Bohr's model of an atom This question is not very mighty. But I'm having a problem wrapping my head around the Bohr's atomic structure. The textbook says electrons revolve around the nucleus in fixed energy shells, with each shell having its different value for energy. What I want to know is, what does it mean for a shell to have an energy level? Where does the electron get the energy to revolve around the nucleus from? And when the electron jumps from a ground state to other shell, after gaining energy, why does it have to return to its original shell? I mean it has satisfied the criteria of going into the next shell, so why does it return.

And my main concern is the energy part. Where does electron get the energy to revolve in a shell.

If possible, please refer me a source to study this in depth.

Think of the planetary system which inspired Bohr to think about his atomic model. The Earth is rotating around the Sun in a fixed orbit. What keeps the Earth rotating around the Sun? The mechanics of the planetary motion, and other electrical phenomena were very well understood in Bohr's time, so much so that by late 1880 to early 1900 a physics professor advised his student, who would become a Nobel laureate later, not to study physics because there is nothing left to do in physics

...here might still be a little dust or bubble at one or the other angle to check and classify, but the system as a whole is quite secure, and theoretical physics is noticeably closer to that degree of perfection that geometry has had for centuries..
"Max Planck: Wege zur Physikalischen Erkenntnis."

So classical electrodynamics (motion of charges) predicted in Bohr's time that a moving negatively charged body should spiral into the positively charged proton. Obviously this was not happening. Secondly, a moving charge should emit electromagnetic radiation because the great Maxwell had already established that.

Bohr had a moment of enlightenment. He just said let us say that the electron does not spiral into the nucleus but stays in a fixed orbit nor it emits radiation. Lo and behold, that assumption helped him explain most the hydrogen atomic spectrum in the UV, Visible, and in the near infrared. This was a big big feat! That is why when you read Bohr's theory, his assumptions are called Bohr's postulates.

What is a postulate in logic as per a dictionary's definition? "A fundamental principle, presupposition, or condition, esp. one assumed as the basis of a discipline or theory; (also) a proposition that is (or is claimed should be) taken as granted; esp. one (to be) used as a basis for reasoning or discussion, a premise."

"Taken for granted" is the keyword, which means when you are reading my theory, accept my assumptions that they are true-don't question them.


Is turning graphite carbon into diamond industrially doable? I've read somewhere on the Internet that heating graphite at 4200°K at a sufficiently high pressure would turn it into diamond.

This chart seems to support this idea:

Is that doable industrially?

It has been done since 1954 and has been a commercial success for some time. Many products now have diamond-like coatings, produced comparatively inexpensively through vapor deposition without need for high pressure.

BTW, in English "burning" means to set on fire, oxidize; I believe you mean "heating graphite", and, as your graph shows, pressure is needed as well as heat.


Can Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together? The following problem I have stated was not given to me as a homework. I have found this one in my textbook. I have solved it in my own way, but I'm not sure about the correctness of my procedure. I have posted to be sure about the procedure that I want to follow to solve such kind of problem.

  A $5~\mathrm{ml}$ sample of $0.2~\mathrm{M}\ \ce{CH3COOH}$ is titrated with $0.2~\mathrm{M}\ \ce{NaOH}$. Calculate the $\mathrm{pH}$ after adding $25~\mathrm{ml}$ volume of base has been added. Where $K_\mathrm{a} = 1.8 \times 10^{-5}$

By applying $v_1s_1=v_2s_2$ formula, I got the concentration of weak acid and it is $0.0333~\mathrm{M}$. Again by applying same formula, for the total volume of $30~\mathrm{ml}$, I got the concentration of base and it is $0.1667~\mathrm{M}$.

Since the concentration of base is more than that of acid, $0.0333~\mathrm{M}$ base will neutralize the $0.0333~\mathrm{M}$ weak acid and the concentration of remaining base will be $0.1334\ \mathrm{M}$.

Then I used this concentration to evaluate the $\mathrm{pOH}$ of this solution and it was $0.875$. So $\mathrm{pH}$ will be $13.125$.

Do you think that my procedure to solve this problem is correct? I am confused. Because my textbook says that $\mathrm{pH}$ will be $8.87$. Has the author of this book made an error in this case? I cannot use Henderson-Hasselbalch equation in this case because it is not a buffer solution. If weak acid ($\ce{CH3COOH}$) and a strong salt of that weak acid ($\ce{CH3COONa}$) stay together in a solution, we can say it to be a buffer solution. But in the problem I stated above we can see that no weak acid exists after reaction takes place, and strong base ($\ce{NaOH}$) and strong salt ($\ce{CH3COONa}$) stay together. We also know that strong base and strong salt of that base cannot form buffer solution.

Now I want suggestion from the experts of this forum. Is my procedure correct? If it is not, which procedure should be followed to solve this problem?

The pH of the solution after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ is $13.12$. Your book or solution manual is wrong. The $\mathrm{pH} = 8.87$ at the equivalence point.

The Henderson Hasselbalch equation only applies in the buffer region - the part of the titration curve before the equivalence point. I have worked out the pH for a number of volumes of titrant and graphed the curve. The buffer region is labeled. You can see that past the equivalence point, the pH jumps to above 12 and stays there. You determined the pH after 25 mL correctly.

There is a point where the $\mathrm{pH=8.87}$ - the equivalence point (labeled with its pH). After $\mathrm{5\ mL}$ of $\ce{NaoH}$ solution is added, all acetic acid has been converted into the acetate ion. We can solve an equilibrium problem to determine the $\mathrm{pOH}$ and $\mathrm{pH}$ at that point. The following equation applies:

$$\ce{CH3CO2- + H2O <=> CH3CO2H + OH-}\ \ \mathrm{K_b=10^{-9.26}}$$

Since all acetic acid was converted into acetate, and the volume has doubled, the initial concentration of acetate is $\mathrm{0.1\ M}$. The initial volume of acetic acid is $\mathrm{0\ M}$ and the initial concentration of hydroxide is $\mathrm{10^{-7}\ M}\approx 0$.

$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{CH3CO2-}] & [\ce{CH3CO2H}] & [\ce{OH-}] \\ \hline
I & 0.1 & 0 & 0\\
C & -x & +x & +x \\
E & 0.1 -x & +x & +x\\ \hline
\end{array}$$

If we plug in the values into the law of mass action, we can solve for $[\ce{OH}]$ and determine $\mathrm{pOH}$.

$$\begin{align}
K_b &= \dfrac{[\ce{CH3CO2H}][\ce{OH-}]}{[\ce{CH3CO2-}]}\\
10^{-9.26} &= \dfrac{x^2}{0.1-x}\end{align}$$

Solving for x gives

$$\begin{align}
[\ce{OH-}] &=x=7.41\times 10^{-6}\\
\mathrm{pOH} &=5.13 \\
\mathrm{pH} &=8.87
\end{align}$$

Even if I had not made the assumption that the initial hydroxide concentration was zero, I would get the same $\mathrm{pH}$.

  Your book has a typographical error. Either you were meant to determine the pH after $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ or the pH after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$. Both would yield a $\mathrm{pH} =8.87$ at the equivalence point.


What's a good resource/book on pH measurements? Most resources on pH and the electrochemistry behind it that I've found are IMHO too superficial. For example, they explain that an electrode with less than 80% slope should be discarded, but not why the decrease in slope happens, or why it's a problem if you can just calibrate it. Or they explain the potential generated at the Ag/AgCl eletrode and the glass membrane potential, but not the interaction between the two - does the H+ combine with the e- from the electrode? And what happens if the current is reversed? After reading for a few days I still feel like I don't understand any of it.

So far I've read a couple PDFs on nico2000.net (the 30 page Introduction to pH measurements), the 160 page book pH Measurements by C. Westcott (1978), a few manufacturer's guides to pH measurement theory and practice (by Radiohm, Emerson, Mettler) and the relevant chapters in the Analytical Chemistry books that my library had.

All help appreciated!

Looks like you understand pH measurement, but you are trying to go deeper into the details of potentiometry and open circuit potential measurements.

There I would recommend basic electrochemistry texts. Here is one simple book that is cheap, easy to read and very informational:

Electrode Potentials; Richard G. Compton and Giles H. W. Sanders


What is reactivity really, and can it be quantified? I understand the concept of a reactive species on a sort of vague and intuitive level: a reactive species is one that has lots of reactions that it can participate in. These tend to have large kinetic constants, be thermodynamically downhill and often have products that are also reactive, leading to knock-on effects. But reactivity isn't the same as energy, and it's possible to have highly reactive species with relatively low free energies and vice versa.

I'm interested in whether there's any kind of formal theory of reactivity, and/or a quantitative definition of the concept. For example, we're often told that free radicals are reactive because they have an unpaired electron and they "want" to fill their outer shell. This makes perfect sense on an intuitive level, but I'd like to know if it can be taken further. Is there some way we can write down an equation that will tell us how reactive a species will be in a given environment, or is there some way we can measure a species' reactivity empirically and use it to make predictions?

To clarify, I'm not looking for quantum-level predictions, although I will look into the theories mentioned in Sam and Greg's answers. Rather, I'm looking for something a bit more heuristic that will help me get a handle on what reactivity really is. In particular, what does it really mean to say that one species is more or less reactive than another? Can we put a number to it and say that a species has a reactivity of 8.2 on some meaningful scale?

ron's answer is a start. It surely must be the case the reactivity relates to the energy landscape. But ron's answer deals with a one-dimensional landscape, corresponding to a single reaction. Reactivity is a concept that involves many reactions, and so a satisfying theory of reactivity would have to involve the properties of a massively multidimensional landscape. If there is anything that has been written about it from that point of view I would be very happy to read it.

I think the term "reactivity" is often used to denote kinetic lability, a proclivity to react.  Let's look at the two reaction coordinates shown below.  They both represent the interconversion of molecules A and B through intermediate I.

Note that the potential well that I exists in is much deeper in the drawing on the left than the drawing on the right.  Therefor the activation energy for I to escape will be much larger for the intermediate on the left than the intermediate on the right.  The intermediate on the right would be said to be more reactive (have a shorter lifetime) than the intermediate on the left, because it requires less energy for it to escape.

An experimentalist can infer the depth of the well through experiments.  Through such experiments the experimentalist learns that free radicals, highly strained hydrocarbons, carbenes, etc. exist in potential wells that must be very shallow because they react so readily.  Using this empirical data the experimentalist can then infer how similar compounds are likely to react. A computational chemist can infer the same thing by calculating the profile of the reaction coordinate for the various reactions under investigation by the experimentalist, or as is becoming increasingly common the computational chemist can predict what should be and let the experimentalist confirm those predictions.


What is the difference between a glassy, amorphous and a vitreous phase? I have been reading about some inorganic materials lately, and these three terms confuse me. If they all mean the same thing, would still one of them have preference over others?

