Racemic mixture or not? Will the following reaction give a racemic mixture or not?

According to me, it should give. But my teacher disagrees, but I cannot understand the reasons he gave. Shouldn't there be equal approaches from above and below the molecules?

I am confused when he says that the adjacent 3° Carbon can change its configuration too.

Update

Original question:-

Solution provided:-

IUPAC advises against using the term racemic mixture because of its ambiguity.

  racemic mixture (obsolete) The term is confusing since it has been used as a synonym for both racemate and racemic conglomerate. Usage strongly discouraged.

The term that applies in this case is racemate.

  racemate An equimolar mixture of a pair of enantiomers. It does not exhibit optical activity.

There are three qualitatively different initial conditions to consider.

1. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is not $\ce{OH-}$

Both enantiomers behave similarly, each yielding a stereocenter at the carboxylic carbon.

Verdict: not optically active but contains two pairs of equimolar enantiomers: $\mathrm{a), d)}$ and $\mathrm{b), c)}$. Even three if you count the reagents. Is a racemate if one relaxes the single pair criterium.

2. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is $\ce{OH-}$

Verdict: not optically active, equimolarity is ensured, a pair of enantiomers. Is a racemate.

3. Either (3S)-methylpentan-2-one or (3R)-methylpentan-2-one, nucleophile is / is not $\ce{OH-}$

Verdict: optically active, does not contain enantiomers. Not a racemate.


Do aldehydes give bromine water test? If they give bromine water test, what do they form?

I think that aldehydes will get oxidised and so decolourise bromine water but I don't know if that happens in normal conditions.

The answer is yes. Aldehydes will decolourise bromine water.

The oxidation of aldehydes to acids is discussed in this answer Does Br2/H2O oxidize all aldehydes?

In addition alpha-bromination of the aldehyde via the acid catalysed bromination of the enol is a well-documented process. https://chem.libretexts.org/LibreTexts/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-1.html


In ESI-MS, why do we assume that the positive charge on the analyte must come from protonation from the solvent? In ESI-MS, we ionise the molecules using a high voltage. Why wouldn't this high voltage cause the ejection of an electron from the analyte itself, generating a M+ ion. Why do we have to assume the the positive charge comes from protonation of solvent molecules?

You might be asking two questions here.

(1) Can the high electric field strength induce electron detachment in the vapor? No. Normal electrospray voltages are not high enough to do so. Air, for instance, has a breakdown voltage of around 30 kV/cm, and liquids and aerosol particles are ionized only at even higher electric field strengths. Electrospray voltages are around one-tenth of that.

(2) Can the positive potential at the capillary walls induce oxidation of $M$ to $M^+$? In principle, yes, but only if there are no other species present with higher oxidation potentials than $M$ -- such as the metal walls of the capillary.


IrO2 (iridium oxide) synthesis How do I synthesize $\ce{IrO2}$ from $\ce{IrCl3}$ ? Please suggest a least expensive method.

One elder reference is Brauer's Handbook of Preparative Inorganic Chemistry, which states for this compound

"In the method of Wöhler and Streicher, $\ce{IrO2}$ is prepared from green $\ce{IrCl3}$ which can readily oxidized in a stream of $\ce{O2}$ at 600$^\circ$C giving blue-black  $\ce{IrO2}$.
The oxidation of fine iridium powder in a stream of air or oxygen does not give $\ce{IrO2}$ quantitatively. [...]"

which is accessible in archive.org's public scanned version here, p. 1590 (book's counting) / p. 1636 (the website's counting) of the 1963-edition by Academic Press.  The age of the edition should neither defer you from using it as a reference, nor inhibit you to look up additional references.


Where's my experimental error coming from? I’m teaching chemistry to a home-ed 15 year old.  Occasionally we do some practical work, since I believe very strongly that physical science learning should have some real lab work to bring it alive, and also to teach things like observational skills and attention to detail.

My background is a PhD in organic chemistry, so I’m familiar with hazard assessment, and all the lab work I do with my student is something I’ve done myself already.  Any time she touches the apparatus it’s under close supervision by myself.

So the practical involved taking hydrated copper sulphate, heating it to drive off the water, weighing before and after, and thus calculating the number of water molecules of crystallisation, based on the respective molecular weights of the anhydrous salt and water.

I have a porcelain evaporating dish to contain the hydrate during heating, electronic scales that will weigh to 300 g in increments of 10 mg, and commercially bought copper sulphate that is allegedly 99.5 % pure.

I started with the hydrate, heated it in the evap dish over a flame (using a gas cooker ring) until it had all turned greyish white, then allowed it to cool and weighed again.

The results were:

Starting wt hydrate: 13.03 g

Wt after heating: 8.22 g

Hence water lost: 4.81 g

So that’s 4.81/18 = 0.267 mol water, and
8.22/(63.5 + 32 + 64) = 0.0515 mol anhydrous copper sulphate

Hence 0.267/0.0515 = 5.18 mol water per mol copper sulphate.

So my question: why not 5.0?

The electronic scales (quite cheap) were an obvious first candidate for a source of error.  I’ve compared its readings at weights of about 2, about 20, and about 50 g, with another similar model, and they agree to within 10 mg (one digit in the final decimal place) at all weights.

I could perhaps have not heated the hydrate enough, and not driven off all the water – but then I’d have less than 5, not more, waters per mol.

Or the hydrate, as supplied, could be a little wet.  I haven’t investigated that yet, but I plan to weigh a sample before and after leaving in a warm airing cupboard overnight.

If you’ve read this far, I’m already grateful, but I’d be even more so, if anyone could suggest where the difference between 5 and 5.18 is coming from.

I did it a second time, and got 5.16.

The difference between 5.00 and 5.16 is a weight error of about 200 mg, and I was really careful, so this seems unlikely.

The weight after cooling of the evap dish is constant.

Any ideas?

Edit (2):

There was no sign of $\ce{CuO}$ after the heating.  No trace of black, and indeed we dissolved the anhydrous salt in water (for growing some seed crystals later), and there was no insoluble residue.
I’ve now heated the dish over a hot blue natural gas flame for ten minutes.  Its weight before cooling was 0.02 g (on my scales) less than its empty weight cold.

Edit (3)

I weighed out some of the hydrate into another dry dish, flattened it out as much as possible within the confines of the dish to expose as much surface area as possible, and left it in the airing cupboard for 24 h at 26 degrees Celsius.  Then I re-weighed.

Pre drying wt: 42.68 g

Post drying wt: 41.98 g

Assumed water content: 42.68 − 41.98 = 0.7 g (!!!)

So assuming that 26 degrees is not warm enough to cause partial loss of water of crystallisation, that suggests that at best my hydrate is only 98.36 % pure, not 99.5 as it says on the bottle (typo corrected above).

So repeating my calculation above:

Wt before: 13.03 × 0.9836 = 12.82 g hydrate
Wt after: 8.22 g
Water lost: 4.60 g

Moles water: 4.60/18 = 0.256

Moles $\ce{CuSO4}$: 8.22/159/5 = 0.052

Ratio: 4.96

I’m much happier with that!

Many thanks.

I would go with a full fledged error propagation analysis on this one, because without systematic approach you might be guessing forever. So if I look at the calculation process you have the following equation to get from known quantities to the answer:

$$X=\left(\frac{W_0}{W_e}-1\right)\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right) \tag{1}$$

where $W_0$ is the start weight, $W_e$ the end weight and $M_i$ is the respective molecular mass. $X$ is the desired answer.

To find the expected error in the answer, $\Delta X$, assuming no error in the molecular masses, we calculate:
$$\Delta X=\sqrt{\left(\frac{\partial X}{\partial W_0}\Delta W_0\right)^2+\left(\frac{\partial X}{\partial W_e}\Delta W_e\right)^2} \tag{2}$$
which works out to:
$$\Delta X=\frac{\sqrt{(\Delta W_0)^2 W_e^2+(\Delta W_e)^2 W_0^2}}{W_e^2}\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right)  \tag{3}$$

So now the final thing left to do is estimate how big $\Delta W_0$ and $\Delta W_e$ are. For $\Delta W_0$ I think it is safe to say that this is only the inaccuracy of the scale, so perhaps $0.01$ g. For $\Delta W_e$ I would go for the worst case scenario: that all of the 0.6% impurity in the original copper sulfate has evaporated, meaning that the weight will be off by 0.6% times $W_0$ i.e. $0.08$ g.

Now plugging in the numbers in Eq.$(3)$ I find $\Delta X=0.14$ mol water per mol copper sulphate. Which is very close to the actual error you find.

This error is almost completely determined by the error in $W_e$. If, like @brinnb suggests, an additional reduction in $W_e$ could be expected due to adhered water say that the error in $W_e$ is $0.1$ g you already get to $\Delta X=0.17$ mol water per mol copper sulphate.

Conclusion

The error you find falls within the worst-case expected error due to impurity and potentially adhered water.


Silver nitrate mirror on 3d printed sla resin For a project, I need to develop special array of reflectors.

Usually this is done through a process of plastic injection and vapor deposition, however the prototyping cost is very high.

Another solution I would like to explore is to 3d print the array of reflector with SLA printer like Form 3 and then deposit silver through the process of silver nitrate on solution.

However, I wonder if that could work and if the bond would be good enough, if the resin part would need prior treatment.

Another method could be by depositing a conductive layer and then do electroplating.

I'm not a chemist and don't really know where to start looking.

Thanks

My experience with glass reduced-silver mirrors is that bonding is poor, and that unprotected Ag rapidly degrades in air (a matter of days, in some cases). The bond to organic surfaces, I believe, might be worse.

