibrates each MFC to a specific gas, so if you only need argon flows, you'll be all set.


What does ingredients being "biologically active" mean? From Wikipedia

  Cosmeceuticals refers to the combination of cosmetics and
  pharmaceuticals. Cosmeceuticals are cosmetic products with
  biologically active ingredients purporting to have medical or
  drug-like benefits.

I was wondering what  ingredients being "biologically active" means?

In cosmetics that are not cosmeceuticals, are their ingredients not "biologically active"?

Chemists usually marks as biologically active compounds, that have clear effect on living being in small amounts. For example, methane is not biologically active, but potassium cyanide clearly is. Of course, the borderline is quite blurry and can move if context changes.


Source for UV cut-off values of buffers When I want to check a $\mathrm{p}K_\mathrm{a}$ value I look it up in CRC Handbook. Is there such source for UV cut-off values of various buffers?
Usually books in chromatography contain tables with that data, but often no source is given.

The primary reference for this type of data is the article by Li [1]. Snyder cites this paper as the primary source for UV cutoff values ("Ref 7" on the screenshot below) [2, p. 299]:

References

Li, J. B. Signal-to-Noise Optimization in HPLC UV Detection. LC GC 1992, 10 (11), 856–864.
Snyder, L. R.; Kirkland, J. J.; Glajch, J. L. Practical HPLC Method Development, 2nd ed.; Wiley: New York, 1997. ISBN 978-0-471-00703-6.


Why is the size of Al3+ less than that of Li+? Why is it so that the size of $\ce{Al^{3+}}$ is lesser than $\ce{Li+}$?

I'm a bit confused about this one as the aluminium(III) ion is composed of 10 electrons and 13 protons, while the lithium ion has only 2 electrons and 3 protons.

Electronic configuration of $\ce{Al^{3+}}$ is $\mathrm{1s^2 2s^2 2p^6}$ and that of $\ce{Li+}$ is $\mathrm{1s^2}$.

Now, it seems obvious that the size of $\ce{Al^{3+}}$ should be greater than that of $\ce{Li+}$, but it is actually the other way around. Why is this? I think that it is a kind of exception, but if anyone can explain the reason behind this, it will help me in my upcoming tests.

Now, it seems obvious that the size of Al3+ should be greater than that of Li+

No it does not. It is true that electrons repulse each other, however, they all are attracted towards the nucleus.  The real 'size' of an atom or ion is a result of compromise between electron-electron repulsion and electron-nuclear attraction.

  I think that it is a kind of exception

Obviously, it is not.

  Why is this?

Commonly, there are three consideration, used to rationalize basic trends in atomic/ionic radii.

electron-electron repulsion (negative charges repulse each other)
electron-nucleus attraction (negative charges are attracted towards positive ones)
electron shielding (electrons in shells closer to nucleus 'shield' outer electrons from some charge of the nucleus.

While electrons repulse each other, increasing nuclear charge more then compensates it; so if not for shell organisation of electrons, atoms would monotonically decrease in size toward the end of the periodic table, as it is observable in each period 1—3.

However, inner electronic shells effectively shield the outer electrons from the nucleus, so they 'see' a nucleus that is a) larger in size and b) has charge equal to nuclear charge minus number of electrons in inner shells. Given that, outer electrons 'see' a nucleus with low charge density, so outer shells progressively increase in size within the same column.

In addition to that, electrons still repulse each other, so positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions.

Now, consider $\ce{Li+}$ and $\ce{Al^{3+}}$. For the former, 2 outer electrons feel the charge of three protons. For the latter, 8 outer electrons feel the charge of 11 protons in the nucleus. Such great charge easily beats all other factors, so the latter cation is much smaller than the former.


Is this an intramolecular SN2 reaction? What's the mechanism of the following reaction? Seems like internal attack occurs, but would be great if someone could provide a step by step mechanism.

Yes, mechanism of this reaction involves internal attack.


what is a difference between outer electrons and unpaired electrons I know outer electrons include (n-1)d...+ all n'ss.. But i don't understand, never been told what unpaired electron is supposed to be? is it the same thing.

so in the following electron configuration:

$4s^2\: 3d^{10}\: 4p^3$:

OuterElectrons: $15$

ValenceElectrons: $5$

from what I remember someone told me unpaired electrons are $15$ too for this...which is the same as  outer electrons so I guess it's the same thing. But I am not sure so please help.

Outer electrons and unpaired electrons are not the same. Outer electrons are, as you said, $(n-1)d$ and all $n$'s. But unpaired electrons are different. To understand what they are, you must know what is electron spin (which, I hope you do). If you don't, read on.
Every electron in an atom has a quantum spin state (either clockwise or anticlockwise, nothing else) denoted by $+\frac{1}{2}$ and $-\frac{1}{2}$. These spin quantum numbers have no classical analogue, i.e., you can't relate them to any phenomenon in real life situations that you can visually perceive.
So, you see, every electron is paired with another electron with opposite spin. Together, the two of them fill one orbital, as for example, $2p_x$, $3d_{xy}$ etc. These orbitals are found out by solving the Schrodinger Equation.
Now, whenever an electron in an orbital does not have another electron with opposite spin, it is called an unpaired electron.

The electronic configuration that is given is $4s^2\: 3d^{10}\: 4p^3$. As, you can see, the $4p$ is filled with $3$ electrons whereas its maximum capacity is $6$. $4p$ has $3$ orbitals and so, each orbital is filled with one electron each, as given (according to Hund's Rule). So, these three are unpaired electrons. All other shells are fully filled and so, there are no more unpaired electrons.
So, number of unpaired electrons is 3 not 15.


Stability of carbocation directly attached to a halogen atom I have two compounds..

How is the first one more stable?
As I have been taught that -I effect of Chlorine dominates its +R effect but to explain this case, My teacher is saying that because of the achievement of the complete octet, chlorine donates its electrons to the carbocation(showing +R effect). If that's the case then the nucleophilic addition of alkanoyl chloride should be the slowest but experimentally proven, they show the fastest nucleophilic addition reaction by increasing the electrophilicity of the carbon atom(showing -I effect). Moreover, on treatment of alkynes with HCl, we get gem dihalides(which also shows the +R effect of chlorine).

Please help me with this because I am really confused between the -I and the +R effect of halogens.

some people say that its because of octet completion of the carbocation
and the NS reactivity that you mentioned is due to the -I effect of chlorine which makes it the most reactive and not the least reactive


Entropy change in adiabatic process Why is entropy not zero for an irreversible adiabatic process though q=0? Ideally entropy is a state function so entropy should be zero.

The 2nd law is

$d S \ge \displaystyle \frac{d q}{T}$

so the best you can do is zero for a reversible process.   Or if you don't like inequalities,

$d S = \displaystyle \frac{d q}{T} + d S_{irr}$

where $d S_{irr}$ is the entropy production due to the irreversible processes within the system.

You can find a intuitive in depth discussion in Prigogine's book:

Modern Thermodynamics: From Heat Engines to Dissipative Structures

http://books.google.com/books?id=6qFxngEACAAJ&dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&hl=en&sa=X&ei=IveUU8HvH8KVqAbDvILoBA&ved=0CCcQ6AEwAA


Why doesn't the solute settle out of a solution, such as brine or sugar water? If I have a solution of salt or sugar dissolved in water, the solute will never settle out or "fall to the bottom", no matter low long I wait. Why is that?

Since salt and sugar (in their pure, solid forms) are denser than water, I would intuitively expect that gravity would eventually pull all the salt and sugar to the bottom. Why doesn't this happen?

Now, I'm guessing that salt and sugar solutions have lower energy than mixtures of solid salt or solid sugar with water. So that may explain why salt and sugar don't spontaneously settle out; reactions tend to proceed from a high-energy state to a lower-energy state, not vice versa. (In "explain like I'm 5" terminology, maybe sugar sticks to water better than sugar sticks to sugar.) Is that accurate?

I've heard that the favorability of a reaction is determined by such things as energy, entropy and enthalpy, but I don't know the details. I'm guessing the answer has something to do with those.