After some reading I have decided to answer this question myself for anyone else with the same confusion. The info is from the downvoted answer of alphonse and the wiki-link, but with a better overview.

Both vitreous and glassy are amorphous. Amorhous is the broadest term to rule them all.

Some amorphous materials have a Glass Transition Temperature (GTT). Above this temperature the material becomes a viscous liquid. note that both below and above the GTT the material is amorphous.

An amorphous solid that has a GTT is called a glass.
The transition (heating) over the GTT results in a glassy phase.
The reverse transition (cooling) corresponding to the GTT is called vitrifcation. The result is a vitreous phase.

If I made mistakes please comment and I will update the answer.


Why are the hydrogen-carbon bonds bent in a graphical depiction of an alkene, but are straight horizontally and vertically in an alkane? In propane, for example, the $\ce{C-H}$ bonds are depicted by straight vertical and horizontal lines. But in propene, the same bonds are bent.
Is it a rule that $\ce{C-H}$ bonds are always bent if the carbon atoms are double-bonded? The image below illustrates what I am asking.

Fundamentally, it depends on how you put a three-dimensional structure onto a 2D surface.
The Valence Shell Electron Repulsion (VSEPR) model can predict the structure of most molecules. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsion. So with the first example (propane), each bond is equally 90 degrees from each other, and in the second (propene), the bonds are 120 degrees from each other. If the bonds were placed 90 degrees away from one another, at least one gap between electron groups would be 180 degrees, thus a less stable arrangement of electrons, regardless of if it is a double bond (c=c) a or a single bond (c-c)


How to remove nitrate from water sample? Are there any simple methods to completely remove nitrate ions from few ml of impure water sample?

One method I am curious about is the following process:

Add one or more reagents to water in test tube, which would crystallize nitrate ions into some solids
Then filter out the solids from water such that the water is now nitrate free.

Is the above process possible? If so, what are some reagents that can be used?

Precipitation won't work

Nitrate removal is a very tough problem in wastewater treatment.  One of the reasons is that nitrate salts of nearly every common cation are highly water soluble.  Unfortunately, I can't think of a way to precipitate nitrate from your water sample.

How commercial nitrate removal processes work

Nitrate is removed from wastewater through a microbiological process called denitrification.  In this process, organic carbon substrates are used as food by special anaerobic bacteria.  In the absence of oxygen, the bacteria use nitrate as an electron acceptor so they can oxidize their food and obtain energy via respiration.  As an example organic food molecule, consider hexose sugar.  Then the reaction could be written as:

$$\ce{5 C6H12O6 + 24NO3- -> 30CO2 + 24 OH- + 18H2O + 12N2} $$

It will be difficult to purify a few mL of water in this way unless you know exactly the BOD of the water and are an experienced microbiologist with access to cultures of the appropriate bacteria.

Another not very good option

Another possibility would be electrochemical removal of the nitrate.  That is a topic which has been researched but which is not practices commercially to any significant extent.  Here is one paper where you can read about the technique.  I don't think it would be very easy for a home chemist either.

The tried and true

That leaves more tried and true methods.  You've simply got to distill your water!  Boil it and collect and condense the steam and you'll have substantially purified water to behold and enjoy!


How would someone in 1917 describe elements 43, 61, 72, 75, 84, 85, 87, 89, and 91? https://books.google.com/books?id=aaELAAAAYAAJ&printsec=frontcover#v=onepage&q&f=false Notice the list of elements on pages 62–63. It includes a short description of each element's "Nature under Ordinary Conditions." I am curious to know what the author might have written for elements 43, 61, 72, 75, 84, 85, 87, 89, and 91 (based on 1917 knowledge; for those which hadn't yet been discovered, some properties could be told from the element's position on the periodic table). Would people be willing to suggest period-appropriate descriptions for them?

Unknown elements were discussed according to Mendeleiev's nomenclature. The prefix eka- was used for the element just under a known element, and it was dvi- for the following one, two lines under. For example, the element 31 under aluminum was celled eka-aluminum, just before it was discovered as gallium. With this same technique, the element 43 was named eka-manganese before it was discovered as technetium, and element 75 was dvi-manganese before its discovery as rhenium.
In your list, the element 61 was not expected. The 72 was eka-zirconium. The radioactive elements (Z > 83) were not expected.
Ref. : M. Fontani, M. Costa, M. Orna, The Lost Elements, Oxford University Press 2015, 530 p., ISBN 978-0-19-938334-4.


Benzotriazole to o-phenylenediamine So I was reading benzotriazole's synthesis is o-phenylenediamine with acetic acid and sodium nitrite.

Is this a reversible process?

Not even specifically to o-phenylenediamine but any derivative.

phenylenedaimne's (usually para) and derivatives and generally aromatic amines form various colour dyes on old colour photographic films that I like to experiment with as a hobby.

Yes, a common synthesis of benzotriazole starts with o-phenylenediamine as shown in the following figure.

Image Source

The first step in the reaction involves straightforward diazotization of one of the amino groups (exactly like what's done in the first part of the Sandmeyer reaction) to produce the ortho-amino diazonium ion. Instead of eliminating nitrogen as in the Sandmeyer case, the molecule loses a proton and then the diazomium ion internally captures the ortho-imino group to yield the triazole product.

The reaction is not reversible, the benzotriazole is very stable and the intermediate diazonium ion is relatively high energy making the back reaction unlikely under normal conditions.  The reaction will also work with substituted o-phenylenediamines as long as the substituent doesn't interfere with the diazotization step.  So for example, alkyl- or halo-substituted o-phenylenediamines will produce the corresponding alkyl- or halo-benzotriazoles.

Edit: response to comment

Intramolecular cyclization to produce a triazole will only occur if the amino groups in the starting diamine are ortho to one another.  If m- or p-phenylenediamines or non-ortho naphthyl diamines are used, then products resulting from intermolecular azo coupling will occur. See here for some examples.


Why do all medicines require storage temperatures below 25 °C? Almost all medicines I buy have storage requirements saying $T\le25^\circ C$.

But do they really spoil at higher temperatures? What if they are stored at e.g. $30^\circ C$, which is common at summertime?

People do not realize that the quality requirements for a medicine is rather high. We are talking about a complex mixture of more or less reactive organic compounds that are potentially sensitive to air, moister and light. Any reaction of the original ingredient is potentially biological active and harmful, and even simple things like recrystallization or coagulation can destroy the effect of the original mixture.

The question "will they really spoil" makes no sense in the given context: if 99.999% of the cases not spoil, but there is a handful cases when things goes bad then the public goes banana. It took only one (purposefully false) paper to convince tens of thousands of people to go anti-vaccine, and decades till it dies out.

If it happens with a medicine, it potentially kills the product, so easier to err (by huge margin) on the safe side.


How do I convert zwitterions to neutral form in Avogadro? I'd like to be able to convert zwitterions to neutral form in Avogadro if possible. How do I do this?

I don't think the neutral protomer of an amino acid can be generated in Avogadro.  What can be done is adjusting the protonation state for different values of pH.  That option is in the Build menu - Build-> Add Hydrogens for pH.


Why is this not an isomer of pentane Given was the fact there are 3 isomers of pentane, two were given: pentane and 2-methyl butane. Then I was asked to draw to other one. I drew up the star like structure and checked it to be ok (coincidental I might add).

Then I drew some more and came up with amore turnstile shaped molecule.

     H
     |
   H-C-H   H   H
     |     |   |
   H-C-----C---C-H
     |     |   |
   H-C-H   H   H
     |
     H

To me this also seems like $\ce{C5H12}$ molecule and all valences seem ok. Why isn't this a valid isomer?

Edit: this is 2-methyl butane according to the book:

        H
        |
   H  H-C-H   H   H
   |    |     |   |
H--C----C-----C---C-H
   |    |     |   |
   H    H     H   H

To clarify Apeiron's answer. You have drawn 2-methyl butane, the same isomer that was given in the question. The rotation of a bond is probably the source of any confusion. A 3D model of your two structures would be the same.

To answer your question:

Neopentane is the third isomer of the pentane series.


What does delta letter mean in some ceramics chemical material formula? Here is a given chemical formula for ceramic material:
La0.6Sr0.4Fe0.8Ni0.2O3-δ

There is a letter δ. What does it stand for? I guess it is something about oxygen vacancies in crystal lattice.

Exactly. In your example, it is related to oxygen vacancy (oxygen deficiency) in the crystal lattice.
It can affect the physical properties of the ceramic material. Take the example of this piezoelectric ceramic material:$\ce{Eu_{0.5}Ba_{0.5}TiO_{3-\delta}} $ Oxygen-vacancy induces antiferromagnetism to ferromagnetism transformation.

In general, $\delta$ denotes deviation from the stoichiometry (exces or deficiency) in a chemical formula


Why does pH change with temperature? Why does pH change with temperature?  I recently read up on some chemistry notes, and found out that the higher the temperature of distilled water, the lower the pH.  Why?  Does this apply to other fluids too?  Does this also mean dipping a litmus paper into 2 beakers of water of different temperature would yield different results?

Water always contains a certain amount of $\ce{OH-}$ and $\ce{H3O+}$ ions. The pH is the negative decadic logarithm of the $\ce{H3O+}$ concentration (pH = - log(c($\ce{H3O+}$)). As chemical equilibria might change with temperature so does the concentration of $\ce{H3O+}$ and therefore the pH.

Although I never tried, I suppose this could affect measurements using pH paper. On the other hand the pH changes might very well be to small to be detectable using this method.

How this affects other fluids / equilibria strongly depends on how each equilibrium is affected by temperature. At higher temperatures equilibria are shifted in the endothermic direction and at lower temperatures they are shifted in the exothermic direction. You will need to look up the details for each equilibrium I suppose.


Do core electrons in bulk material have discrete energy levels or continuous energy band? The energy band of valence electrons gets smaller and smaller when the lattice (bulk) atoms are more and more separated from each other, until discrete energy levels are formed, due to the isolation of the atoms from each other (see image below).

What happens with energy levels of core electrons in the bulk, as the lattice constant is changed? Do they form a continuous energy band in the first place? Why?

(source)

If core electrons form a continuous energy band, how come K,L,M,... shells are differentiable in xray spectroscopy? In other words, if all electrons form one continuous band, how can xray have element specificity?

(source)

What happens with energy levels of core electrons in the bulk, as the
  lattice constant is changed? Do they form a continuous energy band in
  the first place? Why?

Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively. If you continue this chain extension and wind up with a very large polyene, the energy difference between all of the bonding (or antibonding) orbitals becomes smaller and smaller.  In addition to this compression between energy levels, each molecule is in a slightly different environment.  Some are near the surface, some deep in the bulk, some have a structural deviation or packing non-uniformity close by.  This serves to broaden the energy width of each orbital.  Between the energy broadening of an orbital and the compression of the energy separation between orbitals, one can imagine a limiting case where all of the bonding orbitals are so close to one another that one continuous band, the valence band, exists. The same would be true of the antibonding orbitals and they would merge into a continuous conduction band. The only remaining separation is the separation between the valence band and the conduction band.  This separation is referred to as the band gap.

In the real world, at least at normal pressures, this limiting case of a single band comprised of all bonding orbitals seems unlikely. Instead, some orbitals where the energy separation between them is large likely form separate bands.  In this case, the HOMO would still broaden significantly (as would the other molecular orbitals) and transform into the valence band, remaining distinct from the other bands at lower energies.

Still, it is difficult to probe the bulk and to understand exactly just what the situation is there. X-ray spectroscopy probes to a depth of only a micron or so, so it is really examining the surface states.  Certainly at least some of the molecules at the surface retain their individual orbital identity and are not part of the band structure.  Therefore it is not surprising that XPS (since it probes only the surface) can detect individual orbitals and serve to identify materials.


Are there any organic compounds that do NOT contain hydrogen ? And if so, what are they? Organic compounds are typically defined as compounds containing carbon and nearly all contain hydrogen as well. Which organic compounds do not contain hydrogen?

Distinguishing organic and inorganic substances is often very formal. We say CX4 are organic compounds, we should also say carbon does not form carbon halogenides, which are inorganic compounds.

One of more known halogen-hydrogen-free examples is carbon suboxide $\ce{C3O2}$,  $\ce{O=C=C=C=O}$, that is somehere on inorganic-organic border, being often considered as the second anhydride of malonic acide $\ce{HOOC-CH2-COOH}$, as it is prepared from it by $\ce{P4O10}$ aggressive dehydratation.

But the true malonic anhydride has 4-atom cycle.

One of less known examples is the anhydride of the mellitic acid, which could be formally considered as a carbon oxide with the summary formula $\ce{C12O9}$

  ... [mellitic anhydride] is one of the only four [oxides] that are reasonably stable under standard conditions.  It is a white sublimable solid, apparently obtained by Justus Liebig and Friedrich Wöhler in 1830 in their study of Mellite

As an interesting side info:

  Mellite, also called honeystone, is an unusual mineral being also an organic chemical.   Chemically identified as an aluminium salt of mellitic acid; that is, aluminium benzene hexacarboxylate hydrate, with the chemical formula $\ce{Al2C6(COO)6 \cdot 16H2O}$


Change in internal energy of a gas upon expansion against a constant external pressure I have a question about the following problem which I got wrong:

During expansion of its volume from $\mathrm{1.00\,L}$ to $\mathrm{10.00\,L}$ against a constant external pressure of $\mathrm{2.00\,atm}$, a gas absorbs $\mathrm{200\,J}$ of energy as heat. Calculate the change in internal energy of the gas.

First, I used the equation: change in $E = q + w$. I calculated work ($-P \Delta V$) and found it to be
$$-18 = -2 \cdot (10 - 1)$$. Then, I added $-18 + q$, the heat absorbed ($\mathrm{200\,J}$). This gave me a result of $\mathrm{-182\,J}$, which is wrong by a large factor. My textbook gives the answer of $\mathrm{-1623\,J}$, but I am not sure where I am going wrong...

Do you know where I made a mistake? Thank you!

You did everything correctly with the exception of converting the units of your result for work.

First, calculate the work in joules:

$$w = -p\Delta V = (-2\,\mathrm{atm}) \times 9\,\mathrm{L} =
\left(-18\,\mathrm{L\cdot atm}\right)
\left({101.325\,\mathrm{J}\over\mathrm{L\cdot atm}}\right) = -1823.85\,\mathrm{J}$$

Second, add the heat absorbed (already in joules):

$$\Delta E = 200\,\mathrm{J} - 1823.85\,\mathrm{J} = -1623.85\,\mathrm{J}$$

Third, always remember to check your units!


How are the van der waal constants 'a' and 'b' related to molecular size? 'a' represents the measure of attractive forces between the gas molecules so how is it related to the molecular size? if a1> a2 then what does it tell us about the molecular sizes of 1 and 2?

The real gas equation or van der Waals’ equation of state is
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT $$

where the constants a and b are known as van der Waals’ constants.

$a$ is a measure of van der waals forces of attraction existing between the molecules of a given gas. Greater the value of $a$ greater is the strength of the forces. It doesn’t affect molecular size.

On the other hand, $b$ is the incompressible volume per mole for a gas. It is directly proportional to the size of gas molecules.


Is there air in the volume of crystallized sodium hydroxide? I have a container of crystallized NaOH (98%) which has a volume of 500ml. Now, as per wikipedia, Sodium Hydroxide's density is 2.13g/cm3. This should give the weight of a 500ml container of NaOH at 1065g.

However, when measured, the crystalls weighed in at around 560g (minus the weight of the container). Is it because the NaOH is not in liquid or a homogenous crystal? Is the rest of the volume around the crystals air?

Imagine a container with sand; it may seem "full", but surely you may pour in some water, and it will fit in the cavities. Same thing here, except in case of $\ce{NaOH}$ I'd rather not check it directly with water, for the response may be quite intense.

$\ce{NaOH}$ commonly comes in granules; if you'd have one big solid brick inside your container, that would be another story.


Calculating pH of weak base and strong acid: Why is K(a) not required? Question: 35mL of 1.5M $\ce{HCN}$, a weak acid(K(a)= $6.2*10^{-10}$) is mixed with 25mL of 2.5M $\ce{KOH}$. Calculate the pH of final solution.

The solution given in my textbook is as follows:-

$\ce{KOH}$ completely dissociates to give 0.062 mol $\ce{OH-}$(The $\ce{K+}$ becomes irrelevant)
0.052 mol of $\ce{HCN}$

3.The equation is $\ce{HCN + OH- ⇌ CN- + H2O}$

So 0.062-0.052 mol=0.01 mol of $\ce{OH-}$ is left over
so 0.01 mol/(0.035l+0.025l)=0.17M of $\ce{OH-}$

6.pOH=0.76 so pH=13.24

7.Also, $\ce{CN-}$ is a weak base relative to $\ce{OH-}$ so we consider it negligible in the calculation.

Finally, it says that K(a) was not required and it was just to distract.

However, I think K(a) is actually required because $\ce{HCN + OH- ⇌ CN- + H2O}$ is a reversible reaction and I don't understand how all of the 0.052 mol of $\ce{HCN}$ can react with $\ce{OH-}$(and in turn, 0.01 mol of $\ce{OH-}$ is left) when there is also a backward reaction producing $\ce{OH-}$ and $\ce{HCN}$ back.

Please help.

Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water:

$$
\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}
$$
That is derived from this ideal equilibrium:
$$\ce{HCN ⇌ CN- + H+}$$

Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors.
Anything that will interfere with this equilibrium can cause problems to your assumptions.

In your specific case, the conditions are different there is an excess of $\ce{OH-}$ in the solution due to the strong base, so every $\ce{H+}$ produced will react with the $\ce{OH-}$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $\ce{HCN}$ from producing more $\ce{H+}$.

In simple words, if one reaction is favourable the solution doesn't care about your $K_{(a)}$ it will go following the most thermodynamic favourable path to the equilibrium.

Possible practical learning aim of this question?

So if you want to have high products amount (i.e. $\ce{CN-}$) and you have cheap reactant (i.e. $\ce{KOH}$) you can use an excess of reactant to shift the equilibrium where you want.

Bonus question: what can you do if you want to have HCN?


Do 1 mole of both dioxygen gas and its atoms occupy 22.4 litres at STP? The ideal gas law states:

  One mole of an ideal gas will occupy a volume of $\pu{22.4 liters}$ at STP.

Does this mean that both

$\pu{1 mole}$ of $\ce{O}$  would occupy $\pu{22.4 L}$ (or if this doesn't usually occur in nature, say \pu{1 mole }of $\ce{He}$ or any another monoatomic gas)
$\pu{1 mole}$ of $\ce{O2}$ would occupy $\pu{22.4 L}$

I am assuming the answer is yes based on:

reading the ideal gas law definition, it wouldn't make sense if it was true only for $\ce{He}$ but not for $\ce{O}$.
in class when we are finding the volume of e.g. $\pu{100 grams}$ of $\ce{Cl2}$, it was always $\frac{\pu{100 grams}}{(2\times\pu{35 grams} \text{[molar mass of $\ce{Cl2}$]}} \times\pu{22.4L}$  and not $\pu{1\times35 grams}$

But why? Wouldn't $\ce{O2}$ contain twice the amount of $\ce{O}$ matter and hence take up twice the volume?

Does this mean that both

1 mole of $\ce O$ would occupy $22.4~\mathrm L$ (or if this doesn't usually occur in nature, say 1 mole of $\ce{He}$ or another monoatomic gas)
1 mole of $\ce{O2}$ would occupy $22.4~\mathrm L$

Yes, it means exactly that.  And you're right, a stable gas of $\ce O$ atoms is a pretty exotic thing, so $\ce{He}$ is a much better example.

Wouldn't $\ce{O2}$ contain twice the amount of $\ce O$ matter

Yes and no. The $\ce{O2}$ gas would indeed contain twice the mass of the $\ce O$ gas, but it would contain the same number of particles ("molecular entities") as the $\ce O$ gas.

and hence take up twice the volume?

Nope.  There is a lot of empty space in gases. A lot. So much, in fact, that molecules have to get pretty darn large in the gas phase$^\dagger$ before they start to get "cramped" and "insist" on taking up more space (viz., occupying a larger volume).
To give you an idea of just how much room there is, consider what's called the van der Waals volume of $\ce{O2}$.  This is an estimate of the total volume actually occupied by one molecule of $\ce{O2}$, and has a value of about $0.053~\mathrm{nm}^3$ (that's 'cubic nanometers').  For one mole of $\ce{O2}$ molecules, that works out to just under $32~\mathrm{cm^3\over mol}$ of volume actually occupied by the oxygen molecules themselves.
However, as you note, that mole of $\ce{O2}$ gas actually expands to fill a full $22.4~\mathrm L$ of space at STP. So, that space is actually only about ${32~\mathrm{cm^3}\over 22.4~\mathrm{L}} = 0.14\%$ full of $\ce{O2}$.
So, even though one $\ce{O2}$ molecule does take up more space than one $\ce O$ atom, compared to the amount of space available, they're both really, really tiny.  And, thus, the fact that the two $\ce O$'s in $\ce{O2}$ are bound so closely to one another means that they both behave essentially identically in the pressure they exert when they fill up a container of a given size.
$^\dagger$ At ambient temperature and pressure, anyways.