If you have access to a fairly good vacuum pump (less than 10-4 Torr), it's not hard to aluminize. Watch a demo online.

You might also request a quote from an aluminization service, such as those in this list. In 2010, a 200 mm diameter mirror was said to cost about US$70 to aluminize. This British service has an online price list, starting at £20.


Can electrolyzing ice cause an explosion? I know electrolyzing water separates it into hydrogen and oxygen.  However, I am curious, if ice is electrolyzed, could it explode?  My thought process is that oxygen and hydrogen bubbles would build up in the ice, and continued electrolysis would ignite the gasses (forming water again).  However, since it is contained within the ice, it could ignite explosively.  Is this what would indeed happen?  One way to try this might be using a tazer or bug zapper on ice.  I believe this could be dangerous, and appropriate precautions would need to be taken.

Bulk electrolysis of ice would likely not get very far. Electrolysis works in liquids because the liquid is able to transport an ion to complete the circuit. In the case of water electrolysis, water is needed to transport protons (or hydroxide, but we'll stick to just one thing).

The half reactions for electrolysis of water are

Oxidation at anode:

$$\ce{2 H2O (l) -> O2(g) + 2H+(aq) + 2e-}$$

The electrons travel through the wires and the protons through solution to the cathode where the following happens:

Reduction at cathode:

$$\ce{2H+(aq) + 2e- -> H2(g)}$$

Electrolysis requires ion mobility. In solid water this mobility is severely decreased. All water molecules are locked in a rigid hydrogen bond network. A proton can propagate slowly through this network by proton exchange, but it will be slow. Thus electrolysis will be slowed.

Also, electrolysis only occurs at the surface of the electrodes, which must provide a means for the gas to escape (whether they are on the surface of the ice or driven into the ice, gaps will exist for gas to escape).


How to find probability in different basis representations? The matrix representation of the Hamiltonian is given in the position basis: $\{|+a\rangle,|-a\rangle\}$.
$$H = \begin{pmatrix} E_{0} & E_{-A} \\ E_{-A} & E_{0} \end{pmatrix}$$

The initial state of a system is also given in terms of the position basis:

$\mid \psi \rangle  = \frac{1}{5}[3|+a\rangle + 4|-a\rangle]$.

To find the probability of the a particular energy eigenvalue, why is it incorrect to choose the coefficient of $|-a\rangle$ so that it would be $\left(\frac{4}{5}\right)^2$?

Instead, we have to transform to the Hamiltonian basis, using the eigenvectors of the Hamiltonian. Why can't we use the position basis?

In quantum mechanics observables are represented by operators and the outcome of a given measurement can only be an eigenvalue of this operator. After the measurement, the state of the system will be projected to the eigenstate of the operator corresponding to the measured eigenvalue (consider for simplicity the case where you don't have more than one eigenstate corresponding to one eigenvalue).

Now, in you case, the probability $|4/5|^2$ corresponds to the following:
$$
P(\text{-}a) = |\langle \text{-}a|\psi\rangle|^2.
$$
This is the probability of projecting (by measurement) the state of the system into $|-a\rangle$. This corresponds to computing the probability of obtaining the eigenvalue $\text{-}a$ as a result of your measurement. If you want to know the probability of being in a give energy state, you have to compute the following probability
$$
P(E) = |\langle E|\psi\rangle|^2.
$$
where $|E\rangle$ is the state with energy $E$ (eigenstate of the Hamiltonian). Therefore, in order to compute the braket $\langle E|\psi\rangle$ you need to write $|\psi\rangle $ in the basis of Hamiltonian's eigenstates: in this basis, the solution is trivial.


Zinc screw turning black I tried one experiment. I started with ordinal yellow zinc screws.
The screws were sumberged in bath of two parts of $4.7\%$ solution of $\ce{NaClO}$ and one part $\ce{NaOH}$. The screws turned silverish in color. Then, I submerged them into vinegar. Vinegar turned them black. I am now wondering what actually happened. The bath of $\ce{NaClO}$ and $\ce{NaOH}$ is used to blacken the steel, but it doesn't work with zinc. The vinegar can be used to remove zinc.
But why it was black after that bath?
I was expecting that it would need another bath in $\ce{NaClO}$ and $\ce{NaOH}$ to become black.
Another control screw which wasn't in first bath was just stripped of zinc layer and got a steel color.

These screws are steel with thin cover of zinc with colored oxide layer, probably done by anodizing, I think. At first you dissolved zinc oxide with NaOH. This removed the colored layer and exposed zinc, making it look metallic. Then you reacted zinc with vinegar. Zinc itself is more reactive with acids. The zinc acetate that could be forming is transparent, so that doesnt explain the black color. If black color is present everywhere in solution, it could be that zinc was not pure, and as zinc was being chipped away by vinegar impurities colored the solution black. Or, if black was present only on the screw itself it could be that steel formed iron oxide on its surface. In both cases black is likely non zinc metal oxide. Most likely copper or iron.


Could a non-spontaneous reaction occur "on its own"? Does spontaneity really only apply to the net free energy change of a reaction?

I used to be taught that spontaneous reactions could occur on their own BUT without enzymes or a spark they might take thousands of years to occur.

So teachers would say that spontaneity is more describing whether a reaction will occur on its own. Rather than a statement of reaction speed.

Because there is still activation energy to deal with, so technically spontaneous reactions to need an input of energy.

So is a spontaneous reaction more a statement of the reaction energetics?

And could a non-spontaneous reaction occur on its own given enough time?

An analogy may help.  Imagine a rock perched at the top of a hill.  If given a small push, the rock will roll all the way down the hill without any further input of energy (in fact, releasing energy as it goes).  That's what we mean by a "spontaneous" reaction.  The reverse (pushing the rock up the hill) is a "nonspontaneous" reaction into which you have to continuously put energy for it to proceed.

But now let's consider that small initial push.  Just how small is it?  If the rock is small and perched on a very sharp hilltop, maybe a strong breeze or a squirrel jumping on it is enough.  But if the rock is big and perched on a rather broad hilltop, it may need an earthquake or big tree falling on it to get moving.  Of course, earthquakes and big trees falling are much rarer events than a squirrel jumping around or high wind.  So if the rock needs a tiny push, we expect it to come flying down the hill quite soon after it's put up there, and if it needs a big push, it will be quite a long time before it gets rolling.

This is what we mean by the kinetics of the reaction -- how fast it goes.  A fast reaction needs hardly any push to get started, a slow reaction needs a big push, and small pushes are much more common than big pushes.  (The "push" for a real chemical reaction is some fluctuation in energy in the environment, like the coincidence of a bunch of simultaneous high-speed collisions of molecules.)

Hopefully the analogy makes clear that the thermodynamics of a reaction (whether it is spontaneous or not) is completely separate from the kinetics (whether it is fast or not).

Finally, use caution when everyday words (like "spontaneous") are used in scientific language.  The implications may not be the same!  In everyday speech "spontaneous" often has the additional implication of "quick."  If I said "I spontaneously break into song during meetings" you would find it odd if I only did it every 25 years.  There's an implication, in everyday speech, of speed as well as no need for a driving force. But that implication does not carry over to the scientific use of the word!


Electronic configuration of tellurium Below is a copy of the method I am using to determine electron configurations.

Question:

  Write the full and short electron configuration for tellurium.

My final answer was:

Full Configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4

Short: $[\ce{Kr}]~\ce{5s^2 4d^10 5p^4}$

However, checking with google, it seems the answer is actually: $[\ce{Kr}]~\ce{4d^10 5s^2 5p^4}$

Is the silly "follow the yellow brick road" thing faulty? Or is it that order doesn't matter in this situation?

I haven't done that in decades (or so), but I'd bet that you did right!

The result from google only lists the final outcome, not the order in which the orbitals are filled. Except for some examples $\ce{La}$, $\ce{Ac}$, etc., the Aufbau principle is reliable. So: Follow the yellow brick road :)


Why is absolute zero unattainable? We were dealing with the Third Law of Thermodynamics in class, and my teacher mentioned  something that we found quite fascinating:

  It is physically impossible to attain a temperature of zero kelvin (absolute zero).

When we pressed him for the rationale behind that, he asked us to take a look at the graph for Charles' Law for gases:

His argument is, that when we extrapolate the graph to -273.15 degrees Celsius (i.e. zero kelvin), the volume drops down all the way to zero; and "since no piece of matter can occupy zero volume ('matter' being something that has mass and occupies space), from the graph for Charles' Law, it is very clear that it is not possible to attain the temperature of zero kelvin".

However, someone else gave me a different explanation: "To reduce the temperature of a body down to zero kelvin, would mean removing all the energy associated with the body. Now, since energy is always associated with mass, if all the energy is removed there won't be any mass left. Hence it isn't possible to attain absolute zero."

Who, if anybody, is correct?

Edit 1: A note-worthy point made by @Loong a while back:

  (From the engineer's perspective) To cool something to zero kelvin, first you'll need something that is cooler than zero kelvin.

Edit 2: I've got an issue with the 'no molecular motion' notion that I seem to find everywhere (including @Ivan's fantastic answer) but I can't seem to get cleared.

The notion:

  At absolute zero, all molecular motion stops. There's no longer any kinetic energy asscoiated with molecules/atoms.

The problem? I quote Feynman:

  As we decrease the temperature, the vibration decreases and decreases until, at absolute zero, there is a minimum amount of motion that atoms can have, but not zero.

He goes on to justify this by bringing in Heisenberg's Uncertainity Principle:

  Remember that when a crystal is cooled to absolute zero, the atoms do not stop moving, they still 'jiggle'. Why? If they stopped moving, we would know were they were and that they had they have zero motion, and that is against the Uncertainity Principle. We cannot know where they are and how fast they are moving, so they must be continually wiggling in there!