This question is inspired by this question over at the Earth Science site: Why does the salt in the oceans not sink to the bottom?

(I'm surprised that I wasn't able to find a previous question about why solutions don't settle. All I found was a question about why colloids don't settle, which is also an interesting question, but it's not the same thing.)

You discuss two different phenomena, solubility and concentration gradients due to gravity.

Solubility can be explained using the concept of entropy. Solubility requires the the free energy of a solid be higher than the corresponding substance dispersed in solution. Surprinsingly perhaps, additional entropy associated with dispersion of the solute in the solvent suffices to drive the process (interactions can certainly be important, but are not at minimum necessary, although much depends on how much you are attempting to solubilize, and at what temperature).

Gravity can (and does) cause concentration gradients, and can therefore under the right circumstances cause precipitation. The problem is that gravity is a very weak force, so it's effects on light molecules are, well, weak. Gravitational effects require large scales or masses to be significant. Density differences are argued to be critical driving force behind water cycling in the oceans. The air atop Mt Everest is so thin people suffocate without access to an auxiliary supply of oxygen. So these effects do happen, and matter. And you can use a centrifuge to separate components in a mixture, which is analogous to applying gravitational forces, just scaled up in strength.


Error with explaination of shape of weak acid strong base titration? I have read the other posts on the same question posted on chemistry exchange such as this one: Shape of Weak-Strong Acid-Base Titration

and here : What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?

However, I don't see why their explanation is valid. They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution. When a strong base is added to a weak acid, as more $\ce{OH-}$ is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

However, I do not see this is useful. Isn't having the conjugate base only useful if you want to add in more $\ce{H3O+}$ but we are not! As we only add in $\ce{OH-}$ how is having buffer capacity of $\ce{CH3COO-}$ in the other direction useful (as $\ce{CH3COO-}$ and $\ce{OH}$ don't react)? I agree this is a buffer, but how does the reverse reaction help at all? This buffer doesn't seem like it would help to reduce pH changes

This buffer doesn't seem like it would help to reduce pH changes

You have to be clear what you want to compare it with

Adding a strong base to pure water:
If you start with pure water, the concentration of hydroxide will be pretty much equal to the concentration of strong base you add. The pH will drop quickly.
Adding a strong base to a strong acid:
If you start with strong acid, the first few drops of strong base will be neutralized. The pH does not change much because the strong acid is still in large stoichiometric excess, and the pH is a logarithmic scale.

  When a strong base is added to a weak acid, as more OH− is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

In the first approximation, the added hydroxide will be neutralized by the acetic acid, and the pH will not change. However, as you do that, the weak acid/weak base equilibrium will be disturbed, and the reaction will shift to the left, consuming some hydronium and increasing the pH. The more you change the ratio of acetic acid to acetate for a - say - milliliter of added strong base, the more the pH will change. The ratio changes fast at the beginning (where there is no acetate) and near the equivalence point (where there is almost no more acetic acid), so that is when the pH changes are largest because those are the steps where we disturb the equilibrium the most.

  They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution.

If you compare the slope of the dip to that of scenarios 1. and 2., you will find that it is less than in 1. (strong base added to pure water) and more than in 2. (strong base added to strong acid).

  This buffer doesn't seem like it would help to reduce pH changes

The way I explained it was to say the strong base reacts with the weak acid. We could have said it reacts with the hydronium ion instead. Then the shift of the weak acid/base reaction would have been the other way. In the end, we are just rationalizing the result of two equilibria, that of water autoionization (taking a back seat here because hydronium is the dominant species and hydroxide is present at very low concentration) and that of the weak acid dissociation. The shape of the titration curve is what we get experimentally, and the two equilibria explain it fully. Any explanation on top of that is just to convince ourselves.


What is a product of water ionization? I've read that radioactive compounds are dangerous because they emit electromagnetic waves (gamma-radiation etc.) that are dangerous for living organisms because they can ionize atoms in molecules.

So if one water molecule is ionized by using radiation, would you then get:

$\ce{H2O -> H^+ + OH^-}$
or would you get free radicals : $\ce{H2O -> H^{.} + OH^{.}}$?

Water, when hit by radioactive particles, will most likely suffer ionization (or radiolysis). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived species which will tend to react with most other substances nearby, even if neutral. The electron, after being slowed down considerably, becomes hydrated and can participate in reactions itself. This article has several examples of reactions triggered by radiolysis, and also separates them by their relevant timescales.

Some further examples of reactions involving radiolysis of water via beta particles and gamma rays are (from the very well-cited article "Critical review of rate constants for reactions of hydrated electrons, hydrogen atoms and hydroxyl radicals (⋅OH/⋅O−) in aqueous solution"):

Of course, not only are the decay particles damaging, but so are the entities produced by their interaction with matter in the vicinity. If the nuclear decay particle doesn't directly ionize organic matter, reactive species such as $\ce{OH^{.}}$ or $\ce{H^{.}}$ could then react with biomolecules, degrading cells and tissues.

This Wikipedia link has some details on the radiolysis of water. Interestingly, it seems that dissolved hydrogen in water can efficiently suppress the formation of unusually reactive species, presumably as the side reactions eventually cause the formation of hydroxyl anions (if electrons are present in excess) or hydronium cations (if positive charges are present in excess). This image (source) shows a collection of possible reactions in sequence. If you search for articles on water radiolysis, there are many other results, such as these two. It's interesting to note that the effect of radiation on water still doesn't seem to be entirely understood, in part due to our incomplete knowledge of the behaviour of hydrated electrons.

Here's an extra bit of background on radiation and its biological effects (at least for the most common types. There are many ways nuclei can decay).

Depending how far the radiation source is from you, gamma radiation is either the least or the most dangerous type of nuclear decay. If the radiation is coming from well outside your body, then the danger scale goes alpha < beta < gamma. If you happen to have a radioactive nuclide in your pocket, the scale may well become alpha < gamma < beta. However, if the radiation is coming from inside your body (avoid ingesting glowing substances), then the danger scale goes gamma < beta < alpha. Why does radiation behave like that? There are two main effects in play: penetration capacity and ionization capacity.

In collision theory, different types of projectile and target can be characterized by something called an collision cross-section (or absorption cross-section, in nuclear physics), which basically gives you a probability of how likely it is for the projectile to hit a certain target. If the combination of projectile and target have a high collision cross-section, they they are very likely to interact, and so almost any projectile will be stopped by even a thin target. If the collision cross-section is low, then the projectile has a low chance of hitting the target, and therefore will tend to pass through it entirely. To compensate, you need to make a very thick target so there are more chances for the projectile and target to interact.

The different types of radioactive decay have different penetrating capacities. All of matter as we know it is composed of atoms, which contain well-defined oppositely charged regions (the nuclei and the electron shells), and so we can expect them to interact strongly with other charged particles. Alpha and beta(-minus) decays emit particles with a charge of $+2e$ and $-e$, respectively ($+e$ for beta-plus decay). We then expect these particles to interact strongly with matter, being stopped easily. Indeed, most alpha particles are efficiently stopped even by air. Beta particles have half the charge of an alpha particle and are going quite a bit faster (having approximately 1/7000th the mass), so they manage to go further before interacting. A few centimetres of of metal shielding or a concrete wall are usually enough to stop them. However, gamma rays have no charge, so they interact comparatively weakly with matter. These highly energetic photons will go though dozens of centimetres of solid matter. To maximize the odds that they will hit something and be stopped, you want to use a shield made of a very dense substance (which would pack a lot of possible targets in a small space). This is why lead is the substance of choice for radioactive shielding; its high density combined with low cost allows the fabrication of relatively compact shields which are capable of stopping even gamma radiation, when around 10 cm thick.