Tool that generates electrostatic maps of molecules? Do you know any tools that generate electrostatic potential 2d maps of the molecules as an image?

Most computational chemistry software packages geared toward students/education should be able to do this. Examples include Scigress, formerly CAChe, Spartan, which has a 'lite/student' version that can probably do it, etc. Generally, any modeling software, for example, Gaussian or GAMESS should generate a file with the electrostatic potential map provided you give it the correct instructions. The output files of these methods can be visualized using many types of software (like Scigress and Spartan) and open source software like Avogadro.

If you are looking for an inexpensive solution, GAMESS is free for research/education and Avogadro is open source.


Is it possible to have a diatomic molecule of sodium in gaseous state? Already I know that hydrogen, all the halogens, nitrogen and oxygen forms diatomic molecules. But I am confused about $\ce{Na}$? So I would like to know about that.

According to molecular orbital theory, disodium should be stable in the gas phase, with a bond order of one. The molecular orbital diagram is the same for all the alkali metals since they all have one valence electron in an $s$ orbital.

(image source)

NIST chemistry webbook has a small page on disodium and quotes an enthalpy change of formation of $142.07~\mathrm{kJ~mol^{-1}}$ which is moderately endothermic.

This page has a brief discussion of alkali metal dimers and claims that they are all stable in the gas phase, in agreement with the prediction from MO theory.


Painting metal, does it matter if all rust is removed before? In the context of automobile repair work, assume a metal surface has rust on it. Does it matter if all of the rust is removed before the metal surface is painted? If it does matter, then why?

It appears that at least some will claim you can do this, e.g. the company Krylon.

How I've tried to figure this out myself:

I'm assuming that the rust-covered surface area will affect how well the paint will stick to the surface. Maybe this alone can be a problem in terms of durability and sealing the metal from the atmosphere?

If it is governed by a more complex process, what process(es) should be read into further? However, I also assume that (as the aforementioned link suggests) that making the surface as smooth as possible with loose rust and debris removed will yield better results -- i.e. results that approximate that of having no rust at all there? Will it never be as good as simply having all rust removed? I know I'm asking in a way that is difficult to quantify, but "how good" will it get, compared to having all the rust removed?

After studying the Wikipedia article on rust I also imagine that from the dehydration equilibria we have that even a small area of rust can:
$$\ce{Fe(OH)2 <=> FeO + H2O}$$

Even assuming the rust is properly sealed from the atmosphere it appears it can create new $\ce{H2O}$ under this protective layer.

But we need oxygen for the key reaction to occur:
$$\ce{O2 + 4e^- + 2H2O -> 4OH^-}$$

From this how stuff works I have:

  As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.

How exactly is the water broken down into hydrogen and oxygen? From a chemistry textbook (Masterton & Hurley) I have:

  For example, water does not spontaneously decompose to the elements by the reverse of the reaction referred to above
  $$\ce{2H2O(l) -> 2H2(g) + O2(g)}$$ nonspontaneous.

But electrolysis can certainly create these elements. But can it happen in this environment? Sealed under a surface of paint? What does the chemistry look like?

From chemical formula I have that we can get hydroxide from:
$$\ce{4e^- + 2H2O(l) + O2(g) -> 4OH^-(aq)}$$

However still, from where does new oxygen get introduced into the system, assuming the atmosphere is actually completely isolated from this internal system taking place under the surface of the paint?

Should it matter, assume the environment is approximately 1-atmosphere pressure and that the temperature varies between -10 to 40 degrees Celsius.

How exactly is the water broken down into hydrogen and oxygen?

As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.

That's extreme oversimplification. The reaction common for active metals (not including iron) is
$$\ce{Mg + 2 H2O ->  Mg(OH)2 + H2 ^ }$$
Iron does not react this way. However, it does react with acids, say, with hydrochloric acids
$$\ce{Fe + 2 HCl(aq) -> FeCl2(aq) + H2 ^}$$
The reaction is quite slow even for concentrated nonoxidizing acids. However, in presence of oxygen and in diluted acids two subsequent reactions occur, the one noted above is followed by
$$\ce{4FeCl2 + 10H2O + O2 = 4Fe(OH)3 v + 8HCl }$$
The second reaction effectively removes iron from solution, significantly fastening the overall reaction.  However, even without acid present, even in slightly basic conditions iron slowly reacts with water in presence of free oxygen, forming rust. The process is hindered only at pH above 10 or at absence of free oxygen. Since natural water is usually slightly acidic due to presence of dissolved carbon dioxide, iron always rust on contact with water and atmosphere. This, actually, is a reason why concrete layer over steel frame is regulated: concrete always contains water, and surface layers quickly looses basicity due to reaction with atmospheric carbon dioxide, so steel near surface of concrete quickly rusts, increasing in volume and tearing the concrete from inside.

In the context of automobile repair work, assume a metal surface has
rust over it. Does it matter if all of the rust is removed before the
metal surface is painted? If it does matter, then why?

Yes, it does. Rust is weak, so any paint over it will hold weak. As long as the paint film over rusty part is broken, water and atmosphere contacts the metal and start to erode it, resulting in more paint film be destroyed. So it is critical to remove all rust from the surface being painted. Abrasive blasting, if you have the equipment, is probably the best way. While some paints may be applied over rust, it should be a very thin layer, and even then I would prefer traditional paint over cleaned surface anyway.


What does the number of orbitals in a subshell represent? s subshell has 1 orbital and p subshell has 3 orbitals. Does that mean that there is a high probability that the electron is in the p subshell? As orbital are the space where there is a high probability of finding electrons.

Short answer

Number of orbitals does not have direct relation to either probability of  electrons occupying them, either the space they are describing.

The related probabibility is to be understood as:

IF the electron is in the quantum state, belonging to the particular orbital, THEN the wave function of the orbital describes spatial distribution of the electron  occurence probability, what is geometrically illustrated by the orbital 3D shape.

Longer answer
There are 3 meanings of the term orbital:

A complex wave function $\Psi$ being a particular solution of the quantum wave equation
A quantum state of an electron, following 1., described by particular quantum numbers $n$, $l$ and $m$, representing discrete values of electron energy, orbital angular momentum and one of component of angular momentum, respectively.
3D shape, following 1. and 2., representing statistical  probability of the electron presence at given point by$\Psi^2$, determined by the conventional value by residual external probability and by the isoprobability surface.

If there is 1 orbital 2s and 3 orbitals 2p, it means,
for  $n=2$  and $l=0$, there is 1 value $m=0$.
But for $n=2$  and $l=1$,
there are 3 values $m=-1,0,+1$

Electrons are fermions with half integer 4th spin quantum number $m_s$, and every electron in the atom must have unique set of quantum numbers.

Therefore there can be 2 electrons in the orbital 2s, having $n=2$, $l=0$.

But there can be up to 6 electrons in 3 2p orbitals, having $n=2$, $l=1$.


Net direction of dipole moment in oxygen difluoride What is the net direction of dipole moment in $\ce{F2O}$ molecule?
In the case of $\ce{NH3}$ and $\ce{NF3}$ molecules both have net dipole moment with direction reversed due to strong electronegativity of fluorine atoms, whereas in the case of $\ce{H2O}$ the direction of net dipole moment is from oxygen atom towards lone pair with a value $\mu(\ce{H_2O}) = \pu{1.84 D}.$
But in my textbook the value of net dipole moment of $\ce{F_2O}$ molecule is given $(\mu(\ce{F2O}) = \pu{0.3 D}),$ but not its direction.
I have done a lot of research on google but I could not find any related information.

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D)
On the other hand,

In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole (0.3 D)
So, basically, the dipole moment is towards fluorine atoms (which are more electronegative than oxygen atom). The reason why H2O has a higher dipole moment is that the difference between the electronegativity of O and H is higher than that of O and F, so water has a higher dipole moment even tho both water and oxygen difluoride have a similar bent molecular geometry.


Steady State Approximation final integral proof $\frac{d[P]}{d t}=k_{a}[A]$ yielding $[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$ I want to ask a question about the Steady State Approximation integral.

I understand the steady state integral works in Leyman's terms as the intermediate is formed as fast as it is consumed so the intermediate's concentration is really low and nearly always the same.

If I had a reaction:

$$\ce{A->I->P}$$

Then essentially,

$$\frac{d[I]}{d t} \approx 0$$

and therefore using this information and previous knowledge of rate

$$\begin{array}{l}{\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I]} \\ {\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I] \approx 0}\end{array}$$

which results after rearrangement:

$$[I] \approx \frac{k_{a}[A]}{k_{b}}$$

and hence $$\frac{d[P]}{d t}=k_{b}[I] \approx k_{b} \frac{k_{a}[A]}{k_{b}} \approx k_{a}[A]$$

The trouble is, I couldn't understand how:

$$\frac{d[P]}{d t}=k_{a}[A]$$

can be solved to give

$$[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$$

I recognise from previous work that:

$$\frac{d[A]}{d t}=-k_{a}[A]$$ gives

$$A =[A]_{0}e^{-k_{a} t}$$

from my final year at school, but I couldn't seem to apply this understanding to the final step of the integral above.

How can this be shown?

There are a couple of ways to approach this problem. Bear in mind this problem requires some calculus.

$\boldsymbol{1^{st}}$ Way (repeated integration): To start, you were able to recognize the first order integrated rate law.

$$\frac{d[A]}{dt} = -k_a[A] \implies [A] = [A]_0e^{-k_at}$$

This result may be plugged into the rate equation for $P$ to obtain

$$\frac{d[P]}{dt} = k_a[A] = k_a[A]_0e^{-k_at}$$

This equation is directly integrable.

$$\int_0^t\frac{d[P]}{dt}\,dt = \int_0^tk_a[A]_0e^{-k_at}\,dt$$
$$[P] - [P]_0 = -\frac{k_a[A]_0}{k_a}\left(e^{-k_at} - 1\right)$$
$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$

The last step applies the knowledge from the problem that $[P]_0=0$.

$\boldsymbol{2^{nd}}$ Way (conservation law): orthocrestol in the comments above pointed out that the sum of the concentrations of all chemical species should equal to some constant in this problem. Call this constant $C$ and ignore $I$ due to the steady state approximation. Thus,

$$[A] + [P] = C$$

Now substitute this into your rate law for $P$.

$$\frac{d[P]}{dt} = k_a[A] = k_a(C - [P])$$

This differential equation is separable so we can use the neat tricks to integrate this equation.

$$\int_0^{[P]}\frac{d[P]}{C-[P]} = \int_0^tk_a\,dt$$
$$-\ln\left(1-\frac{[P]}{C}\right) = k_at$$
$$[P] = C\left(1-e^{-k_at}\right)$$

Almost done. Now recognize that at time zero, we can write the conservation law as

$$C = [A]_0 + [P]_0 = [A]_0 + 0 = [A]_0$$

This makes physical sense because since $C$ is a constant, this value cannot change at any time. Moreover, the total number of molecules at any time should be the same as the amount of molecules that you start with due to stoichiometry (everything is 1:1 in this particular problem). So make the substitution to get your answer.