So, can anyone account for Feynman's claim as well? To the not-so-hardcore student of physics that I am (high-schooler here), his argument seems quite convincing.

So to make it clear; I'm asking for two things in this question:

1) Which argument is correct? My teacher's or the other guy's?

2) At absolute zero, do we have zero molecular motion as most sources state, or do atoms go on "wiggling" in there as Feynman claims?

There was a story in my days about a physical chemist who was asked to explain some effect, illustrated by a poster on the wall. He did that, after which someone noticed that the poster was hanging upside down, so the effect appeared reversed in sign. Undaunted, the guy immediately explained it the other way around, just as convincingly as he did the first time.

Cooking up explanations on the spot is a respectable sport, but your teacher went a bit too far. What's with that Charles' law? See, it is a gas law; it is about gases. And even then it is but an approximation. To make it exact, you have to make your gas ideal, which can't be done. As you lower the temperature, all gases become less and less ideal. And then they condense, and we're left to deal with liquids and solids, to which the said law never applied, not even as a very poor approximation. Appealing to this law when we are near the absolute zero is about as sensible as ruling out certain reaction mechanism on the grounds that it requires atoms to move faster than allowed by the road speed limit in the state of Hawaii.

The energy argument is even more ridiculous. We don't have to remove all energy, but only the kinetic energy. The $E=mc^2$ part remains there, so the mass is never going anywhere.

All that being said, there is no physical law forbidding the existence of matter at absolute zero. It's not like its existence will cause the world to go down with error 500. It's just that the closer you get to it, the more effort it takes, like with other ideal things (ideal vacuum, ideally pure compound, crystal without defects, etc). If anything, we're doing a pretty decent job at it. Using sophisticated techniques like laser cooling or magnetic evaporative cooling, we've long surpassed the nature's record in coldness.


Why do we need acid when making acetone peroxide? Acetone peroxide is a very dangerous explosive, easily detonated by mild heating, friction or shock. It appears in the form of small white crystals.

This compound forms from the mixture of hydrogen peroxide, acetone and hydrochloric or sulphuric (and maybe any other) acid. The acid, however, is only a catalyst in the reaction.

I'm curious why the acid is required for the reaction to happen. Is it actually necessary? Does the acid change during the reaction?

The role of the acid in this reaction (really a polymerization) is two-fold.  First, it protonates the acetone carbonyl oxygen making the acetone carbonyl carbon more electrophilic.  This speeds up the rate of attack by the nucleophilic hydrogen peroxide (or other hydroperoxides formed later in the reaction) on this carbon.  Second, the acid also protonates hydroxyl groups in intermediates formed later in the reaction.  This makes them better leaving groups and allows nucleophiles like hydrogen peroxide or other hydroperoxides formed in the reaction to displace them.

The acid is a catalyst, the more acid the faster the reaction, but the reaction will proceed, albeit more slowly, without the acid.  Any proton acid can be used, but as expected, the stronger the acid, the faster the reaction.

Here are links to two articles on the subject (reference 1, reference 2) along with a diagram of the proposed reaction mechanism.


What guarantees that there is a thermodynamic equilibrium in a closed chemical reaction network independently of the mass-action kinetics? I would like to know according to which we can state that there is a unique equilibrium of a closed chemical reaction network (at constant circumstances) without supposing mass-action kinetics if time is increasing beyond all limits. Can we derive it from a non-equilibrium thermodynamic law or is it a postulate? Thank you for your answer!

The equilibrium is determined by the thermodynamics, not by the kinetics.  The rate constants for the reaction kinetics must be consistent with the equilibrium determined by the thermodynamics.  For example, for a single reversible reaction, the reverse reaction rate constant must be equal to the forward rate constant divided by the equilibrium constant.


Decomposition of potassium permanganate under prolonged heating When we heat potassium permanganate, the equation of the decomposition of it is:
$$\ce{2KMnO4 ->[\Delta] K2MnO4 + MnO2 + O2}$$

So  what will happen to potassium manganate ($\ce{K2MnO4}$) when we continue heating?

Potassium manganate is produced by reacting manganese dioxide with potassium hydroxide and oxygen gas :

$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$

So I would suspect that the reverse process will occur when further heating. (Of course when there is no water available you might get other products, maybe potassium oxide instead of potassium hydroxide)


Is oxygen necessary for burning? Is oxygen really necessary for burning things?

What if there was no oxygen around? Would there still be a way to burn things?

Yes, absolutely. Burning refers to oxidation accompanied by a flame and there are numerous reactions in which oxygen takes no part. Most of these are either with elementary halogens, or compounds thereof.

Here are some examples.

\begin{align}
\ce{2K + F2 &-> 2KF}\\
\ce{2Fe_\text{wool} + 3Cl2 &-> 2FeCl3}\\
\ce{H2O + ClF3 &-> HF + HCl + OF2}\\
\ce{2H2O + ClF3 &-> 3HF + HCl + O2}\\
\ce{2Cl2 + 2OF2 &-> 4ClF + O2}
\end{align}

So in fluorine chemistry, you don't oxygenise it; it flourises you.


Is there always a need for a salt bridge to let a galvanic cell work continuously? I read in a book that "Sometimes, both the electrodes dip in the same electrolyte solution & in such cases we do not require a salt bridge".

How is this possible?

It depends where and how the electrodes are positioned. The UC Davis ChemWiki page Electrochemistry 2: Galvanic cells and electrodes describes the possible arrangements.

For example, if the electrodes are placed in the same vessel, a salt bridge is not required (as shown below). Ions can pass through the porous barrier.

If, however, the electrodes are are placed in separate vessels (such as in the diagram below), then a salt bridge is required to facilitate ion movement.


Do orbital models derive from the solution of the stationary Schrödinger equation? I know that hydrogen-like atomic orbitals can be seen as a solution of the stationary Schrödinger equation - but what about atoms with more electrons? As I am a pupil and quite uneducated about this subject, I would be very grateful for an answer that is easy to understand.

I know that hydrogen-like atomic orbitals can be seen as a solution of
  the stationary Schrödinger equation - but what about atoms with more electrons?

Orbitals are one-electron wave functions, i.e. they are wave functions that describe the state of only one electron. For a hydrogen-like atom, which is a one-electrons system, if treating its nucleus as a classical positively charged point particle, the wave function is indeed an orbital, since there is just one quantum particle out there and it is the single electron. So, right, orbitals are the solutions of the (non-relativistic) time-independent Schrödinger equation if the nucleus is assumed to be a classical point particle.

For any many-electron atom, though, even when its nucleus is though of as a classical point particle, the wave function describes more than one electron, so by the very definition it is not an orbital. It can be written in terms of the one-electron wave functions (orbitals) as their product, but this can be done only approximately due to interactions between the electrons. It is in this sense we say that orbitals aren't real.

Note though that, strictly speaking, even for a hydrogen-like atom orbitals aren't real since they arise in the non-relativistic treatment which is also an approximation. The reality is even more weird than its non-relativistic description, but for atoms from the first and second rows of the periodic table the non-relativistic description is usually a good one in a sense that it introduces relatively small errors that can be neglected for the most of the chemical purposes. The orbital approximation on the other hand might be the source of much larger errors. For instance, for a $\ce{He}$ atom in its ground electronic state the orbital approximation introduces the error of the $10^{-2} E_{\mathrm{h}}$ order while the relativistic correction is about two orders of the magnitude smaller, $10^{-4} E_{\mathrm{h}}$.


Modelling planar molecules and their evolutions I am trying to simulate a chain of molecules which are planar.

Can anyone please give me an example of the molecular chain which can stay as planar with it's minimum energy state is planar. A published study on the evolution of the geometry of the molecule will be highly helpful.

One-dimensional Fabre and Bechgaard salts, TTF-TCNQ,

http://www.chem.tamu.edu/rgroup/dunbar/Publications/2002-147.pdf
platinum blues

DOI:10.1002/asia.200700370
hexaphenylbenzene discotic liquid crystals
http://www.crcpress.com/product/isbn/9781439811436
"Chemistry of Discotic Liquid Crystals: From Monomers to Polymers," Sandeep Kumar (2010)

http://link.springer.com/chapter/10.1007/978-94-017-2319-0_16
middle
http://physics.aps.org/synopsis-for/10.1103/PhysRevB.82.180401
  Organic magnets

http://pubs.acs.org/cen/topstory/8201/8201notw8.html
Ladderanes

stacks
http://www.google.com/patents/WO2011017711A2?cl=en
http://www.halcyon.com/nanojbl/Images/ConPoly3.gif
and a central hexacoordinate silicon with oxygen bridges


Why does the aqueous solution of sodium peroxide turns red litmus into white? A white solid is either $\ce{Na2O}$ or $\ce{Na2O2}$. A piece of red
  litmus paper turns white when it is dipped into a freshly made aqueous
  solution of white solid.

  Identify the substance and explain with the help of balanced
  equation.
  Explain what would happen to the red litmus if the white solid
  were the other compound.

From the solution and theoretical knowledge, I know the following:

$$\begin{align}
\ce{Na2O2 + 2 H2O &-> 2NaOH + H2O2}\\
\ce{Na2O + H2O &-> 2NaOH}
\end{align}$$

The aqueous solution of $\ce{Na2O}$ will convert red litmus into blue (basic property), so, the substance is $\bf{not}$ $\ce{Na2O}$ for sure.

But, how will it be $\ce{Na2O2}$? Why does the aqueous solution of sodium peroxide turns red litmus into white?

Your analysis seems right and you remember correctly that colour change of litmus of litmus paper will indicate the pH value of the solutions into which it is dipped.