Now, for the mechanism of how particles from radioactive decay are stopped. For uncharged particles, the absorption cross-section is low because interactions only happen when the uncharged particle comes very close to a target particle. For gamma rays (very high energy photons), when there is an interaction, it is usually due to Compton scattering, where an energetic photon interacts with an electron, giving it a large amount of kinetic energy, while the photon is re-emitted with a smaller amount of energy. The target electron (not necessarily a valence electron) often gains enough energy to be completely removed from the atom, causing ionization.

For collisions between charged particles (both same and opposite charges), interactions happen much more easily, as charges interact strongly even at comparatively much larger distances. Beta-minus particles (very fast electrons) can more efficiently repel electrons and shove away from atoms, creating cations. Alpha particles (bare helium-4 nuclei) are interesting because their positive charge means they can tear electrons from any atom or ion. In fact, it's defensible that alpha particles are the strongest Lewis acid one can commonly find; they are effectively free protons with twice the charge.

Notice that penetration capacity and ionization capacity are inversely correlated; if a radioactive decay particle is strongly ionizing, it interacts well with matter and is stopped in short order. If it is weakly ionizing, one of the main interaction mechanisms is suppressed, and the decay particle can go further before it is stopped. In order to not damage you, you want the nuclear decay products to not dump their energy in your body, i.e. you want them to pass right through you. Here, it is clear that alpha particles are the most dangerous, followed by beta and gamma radiation. However, alpha particles interact so strongly with matter that they literally need to be released closer than your skin to dump energy into living tissue, so you need to ingest, breathe or inject alpha emitters. When this happens, they are extremely deadly. It is a very rare occurrence, however, since radiation usually comes from far outside. In that case, gamma radiation has a larger chance of reaching you, so it is said to be the most dangerous.


Investigating a factor affecting buffer capacity I'm preparing for an experiment to investigate some factor affecting buffer capacity. At the moment, I'm looking at using a $\ce{NH_3}$ ($2\,\mathrm{M}$) $+$ $\ce{NH_4Cl}$ buffer, but I'm not sure which factor to test. My research tells me that only the concentrations of each will affect buffer capacity, and that the ratio of their concentrations will affect the $\mathrm{pH}$ of the buffer (by the Henderson-Hasselbalch equation). Are there other factors which I can test, because I'm trying to make it a bit more interesting? As far as I can tell, something like temperature won't affect buffer capacity.

As far as the actually experiment goes, what would be the best way to measure buffer capacity? I could just do a titration and record the $\mathrm{pH}$ as I go, but I'm thinking of using an indicator and measuring absorbance. I'm open to any ideas here, because I'm struggling to think of anything "out of the box"?

As far as I can tell, something like temperature won't affect buffer capacity.

In this particular case, I wouldn't be too sure about it. Isn't it conceivable that your buffer will lose ammonia upon heating?

Did you already decide about the pH indicators? For the lower end, thymol blue might be a good choice. The supplementary data
to J. Mater. Chem. B, 2015,3, 2801-2807 contains a UV/VIS spectrum of this indicator.


How to be able to tell if a molecule will react via SN1 or SN2? I've been having a lot of issues with a question.

2 compounds, (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene react with EtOH in water.

It asks to classify the reaction mechanism for each and to then go on and explain why (1-chloro-2-methylpropyl)benzene has a much faster relative rate.

I am having major issues with the question, because I just cannot tell where there are any major differences between molecules which could make a difference. Both of them have a benzene ring, but this stabilises both the transition state and the carbocation intermediate so will favour both SN1 and SN2. Both reactions are done in the exact same solvent with the same nucleophile.

The only differences I can spot is that in (1-chloro-2,2-dimethylpropyl)benzene, there will be more stabilisation due to the inductive effect and hence it will favour SN1. But this does not help with determining the mechanism for (1-chloro-2-methylpropyl)benzene.

Any help will be much appreciated. Many thanks :)

Factors affecting SN1 mechanism:

  Weak nucleophile / base and Good leaving group.
  More steric hindrance.
  Polar protic solvent should be used.

Factors affecting SN2 mechanism:

  Strong nucleophile / base and Good leaving group.
  Less steric hindrance.
  Polar aprotic solvent should be used.

Coming back to your problem. Actually, both of the compounds can go through SN1 or SN2 mechanism as the carbon linked with $\ce{Cl}$ is a secondary carbon. But, as you mentioned that you're using $\ce{EtOH}$, and you should know that $\ce{EtOH}$ is little basic, almost neutral, therefore, we can say that now, both of the compounds will go through SN1 mechanism (Remember, the first point of SN1 mechanism, i.e., weak base).

Now, the question is if both the compounds will follow SN1 mechanism, then which will react faster. For this, we will have a look at the SN1 mechanism:

  Step 1: Leaving of the (-X) group

  Step 2: Weak nucleophile attacks from either side

  Step 3: Substitution product formed (Inversion or retention, based on
  the step 2)

So, our concern is only the step 2.

Note that, rate of reaction in step 2 depends on steric hindrance in your case. So, let's have a look at the structure of (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene:

The Carbon with red dot is the point where $\ce{EtOH}$ will attack after the (-Cl) will leave.

Now, on comparing the both the compounds, you will notice that in the first compounds there is greater steric hindrance as compared to the second compound. And, that's the reason why second compound will react faster than the first one.

Conclusion:

  (1-chloro-2-methylpropyl)benzene will react faster than
  (1-chloro-2,2-dimethylpropyl)benzene due to less steric hindrance
  around the carbon that is directly attached to aromatic ring (i.e.,
  the carbon with red dot).

  And, both of the compounds will go through SN1
  mechanism due to presence of $\ce{EtOH}$.


Why density functional theory favors a uniform electron density? Why density functional theory (DFT) favors a uniform electron density when we use it to calculate ground state electronic structure self-consistently?

For example, when we use DFT for calculating the (ground state) electron density of two $\text{H}$ atoms with a separation somewhat larger then the bond length of an $\text{H}_2$ molecule, there will still be a non-negligible electron density between the two atoms as if there is still a covalent bond. As a result, the calculated electronic ground state is essentially an excited state of the system.

I cannot remember where did I read this, and I don't understand why DFT has such a problem. I think the problem resides in the fact that we use a local exchange-correlation functional $E_{XC}$ like LDA and GGA. But I don't understand why a local $E_{XC}$ will lead to a uniform electron density.

I think I can see the confusion, and I will do my best to explain.

As an initial attempt to use the electron density, the uniform electron gas (UEG) model was developed. This fictitious model assumes: an infinite number of electrons, a constant non-zero density, an infinite volume, and a uniformly distributed positive charge. IMPORTANT: The electron-electron interactions (correlation and exchange) are fundamentally ignored.

Two main theorems comprise modern DFT: Hohenberg-Kohn (HK) and Kohn-Sham (KS).
In principle DFT is exact if we knew the universal HK functional (which contains electron-electron exchange-correlation terms); however, no mathematical description has been developed. As we do not know the exact form of the exchange-correlation term, a number of exchange-correlation functional have been developed. These can be derived from experimental thermochemical date or high-accuracy quantum calculations.

As exchange-correlation functionals are approximations, the electron density is not fully localised to a specific area. I hope this helps.


What are the possible products in allylic bromination of 3-methylcyclopentene? Bromination of an alkene by N-bromosuccinimide (NBS) in the presence of light or peroxide is a radical reaction and produces an allylic bromide. For the following bromination of 3-methylcyclopentene, which of the following allylic bromides would be products of the reaction?

In some other reactions such as elimination, there is a tendency to form the most substituted double bond, since it is the most stable. If that were the case, then the products should be A, E, F, and G, since these are all trisubstituted alkenes (the others are disubstituted). Is that a valid consideration here, or is there a different overriding rule?

Your comment shows that you are on the right track.

You have chosen A and C because they are formed when the initial allylic radical instantaneously reacts with $\ce{Br2}$ in the propagation step of the reaction, isn't it? (Hint: Just say YES!)

Now give the allylic radicals corresponding to A and C a bit of time, just enough to flip the dot.

How do these other radicals look like? There you have the other two products.

For an extra point: Which isomer of methylcyclopentene could lead to G?