$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$


Acidic strength order in gaseous phase I got this question in my exam:

Arrange the following in increasing order of acidic strength in gaseous phase
(i) $\ce{CH3CH2COOH}$ (ii) $\ce{CH3COOH}$

I felt that increasing order must be (i) $<$ (ii). But answer given is (ii) $<$ (i). Why will the order change in gaseous phase?

Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists.

In aqueous solution, $\mathrm{p}K_{\mathrm{a}(\ce{CH3CO2H})} = 4.76$ while  $\mathrm{p}K_{\mathrm{a}(\ce{CH3CH2CO2H})} = 4.87$, thus acidity of acetic acid is greater than that of propanoic acid. Keep in mind that solvent plays a huge role in $\mathrm{p}K_{\mathrm{a}}$ measurements. For example, compare $\mathrm{p}K_{\mathrm{a}}$ measurements in $\ce{H2O}$ and in $\ce{DMSO}$:

$$
\begin{array}{c|cc}
\hline
\ce{HA} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{H2O} & \mathrm{p}K_{\mathrm{a}} \text{ in } \ce{DMSO}  \\
\hline
\ce{CH3CO2H} & 4.76 & 12.3  \\
\ce{H2O} & 15.7 & 31.2 \\
\ce{CH3OH} & 15.5 & 27.9 \\
\ce{C6H5OH} & 9.95 & 18.0 \\
\hline
\end{array}
$$

However, the proton affinity (PA) measurements of the conjugate bases of these acids in gas phase tell a different story. Proton affinity is a thermodynamic measurement conducted in the gas phase. These measurements give a pure view of base/acid strength, without complications from solvent effects such as salvation, thus take the solvent out of the equation.

Cardwell, et al. (Ref.1) has measured the proton affinities of various $\ce{R-CO2H}$ acids, including ethanoic and propanoic acids ($\ce{R}=\ce{CH3}$ and $\ce{CH3CH2}$, respectively). They conclude that:

  The acidity increment (with the size of $\ce{R}$-group) is relatively small and can be attributed to the stabilizing effect of the polarizability of the alkyl group on the negative ion.

References:

Gary Caldwell, Richard Renneboog, Paul Kebarle, “Gas phase acidities of aliphatic carboxylic acids, based on measurements of proton transfer equilibria,” Canadian Journal of Chemistry 1989, 67(9), 611-618 (https://doi.org/10.1139/v89-092).
Also read: Diethard K. Bohme, Edward Lee-Ruff, L. Brewster Young, “Acidity order of selected Brønsted acids in the gas phase of $\pu{300  K}$,” Journal of American Chemical Society 1972, 94(15), 5153-5159 (https://doi.org/10.1021/ja00770a002) and references there in.


Why are nitro compounds with at least one alpha-H soluble in NaOH? Are all the nitro-compounds having at least one alpha-$\ce{H}$ soluble in $\ce{NaOH}$? And what does solubility has to do with alpha-$\ce{H}$?

I was reading about tautomerism today and got stuck on this thing. Is it due to its conversion to the enol form? Does the keto form also show this? What could be the possible reasons?

The compounds containing nitro groups when dissolved in basic medium, the following things can happen.The base can take the $\ce{$\alpha$-H}$ of the nitro-group, which will create a negative charge on that $\ce{$\alpha$-C}$, which can further delocalise with the Nitro-group (similar to Nitro-Acinitro tautomerism).
So, the whole compound will be in the state of a Resonance Hybrid where there will be partial double bond character between $\ce{$\alpha$-C, N }$ and two $\ce O$ atoms. and also there will be a $\delta$- charge on the $\ce{$\alpha$-C }$ atom as well as on the $O$ atoms. , and $\delta$+ charge on the $N$ atoms.Due to such salt like compound formation, it becomes soluble. This $\delta$- charge created on the Carbon atom is stable if the side chain is short, but starts to get destabilised as the side chain grows longer. So, I think the solubilities of the Nitro-compounds decrease as the side chain grows longer, and the compound with $10-12$ C atoms may not be at all soluble in base because of such large Hydrophobic part and destabilisation of the $\delta$- charge.


Does the prefix Ortho have some significance? Are orthoboric acid and boric acid one and the same thing or we can not write H3BO3 as B(OH)3?

The reference given by Andselisk does not bring an answer to the question about the origin of the prefix "ortho" in inorganic nomenclature.

The prefix "ortho" is added before the name of an acid $H_aX_bO_c$ if first $b = 1$ and if it corresponds to the highest possible number of $OH$ groups attached to the central X atom. For example $H_4SiO_4$ is the orthosilicic acid, $H_3BO_3$ is the orthoboric acid. $H_4CO_4$ is the orthocarbonic acid (which only exists in esters). $H_3PO_4$ is the orthophosphoric acid (where the prefix is often skipped). $H_3AsO_4$ is the orthoarsenic acid. The use of the prefix does not depend on the oxidation number of $X$. For example, if $P$ and $As$ are taken with a lower oxidation number, one obtains $H_3PO_3$ as the orthophosphorous acid, and $H_3AsO_3$ as the orthoarsenous acid.

The prefix "meta" is added to the name of an acid, if its formula is formed by removing one molecule $H_2O$ from the formula of the ortho-acid. $H_3BO_3$ - $H_2O$ = $HBO_2$ and this is the metaboric acid. $H_3PO_4$ - $H_2O$ = $HPO_3$ which  is the metaphosphoric acid. Repeating this process, one gets :$H_2CO_3$ is the metacarbonic acid (often written without the prefix meta). $H_2SiO_3$ is the metasilicic acid. As with ortho-acids, the use of "meta-"  does not depend of the oxidation number. $HPO_2$ is the metaphosphorous acid. $HAsO_2$ is the metaarsenous acid.

It should also be mentioned that removing one $H_2O$ out of two molecules of an ortho- acid gives a pyro- acid. Example :
$2 H_3PO_4 - H_2O = H_4P_2O_7$ and this is the pyrophosphoric acid. $H_4As_2O_7$ is the pyroarsenic acid. This prefix pyro- is even used for acids existing only under the ortho- definition. For example, a pyro derivate from the sulfuric and sulfurous acids $H_2SO_4$  and $H_2SO_3$ exist. They are the pyrosulfuric acid $H_2S_2O_7$. and the pyrosulfurous acid $H_2S_2O_5$

It should also be mentioned that the prefixes ortho-, meta-, and pyro- are never used with acids having already a prefix like hypo- and per-.

All these rules have been edited in the middle of the 19th century.


What is Drago's rule? Does it really exist? My textbooks states the Drago's rule in inorganic chemistry as follows:

  The more electronegative atom prefers the orbital having more p
  character, and lone-pairs or less electronegative elements prefer such
  orbitals, as have more s character.

  Now, according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$,
  $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

  So, in $\mathrm{sp^3d}$ more electronegative elements like Fluorine go
  to axial position, but such elements go to equatorial position in
  $\mathrm{sp^3d^2}$ or $\mathrm{sp^3d^3}$.

  Also, as a result of Drago's effect when atoms of group $14, 15$ and
  $16$ (belonging to third period and above) are attached to atoms having Electronegativity less than
  $2.5$ then it has valence electrons in pure s-orbitals. And apparently
  no hybridization effect is observed in such cases. That is why in
  $\ce{NH3}$ the bond angles are $107.8^\circ$ but for
  $\ce{PH3},\ce{AsH3},\ce{SbH3},\ce{BiH3}$ the bond angles are very
  close to $90^\circ$.

I couldn't make much sense out of the above rule. My questions are as follows:

What does the following part mean?

  according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$,
  $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

Does it mean that the hybridized orbitals split in this way? If so, why should they split?
Why would the more electronegative atom prefer the orbital with more p character?
Even if Drago's rule holds, why is the statement in the paragraph beginning with "Also, as a result of Drago's effect when atoms …" true? I could not understand why the same won't occur for atoms with electronegativity $>2.5$.

I could not find much information regarding Drago's rule on the internet. Not even on Wikipedia. One of my acquaintances (who is a chemist) said that no such rule exists. But I didn't get much time to discuss the matter with him.

So does such a rule really exist? And, in that case, how can we answer the above two questions given as bullets  using the existing theories in chemistry?

Going through what you posted, I think ‘Drago’s rule’ (which I never encountered, either at school or at university) gives good predictions but uses a largely terrible set of arguments.

For any atomic system with more than one electron, quantum chemistry predicts the energies of the $\mathrm{s}$ and $\mathrm{p}$ subshells of a shell to differ — while for hydrogen-like systems (one-electron systems) all subshells of a given shell have the same energy. The $\mathrm{s}$ subshell of a certain shell in multi-electron systems has a lower energy than the $\mathrm{p}$ subshell, which in turn has a (far) lower energy than the $\mathrm{d}$ subshell and so on and so forth. Since $\mathrm{s}$ subshell is determined to be more stable by this method, it is not surprising that the first two electrons of a given shell are added to said $\mathrm{s}$ subshell; the next six electrons are added to $\mathrm{s}$ subshells whereafter a new shell is used.

This immediately explains that it is most stable for any atom to treat the $\mathrm{s}$ electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the $\mathrm{s}$ electrons to be energetically favourable with respect to any configuration that does a priori. (You can consider that part of the quoted statement to be correct.) Therefore, the a priori predicted most stable configurations of certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 $\mathrm{p}$ orbitals to form bonds; keep one lone pair in an $\mathrm{s}$ type orbital. (Predicted bond angle: $90^\circ$.)
tetracoordination (no additional lone pairs): since the former is not possible, hybridise $\mathrm{s}$ and $\mathrm{s}$ orbitals to form $\mathrm{sp^3}$ hybrid orbitals. (Predicted bond angle: $109.5^\circ$.) Actually, in MO terms you should consider the four ligands to supply four group orbitals whose symmetry always fits one of the three central atom’s orbitals, resulting in one lower bond energy and one higher bond energy corresponding to triple-degenerate bonds. The $\mathrm{sp^3}$ description is mathematically equivalent, though.
penta- and hexacoordination; tetracoordination with additioal lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an $\mathrm{s}$ orbital if possible. Use remaining $\mathrm{p}$ orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^\circ$ going from 2e2c bonds to 4e3c bonds; $180^\circ$ between a pair of coordinating atoms contributing to the same 4e3c bond.)