That is, unless the dye (litmus) is irreversible transformed to another compound which no longer shows a pH dependent colour.

As the experiment states:

  A piece of red litmus paper turns white [...]

With other words: Litmus was bleached.

This bleaching typically is the result of an oxidative transformation of a (former) dye.

Here, the only conceivable oxidant in solution is hydrogen peroxide. Consequently, the initial compound in question must have been $\ce{Na2O2}$.


How do d-orbitals interact with molecular orbitals? I  recently stumbled upon a question asking how iron destabilizes the bonds in a nitrogen molecule. I have searched the web and my textbook for quite a while now and I cannot find a satisfactory answer. The only explanation I could muster was that the d-orbitals of the Iron atom interact with the HOMO and LUMO of the nitrogen molecule, due to the overlap of the molecular orbitals with d-orbitals.
I am not sure if this makes sense. I'd really appreciate any answer or advice!

You are on the right track with your answer.

Notice that the LUMO is a $\pi^*$ orbital, meaning that filling it with electrons would weaken the bonding. When nitrogen and iron interact, electrons from the d-orbitals on iron are well aligned, in terms of symmetry and energy, to fill into $\ce{N_2}$'s $\pi^*$ orbital. This forms a bond between $\ce{Fe}$ and $\ce{N_2}$, but at the cost of weakening the bond between the nitrogens. This is a very important concept in inorganic chemistry referred to as Pi Backbonding.

The classic example given of a ligand that exhibits $\pi$ backbonding in metal complexes is $\ce{CO}$, which is isoelectronic to $\ce{N_2}$ and has the same molecular orbital diagram (though the orbitals themselves look different due to the asymmetry of $\ce{CO}$).


Why are complex basis functions not used in electronic structure calculations? I've noticed while reading about Hartree-Fock in Modern Quantum Chemistry (Szabo and Ostlund) that most basis sets use only real basis functions. What I'm wondering is why aren't complex basis functions used? I might be mistaken, but from my experience it seems as though integration on the complex plane would have a lot of convenient properties that would make computations easier. Put another way, are complex basis functions not used because of some difficultly in implementation or is it because of difficulty in interpretation of properties like the molecular orbitals?

The basic answer is rather simple: In the non-relativistic limit it is just not necessary. There is no complex operator and hence nothing which could change the complex component of your basis functions. Using complex basis functions would only double the required work (i.e., computer time). There are very efficient schemes in order to calculate the standard integrals (McMurchie-Davidson, Obara-Saika, ...). I don't think that the representability of MOs is a reason for using real basis functions, since, strictly speaking, you can interprete only the square of the MOs and not their amplitude.

However, there are situations, where your MOs can be complex. This can happen when you have a complex operator in your equation. An example are the Pauli spin matrices of which $\sigma_y$ has only an imaginary compound. They occur, e.g., the Dirac equation (see here one formulation where they explicitly state the incorporation of the Pauli matrices). Complex MOs can be obtained either by using complex basis functions and real MO coefficients or by using real basis functions with complex coefficients. Starting from a non-relativistic implementation it might be simpler just to keep your basis functions real and to choose the MO coefficients complex.


Which of the following is the more stable carbocation? Which of the following is the more stable carbocation?

I thought the 1st carbocation would be the more stable one as the pi-electron density of phenyl group can overlap with the vacant orbital on the sp-hybrid carbocation (1st one). But my tutor pointed out that the vacant orbital on carbon can never be parallel to the p-orbitals of the phenyl group carbons and overlap cannot occur. So only negative inductive effect of phenyl group has to be considered, and so the 2nd one will be more stable.

Please explain. Please also mention about some sources where I can read about these type of carbocations.

Vinyl cations can also be generated by the protonation of acetylenes in strong acid.  The vinyl cations are captured by a nucleophile to produce a stable alkene with the nucleophile attached to the double bond.

  Which of the following is a more stable carbocation?

This route to vinyl cations has been well studied.  As shown in the following diagram, protonation of phenyl acetylene can proceed in two different ways.  The acetylenic carbon attached to the phenyl ring may be protonated or the terminal acetylenic carbon may be protonated.  As you can see, these two routes produce the two vinyl cations you want to compare.

Upon examining the product(s) found in this reaction, we find that only alkene A2 is produced.  Therefore, the reaction proceeds through the intermediacy of vinyl cation A1, not B1.  Consequently, the pathway involving A1 must be lower in energy, or said differently, vinyl cation A1 is lower in energy than vinyl cation B1.

Reasons Why Vinyl Cation A1 Is More Stable Than Vinyl Cation B1

The barrier to rotate the phenyl group is very low. Hence, in A1 the phenyl group can readily adopt a conformation where the aromatic p-orbitals are aligned with the empty p-orbital on the adjacent carbocation center. Additional resonance structures can now be drawn with the positive charge delocalized into the aromatic ring.  There is no such resonance stabilization with the aromatic ring possible for vinyl cation B1. This suggests that vinyl cation A1 is more stable than vinyl cation B1.
In addition to the resonance structures involving the aromatic ring, as shown at the bottom of the diagram, one can draw 2 more hyperconjugated resonance resonance structures for vinyl cation A1, but only 1 hyperconjugated resonance structure can be drawn for B1; again suggesting that vinyl cation A1 is more stable than vinyl cation B1.
Resonance effects are generally stronger and more controlling than inductive effects.  Therefore, the inductive arguments presented in @orthocresol 's answer play only a minor role, being outweighed by the stronger resonance effects described above.


Why do XeO and XeF8 not exist? Since Neil Bartlett's 1962 discovery that xenon was capable of forming chemical compounds, a large number of xenon compounds have been discovered and described. Almost all known xenon compounds contain the electronegative atoms fluorine or oxygen. The oxidation state of xenon in its compounds is generally +2, +4, +6, or +8.

Xenon is known to form three fluorides: $\ce{XeF_{$n$}} \;(n = 2, 4, 6)$ where the oxidation states of xenon are +2, +4, and +6. $\ce{XeF8}$ is not known to exist even though the oxidation state would be +8. Why is this so? Could it possibly be due to the fact that 8 fluorine atoms can't fit around a xenon atom?

Xenon is known to form three oxides: $\ce{XeO_{$n$}} \;(n = 2, 3, 4)$ where the oxidation states of xenon is +4, +6, and +8. $\ce{XeO2}$ was not known until 2011. Why did it take so long to be discovered? Moreover, the lowest oxide of xenon $\ce{XeO}$ is not known though the oxidation state of xenon would be +2. Why?

$\ce{XeF8}$ is not known to exist though O.N is +8. Why is this so?

At least 2 compounds have been reported that contain the $\ce{XeF8^{2-}}$ unit. See, for example:

$\ce{(NO^+)2[XeF8]^{2-}}$ (reference)
Metal salts of the form $\ce{(M^{+})_2[XeF8]^{2-}}$ where M is a metal salt such as $\ce{Cs, Rb}$ (see the above reference) or $\ce{Na}$ (see p. 62 in Advances in Inorganic Chemistry, Volume 46, A. G. Sykes editor; link

The $\ce{XeF8}$ portion of the molecules approximates a square-antiprism geometry.

(image source)

  the lowest oxide of xenon $\ce{XeO}$ is not known though O.N is +2. Why?

Perhaps the molecule is unstable, favoring disproportionation to other xenon oxides plus oxygen. For example, Andreas Hermann and Peter Schwerdtfeger suggest the following pathway at high pressure:

$\ce{3XeO → Xe3O2 + 1/2O2}$

The authors go on to note "we do not find $\ce{XeO}$ to be stable at any pressure" (link to full paper).

Edit: See orthocresol's comment below. He argues that at ambient pressure the following decomposition pathway might be more likely:

$\ce{XeO -> Xe + 1/2 O2}$


How to determine the composition of a gas mixture after a combustion? The problem is as follows:

  A mixture of gases consisting of $\ce{CH4}$ and $\ce{C2H4}$ is let to pass over red hot
  $\ce{CuO}$. Then $\pu{0.6g}$ of $\ce{H2O}$ is collected along $\pu{1.185g}$ of
  $\ce{CO2}$. What will be the composition of the mixture if it is known
  that the combustion was complete?.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.& \ce{CH4}=51.8\%\,and\,\ce{C2H4}=48.2\,\%\\
2.&\ce{CH4}=50\%\,and\,\ce{C2H4}=50\,\%\\
3.&\ce{CH4}=33.3\%\,and\,\ce{C2H4}=66.6\,\%\\
4.&\ce{CH4}=38.8\%\,and\,\ce{C2H4}=64.2\,\%\\
5.&\ce{CH4}=36.3\%\,and\,\ce{C2H4}=63.7\,\%\\
\end{array}$

I'm not sure exactly how to proceed with this question. Should it be treated as if it is an analysis by combustion?.