How to increase the velocity of a reaction? I'm doing relatively simple Chemistry but I've been given a formula for the velocity of a reaction.

$$v= k \cdot c_\mathrm A^a$$

and

$$k= A \cdot \mathrm e^{E_\mathrm a/RT}$$

As a result, from that formula, surely decreasing the temperature $T$ would increase the velocity of the reaction? As that would increase the value of $k$ and therefore drive up the reaction speed? Also, same thinking, increasing the activation energy $E_\mathrm a$.

I know that's wrong, because common sense. But the formula is lying to my face.

It should be $-E_a/RT$

Therefor, increasing the temperature, lowering the activation energy (catalysis), or increasing the concentration of reactant(s) would increase the reaction velocity.


How this tricky Wagner-Meerwein rearrangement occurs? In 1979, Karel Frantisek Wiesner performed a tremondous total synthesis in $59$ steps of the 13-desoxydelphonine. In one of those steps there is a very tricky Wagner-Meervien rearrangement. I tried to find something by myself, I also look on the net, I found this and you can see page $8$ they have the same problem than me.

What is the mechanism of this rearrangement ?

Just to be a bit more clear, the acetate which looks to appear magically obviously don't, it comes from the previous step because of the solvant, they just heat several days after so the solvant is still the same as you can see in the picture below.

If someone has an idea, I would be glad! Thank you in advance :)

My literal back of the envelope solution. Notable points:

I invoked a nonclassical carbocation in the first step of the rearrangement. This was the easiest way to explain the stereochemistry of the new acetate group.
This is a reaction from 1959-1960. It's almost certainly racemic. I think the product as drawn is actually the enantiomer of the one that you would expect from the reactant.


Can ferrocene type sandwich compounds be made with fluorene? Ferrocene will be well known to made members of this site. It consists of two cyclopentadienyl anions sandwiching an iron-II ion.

During my first year organic chemistry course (about 40 years ago) I remember a friend suggesting that the fluorene anion had a similar stability and it might be possible to make ferrocene type compounds with an iron-II ion sandwiched between two fluorene anions.

A group of us persuaded a sympathetic professor to let us have a go, and as I recall we prepared the anion by adding lithium aluminium hydride to solution of fluorene in ether cooled with dry ice then adding iron-II chloride. The iron chloride immediately went into solution, so some form of complex was formed, but we couldn't get much further as on warming to room temperature we got a hideous mess of many different products and we didn't have time to analyse it thoroughly.

So my question is whether these types of fluorene compound are known, or alternatively whether it is known that they cannot be prepared? I have attempted to Google this, but there is such a vast literature on ferrocene type compounds that I have not been able to find anything relevant.

Yes there are, many in fact.  One of the examples is having a complex consisting of a fluorene molecule and of cp altogether with iron, described by Decken in Acta Cryst. E (doi: 10.1107/S1600536804028077):

(source)

You are not limited to complexation with iron, though.  Already by the title «Synthesis and characterisation of Ti, Cr, Mo and W bis(fluorene) complexes» by Cloke et al. in J. Organomet. Chem. (source, paywall) you may see that there is a lot of variation possible by either varying the metal, or the second molecule to build the complex.


Spontaneity of reversible chemical reactions - Both sides? Does the reversibility of a chemical reaction make it spontaneous from both sides also? (i.e.: forward & backward reactions) If not then is there any relationship between $\Delta G$ of forward and backward reactions. Am I right to assume that if $\Delta G$ of one of the process is positive then $\Delta G$ for the reverse process must be negative. So at equilibrium $\Delta G = 0$ ?

You may look here for more information.

We know the formula $\Delta G=\Delta H-T\Delta S$ and also know that if for the left reaction $\Delta H_\text{r}=c$, then for the other-way reaction $\Delta H_\text{r}=-c$ (because $\Delta H_\text{r}=\sum\Delta H_\text{f} \text{(product)}-\sum\Delta H_\text{f}\text{(reactant)}$)

Same for the entropy change $\Delta S$: If for the left reaction $\Delta S_\text{r}=c$, then for the other-way reaction $\Delta S_\text{r}=-c$ (because $\Delta S_\text{r}=\sum\Delta S_\text{f}\text{(product)}-\sum\Delta S_\text{f}\text{(reactant)}$).

So, for the reverse of a reaction, since both $\Delta H_\text{r}$ and $\Delta S_\text{r}$ are opposite, the whole $\Delta G_\text{r}$ will be the opposite, hence at equilibrium $\Delta G=0$, so that neither way is thermodynamically favoured.

In fact, how much a reaction (warning: reversible reactions isn’t the same as reversible processes, not to be confused) is reversible and the direction it proceeds is determined not only from thermodynamics (I would say much more from other factors), but from Le Chatelier’s principle (temperature, pressure, concentration), catalysts and of course activation energy.

So, we have seen the relationship of $\Delta G_\text{r}$ of a reaction and its reverse and the fact that at equilibrium $\Delta G=0$.

It must be noted that:

$\Delta G_\text{r}$, $\Delta H_\text{r}$, $\Delta S_\text{r}$ refer to “reaction” (r)

and

$\Delta G_\text{f}$, $\Delta H_\text{f}$, $\Delta S_\text{f}$ refer to “formation” (f) (of chemical compounds and elements)


Why does O2NCl only have two resonance structures? Why is it that $\ce{O2NCl}$ has only two resonance structures? More specifically, why is the Lewis drawing where $\ce{N}$ is double bonded to $\ce{Cl}$ not considered a resonance structure along with the other two where $\ce{N}$ is double bonded to an oxygen?

$\ce{O2NCl}$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if weighted accordingly.
This last expression in italics is important. Not all resonance structures are equal. People teaching resonance structures usually use expressions like most probable resonance structures. To determine whether a resonance structure is more probable than another one, check the following list:

does it give all main group atoms a valence electron octet?
are formal charges minimised?
are formal charges distributed in accordance with electronegativity (i.e. do electronegative atoms have more formal negative charge than electropositive ones)?

The list is ordered from most important to not so important although traditional teaching sometimes weighs maximised amount of bonds higher and thus arrives at hyperoctet structures.
Now let’s check the structures of $\ce{O2NCl}$ that you are talking about:

We see that all structures fulfill rule 1. However, rule 2 clearly shows a difference: The two structures on the left contain $+1$ and $-1$ formal charges each while the rightmost structure has $+2$ and $-2$. We don’t need to check rule three, we can see that the rightmost structure is unfavourable and thus less close to the truth. Finally, the formal charges are distributed correctly on all three structures, because oxygen is more electronegative.
Therefore, the left two structures are equally probable while the rightmost one is not so; oftentimes in exams, only the left two will be considered ‘correct’. But as I said, the true picture (or something really close to the truth) can only be determined by quantum chemical calculations.


Which form of chromium oxide is formed naturally in the atmosphere? I'm trying to figure out which form of chromium oxide forms naturally in our atmosphere, i.e., if you take a piece of elemental $\ce{Cr}$ from vacuum to a normal, room temperature environment with air, does it form $\ce{CrO}$ on the surface? Or $\ce{Cr2O3}$? Etc.

This site says:

  Chromium (III) oxide, $\ce{Cr2O3}$ is the main oxide of chromium.

So it seems like it might be that one, but "main" is pretty vague. Does anyone have a more definitive answer, with a source?

thank you!

An investigation of the phase diagram reveals that the only stable chrome oxide near room temperature is $\ce{Cr2O3}$.  $\ce{Cr3O4}$ appears, but only near 2000K.  There is no stable chrome analogue to $\ce{FeO}$ on the phase diagram.

Thus, the one stable oxide that will form under normal atmospheric conditions is $\ce{Cr2O3}$. However, note that it is easy to find mixed metal-oxygen compounds with high mutual solubility.