Using the aforementioned 4e3c bonds nicely explains why electronegative atoms prefer to occupy these positions. You should always remember that 4e3c bonds can be described by the following resonance structures (drawn for $\ce{ClF3}$):
$$\ce{F-Cl^+\bond{...}F- <-> F^-\bond{...}Cl^+-F}$$
In this resonance, each outer atom taking part in a 4e3c bond has a formal average charge of $-0.5$; it is generally better to assign charges to electronegative atoms.

The theories I have outlined so far do not adequately predict the bond angles of water, ammonia and related compounds. Indeed, these compounds are remarkably complicated although they seem simple. The gist, however, is that a bond angle of $90^\circ$, while being most favourable on a pure orbital basis, introduces steric strain between the outer atoms which approach each other too closely if the central atom is relatively small. To combat this destabilisation, the bond angle is extended by mixing $\mathrm{s}$-contribution into the bonding orbitals — which in turn means that the lone pair will be less $\mathrm{s}$-like. The ideal bond angle for dicoordinated systems based on sterics is $180^\circ$, showing that the electronic contribution ($90^\circ$) is more important than the steric one ($180^\circ$). Larger central atoms than oxygen and nitrogen, such as phosphorus, antimony, selenium, etc., allow for smaller bond angles since the bond lengths are larger increasing the spacing between the ligands.

The reason for the special behaviour of nitrogen and oxygen is not their high electronegativity but their small size. Chlorine, which has an electronegativity similar to nitrogen, behaves as predicted for large atoms.

It has been commented that you should not invoke $\mathrm{d}$ orbital participation for main group elements. Remember what I said in the second paragraph: a shell’s $\mathrm{d}$ subshell has a higher energy than the next shell’s s subshell. Thus, before any same-shell $\mathrm{d}$ subshell is added to hybridisation, the next shell’s $\mathrm{s}$ subshell should be. Since hybridisations involving two different shells’ $\mathrm{s}$ subshells are not observed, we should reject the possibility of using $\mathrm{d}$ orbitals for hybridisation. Therefore, we can also flat-out reject the central two paragraphs of your quotation.


Why 4s lose electron to 3d? In chromium and copper, the are the configuration fill up like
$$\mathrm{3d^5  4s^1}$$
but not $\mathrm{4s}$ first as it has lower energy level. and why only these 2 elements.

In any element after Argon [1s22s22p63s23p6],
the repulsion force between the elctrons increase at a great extent and also the atom becomes very stable. So the elctrons tend to move in far 4s orbital first and after filling it completely , fill the 3d orbital. About chromium and copper, they are exceptions. Any half filled or fully-filled orbitals are more stable then rest of the configurations, so in Chromium one electron of 4s comes to 3d orbital and a stable 3d5 is obtained, similarly in Copper one electron from 4s comes to 3d orbital and forms stable 3d10.

I hope this answers your question,
If you have any further doubts you can ask me :)


Microscopic and mechanical approach to boiling point I am confused as to how I should visualize the boiling process throughout a liquid.

From my understanding, the boiling point of a liquid is when its vapor pressure reaches atmospheric pressure. This makes sense at the surface of the liquid where the pressure of the atmosphere no longer overcomes the pressure of the gas molecules escaping the surface of the liquid, so in some sense it can be thought of, on a conceptual level, as becoming a mixture of "air" molecules and "liquid" molecules.

BUT, how does this account for the formation of bubbles on the bottom of container holding the liquid? Wouldn't the gas that is trapped in these crevices need to overcome to pressure of the atmosphere AND the pressure of the liquid itself (looking at it from a physics perspective as well)? How would this phenomena be explained using vapor pressures?

Furthermore, when gas bubbles form within the liquid, should I think of it as the kinetic energy of those particular liquid molecules are so great that it's "pushing" the surrounding liquid molecules with less kinetic energy?

EDIT: Part of my confusion was that I wasn't sure what the composition of these gas bubbles would be exactly—and yes the assumption is that it is being heated from the bottom. Although now I'm curious what would happen if it was heated uniformly somehow.

Let’s ignore the air, first of all, and put it back later.

When you heat the bottom of the saucepan, the first thing that happens to the water at the top is nothing. It stays cold, because water is not very conductive.

The next thing that happens is that the water at the bottom becomes so hot that it is less dense than the water above it. The first thing that happens then is that nothing happens; the second thing is that the situation is unstable and the tiniest irregularity in the layers makes a blob of less dense water float upwards. The colder denser water sinks to the bottom to replace it, and then is heated in its turn.

(The long thin tube mentioned in the link finds it far harder to convect in this way, which is why the boiling, when it happens, is so explosive).

The rising blob of hotter water can cool by mixing with the water it meets on the way up. Mixing is more or less efficient depending on various things (such as how big the difference of temperature is, and so how buoyant the blob is), so the blob may still be travelling upwards when it hits the surface. Look sideways at a saucepan near to boiling and you will see this.

The boiling point of water is higher at the bottom than at the top, so at a certain moment you will have a blob that is below boiling point when it leaves the bottom finding itself above boiling point when it gets higher up. So boiling can start above the bottom of the vessel in that case; and indeed once a saucepan starts boiling, you see it boiling at the surface and only then does the boiling spread throughout the liquid.

Alternatively the blob may boil on the way up, and the bubble of steam then float up into the cooler layers and condense suddenly. This is the noise a kettle makes when it is near to boiling, and why it goes suddenly quieter when it really starts to boil.

An extra detail: water heated a little above boiling point does not boil. It stays a superheated until some irregularity or impurity acts as a nucleus for the change of state. This is why if you get a saucepan at the late “hot convection” stage and throw salt into it, it fizzes violently. The salt nucleates the superheated water and a bubble of steam forms round each grain.

Now air. Air is less soluble in hot water than in cold, so as the water is heated the air comes out of solution. First, bubbles form nucleated by any roughness of the surface of the base of the pan (if your pan is too shiny, scratch it). Eventually they grow big enough to detach themselves and float to the surface. The air bubbles stir up the liquid they float up through, so that convection happens quicker and in smaller blobs. That is why when you boil already-boiled water the second time, it makes such explosive noises: you get much bigger blobs (because no air bubbles intervene) making bigger steam bubbles which then implode more violently.

Finally, not heating from the bottom. If you grill water, it doesn’t boil at all, it evaporates in a quiet and genteel way. If you microwave water then there is little or no convection and most of it is superheated. This when you take your mug out of the microwave, it is still and innocent and unboiling. When you put a tea-bag into it then its rough surface nucleates the boiling everywhere (like the thrown salt) and you get an eyeful of boiling water.


Is there a difference between negative/positive controls and control/experimental groups? First of all, just so I can fully understand, what is the correct definition of a control regarding science experiments?

Is there a difference between the definitions of a control/experiment group and negative/positive controls?

I know that the control group is the group that will not be tested on while the experimental group is the group that will be experimented on, but that seems very similar to negative and positive control.

Is it true that negative controls are the controls in the control group while positive controls are the controls in the experiment group?

Consider a chemical test, such as a test for the presence of lead in drinking water. You would use both experimental and control samples. Whenever you run the test on experimental samples, you also need to run it on both positive and negative control samples--positive samples are prepared deliberately with a certain known concentration of lead, and negative samples are known not to contain any lead (you have to be careful here!). If you get the incorrect results for the control samples, then you know that the results for the experimental samples may be off, so you have to repeat the test. You also might want to periodically have the analyst run a proficiency test with the standard positive and negative controls as well as samples containing various known amounts of lead.

Experimental and control groups might more typically refer to situations such as drug trials, where you are comparing a new drug to another drug that is already approved and used to treat an illness. The experimental group would be the group of subjects having the illness that are given the new drug and the control group would also have the illness but would be given the approved drug (preferably in a double-blind study design--where neither the subjects nor the people administering the drug know which treatment is which). Note that both the experimental group and control group are "experimented on" - otherwise they could tell whether they received the treatment or not, and the results could be erroneous due to the placebo effect.


Phenol smell from hands It is a question which I faced when I was in Senior year in high school, I used my hands to handle phenol which has turned pink due to oxidation, I washed my hands with all kind of soaps but smell in hands didn't go for 3-4 days. Why was it, did it bond to my skin? What I could have done to remove smell?

As an acidic compound, phenol started to degrade the protective layers of your skin, and partially diffused into deeper layers.  This, and because of the moderate solubility in water are the reason the scent of phenol was still recognisable after some time.  While phenol is better soluble in ethanol and methanol, one refrains from these solvents to clean the skin because these actually would help phenol to reach even deeper tissues.
 Your writing that it was recognisable even after 3-4 days, even after extensive rinsing with water and soap, hints to an extended exposure of the skin towards pure phenol, which however should be avoided.

Not visible to human eye, but thanks to antiseptic power, it killed some germs, too.  Earlier in times, it was known and used as carbolic acid.


Thermal equilibrium in reversible processes In whatever resources I have consulted to study reversible thermodynamic processes, a common statement seems to be :

The system and surroundings (if the boundary allows) are in mechanical ($P_{sys}=P_{surr})$ and thermal $(T_{sys}=T_{surr})$ equilibrium at all instants.

However, in various answers on the Chemistry stack exchange, and in physical chemistry by peter atkins, it is claimed that the "surrounding" is an infinite reservoir: so the temperature can be assumed to be constant.
1.Is this not contradictory to the "thermal equilibrium" I mentioned in the beginning?
2.Thermal equilibrium of a system with its surroundings means that the entropy change of the surroundings , should be given by $\int\dfrac{-dq_{sys}}{T_{surr}}$. However, everywhere, this equation is simply given as $\dfrac{-Q_{sys}}{T_{surr}}$, which seems to be explained by the claim that the Temperature of the surroundings doesnt change. This equation, according to me should be applicabe only to Irreversible processes happening quickly. (or to isothermal processes).
so what exactly should we consider: $T(sys)=T(surr)$ at each instant, or $T(surr)$=constant?

In a cases where where the temperature is changing during the reversible process, instead of just using a single reservoir at a constant temperature, what you do is use a continuous sequence of reservoirs, all at slightly different constant temperatures (over the range from the beginning to the end of the process), with a tiny (differential) amount of heat exchanged with each.  So, for each reservoir, you have $$dS_{surr}=\frac{dq_{syst}}{T_{surr}}$$ and, if you add all these differential changes in entropy up, you end up with $$\Delta S_{surr}=-\int{\frac{dq_{syst}}{T_{surr}}}$$


'In MO theory, electrons are treated as spreading throughout the entire molecule'- isn't it exhibited in VB theory also? First statement:

  In the molecular orbital theory, the valency electrons are considered to be associated with all the nuclei in the molecule. - Concise Inorganic Chemistry by J.D.Lee.