What I did to solve this was to treat each gas in the mixture as reacting independently for the oxygen from interacting with the red hot $\ce{CuO}$ as follows:

$\ce{ CH4 + 2 O2 -> CO2 + 2 H2O}$

With: $x=\ce{ CH4}$

$x\,g\times\frac{1\,mol\,\ce{ CH4}}{16\,g\,\ce{ CH4}}\times \frac{2\,\,mol\,\ce{ H2O}}{1\,\,mol\,\ce{ CH4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{18}{8}x\,g\,\ce{H2O}$

$x\,g\times\frac{1\,mol\,\ce{CH4}}{16\,g\,\ce{CH4}}\times \frac{1\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{CH4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{44}{16}x\,g\,\ce{CO2}$

For:

$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O}$

With $y=\ce{C2H4}$

$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{H2O}}{1\,\,mol\,\ce{C2H4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{36}{28}y\,g\,\ce{H2O}$

$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{C2H4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{88}{28}x\,g\,\ce{CO2}$

This is reduced to a system of equations as follows:

$\frac{44}{16}x+\frac{44}{14}y=1.185\,g\,\ce{CO2}$

$\frac{18}{8}x+\frac{18}{14}y=0.6\,g\,\ce{H2O}$

Solving this system yields:

$x=0.1024\,g\,\ce{CH4}$

$y=0.2874\,g\,\ce{C2H4}$

Then to calculate the composition of the mixture I would use the molar fraction:

$n_{\ce{CH4}}=\frac{0.1024}{16}=0.0064\,mol$

$n_{\ce{C2H4}}=\frac{0.2874}{28}=0.01026\,mol$

Then the percentage for composition for $\ce{CH4}$ would be:

$\frac{n_{\ce{CH4}}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.0064}{0.0064+0.01026}=0.3841$

For $\ce{C2H4}$:

$\frac{\ce{C2H4}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.01026}{0.0064+0.01026}=0.6159$

To which correspond each one as roughly $38.41\%$ and $61.59\%$ respectively for $\ce{CH4}$ and $\ce{C2H4}$. But none of this seem to check with any of the answers.

Could it be that my method was wrong or did I overlooked something?.

The only answer which is closer to mine is the forth alternative but the second number is off by a big margin thus I'm not very convinced on my result. What would be the recommended procedure in this case?

Can somebody help me here?.

This is clearly a homework question. Yet, since OP has given good effort to solve the question, I'd like to give OP some insight to solve these kinds of questions, taking current question as an example.

Suppose the mixture contains $x\;\pu {g}$ of $\ce{CH4}$ and $y\;\pu {g}$ of $\ce{C2H4}$. The combustion reactions of $\ce{CH4}$ and $\ce{C2H4}$ can be given as:

$$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$
$$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \tag{2}$$

Thus, the amount of $\ce{CO2}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{1 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{11}{4}x\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{4}x\,\pu{g} \,\ce{H2O}$$

Similarly, the amount of $\ce{CO2}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{22}{7}y\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{7}y\,\pu{g} \,\ce{H2O}$$

Since the mixture produced $\pu{1.185 g}$ of $\ce{CO2}$ and $\pu{0.6 g}$ of $\ce{H2O}$, we can derive two equations with two variables, $x$ and $y$:

$$\frac{11}{4}x+\frac{22}{7}y=1.185$$
and,
$$\frac{9}{4}x+\frac{9}{7}y=0.6$$

Up to this point OP has taken correct path. Now, let's consider these two questions. These equations can be simplify as:

$$77 x + 88y=1.185\times 28$$
or
$$ 7x + 8y=\frac{1.185\times 28}{11}=3.016 \tag {3}$$
and,
$$ 7x + 4y=\frac{0.6\times 28}{9} = 1.867 \tag {4}$$

Thus, $(3)-(4)$ gives you:

$$4y = 3.016 - 1.867 = 1.149 $$
$$\therefore \; y = \frac{1.149}{4} = \pu{0.2873 g}\,\ce{C2H4}$$

And $(4)\times 2-(3)$ gives you:

$$7x = 1.867 \times 2 - 3.016 = \pu{0.718 g}\,\ce{CH4}$$
$$\therefore \; x = \frac{0.718}{7} = \pu{0.1026 g}\,\ce{CH4}$$

Thus, it is clear that you were still in the correct path on solving the same two equations for $x$ and $y$.

According to my solution, mass percentage of $\ce{CH4}$ is:

$$\frac{0.1026}{0.1026 + 0.2873} \times 100 = 26.31\% \; (w/w)$$

And mass percentage of $\ce{C2H4}$ is:

$$\frac{0.2873}{0.1026 + 0.2873} \times 100 = 73.69\% \; (w/w)$$

Since this is not the one of the given answers, let's check for mol% like what you have done:

$$n_{\ce{CH4}}=\frac{0.1026}{16}= \pu{0.0064 mol}$$

$$n_{\ce{C2H4}}=\frac{0.2873}{28}= \pu{0.0103 mol}$$

Thus, the percentage of composition for $\ce{CH4}$ would be:

$$\frac{0.0064}{0.0064+0.0103}\times 100 = 38.32\%\: \pu{mol}$$

Similarly, the percentage of composition for $\ce{C2H4}$ would be:

$$\frac{0.0103}{0.0064+0.0103}\times 100 = 61.68\%\: \pu{mol}$$

Accordingly, the most closest answer is (5), although both given values are $\approx 2\%$ off from calculated values. It is noteworthy that OP's calculations are correct to the end.

Note: This question is not well written. That's why I have to go through two calculations to see which $\%$ is they are talking about. Also, the sum of compositions given in answer choice (4) is not $100\%$.


Why aqueous hydroxide is slippery (beyond just saponification) A recent question mentions:

  Bases are soapy to touch because of soap formation on contact with lipids present on our skin.

This is always the reasoning I hear, as well, and searching the internet gave the same results. However, when working with nitrile gloves, glass with aqueous sodium or potassium hydroxide on the outside always feels slippery. I believe that even nitrile-on-nitrile (rubbing the fingers together) gives the same slippery feeling. This effect is even stronger for the liquid in the base bath, but I assume that's because alkoxides would be detergents similarly to soap molecules in the absence of water.

Point is, if saponification is the culprit, why does the exterior of the glove experience a similar soapy effect to that my fingers feel? Could it be negative charges on the exterior of the glass? Is it possible for hydroxide to react with nitrile gloves to cause some similar kind of electrostatic repulsion, maybe by deprotonation or nucleophilic attack of the nitrile? This seems like a reasonable idea as far as glass goes, but I don't think that reasoning crosses over to gloves. Of course, I am not sure it really happens with glove-on-glove contact. I will safely test that theory again and edit if it's incorrect.

Glass itself is soluble in strong alkalis, forming sodium silicate solution, which is a syrupy, somewhat slippery-feeling substance. You can use sodium silicate to make a fire-resistant coating or high-temperature cement.

As you state, it's also possible the glove itself is being attacked. If the glass feels slippery in $\ce{NaOH}$ and $\ce{KOH}$ solutions, but not in $\ce{NH4OH}$, then it is most likely a silicate forming, though.


How does a fire start? I know that fire in a few words is the exothermic reaction of a fuel with an oxidizing agent, but I can't fully understand what exactly happens to piece of wood when it is ignited. How do molecules start producing a flame? In other words, what is the chemistry behind the production of flame?

https://www.youtube.com/watch?v=B0E4PX3e3RE It feels somewhat weird to answer my own question but I think this video describes exactly what I wanted. As it supports, when heat is applied to a piece of wood, some bonds of the molecules that make up wood, break and thus different compounds are formed. These compounds are not held back by some force and so they are released in the air. When these compounds meet atmospheric oxygen, under heat (=energy), they burn and thus more heat is released along with carbon dioxide and water. That stage can be described as ignition. Finally, this produced heat is able to preserve the fire.


Is benzene harmful to human health? I have heard that benzene is a dangerous chemical for humans; it causes cancer and so it's use is restricted/regulated. But many organic compounds that we use contain benzene in some form such as oils, foods, even medicines (paracetomol). So is benzene really harmful to people's health?

As Ivan Neretin commented, parts of a compound can never dictate its own properties to the entire molecule's properties. Additionally benzene is not a very effective functional group by itself.

In short, a restricted use is only applied when that certain chemical is exposed by itself. In paracetamol or any other thing including benzene, you will not see any benzene vapor emanating from them or something. It would also be meaningful if benzene containing compounds degradation products included benzene, like in environmental degradation of chlorobenzenes, though this is not a usual case as I see.


Response factor calculation for gas chromatography While dealing with gas chromatography, how do you calculate the response factor with only the area known?

I am given the equation

$${k_1\over k_2} = {f_1 \times O_1 \over f_2 \times O_2}$$

where $O_1 = 21.28$, $O_2 = 178.35$, and $f_1 = 1.00$. $k_1$ and $k_2$ are the weights (in moles) that we added into the mixture. I calculated $0.05\,\mathrm{mol}$ for each component.

Using the formula (above) and the given quantities, I get

$$1 = {1.0 \times 21.28 \over f_2 \times 178.35}$$

Solving for $f_2$, I get 0.119, and I do not believe that is right. Is my approach valid?

This is a bit convoluted.

A simply dimensional analysis shows that your approach doesn't work. The units of $O$ are unit area. The conversion factor $f$ has units moles per unit area. The variable $k$ is in moles. If you divide the two equations you get a relative sensitivity of $f_2$ as compared to $f_1$, but the value is dimensionless. So each compound has its own response factor. So for two unknowns you have:

$f_1 = \dfrac{O_1}{k_1}$
$f_2 = \dfrac{O_2}{k_2}$

The twist here is the GC typically uses a standard either on a run before the unknown or spiking the unknown sample. So the standard is known to be $x$ ppm and you'll get some area from the run. The hope is that if the gain of the detector drifts them the area of the standard will change too by the same amount. So a 10% gain in area means that the detector gain has drifted up by 10%. Thus measuring area relative to the standard compensates for the drift.

The sensitivity changes a bit from day to day because the ionization in the detector changes due to aging, the gas flow rate isn't 100.0000% the same each time, the column ages and so on.

So the ratio of two unknowns is meaningless. You ratio each of the unknowns to a standard.