If you are interested in the thermodynamics of the $\ce{Cr-O}$ system, you could start with M. Kowalski and P.J. Spencer, Calphad 19(3) 229-243 (1995), which covers the $\ce{Cr-O, Fe-O}$ and $\ce{Ni-O}$ systems


Flaw in my logic on Henry's Law? It states mathematically $ p=Kx $ where $x$ is mole fraction of dissolved gas and $p$ is partial pressure of the gas.
Dissolution of gases is exothermic, so with increase in temperature, solubility $x$ decreases.
But increase in temperature also means an increase in partial pressure $p$ which in turn means an increase in solubility $x$. Where am I going wrong?

Henry's law is:

  At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

If temperature changes, the Henry's law constant itself changes.


Does carbonated water change over the course of several days, or was my container leaky? I have a Klean Kanteen flip-top growler, allegedly leakproof at up to 35psi of pressure, that I filled with some freshly-carbonated water 4 days ago — just to see how long it would last. I used a Soda Stream for carbonation. The bottle was filled to the brim.
When fresh, the carbonated water fizzed and sparkled. You could see the bubbles constantly rising to the surface. It felt like pop rocks when you put it in your mouth.
Now, 4 days later, the water is still clearly carbonated, but it's a lot more subdued. You can taste the fizz when you drink it, but there's no more surface bubbling. When you pour it into a cup, you see bubbles for about a second before the water clears up.
The obvious theory is that some CO2 escaped over the course of the week, but I'm fairly certain there were no liquid leaks. I also couldn't hear any hissing. The bottle was stored on its side in the fridge.
Is it possible for a container to be liquid leak-proof, but not gas leak-proof?
If not, does the nature of carbonated water change over time? Does the water, perhaps, absorb more of the CO2 the longer it sits? If so, why does this not happen with soda cans and such?

Update 2015-11-3:
I repeated the experiment with a 20oz screw-top Klean Kanteen I had lying around, ensuring that the water was as close to the cap as possible and that the lid was screwed on as tightly as possible. The results after 4 days were more carbonated than my initial trial with the flip-top Kanteen, though still not as much as fresh. Currently repeating the experiment once again with the flip-top, this time making 100% sure that the water is touching the lid.

The difference is in what is leaking. Gas leaks out a lot easier than fluid. The container is rated not to leak liquid, not that it won't leak air. So if you turned it upside down with 35 psi inside (a carbonated drink) the fluid won't leak out.

So turn the Klean Kanteen flip-top growler upside down. ;-)


Getting the pressure of a system by series expansion An average number, $N$, of bosons of spin $S = 0$ is conned to a two-dimensional domain
  with surface $A$. The gas is ultrarelativistic with a single particle energy $\epsilon = cp$, where $c$ is the speed of light in vacuum and $p$ is the absolute value of the momentum. Compute the pressure, P, of this system as a function of z, A and T. While still being
  in the high-temperature regime, use the result for $N(z, A, T)$ to find $P(N, A, T)$ (keep up to quadratic terms in N). Discuss your results and the relation
  to the ideal gas law.

$N(z, A, T)$ in function of $z$ has been previously calculated:

$$N = \frac{2\pi A}{(h \beta c)^2}(z + \frac{z^2}{4})$$

Where:

$$\beta = \frac{1}{K_B T}$$

$$z = e^{\beta \mu}$$

The following equation also holds:

$$\frac{P}{K_B T} = \frac{2\pi }{(h \beta c)^2}\sum_{n=1}^{\infty} \frac{z^n}{n^3}$$

With this information we should be able to get $P(N,A,T)$ (keeping up to quadratic terms in N), but I am not getting the stated pressure.

The provided solution is:

$$P \approx \frac{NK_BT}{A} (1 - \frac{N(h \beta c)^2}{16 \pi A})$$

Since this is clearly an exercise, I don't think it's appropriate to give the full solution, but I see no harm in pointing you in the right direction.

You are aiming to convert an expansion in $z$ into an expansion in the number density $\rho=N/A$. These are often called virial expansions. Your first equation can be written
$$
\rho = q\left[z+\frac{z^2}{4}\right] \qquad\text{where}\qquad
q = \frac{2\pi}{(h\beta c)^2} .
$$
I have defined $q$ just to cut down on needless clutter.
The brute force approach to this would be to solve this equation, to give $z$ in terms of $\rho$,
and then substitute this expression into your series formula for $P/K_BT$ as a function of $z$. You can do this; it's a bit fiddly, since it is a quadratic equation in $z$, but it could be done.

For your purposes, a slightly more straightforward approach is to assume that you will get a result of the form
$$
z = c_1 \rho + c_2 \rho^2 + \ldots
$$
where you need to determine the coefficients $c_1$, $c_2$ etc.
Hopefully you can see why there is no need for a $\rho$-independent term $c_0$, when you look at the equation for $\rho$.
Also, it should be obvious that you don't need to consider more terms like $c_3$ etc, if you are only interested in the final result having the first couple of powers of $\rho$.

So you determine the coefficients $c_1$, $c_2$ by substituting that expression for $z$ into the right hand side of the equation for $\rho$,
and equating the coefficients of corresponding powers of $\rho$ on each side.
The term linear in $\rho$ will give the coefficient $c_1$ immediately:
$$
[\rho^1]:\qquad 1 = q \, c_1 .
$$
The quadratic term in $\rho$ will give an equation for $c_2$, which
also involves $c_1$,
$$
[\rho^2]:\qquad 0 = \text{I'll leave you to work this out.}
$$
Since you just determined $c_1$, this will be easy to solve to get $c_2$.

So now you have $z\approx c_1 \rho + c_2 \rho^2$ and you can insert this expression into the given series expansion for $P/K_BT$,
only keeping the terms in $\rho$ and $\rho^2$.
If you do all this carefully, you will get the answer provided.


Why is Cr(2+) a stronger reducing agent than Fe(2+) in water? According to me, $\ce{Fe^{2+}}$ should be a better reducing agent because $\ce{Fe^2+}$ - after being oxidized - will attain a stable $\ce{d^5}$ configuration, whereas $\ce{Cr^2+}$ will attain a $\ce{d^3}$ configuration. I think the half filled $\ce{d^5}$ configuration is more stable than the $\ce{d^3}$ configuration. Why is this not so?

The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept.

It is tricky to compare the "stability" of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the change in (free) energy that occurs from start to finish.

To put it another way, it may be that $\ce{Fe^3+}$ is more stable than $\ce{Cr^3+}$ on an absolute scale, but what we really care about is how much more stable $\ce{Cr^3+}$ is to $\ce{Cr^2+}$ compared to how much more stable $\ce{Fe^3+}$ is to $\ce{Fe^2+}$.

Let's examine all four using electron configuration as you have done:

$\ce{Fe^2+}$ is $\ce{d^6}$ or more probably $\ce{[Ar] 4s^1 3d^5}$ - two half-filled half shells!
$\ce{Fe^3+}$ is $\ce{d^5}$, which is $\ce{[Ar] 3d^5}$ - one filled half shell.

The difference between the two iron ions might not be that large.

$\ce{Cr^2+}$ is $\ce{d^4}$, which is $\ce{[Ar] 3d^4}$ or $\ce{[Ar] 4s^2 3d^2}$
$\ce{Cr^3+}$ is $\ce{d^3}$, which is $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^2 3d^1}$

The energy difference between chromium ions might be larger. Represented graphically, the reaction coordinate energy diagram for the two process might be:

  Now the number part.

We can go looking for some standard thermodynamic data to help make our case. The best data are for standard reduction potentials, because these data are for exactly what we want!

Taking the reduction potential data and writing the equations in the direction we care about, we have:

$$\begin{align}
&\ce{Fe^2+ -> Fe^3+} &&&E^\circ=\pu{-0.77 V}\\
&\ce{Cr^2+ -> Cr^3+} &&&E^\circ=\pu{+0.44 V}
\end{align}$$

Spontaneous reactions produce positive potential differences, so we can see right now that $\ce{Cr^2+}$ is a better reducing agent.

Let's go a step farther to free energy.

$$\Delta G^\circ =-nFE^\circ$$

However, to deal with free energy, we need a full reaction. Technically, comparing the two half-reactions in isolation is just as bad. However, the definition of the standard electrode potential and the standard free energy come with a common zero point in terms of half-reaction: $\ce{2H+ + 2e- -> H2}$.