Second statement:

  In molecular orbital theory, electrons are treated as spreading throughout the entire molecule: every electron contributes to the strength of every bond. - Elements of Physical Chemistry by Peter Atkins.

Two different sources are actually saying the same thing in regard to MO theory: electron density is spread over the entire molecule, okay that is quite right & agreeable.

But the way the fact is presented as if it were the sole property of MO theory. Doesn't the electron density span over the entire molecule in VB theory also?

Check this pic taken from Peter Atkin's above mentioned book:

This is the electron density in $\ce H_2$ as predicted by VB theory; notice how the resultant wavefunction after bonding is spread over the entire molecule & this implies the electron density is spread over the entire molecule. So, even in VB theory, the electron density is spread over the nuclei of the molecule. So, why those special writings above?

Why is it that the spreading of electron density over the nuclei of the molecules specially in MO theory? Is not this exhibited even in VB theory also? Or am I missing something?

Can anyone please help me whether the spreading of the electron density is the sole property of Mo theory as is put up by the two authors? Is not it exhibited in VB theory also?

In MO theory, you construct a molecular orbital by a linear combination of atomic orbitals. Because those atomic orbitals are part of different atoms, in this way, the valence electrons are associated with all the atoms in the molecule. So even if you consider a molecular orbital that is bonding between, let's say, atom 1 and atom 2 in the molecule, that molecular orbital has also contributions from all the other atomic orbitals (besides those that arise from atom 1 or atom 2).

In contrast, in VB theory, a bond between atom 1 and atom 2 can only have atomic orbitals of atom 1 and atom 2. The bond between two atoms is then considered to arise from two weakly coupled atomic orbitals. I have to point out though that modern valence bond theory now complements molecular orbital theory and no longer adheres to the valence bond idea that electron pairs are localized between two specific atoms in a molecule but that they are distributed in sets of molecular orbitals which can extend over the entire molecule.

Given the discussion below, I would like to add to my answer a simple example. Let us consider the $\ce{H2O}$ molecule. Within the framework of Hartree-Fock and using a STO-3G basis set, the $2\sigma$ orbital of water can be expanded in terms of the atomic orbitals as follows:
$$
\psi_{2 \sigma} = (-0.4330)\; \phi_{\textrm{O},1s} + (-0.4686)\; \phi_{\textrm{O},2s} + (0.2369)\; \phi_{\textrm{O},2p_y} + (0.2685)\; \phi_{\textrm{H1},1s} + (0.2685)\; \phi_{\textrm{H2},1s}.
$$

From this expansion you can see that multiple atomic orbitals contribute to the same molecular orbital. This is a molecular orbital that forms, in a way, a multi-atom bond between O and the two H atoms. You could also say that this molecular orbital contributes to both of the O-H bonds (in fact, more molecular orbitals than just the $2\sigma$ do).


What is the reason for the anomalous electronic configuration of palladium? The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.

Electron behavior is elaborated by other principles of atomic physics, such as Hund's rule and the Pauli exclusion principle. Hund's rule asserts that even if multiple orbitals of the same energy are available, electrons fill unoccupied orbitals first, before reusing orbitals occupied by other electrons. But, according to the Pauli exclusion principle, in order for electrons to occupy the same orbital, they must have different spins.

The order in which these orbitals are filled is given by n+l rule known as Madelung Rule or diagonal rule.Orbitals with a lower (n+l) value are filled before those with higher (n+l) values where n represents principal quantum number and l represents azimuthal quantum number.

The fact that most of the ground state configurations of neutral atoms fill orbitals following this n + ℓ, n pattern was obtained experimentally, by reference to the spectroscopic characteristics of the elements.

The Madelung energy ordering rule applies only to neutral atoms in their ground state,even in some cases there are examples which show different electronic configuration than those of experimentally determined.Copper,Chromium and Palladium are common examples of this rule.

For Palladium,according to Madelung's rule, 5s^2 has lower (n+l) value i.e. (5+2=7) and 4d^8 has higher (n+l) value i.e. (4+8=12),so the 5s orbital  fills up first before 4d orbital.

Hence, Palladium shows anomalous electronic configuration.


How do I heat my borosilicate reflux column? I am gonna use a ceramic tape heater, but cannot wrap it around evenly. Is the glass column gonna be okay?

I am running some tests on plastic pyrolysis. I am using the reflux condenser to make the heavier hydrocarbons flow back into the boiling flask. I am also trying out various catalyst in the reflux condenser which will facilitate catalytic cracking.

I have decided to maintain the temperature at top of the reflux column to be of 150C and it will allow the lighter fraction to be removed off and the heavier fraction to be condensed back into the boiling flask.

The boiling flask is made up of quartz and the rest of the glassware is borosilicate.

I am gonna use the glass setup to test catalysts. If results are satisfactory, will get a microreactor made up of steel for further testing. I can better monitor and control temperature in a metal setup I think. Gonna try on only PE and PP types of plastic.

I am a structural engineer by profession and am new to organic chemistry. Thanks for you help in advance.

That should pose no issue regarding the glass

Without images to see how unevenly you will apply the heat, I have to modify my instinctual reaction "That cannot possibly be an issue" to "Most likely it will be OK". I'd be much more worried about variation in the distilling if you are at some point drawing a distillate. If you cannot control the heat flux, the yield and temperature profile will be offset from the theoretical and can (if the process is delicate) lead to variation in the results.

But borosilicate glass is sturdy, you can probably heat it with a naked propane flame - it is designed to have a low thermal expansion coefficient. Directly from that, it will have a high thermal shock resistance.


Are there any other (meta)stable oxygen molecules beyond O3? Apart from Ozone, are there any longer oxygen molecules that are at least stable for short amounts of time? What configurations would that be?

According to my German version of Holleman-Wiberg: Inorganic Chemistry there is one additional, very short lived species, the so called tetraoxygen. A short article is also available on Wikipedia.

Wiberg describes it's synthesis from dioxygen in a mass-spectrometer employing the "neutralisation-reionisation-technique" (I am translating and paraphrasing at the same time, there might be something lost in translation.) It's stable time frame is in the micro second area. The decomposition barrier of $\ce{O4}$ is about 25 kJ/mol. The structure is not known, but most likely an aggregate of two oxygen molecules, i.e. $\ce{(O2)2}$. In lower concentrations $\ce{O4}$ might be found in liquid oxygen and they may be a template for the  ε-high pressure modification of solid oxygen forming at 0.6 Mbar.

Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XIII 1.1.4 Kurzlebige Sauerstoffspezies. S. 513f.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.


Reaction of glucose acetal with acetic anhydride The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

Consider the following reactions:
$\ce{ (i) Glucose + ROH ->[dry HCl] Acetal ->[$x$ eq. of (CH3CO)2O] acetyl derivative }$
$\ce{ (ii) Glucose ->[Ni/H2] A ->[$y$ eq. of (CH3CO)2O] acetyl derivative }$
$\ce{ (iii) Glucose ->[$z$ eq. of (CH3CO)2O] acetyl derivative }$
What are $x, y,$ and $z$ in these reactions, respectively?

I know that the reaction in (i) is acetal formation (used to protect aldehyde), and that alcohol reacts with anhydrides to form acids. As, there are only 5 hydroxyl group in acetyl deriv. of (i), so $x=5$.
In (ii), carbonyl part of glucose reduces to hydroxyl group, so $y=6$. And, in (iii), glucose directly reacts with anhydride, so $z=5$.
But, my answer was wrong (x only), as per the key. What am I doing wrong?

The answer given is,

 x = 4, y = 6, z = 5

I think there's nothing to discussed about this further as all comments directing to the correct answer: $x = 4, \ y = 6,$ and $ z = 5.$ I think it's better show in the structures:

Glucose is naturally occurring in pyranose hemiacetal form. When subjected to react with $\ce{H+/R-OH}$, the remaining acetal hydroxy group (anomeric hydroxyl group) will undergoes alkylation. Thus, the first scheme is showing alkylation of this anomeric hydroxyl group by methanol (Ref.1). The resultant methylglucose has four free hydroxy group, hence, $x = 4$.
When subjected to reduction with suitable conditions (e.g., $\ce{Ni/H2}$, $\ce{Ru/C/H2}$, etc.), the hemiacetal group would be reduced to a $\ce{-CH2OH}$ group (second scheme in the image). The resultant product is sorbitol (Ref.2), which has six free hydroxy groups, hence, $y = 6$.
Finally, pyranose glucose structure has five free hydroxy groups, hence, $z = 5$.

References:

B. Helferich, W. Schäfer (Checked by H. T. Clarke and M. R. Brethen), "$\alpha$-methyl d-Glucoside," Org. Synth. 1926, 6, 64 (DOI: 10.15227/orgsyn.006.0064).
J. Wildschut, J. Arentz, C.B. Rasrendra, R.H. Venderbosch, H.J. Heeres, "Catalytic Hydrotreatment of Fast Pyrolysis Oil: Model Studies on Reaction Pathways for the Carbohydrate Fraction," Environmental Progress & Sustainable Energy 2009, 28(3), 450-460 (https://doi.org/10.1002/ep.10390).


why maximum (w/w) percent concentration of commercial acids is same to their molar mass? i want to know why maximum (w/w) percent concentration of commercial acids is same to their molar mass, for example for HCl the maxmim concentration is 37 and it's molar mass is 36.5 and maximum concentration of H2SO4 is 98%(w/w) and it's molar mass is 98.
thanks alot!

It's a coincidence as HBr is 48% (MW ~81), phosphoric acid is 85% (~98), hydrofluoric acid 97% (MW ~20).Look here for more http://www.csudh.edu/oliver/chemdata/acid-str.htm


Why is mercury used in thermometers? I know it expands when hot and grows compact when cool (that's about all I know), but I suppose many other substances do, too.

Also, and excuses for my ignorance, what the term used to describe the function of mercury in a thermometer? I mean, would you call it an "indicator" (insofar as it indicates the temperature), or is there some technical term?

Thanks!

In addition to the answer and the comment given, the use of mercury in thermometers had another advantage frequently used in the lab:

Mercury, as a metal, has a high electrical conductivity. Imagine two thin wires inside the capillary of your thermometer. One of them is fixed.
The other wire is can move along a threaded rod. Typically, this was realized via magnetical coupling to a freely rotating knob on top of the (hermetically sealed) thermometer. Upon heating, the mercury expands and finally reaches the upper wire previously adjusted to a particular working temperature. At this point, an electrical current may flow through the thermometer.

Source: http://upload.wikimedia.org/wikipedia/commons/4/48/Kontth.jpg

These contact thermometers, nowadays replaced by mercury-free electronic thermoswitches, once were the working horses in the lab when it came to controlled heating of oil baths or monitoring cooling circuits for uv lamps.