How to derive van 't Hoff equation for Henry's law constant? While studying about Henry's law, I encountered this equation:
$$\frac{\mathrm d\ln H}{\mathrm d(1/T)}= \frac{-\Delta_\text{sol}H }{R},$$
where $\displaystyle\mathrm d(1/T) = -\frac{\mathrm dT}{T^2}$ and the $\displaystyle\frac{\Delta H}{RT^2}$ reminds me of some kind of $\Delta G$ equation.
I didn't know how to derive it, so I searched about it and got nothing. I tried to substitute $H$ (Henry's law constant) by $\gamma\cdot f_i$ where $\gamma$ is the activity coefficient at infinite dilution and $f_i$ is the fugacity of pure species.
But still I have no idea what to do to obtain the formula that I'm familiar with, so I welcome any idea about how to derive or how to reverse engineer the formula.

Assuming Henry's coefficient $\mathcal{H}_i$ is defined by the following relation:
$$
\mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat}
$$
Where $\gamma_i^\inf$ is the activity coefficient for an infinite diluted solution and $p_i^\mathrm{sat}$ is the vapor pressure of pure component $i$.
One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)}
$$
The RHS maybe rearranged in order to have the well know form of ratio of  mixed equilibrium constants $K_i$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)}
$$
That is, we are stating that ratio of Henry's constants approximates to the ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.
Where the concerned reaction is the converse of  gas solubilization:
$$
\ce{X_{i,(aq)} <=> X_{i,(\mathrm{g})}}{\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H}
$$
And its mixed equilibrium constant:
$$
K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i}
$$
Then one may apply van 't Hoff relation using enthalpy $\Delta_\mathrm{sol}H$ of the concerned reaction, from its integrated form, it comes:
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right]
$$
This formulae is generally a good approximation around $\pu{20°C}$ over a small range of temperature. Do not forget that $\Delta_\mathrm{sol}H$ is not a constant but rather a function of temperature.
If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2}
$$
Or equivalently:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R}
$$


What is this device? (Constant level device?) Found this in my lab on a random shelf (water quality lab). It is old enough that nobody knows where it came from. I think it's a constant level device. It has in- and out-flow connections, as well as electrical plugs for some kind of monitoring equipment. The only writing is on the glass, which says "FSTREEM [ASTREEM?] MADE IN ENGLAND". The inside float looks like it is meant to float up and down, while the thin metal wire keeps track of its level via a small electric recorder. What is this thing called, and where can I find more info?

Thanks!

Thanks Joseph. Knowing that it is a Fi-Streem product, I called/emailed them and got an answer from their friendly support tech. From his email response: [This] is a "Switch Level Device, front" that is used in conjunction with an older series Fi-Streem (Fisions) Still, which is no longer manufactured. The part number for this item is, 06756."


Diffuse functions for Ahlrichs/Karlsruhe basis sets in Gaussian? As the question states, can one use the basis sets of Ahlrichs and coworkers (e.g. def2-SVP, def2-TZVP, ...) with diffuse functions (e.g. def2-TZVPD) in Gaussian?

The regular basis sets have been implemented, but not the diffuse functions according to their homepage. Is there some other way to access diffuse functions for these basis sets in Gaussian?

If they are not in Gaussian yet, there won't be keyword to include these diffuse functions, but you can add them manually to your jobs.

You can find the basis set specification for def2-TZVPD for the atoms you are interested from the Basis Set Exchange. From there, you can either just insert the diffuse functions after your molecule specifications using the keyword Extrabasis or just insert the whole basis manually using the gen keyword where you would normally put the name of the basis set.


Cleaning silver - which method is better? What is the advantage of this approach over polishing the silver with an abrasive?

"Silver tarnish is an outer coating of silver sulfide, $\ce{Ag2S}$, formed when silver reacts with trace quantities of air-borne hydrogen sulfide, a smelly gas produced naturally by the digestion of food in living animals.  Silver atoms in silver sulfide have lost electrons to sulfur atoms in creating ionic bonds.  You can remove this tarnish with home electrochemistry.  Convert silver atoms back to their elemental state by restoring their electrons.  Sulfur atoms won't relinquish electrons to silver, but with the proper connection, aluminum atoms will.  Add about a liter of water with several heaping tablespoons of baking soda to an aluminum pan that has been scoured clean.  If you don't have an aluminum pan, use several large strips of aluminum foil at the bottom of another pan.  Bring the water to a boil and remove it from the heat source.  Slowly immerse a tarnished piece of silver; you'll see an immediate effect as the silver and aluminum make contact.  (Add more baking soda if you don't see this.)  Also, as silver is brought back to its elemental state, hydrogen sulfide is released back into the air.  You'll smell it!  The baking soda serves to remove a thin and transparent coating of aluminum oxide, which normally coats aluminum and thus prevent direct contact.  The baking soda also serves as a conductive ionic solution to permit the passage of electrons from the aluminum to the silver."

Abrading the silver sulfide removes it (and some silver underneath the tarnish). Electrolytic cleaning actually restores the silver previously part of the tarnish. In addition, it cleans crevices inaccessible to manual polishing.

There are disadvantages, though:

If the article has niello inlay, it could be damaged.
If other materials, such as parchment or glue, are part of the article, they would be damaged.
If the article is a valuable antique, cleaning may reduce its value.


Confused about writing out electron configurations I watched the Crash Course chemistry video about the electron. In the video Hank showed an example configuration of iron as $\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6}$ and explained that due to energy constraints, electrons go to the s orbital of the 4th shell before going into the d orbital of the 3rd shell and so they are written in this order.

I looked up iron in Wikipedia and there the configuration is given as $\mathrm{[Ar](3d)^6 (4s)^2}$. I understand how it's shortened with argon, but why in Wikipedia is the d orbital of the 3rd shell written before the s orbital of the 4th shell?

Which one is preferred?

For a given configuration, the order of writing the orbitals is not
  completely fixed since only the orbital occupancies have physical
  significance. For example, the electron configuration of the titanium
  ground state can be written as either $\mathrm{[Ar] \, 4s^2 3d^2}$ or $\mathrm{[Ar] \, 3d^2 4s^2}$.
  The first notation follows the order based on the Madelung rule for
  the configurations of neutral atoms; $\mathrm{4s}$ is filled before $\mathrm{3d}$ in the
  sequence $\ce{Ar, K, Ca, Sc, Ti}$. The second notation groups all orbitals
  with the same value of $n$ together, corresponding to the
  "spectroscopic" order of orbital energies that is the reverse of the
  order in which electrons are removed from a given atom to form
  positive ions; $\mathrm{3d}$ is filled before $\mathrm{4s}$ in the sequence $\ce{Ti^4+, Ti^3+, Ti^2+, Ti+, Ti}$.

Source: Wikipedia article on electron configuration.


Can rearrangement in carbocations cause inversion? Can rearrangement reactions in carbocations cause inversion?

If you have a carbocation rearrangement (i.e. a methyl shift or a hydride shift), is it possible for inversion of stereo chemistry to take place?

The carbocation has the free p-lobe, so is it possible for the hydride/methyl group to bond to either lobe, such that one can cause retention of stereochemistry while the other can cause inversion of stereochemistry? Does this logic even make sense?

Originally this was a comment, but I figure it's good enough for an answer.

As far as I can tell, you seem to be confusing the notion of suprafacial/antarafacial with retention/inversion.

You're asking whether the migrating group can loop around to the other side of the carbocation, right? That's known as an antarafacial process, not an inversion.

Retention, or inversion, refer to the stereochemistry at the migrating centre. If it is a hydride shift, then the migrating centre is a hydrogen, and there is no stereochemistry to speak of. The terms retention or inversion are therefore not applicable to hydride shifts.

In any case, the crux of the matter is that this migration is a concerted process (it occurs in one step), and an antarafacial migration cannot occur due to geometric reasons. The lobe of the p orbital on the other side is way too far away, and the $\ce{C-R}$ $\sigma$ orbital (which contains the electrons that you are trying to migrate) is not going to be able to reach it.

[1,2]-alkyl (or hydride) shifts are always going to be suprafacial!

(As a side note, the suprafacial [1,2] shift has to proceed with retention, as per Woodward-Hoffmann rules.)


What is the difference between Molecularity and Order Of Reaction? In the field of Chemical Kinetics, terms like Molecularity and Order Of Reaction are often used. I am a bit confused between these two terms.

Can someone explain to me about these two terms and their origins?

In fact, molecularity and order of reaction are identical if the reaction is elementary. They may become different when the reaction is complex and has a mechanism.

The molecularity of a reaction is defined as the number of molecules or ions that participate in the rate determining step. A mechanism in which two reacting species combine in the transition state of the rate-determining step is called bimolecular. If a single species makes up the transition state, the reaction would be called unimolecular. The relatively improbable case of three independent species coming together in the transition state would be called termolecular.

The order of a reaction is an experimentally derived number. Let's consider the general reaction: $\ce{A + B-> C + D}$
 The reaction rate$=\ce{k[A]^n.[B]^m}$ where the rate constant $k$ is a proportionality constant that reflects the nature of the reaction, $\ce{[A]}$ is the concentration of reactant $\ce{A}$, $\ce{[B]}$ is the concentration of reactant $\ce{B}$, and $n$ & $m$ are the exponents used to fit the rate equation to the experimental data. Chemists refer to the sum $n + m $ as the kinetic order of a reaction.

In a simple bimolecular reaction $n$ and $m$ would both be $1$, and the reaction would be termed second order, supporting a mechanism in which a molecule of reactant $\ce{A}$ and one of $\ce{B}$ are incorporated in the transition state of the rate-determining step. A bimolecular reaction in which two molecules of reactant $\ce{A}$ (and no $\ce{B}$) are present in the transition state would be expected to give a kinetic equation in which $n=2$ and $m=0$ (also second order).