The two full reactions (net ionic equations anyway) are:

$$\begin{align}
&\ce{2Fe^2+ + 2H+ -> 2Fe^3+ + H2} &&E^\circ=\pu{-0.77 V}&&&\Delta G^\circ=\pu{+74kJ/mol}\\
&\ce{2Cr^2+ + 2H+ -> 2Cr^3+ + H2} &&E^\circ=\pu{+0.44 V}&&&\Delta G^\circ = \pu{-42 kJ/mol}
\end{align}$$

And the winner is chromium, by a whopping $\pu{116 kJ/mol}$. In fact, the data might show that $\ce{Fe^2+}$ is more stable than $\ce{Fe^3+}$, while $\ce{Cr^2+}$ is less stable than $\ce{Cr^3+}$.


Determining exponents in a rate law with different data Here is the experimental data: Note all concentrations are the initial amounts

\begin{array}{cccc}
\text{Experiment} & \ce{[C2H4Br2]} &\ce{[I-]} & \text{Initial rate of formation of }\ce{I3-}\\\hline
1 & 0.127 & 0.102 & 6.45\times 10^{-5}\\
2 & 0.343 & 0.102 & 1.74\times 10^{-4}\\
3 & 0.203 & 0.125 & 1.26\times 10^{-4}\\\hline
\end{array}

$\mathrm{Rate}=k\ce{[C2H4Br2]}^m\ce{[I-]}^n$

We can determine the exponent $m$ easily because experiment one and two have the same initial $\ce{[I-]}$. However, when determining $n$, there is no data that has the same initial concentration. What the solution manual does is the following:

$$n=
\dfrac
{\ln \left(\dfrac{\mathrm{Rate}_{3}\cdot [\ce{C2H4Br2}]_2}
{\mathrm{Rate}_{2} \cdot [\ce{C2H4Br2}]_3}\right)}
{\ln \left(\dfrac{[\ce{I-}]_3}{[\ce{I-}]_2}\right)}=\dfrac{\ln \dfrac{(1.26\times 10^{-4})(0.343)}{(1.74\times 10^{-4})(0.203)}}{\ln \left(\dfrac{0.125}{0.102}\right)}=1$$

My question is, why are they multiplying the rate of the third experiment with the initial concentration from the second experiment and then multiply the rate from the second experiment with the initial concentration of the third experiment. I don't see any correlation with these numbers and how they can be used to determine the exponent.  Any thoughts?

If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):

$$
\dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Rearranging this equation, one obtains:

$$
\dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:

$$
\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2}
$$

Solving for $n$, one finds:

$$
\frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n
$$

Which is the final equation from your textbook's solutions manual.


Why do two atomic orbitals form two molecular orbitals? According to molecular orbital theory, two atomic orbitals form two molecular orbitals analogous to waves combining constructively or destructively but how can a wave combine destructively and constructively at the same time to produce two molecular orbitals?

Why don't two atomic orbitals combine to form a molecular orbital that can be BMO or ABMO?

Well, one important thing to keep in mind is that atomic and molecular orbitals are not "real" things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules.  They are like a set of rules that tell us all of the ways in which it is possible for the electrons with a particular set of quantum numbers to behave.  For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\mathrm p_z$ orbital (quantum numbers: $n=3$, $l=1$, $m_l=0$).

Now, when you are dealing with molecular orbital theory, you are talking about the possible ways in which electrons in atomic orbitals can combine in a covalent interaction.  So it's not that the two atomic orbitals combine "constructively and destructively at the same time", it's just that those are the two ways in which its possible for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them.  The "constructive" case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the "destructive" case, so by the aufbau principle, electrons will occupy the lower energy "bonding" orbital first.  In a case like $\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\ce{He2+}$), it will occupy the higher-lying "anti bonding" orbital.


Producing ketones from secondary alcohols using hydrogen peroxide Usually, to produce ketones from secondary alcohols (e.g. propanone from 2-propanol), people use a solution of potassium or sodium dichromate with a small amount of sulfuric acid. Can other oxidizers like hydrogen peroxide be used, and will the use of hydrogen peroxide produce organic peroxides?

$$\ce{C3H8O + H2O2 -> C3H6O + 2 H2O}$$

This is not for any particular purpose.

I think it is possible to use hydrogen peroxide. This paper discusses it: http://greenchem.uoregon.edu/PDFs/ResourceID88.pdf

Other alternatives:

Pyridinium Chlorochromate (or PCC for short)
Dess-Martin Reagent
Swern Oxidation: oxalyl chloride in DMSO with triethyl amine

Fétizon Oxidation: silver(I) carbonate absorbed onto the surface of celite (Fétizon's reagent). This can be prepared by adding silver nitrate to aqueous sodium carbonate, in the presence of purified celite.

Oppeneaur Oxidation:  excess ketone reagent (such as acetone) and an aluminium triisopropoxide catalyst

Pfitzner–Moffatt oxidation: dimethyl sulfoxide (DMSO) activated with a carbodiimide, such as dicyclohexylcarbodiimide (DCC)

Corey-Kim Oxidation: Dimethyl sulfide is treated with N-chlorosuccinimide (NCS), and then addition of triethylamine


The effect of NaCl on the ability of an acid to protonate benzoate I've been doing several experiments using soda to quantify the amounts of caffeine and benzoate. Because of our methods of analysis, we had to decarbonate our soda. The experiment offered two ways for decarbonating the soda: shaking it or heating it. Then, to make sure all the benzoate in the soda is converted to benzoic acid, we add a small amount of $\ce{HCl}$ (to protonate it.)

Several years ago, I did a similar experiment to quantify the amount of caffeine. To decarbonate the soda, I added salt ($\ce{NaCl}$.)

I'm wondering if adding salt would be inappropriate in the more recent experiment. None of the lab procedures mention adding salt and, to me, this is the easiest way to decarbonate soda. Therefore, I'm thinking there may be a specific reason the salt method was left out of the procedure.

My thought is there could be some interaction between $\ce{Na+}$ and $\ce{Cl-}$ that could "neutralize" the $\ce{HCl}$ so to speak, making the protonation of benzoate less effective.

Perhaps it would make sodium benzoate instead of benzoic acid? Perhaps they just didn't think to add salt. Any thoughts on this?

Some salts added obviously will alter the pH value of the soda ($\ce{NaHCO3}$, for example), and others not (like the example provided by you, $\ce{NaCl}$). Changing the ionic strength by addition of salts may render your mathematical model to describe the composition of the analyte more complex.

Hence, let me I suggest an alternative to outgassing the soda: immerse the open container into an ultrasound bath.  This works rapidly, however ensure the container is widely open and may accommodate the foam briefly generated.  Otherwise, it may surprise you as a sticky geyser.  Second hint: gently approach the active regions in the ultrasound bath to moderate the gas evolution.

(Degassing solvents with ultrasound is a common preparation for HPLC, too.)


What is the mechanism of this reaction? I've been stuck with this for some time. I just don't see it. As the reactant is Pd(0), I think oxidative addition will happen to the bromine. Then I would say an insertion reaction happens. However this would form a 7-membered ring which isn't as stable as a 5-or 6-membered ring and isn't even in the final product.

Then I tried complexating it with the double bond next to the ether. However I don't see what use the Pd could have from there. Maybe deprotonate the alpha-carbon which would give a hapto-3 complexation? But what next, I wouldn't know. Any help would be appreciated!