Clemmensen reduction of α-, β-keto acids The following is from Organic Chemistry-2 for JEE Advanced by KS Verma:

α- or β-keto acids do not undergo Clemmensen reduction.

My reasoning was that the double bonds are in conjugation which makes them reluctant to participate in such reactions. However, this doesn't seem to be the case with β-keto acids which makes me suspect that my line of reasoning might be incorrect.
What is the reason for this behavior?

The likely reason for this is a steric hindrance. The Clemmensen reduction is very vulnerable to hindrance. See the image below:
There was also another reaction conducted that had a 67% yield for a gamma keto acid with resonance. See the image below:

Therefore, the main factor that makes α- or β-keto acids not undergoing Clemmensen reduction is likely not because of resonance, but primarily of steric hinderance. But this also may raise another question, why doesn't the carboxylic acid group get reduced? The reason for this is that the carbon in a carboxylic acid is not quite partially positive because OH is an electron-donating group. The mechanism involves a transfer of an electron from zinc to the partially positive carbon atom, which, as mentioned, carboxylic acid doesn't have.
However, even for diketones, the Clemmensen reduction doesn't often yield what one would expect. See the image below:

The reason is because of the mechanism. In the mechanism, radicals are formed on the carbonyl carbon. However, if you have a diradical, they can come together and form a bond if they are close enough. This is just some extra information.
The source I took the images from: https://www.sciencedirect.com/topics/chemistry/clemmensen-reduction#
Extra papers I found that may be helpful:
https://d.lib.msu.edu/etd/37284/datastream/OBJ/view
https://pubs.acs.org/doi/abs/10.1021/jo01017a512


Carbon in Glass - XPS Analysis I was reading a paper and found this elemental XPS analysis of an unmodified microscope glass slide surface.

C (1s): 17.7%

Si (2p): 28.47%

O (1s): 53.83%

It seems strange for a piece of normal glass to have almost 20% of Carbon, since most of it is silicon dioxide and other metal oxides in smaller amounts.
I am not used to XPS analysis so there might be something I am overlooking (maybe adsorbed CO2 on the surface ?), can someone tell me if this is a normal result for an XPS analysis of glass, and if it is, where does the carbon come from?

Well, XPS analysis should detect atoms as light as Carbon, but not atoms as light as Hydrogen.

Far be it from me to claim any expertise on microscope sliders, but I have used one or two in my career, and I am hazarding a guess. It is likely a pure glass ($\ce{SiO2}$) sheet laminated with some acrylic glass. The hydrogen in the acrylic polymer are too light to appear on the XPS.

There is a slight possibility, although I consider it very unlikely, that it is glass with a substrate of moissanite on top, which would in theory make it scratch proof, due to the hardness of $\ce{SiC}$.

If you would include whether the %'s are weight percent or atomic percent, it would be easier. If it is atomic, then the case for $\ce{SiC}$ is more likely, since the ratio ($\ce{Si:O} > \ce{1:2}$) and the extra Si would be in SiC. If it is weight percent then there is some math to be done. I still think it is just glass and acrylic glass.


Solubility of n-butanol and n-butanamine I ran across something saying that n-butanamine is miscible in water while butanol is only partially soluble. Is this correct? Can anyone offer an explanation? I would have guessed that the decreased polarity of the amine relative to the alcohol would have resulted in lower solubility for n-butanamine.

Out of butanamine and butanol, butanol is more soluble in water due to higher electronegativity of oxygen . And as butanol  have more polar oxygen so it will be more attracted towards water and form H bond.so it is soluble but amine are miscible due to enthalpies that will release out because of electron donating nature of nitrogen.


Typography of Orbitals For mhchem, I need to decide on a typographical representation of orbitals.

Wikipedia tells me, there are atomic orbitals with letters s, p, d, f and g. In a later table, there is even h.
Then, there are molecular orbitals with greek letters.
Orbital hybridisation uses the letters s, p and sp, sp², sp³.

The IUPAC green book
on (printed) page 32 uses upright characters for 1s, 2s, 2p etc.
However, in the footnotes on page 19, ‘orbitals r and s’ are in italic. (r? Wikipedia did not tell me about this.)

That is the only occurrence in an IUPAC document I could find via search. Are there more?

There are other occurences in the Blue Book in italic, but they are stereodescriptors. (1s,3s)-cyclobutane-1,3-diol.

A quick search at Google Books revealed the convention of italic font for orbitals,
e.g. Organic Chemistry, 2008
or Essentials of Organic Chemistry, 2006,
or Orbital Interactions in Chemistry, 2013,
all published by Wiley.

However, there are also
Molecular Orbitals and Organic Chemical Reactions, 2011
and
Discovering Chemistry With Natural Bond Orbitals, 2012
by the same publisher that use an upright font.

Anyhow, I need to decide for one way or the other. Is there any IUPAC document, I could base my decision on? One, that defines the meaning of the letters and not just uses them.

Second question: I need to distinuish orbitals form other notations. Are my following assumptions correct to reliably detect orbital notations?

Orbitals are always written as number + lower-case letter
Letters can be s, p, d, f, g, h.
There is no element entity in chemical notations with these (lower-case) letters. But there is p for proton.
The letters for abbreviations for organic structure fragments such as ‘c’, ‘m’, ‘p’, ‘r’, ‘s’, and ‘t’ overlap, but they are never written with a number in front of them.

Disclaimer:
I am not a chemist and am not familiar with the meaning of all of this.
I am interested in the typography, though.

In guidelines published by the International Union of Pure and Applied Physics or IUPAP$^{[1]}$, upright letters are recommended. Whether these letters are capitalised depends on their use. Lowercase is for quantum numbers of a single particle, and uppercase for describing a system.

A letter symbol indicating a quantum number of a single particle should be printed in lower case upright type.  A letter symbol indicating a quantum number of a system should be printed in capital upright type.$^{[1]}$

Almost identical text is found in the IUPAC Green Book.$^{[2]}$
The letters themselves ignoring capitalisation are $\mathrm{s}$ – sharp, $\mathrm{p}$ – principal, $\mathrm{d}$ – diffuse, $\mathrm{f}$ – fundamental, $\mathrm{g}$, $\mathrm{h}\ \ldots$
Continue according to the English alphabet skipping letter j.$^{[1]}$
In summary, for orbital angular momentum quantum numbers$^{[1]}$

\begin{align}\begin{split} l\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{s}\ \ &\mathrm{p}\ \ &\mathrm{d}\ \ &\mathrm{f}\ \ &\mathrm{g}\ \ &\mathrm{h}\ \ &\mathrm{i}\ \ &\mathrm{k}\ \ &\mathrm{l}\ \ &\mathrm{m}\ \ &\mathrm{n}\ \ &\mathrm{o}\end{split}\tag{for a single particle}\end{align}
\begin{align}\begin{split} L\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{S}\ \ &\mathrm{P}\ \ &\mathrm{D}\ \ &\mathrm{F}\ \ &\mathrm{G}\ \ &\mathrm{H}\ \ &\mathrm{I}\ \ &\mathrm{K}\ \ &\mathrm{L}\ \ &\mathrm{M}\ \ &\mathrm{N}\ \ &\mathrm{O}\end{split}\tag{for a system}\end{align}

As an example, here are two ways$^{[1]\ [2]\ [3]}$ of writing the predicted ground state electronic structure$^{[4]}$ of oganesson, element $118$:
$$\ce{Og: [Rn] {5f^{14}}{6d^{10}}{7s^2}{7p^6}\\ Og: [Rn] (5f)^14(6d)^10(7s)^2(7p)^6}$$
Typography regarding spacing is not explicitly stated, but is clear from the examples provided. Nevertheless, if you prefer to use
$$\ce{Og: [Rn] 5f^14 6d^10 7s^2 7p^6},$$
it should be fine as well.

$[1]$ E. Richard Cohen, Pierre Giacomo. $(1987)$. Symbols, Units, Nomenclature and Fundamental Constants in Physics. ('IUPAP Red Book'). International Union of Pure and Applied Physics. Commission C2 – Sunamco. ($2010$ reprint) (pp 12$-$13)
$[2]$ E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. ($2007$). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. ($2008$ IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. (p 32)
$[3]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)
$[4]$ 'The Element Oganesson'. $(2016)$. Steve Gagnon. Thomas Jefferson National Accelerator Facility – Office of Science Education. (February 1, 2017)


Question regarding chemical equilibrium when adding a salt to acid solution Having read the following was wondering something:
$$\mathrm {p}K_\mathrm{a} = -\log\frac{[\ce{H+}]_\mathrm{eq}[\ce{A-}]_\mathrm{eq}}{[\ce{HA}]_\mathrm{eq}} $$
$$\ce{HA <=> H+ + A-}$$
$$\mathrm{pH = p}K\mathrm{_{a}} +\mathrm{log\frac{[\ce{A-}]}{[\ce{HA}]}} $$
This rmgd
$$ \mathrm{[\ce{A-}] = [salt] }$$
when acid is weak.
Now my question is like this: take a beaker of $\ce{CH3COOH}$. Now add $\ce{CH3COONa}$. Now here using the a over formula as salt concentration increases $\mathrm{pH}$ should decrease when take in $\ce{A-}$ as salt concentration. However this same $\ce{A-}$ was there in the $\mathrm{p}K_\mathrm{a}$ formula so shouldn't the $\mathrm{p}K_\mathrm{a}$ change to compensate? If so then there will be no $\mathrm{pH}$ change but I have read that there is $\mathrm{pH}$ change on adding $\ce{CH3COONa}$ in a solution. What is wrong with this?

the second relation is wrong to : $pK_{a} = -\log\frac{[\ce{H+}]_{eq}[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}=-\log[\ce{H+}]_{eq} -\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$ so $pH=pK_a+\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$
the acidity constant is the relation which links the concentrations of the species present (and not the reverse): if we add the salt $ \ce {A^-} $, then according to the relation you see that the pH increases. You can also see it by reasoning thus: you add a base so the pH can only increase.


How much electricity could be generated by a person's stomach acid? Motorola has a device that is powered by stomach acid. Given that a battery consumes the electrolyte as it generates a charge, how much power could be generated using a device such as this? Let's assume the electrodes are large enough that the electrolyte is the limiting factor in this reaction. Would the body just generate more acid to compensate? Would there be any physiological side-effects from using such a battery?

Well a battery works on the principle of dissolving one of the electrodes with the electrolyte. This means that you would have a solution of the metal and electrolyte in your stomach. If consumed in large quantities or in quick succession, then you most probably would notice some physiological effects, however, I would assume they'd be very minor given the mass of the tablet in question. Other than under those circumstances, I would expect the physiological effect to be non-existant. Given this has been approved by the FDA, you can be pretty confident that it's safe.