Product/(s) of the following reaction I was wondering what the product will be in the following reaction:
$$\ce{H3COOCH3 ->[\ce{H+}]} \:?$$

I was thinking about $\ce{CH3COOH}$  being a product but I do not know the mechanism.

If you're sure that the initial formula is $\ce{H3COOCH3}$ (and that's a big if), then I disagree with Richard's answer in comments. $\ce{H3COOCH3}$ is dimethyl peroxide. While peroxides are not the most stable of molecules, and decompose rapidly (often by oxidizing other compounds), they do not spontaneously transform in acid conditions.

If the formula were $\ce{H3C–COOCH3}$ (note the added C), then yes, it's be acid-catalysed ester hydrolysis.


Cathode ray discharge tube I am learning about 'cathode ray discharge tube'. I know when electrons are given out from cathode, they travel to anode, because anode being positively charged attract those electrons and continue to accelerate those electrons. When the electron hit any of the gas atom in tube, it helps to knock out electron from it and therefore, electron knocked also start moving towards anode. But i think that one stage would come when there are no electrons left with gas atoms those can be knocked out and when electron hits the gas ions, it would just slow down the electrons(emitted by cathode) and eventually leading to no ray at the end.

For a cathode ray tube (old-style TV), you want the electrons traveling collision-free in a straight line from the cathode to the screen (anode), where they generate light hitting a phosphorescent layer. For a neon light, you want the electron hitting gas particles so that the entire tube glows.

This is further explained here: Why does the cathode ray tube only start glowing at low pressures?


Finding the initial concentration The solubility product of $\ce{AgBr}$ is $7.7 \cdot 10^{-13}\:\mathrm{mol^2/L^2}$.
What was the initial concentration of $\ce{AgNO3}$ solution, if the precipitation of $\ce{AgBr}$ appears after the addition of $20\:\mathrm{mL}$ of a $0.001$ molar solution of $\ce{NaBr}$ to $500\:\mathrm{mL}$ of the $\ce{AgNO3}$ solution.

I got the solution as $0.054\:\mathrm{M}$. I'm confused with the procedure. This is what I have done.

Precipitate occurs at $K_{sp}=Q$ and $Q=[\ce{Ag+}][\ce{Br-}]$
$[\ce{Ag+}] = [\text{(Vol of $\ce{AgNO3}$)}\cdot\text{Molarity} ] / \text{Total Volume of the mixture}$
Similarly for $\ce{Br-}$
$[\ce{Ag+}][\ce{Br-}] = 2\cdot10^5$

The answer I got it $0.054\:\mathrm{M}$ ? Is that correct ?

This is a titration problem for quantitatively determining the concentration of a solution.

What reaction does occur?

   $\ce{AgNO3 (aq) + NaBr (aq) <=> AgBr v + Na+ (aq) + NO3- (aq)}$
 or essentially $\ce{Ag+ + Br- <=> AgBr v}$

Why is the solubility product important?

   The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.

What can you tell about the equilibrium state at the point when the first precipitate falls?

   The solubility product is matched, hence $\ce{[Ag^+][Br^-]}<K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$

What is the amount of bromine ions added to the solution?

   $n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$

What can you tell about the concentrations in the final mixture?
First, what is the concentration of bromide ions in this mixture?

   $V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$
 $c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$

Second, what can you tell about the concentration of silver ions in the final mixture?

   $c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$

What is the number of moles of silver ions in the final mixture?

   $n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$

What is the initial concentration of the silver nitrate solution?

   $c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$


How much energy is there in 1 mol of diamond and how can you get it? Diamond is made of carbon which is during maybe 3 billion years under pressure is formed in the earth. Imagine you found a big source of diamonds such as coal...how much is the energy of the diamond compared to the coal, and how should you get the energy out of it?

Diamond is just one of the allotropes of carbon. There is nothing unique about diamond that allows it to store significantly more energy than any other compound. It is true, however, that much energy is required to separate carbon atoms in a diamond lattice. This is due to the high bond dissociation energy of a carbon-carbon covalent bond.

Although bond dissociation energy in a C—C bond is relatively high, it is dwarfed by the energy predicted to be released by the mass-energy equivalence theorem.

$E = mc^2$

The nuclear mass of 1 mole diamond is about 12 grams. So the theoretical energy released in a fusion reaction involving one mole of diamond is around 1*$10^9$GJ.

I don't think this is the answer you were expecting. But this is the amount of energy in 1 mole of carbon. The method of releasing all this energy is a different story. Nuclear fusion would release a significant portion of this energy but not all of it.


Are the best buffers those that are in a 1:1 ratio with its conjugate acid/base? I wanted to comment this, but I don't have enough reputation, so I just made this a question. I know that buffers and their conjugate base/acid must exist in similar amounts for the buffer to be effective, so would the buffer be most effective when the ratio is close to 1:1 (of the buffer and its conjugate acid/base)?

Also I'm confused about how much of a weak base/acid buffers should be. If you have a very weak acid, then wouldn't you have a very strong base. In this case will the buffer be ineffective? So the upshot of this paragraph is asking will buffers still work if you have a very weak acid?

Yes, the best buffers are those that are in a 1:1 ratio with its conjugate acid/base, but in a somewhat indirect way; it's because buffer capacity is maximum when $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$, and for that value of $\mathrm{pH}$, the proportion of the acid/base conjugates is 1:1.

The contribution of an acid/base conjugate pair to buffer capacity is

$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{a}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{a}} \right)^2} \ln{10}$

(For derivation, check this response to another question)

The maximum of buffer capacity is achieved when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, and is

$\beta_\mathrm{max}= \cfrac{1}{4} C_\mathrm{A} \ln{10}$

As you deviate from $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ in either direction, concentrations also deviate from 1:1 (as per Henderson-Hasselbalch) and buffer capacity will fall.

Note that the maximum buffer capacity does not depend on the strength of the acid or base, only on the total concentration of the conjugate system $C_\mathrm{A}=\ce{[HA]}+\ce{[A-]}$.

What $\mathrm{p}K_\mathrm{a}$ will determine is at which $\mathrm{pH}$ value will this maximum buffer capacity be achieved, so you will choose an acid/base pair with a $\mathrm{p}K_\mathrm{a}$ that will allow the system to act as a buffer around the $\mathrm{pH}$ that you want to preserve; ideally by choosing a pair with a $\mathrm{p}K_\mathrm{a}$ equal to the target $\mathrm{pH}$, which is when you will obtain the maximum buffer capacity.


Does the composition of a solution keep changing when being boiled? The mole fractions of the components of vapours aren't the same as those of the liquid-liquid solution (unless the solution is azeotropic), this implies a change in the composition of the liquid solution when vapours arise. Consequently, the partial vapour pressures of the components would also change and there would be a messed up cycle. There has to be something that I'm understanding wrongly and I'd like to be told what it is.

It is not a messed up cycle. Let's consider what happens with an example.
If you heat up a mixture containing $10$% ethanol + $90$% water, the mixture boils at $92$°C, producing a vapor containing $50$% ethanol + $50$% water. With this operation the liquid loses more ethanol than water. So its concentration in  ethanol decreases, and it is necessary to heat it up to a higher temperature to maintain the process of evaporation. When about one tenth of the liquid has distilled, the residual liquid contains still $5$% ethanol, and it is necessary to heat it up to $95$°C to obtain and continue boiling. At $95$°C, the vapor emitted contains $35$% ethanol. At $98°$C, the liquid contains still $2$% ethanol and the vapor $20$% ethanol. When the last drops are vaporized, the temperature is $100$°C and the last drops are made of pure water.


Why doesn't tetraiodine exist? I've run into a question that says after doing the Lewis structure of $\ce{IF3}$, which I did, explain why $\ce{I4}$ doesn't exist.

I am confused as to how it can be explained. I've thought that maybe it has to do with iodine's valence orbitals, both hybridised or not being too large and as a result the different orbitals would have to overlap and superimpose. But again I'm not sure.

While it's true that there are polyiodides, the $\ce{I4^2-}$ ion has a completely different geometry from $\ce{IF3}$ molecule. The ion is most easily regarded as two iodine molecules stuck end on and the ensemble is bound because two additional electrons are popped into the in-phase combination of each side's $\ce{I(p_z)-I(p_z)}$ antibonding orbital. This is bonding between the fragments and antibonding within the fragments. Since the individual molecules can't do anything with the extra two electrons, it's best just to stay together and stabilize them together.

As to why there isn't a $\ce{I4}$ that is isostructural to $\ce{IF3}$, this is because the stability of interhalogens is heavily influenced by the ionic character of bonds. Recent work[1,2] has shown that the bonding in halogens is dominated by "charge-shift" interactions in which the ground covalent state is stabilized by ionic states in which both electrons of the bond have been transfered to either atom (e.g. $\ce{F-F}$ is stabilized by $\ce{F+-F-}$ and $\ce{F^--F+}$ states). This stabilization is due to highly electronegative atoms. The stability of higher interhalogens relies on this "charge-shift" stabilization. Since I is a lot less electronegative than F, this charge-shift stabilization is much weaker (and this is also the reason why nearly all of the higher interhalogens have fluorine as a bonding partner). Since this charge-shift stabilization is absent, the 4 iodines are much happier to break apart into 2 $\ce{I2}$ molecules.

Sason Shaik, David Danovich, Wei Wu, and Philippe C. Hiberty, Nature Chemistry 2009, 1, 443 - 449. DOI: 10.1038/nchem.327
Daniel S. Levine, Paul R. Horn, Yuezhi Mao, and Martin Head-Gordon, J. Chem. Theory Comput. 2016, 12 (10), 4812–4820. DOI: 10.1021/acs.jctc.6b00571


Reverse Image Searching a ChemDraw Figure Is there any image-recognition software out there that can read in a high-resolution image of a ChemDraw-generated figure and back out the structure (such as the SMILES or INChI)? I have a bunch of review papers where there are tables upon tables of ChemDraw images of molecules and would really like to put them into some useable database.