Nicco: Of course you can't "see it". If you were given the correct structure, the solution to your post would be straightforward. There should be a double bond in the lactam ring as shown in compound 1. The reaction is conducted in acetonitrile (bp. 82oC) at reflux. These conditions allow for a Claisen rearrangement to occur to form  ketolactam 2. Subsequent oxidative addition of palladium(0) [ligands ignored] to the C-Br bond creates the Pd(II) species 3 which adds to the double bond to afford cyclized species 4. Reductive elimination of Pd(0) regenerates the catalyst as alkene 5 is formed. The role of $\ce{Et3N}$ is to neutralize HBr that is formed. Addendum: Just found the original paper: https://doi.org/10.1016/0040-4039(94)88380-7


How to find out the state symbol of a compound by looking at its chemical formula? Like, in the equation between rubidium and water:

2Rb(s) + 2H2O(l) → 2RbOH(?) + H2(?)

How do we know what state symbol to use in 2RbOH and H2? I'd like to do this just by looking at the chemical formula, but it's OK if I have to memorize a list of stuff to be able to do this.

You can't do this by:

  just by looking at the chemical formula

But instead look up the properties of the compounds in question (Wikipedia is a good source of data).

$\text{RbOH}$ is highly soluble in water, so assuming there's an excess of water present, it will end up being dissolved in the water. So its state symbol will be $\text{(aq)}$.

Hydrogen on the other hand is a gas (at room temperature and above) and highly insoluble in water: it will leave the reaction as gas and its state symbol will be $\text{(g)}$.

So without knowing the physical/chemical properties of the reagents/reaction products, the state symbols cannot be determined.

Note that the state symbols are often idealisations: solid or liquid compounds are rarely completely soluble or completely insoluble. The same is true for gases. Even water is a case in point: liquid, $\text{(l)}$, at room temperature but with significant vapour pressure.


How do I visualize an atom? I have searched and searched, oh how I have searched.

I am looking for a 3-dimensional visualization of a whole atom, one that that includes all the orbital geometry. A proper "layered" view of the orbitals.

What I am not looking for:

Individual orbital geometry.
A "picture" of an atom (i.e. what looks like television static to me).
Little spherical balls.
Dot-and-cross diagrams.

I think I understand that this cannot be expressed in just one picture alone, so I look forward to discovering software, or a collection of pictures, that could help express this. Maybe something similar to an exploded view, for atoms?

Please. The universe is weird and I want to see.

I have searched and searched, oh how I have searched.

Do you know what I always tell my mom when she asks me to find something in the Internet she was not able to find herself? I ask her: "Are you sure that the thing you are looking for even exists?"

  I am looking for a 3 dimensional visualization of a whole (moderately
  complex, hydrogen is just a ball) atom that includes 3 dimensional
  orbital geometry.

Hydrogen atom is not "just a ball".
There is no orbitals. In short, except for one-electron systems (such as the hydrogen atom) orbital description is just an approximate model of the reality.
Usually when one asks to visualize a physical object, he/she wants to visualize its physical "boundaries" with respect to some medium. The notion of such "boundary" simply looses its meaning in the microscopic world: atoms do not have "boundaries" like macroscopic objects do.

  Please. The universe is weird and I want to see.

Sorry, but you can not do this: you can not visualize some things in the Universe, you can not even imagine them. Our visualization & imagination abilities (unlike an atom) has some boundaries, because they are based ultimately on our senses which in turn has their boundaries.

As Paul Dirac said on that matter:

  […] the main object of physical science is not the provision of
  pictures, but is the formulation of laws governing phenomena and the
  application of these laws to the discovery of new phenomena. If a
  picture exists, so much the better; but whether a picture exists or
  not is a matter of only secondary importance. In the case of atomic
  phenomena no picture can be expected to exist in the usual sense of
  the word 'picture', by which is meant a model functioning essentially
  on classical lines. One may, however, extend the meaning of the word
  'picture' to include any way of looking at the fundamental laws which
  makes their self-consistency obvious. With this extension, one may
  gradually acquire a picture of atomic phenomena by becoming familiar
  with the laws of the quantum theory.


Finding mathematically the ground state density in DFT To find the ground state density in DFT, you set the following Lagrangian:

$$L = E[\rho(\vec r)] - \mu\left(\int \rho(\vec r) \mathrm{d}\vec r - N\right)$$

While minimising with respect to the density, we get:

$$\frac{\partial L}{\partial \rho} = \frac{\partial E[\rho(\vec r)]}{\partial \rho} - \mu = 0$$

My problem is with the explicit derivation of the energy functional. Let's consider the Thomas-Fermi-Weizsacker kinetic energy functional:

$$T_s(\rho) = C_F \int \rho^{5/3}(\vec r)\mathrm{d}\vec r + C\int \frac{| \nabla \rho (\vec r)|^2}{\rho (\vec r)}\mathrm{d}\vec r$$

For example, how do you derive this with respect to the density?

$$\frac{\partial T_s(\rho)}{\partial \rho} = C_F \frac{\partial}{\partial \rho}\int \rho^{5/3}(\vec r)\mathrm{d}\vec r + C\frac{\partial}{\partial \rho}\int \frac{| \nabla \rho (\vec r)|^2}{\rho (\vec r)}\mathrm{d}\vec r$$

For the first term, you could use Leibniz integral rule, since the density very far away from the nucleus is zero. However, I really have no idea how to tackle the second term. Moreover, what is the algorithm afterwards to find the ground state density? How does one figure out this density from the equation?

What we have to do here is a functional derivative.
Lets consider the following method of finding functional derivartives:

$$F[\rho+\delta \rho]-F[\rho] = \int_\Omega \delta\rho \frac{\delta F[\rho]}{\delta\rho}\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{1}\label{FuncDer}$$

Lets now consider the Thomas-Fermi kinetic energy functional using the above equation.
The Thomas-Fermi functional is given as:

$$T_\mathrm{TF}[\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}\tag{2}$$

Lets now try to evaluate $T_\mathrm{TF}[\rho+\delta\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\left(\rho+\delta\rho\right)^{5/3}\mathrm{d}\mathbf{r}\tag{3}$$

Now lets do Taylor expansion of the integrand:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}+\frac{5}{3}\rho^{2/3}\delta\rho+\mathscr{O}(\delta \rho^2)\mathrm{d}\mathbf{r}\tag{4.1}$$
$$T_\mathrm{TF}[\rho+\delta\rho]= C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}+\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{4.2}$$

Now subtracting $T_\mathrm{TF}[\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] - T_\mathrm{TF}[\rho]=\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{5}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{TF}[\rho]}{\delta\rho}=C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\tag{6}$$

Now for the second part of the kinetic energy functional in your question, i.e. the Weizsacker kinetic energy functional:

$$T_\mathrm{W}[\rho]=C\int_\Omega\frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r}\tag{7}$$

Again lets evaluate $T_\mathrm{W}[\rho+\delta\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho]=C\int_\Omega\frac{\nabla(\rho+\delta\rho)\cdot\nabla(\rho+\delta\rho)}{\rho+\delta\rho}\mathrm{d}\mathbf{r}\tag{8}$$

Now doing a Taylor expansion again:

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho} + \frac{-2\nabla^2\rho+\frac{\nabla\rho\cdot\nabla\rho}{\rho}}{\rho}\delta\rho+\mathscr{O}(\delta\rho^2)\mathrm{d}\mathbf{r}\tag{9.1}$$

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r} + \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{9.2}$$

Now subtracting $T_\mathrm{W}[\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho] - T_\mathrm{W}[\rho] = \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{10}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{W}[\rho]}{\delta\rho} = C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\tag{11}$$

Now for the second part of your question.
For molecular calculations involving kinetic energy functionals you have to look into orbital-free density functional theory (OF-DFT).
The kinetic energy functionals are not used in Kohn-Sham density functional theory (KS-DFT).
In KS-DFT the kinetic energy are calculated from the Kohn-Sham orbitals instead of the density.

Often when working with OF-DFT the equations can be solved numerically using finte element method (FEM) for a nice introduction to this take a look at Joel Davidsson master thesis [1].
Keep in mind that doing OF-DFT calculations is not a trivial task!

[1] : https://www.diva-portal.org/smash/get/diva2:864857/FULLTEXT01.pdf


Why is sulfur hexafluoride more stable than selenium or tellurium hexafluoride? I was just reading about p Block elements from two different books. Both books say that $\ce{SF6}$ is extremely stable in 16th group due to steric reasons but my question is why is it more stable than $\ce{SeF6}$ and $\ce{TeF6}$ when $\ce{S}$ is more electronegative and smaller than $\ce{Te}$ and $\ce{Se}$ and so it would bring the electrons in bonding with fluorine closer to the central atom i.e $\ce{S}$ and thus it would bring the electrons closer and there would be more repulsion between the electron pairs.

Maybe this can be explained by Fajans rules that due to such conditions $\ce{SF6}$ has more covalent character than others and thus is more stable. But if this is the reason, where does "steric" come in here?

$\ce{SF6}$ is extremely stable for purely steric reasons, because S is completely blocked by fluorine atoms from all directions, so the reactions starting with an attack on S that otherwise would readily occur (hydrolysis, etc.) never have the chance to occur. This has nothing to do with electronegativity. For another similar example, look at $\ce{CCl4}$.

Now, the word "stable" is commonly used in two different senses. A compound may be stable by itself, i.e. not prone to spontaneous decomposition; in that sense $\ce{SF6}$, $\ce{SeF6}$, and $\ce{TeF6}$ are all totally stable under normal conditions, so there is nothing to compare. On the other hand, we may consider "stability" as inability to react with common reagents, e.g. water. In that sense $\ce{SF6}$ is indeed far more "stable" than the analogs, and that for the steric reasons (see above).

As for the thermal stability, that's another story altogether, so I don't really know how it would turn out. I wouldn't be surprised if $\ce{TeF6}$ would indeed happen to be the most stable of them all.


Does the dipole moment increase or decrease by increasing the bond length? It is established the dipole moment is a result of multiplication of the magnitude of charges (Q) and the distance between them (r). What I understand is that when an electron and a proton get closer (bond length decreases), polarity decreases hence the dipole decreases. When they move further away from each other(bond length increases), the polarity increases and hence the dipole moment increases. However, in case of halides (H-X), let's take for example HF and HI, HF has a bigger dipole moment than HI, it is stated that the bond length of HF is smaller than HI. Why do I find this contradictory?
Can you help me with What I am missing here?

Strictly speaking, you can neither decrease nor increase the bond length a slightest bit.$\dagger$ All molecules of a given compound are equal. It is not like you may grab one with tiny pincers of some sort, stretch it a little and see what happens.

You may, however, consider a molecule of a different compound; indeed, that's what you do in your question. But that's not just a different bond length; it is different atoms with different properties, and hence different everything.

Most importantly for the question at hand, iodine has much lower electronegativity than fluorine, so the effective charges on $\rm H$ and $\rm I$ in $\ce{HI}$ will be much lower than those on $\rm H$ and $\rm F$ in $\ce{HF}$. It just turns out that the increase in bond length can't compensate for this effect, hence the lower dipole moment.

$\dagger$ For those who might want to point out possible vibrational excitation or some other subtle (or not-so-subtle) effect: don't. Let's keep it simple.


Coloured compounds on reacting with oxygen In the reaction of chlorine gas with water, Hydrogen chloride and nascent oxygen is produced.The reaction of any coloured substance with this nascent oxygen makes it colourless.

Is it true?if yes, how?

The feature that makes colored compounds colored is the one that makes the compounds easily oxidized. When the compound is oxidized, something does not exist anymore. I'm not giving you a straight answer since you will experience a nice enlightenment when you figure that out. Take a look at the structure of beta-carotene, for example. Where would the oxygen go and what would the structure look like after that?


Possible to grow optical calcite? I notice that most optical calcite for sale seems to be from natural (mined) sources.

Also, I know in World War 2, mining optical calcite was a strategic objective.

Is there some reason why high-grade optical calcite cannot be lab grown?

The crystal structure is trigonal, so has very low symmetry and high chance of twinning or other line defects forming during, say, Czochralski type growth methods.  My guess would be the good optical quality stuff sat and annealed in situ for a long time to clear most of the twins/dislocations out.

Support for this comes from A.J. Gratz et al., Geochimica et Cosmochimica Acta 57 491-495 (1993) where they observed, with AFM, calcite growth occuring primarily on screw dislocations.


Are there any fluids which are both viscous and volatile? I'm looking for a viscous fluid which will evaporate (completely) rather quickly.

If propylene glycol evaporated like acetone, that would be the dream ticket, but neither the viscosity nor the volatilty need be quite that high.

Non-toxic would be a plus!

Volatility and viscosity are 2 properties going each against the other.

A volatile liquid has rather small molecules a/o  rather weak intermolecular bonds.

A viscous liquid has rather big molecules a/o  rather strong ( often multiple ) intermolecular bonds.

A partial solution to the task is to have composite liquid, a volatile liquid mixed with either viscous liquid ( e.g  glycerine),
either with dissolved thickener ( like polyvinylalcohol or vaseline/wax )


Which functional group is more polar in a drug: primary amide or a carboxylic acid? There is a lot of speculation as to which functional group is more polar, I've read a lot around this on the web and I can't decide which one is actually more polar? Can someone help because I need to know how to identify the most polar groups in drugs to create some drug monographs.

It depends on your exact definition of polarity
The measured dipole moments of formamide (3.7 D)1 and formic acid (1.4 D)2 in the gas phase indicate that the amide is intrinsically more polar.

However, this neglects additional factors relevant to the discussion: ionization and solvation. Under physiologic conditions, carboxylic acids are generally deprotonated resulting in a formal charge separation with a -1 charge delocalized across two oxygens. Depending on what parameters you are concerned with, this may be a more profound effect than the difference in gas phase polarity.


Roothaan matrix equation equation orbital coefficients To go from the Hartree_Fock

$f\psi{_i}=\epsilon\psi{_i}$

to the  Roothaan equation

$FC=SC\epsilon$.

equation we expand the orbitals as

$\psi{_i}=\sum C_{\mu i}\phi_i$

But for Helium atom we have just one orbital function so the coefficient $C_{\mu i}$ should be a vector not a matrix.
My question is  what is the interpretation of the matrix coefficient $C$ in the Roothaan equation  $FC=SC\epsilon$ for the helium atom since we've got just one spatial orbital function?

The $\mathbf{C}$ matrix will always be $K\times K$, where $K$ is the size of the finite basis set you are using. That is to say, with a basis set of size $K$, you will produce exactly $K$ MOs. This means you will never wind up with $\mathbf{C}$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $\mathbf{C}$ matrix is $2\times 2$. While helium has only one occupied spatial orbital, it will have $K-1$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.


Number of unpaired electron and total spin angular moment IIT-JEE advanced 2013 has a question:

  Give the correct order of the complex ions given below according to their spin-only magnetic moment:
  A. $\ce{[FeF6]^{3-}}$
  B. $\ce{[V(H2O)6]^{2+}}$
  C. $\ce{[Fe(H2O)6]^{2+}}$

So, my reasoning is more unpaired electron results in greater spin-only magnetic moment. So, the order should be B>A>C

But the answer is given A>C>B.

I don't know where I'm wrong.

The first choice lists $\ce{Fe^3+}$ as the central metal atom (CMA), so you have the configuration: $\mathrm{3d^5 4s^0}$. Remember that fluoride is a weak field ligand and would not cause pairing of electrons. This gives you 5 unpaired electrons.

The second choice has $\ce{V^2+}$ as the CMA, the configuration being $\mathrm{3d^3 4s^0}$. This gives you 3 unpaired electrons.

Finally, the third choice has $\ce{Fe^2+}$ as the CMA. Yet again water is a weak field ligand and does not cause pairing of electrons. The isolated configuration although is $\mathrm{3d^6 4s^0}$, but now you'd have 4 electrons in the $\mathrm{t_2g}$ set and 2 electrons in the $\mathrm{e_g}$ set. So now you've 4 unpaired electrons.

  more unpaired electron results in greater spin-only magnetic moment.

This is true and I can see it being followed as $\mathrm{A>C>B}$.

Also see my comment for the strength of ligands.