The best tool I have come across so far to do this is this one: https://cactus.nci.nih.gov/osra/
I was also reading about this other one, but can't find the link to the tool itself.


Nomenclature of Acetone Why is the preferred IUPAC name of acetone "propan-2-one" and why not simply propanone? It doesn't make sense to specify the position of the carbonyl group as it can occupy only one position in acetone, i.e. on the 2nd carbon.

The IUPAC Blue Book defines rules where locants can be ommitted in IUPAC preferred names.

Unlike terminal locants in names for aldehydes that are omitted, there is no rule for homogeneous chains with three or more identical atoms that would allow for the ommission of the locant $2$ in propan-2-one.

This leads to the result that propanal is a IUPAC preferred name while propanone isn't.


Why doesn't tetrachloridotitanate(II) obey the 18-electron rule? I have worked out that the complex comes to 10 electrons, however I do not understand why it does not obey the 18-electron rule? What are the factors that account for this?

There is more to complex chemistry than the 18-electron rule. It states one condition when complexes may be particularly favourable, namely when all metal d-orbitals are fully populated and the ligands favourably interact with all metal s- and p-orbitals leading to all paired spins and hopefully a large energy difference between populated and unpopulated orbitals. The list of complexes that do not obey the 18-electron rule may well be larger than that of those that obey it. In no particular preference that includes $\ce{[MnO4]-}, \ce{[CrO4]^2-}, \ce{[Fe(CN)6]^3-}, \ce{[Mn(H2O)6]^2+}$ and many more. (A few of those are not typically described as complexes but they are nonetheless.)

Further constraints that you need to be aware of include, but are not limited to:

steric stress, i.e. too many large ligands around too small a metal centre,
charge density, i.e. surrounding a weakly charged cation with too many anions,
the actual electronic configuration, which can render octahedral, Jahn–Teller-distorted octahedral or square planar complexes more favourable.

In the case of a chloridotitanate(II), we first need to realise that we are surrounding a moderately small cation (titanium(II)) with rather large anions (chloride) which may generate steric stress. We also need to observe that titanium(II) is only weakly charged and attempting to generate an octahedral configuration would give us an anion with $-4$ charge — not exactly favourable (exceptions may apply). Also, chlorido ligands are known to be weak-field ligands, so one would not expect an overly large crystal field split.

The two greatest factors are probably sterics and electronics. Consider tetrachloridocobaltate(III) $\ce{[CoCl4]-}$: while potentially six chlorides may fit around the rather small cation, even though it is $+3$ the electronic situation favours only tetracoordination. This results in a 14-electron complex rather than an 18-electron one. In analogy, I am tempted to say that the even lesser charged titanium(II) ion cannot accommodate for a total of six chlorides, allowing only four of them in its vicinity. Taking electronics into account, square planar coordination is out of the question due to lack of stabilising interactions and the complex would be tetrahedral.

In fact, the tetrahedral geometry allows for a mildly stabilised triplet state: both $\mathrm{e_g}$ orbitals are semipopulated which is generally a tad better than two of three $\mathrm{t_{2g}}$ orbitals as it would be in the octahedral case.


Why is the effect of competitive antagonist less than the effect of non-competetive antagonists? Drugs can  be provided as a competive or non-competitive form of agonists. But the effect of non-competitive antagonists is much more than the other, so what is the cause of that? And are there some exceptions?

Personally, I think the statement "the effect of a competitive antagonist is less than the effect of a non-competitive antagonist" is too general to have much meaning. There's a large number of systems, each with a potentially large number of competitive and non-competitive antagonists, each with a potentially large range of activities. Saying a priori that a non-competitive antagonist is more effective than a competitive one is presumptuous.

However, there is a bit of sense in there, for a particular defined sense of circumstances.

By definition, a competitive antagonist is only active against the non-liganded state of the receptor. The native ligand (agonist) and the competitive antagonist compete for the receptor. In contrast, a non-competitive antagonist can generally work against both the liganded (with agonist bound) and ligand-free versions of the receptor. (How active it is against each depends on the particulars about the antagonist and receptor.)

So in a situation where the antagonist isn't at saturating concentrations (say at a concentration near it's IC50), saturating amounts of native ligand (agonist) will completely reverse the effects of a competitive antagonist, as the agonist will compete off the antagonist. However, the non-competitive antagonist will still have some remaining inhibitory function, even at saturating concentrations of native ligand, as it can bind to and inhibit the ligand-bound (agonist-bound) form of the receptor.

However, there's an apples-to-oranges comparison embedded in there. That comparison assumes that the competitive and non-competitive are each near their respective IC50s. This value can change greatly between drugs, so you can have a competitive inhibitor with a really low IC50, and a non-competitive inhibitor with a really high one, such that if you used identical molar concentrations of each antagonist, the competitive antagonist would be successful in almost completely shutting off receptor activity, whereas the non-competitive antagonist would show barely noticeable activity.


Why doesn't chocolate ever feel cold? I noticed that every time I put chocolate in the freezer, then I touch it or eat it, it somehow doesn't feel cold at all.

I figured that it must be that it heats up extraordinarily fast from body heat, because if it had a super high coefficient of heat then it would be just as hard to warm it up as it would appear to get cold.

What property would give chocolate that capacity? I've never seen any other substance be put in a freezer and somehow not feel cold.

One possibility is that chocolate has a lower thermal conductivity than water-based foods, which make up much of the food in our freezers.

If you open your freezer and touch a cardboard food container, then touch a metal ice tray, which one feels colder? They are both the same temperature, but the metal tray feels colder because it conducts heat better than the cardboard box, so the heat gets transferred out of your hand quicker and it feels colder.

Water is a relatively good conductor of heat and is present in large quantities in many foods. Chocolate is largely made up of fatty organics, which do not conduct heat as well. It is thus very possible that chocolate has a lower thermal conductivity than many frozen foods and therefore doesn't feel as cold to the touch in comparison.


Charge separation and stability in resonating structures I know that when there are like charges on a resonating structure, the farther apart they are, more stable is the structure due to reduction in repulsive forces.Is the same true for opposite charges? which structure is more stable -
1)with positive charge on one C and negative on adjacent C
2)with positive charge on one C and negative charge on C separated by 4 carbons between them?

The answer is 1) when positive and negative charges on C atoms are adjacent.This is because in the second Resonating Structure the charges have to be seprated and as both charges are opposite in nature they will attract each other and make the process of seprating them difficult making it unstabilized.


How to find the strongest base using stability arguments Answers: The most acidic is D and the most basic is A.

So first off I compared A with B:

I'm kinda confused on this. I looked at the base NH one and saw that when it was protonated it lost its aromaticity. So as a result, the NH²⁺ one must be an unstable acid, thus it is the stronger acid. The stronger acid must have the stronger base on the same side, thus I concluded that A is stronger than B. Is this the correct way to think about it?

Then, I compared A with D:

I'm unsure about this as well.

When I compare the two bases on each side (the O⁻ and OH⁺ one), I see that the O⁻ one is more stable (on the left side), thus it is the weaker base, meaning the OH⁺ one (on the right side) is the stronger base. However, that is not right as A is the strongest.

However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O⁻ one must be stronger. This way of thinking got me the correct answer. However, I'm still unsure if this is the reason why as I was always told to compare the charged species.

Also E doesn't have a lone pair to donate, so I neglected that in the conversation of the strongest base. For C, it just didn't seem likely to me.

A vs. B

I will label your first reaction scheme, using the letters A-E for the deprotonated (i.e. as shown in the problem) form and the symbol $\ce{AH+}$ etc for the protonated form (after acting as a base and accepting a proton):

$$\ce{A (stronger base) + BH+ (stronger acid) <=> AH+ (weaker acid) + B (weaker base)}$$

Your argument about aromaticity is great; imidazolium ($\ce{BH+}$) is the stronger acid compared to cyclohexanol ($\ce{AH+}$), so the alkoxide ($\ce{A}$) is the stronger base compared to imidazole.

A vs. D

The protonated tetrahydrofuran (D) is highly reactive ("unstable") in water because it would want to get rid of a proton. That does not make it a stronger base. In your reaction scheme, you add another proton to a species that already is a strong acid (not sure if it even exists, see Why doesn't H₄O²⁺ exist?).

$$\ce{A (stronger base) + DH+ (stronger acid) <=> AH+ (weaker acid) + D (weaker base)}$$

The cyclohexoxide ($\ce{A}$) is the stronger base than protonated tetrahydrofuran ($\ce{B}$) because we know phenol ($\ce{AH+}$) exists, but we don't think twice-protonated tetrahydrofuran ($\ce{BH+}$) exists.

  the O- one is more stable (on the left side), thus it is the weaker base, meaning the OH+ one (on the right side) is the stronger base

You have to specify stable in terms of what reaction. Protonated tetrahydrofuran is very stable against protonation, so it is not a base.

  However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O- one must be stronger.

Now you are comparing the two in terms of either one losing a proton, which is the right thing to consider when comparing two acids.

C and E

You are right about E, the nitrogen already makes four bonds, so it can't accept another proton. I'm not sure where a proton would go in C. In any case, it is not a good base.


Is selenocysteine C3H6NO2Se or C3H7NO2Se? I'm a beginner in chemistry, and recently I am interested in selenocysteine.

While most websites (Wikipedia, ChemSpider) say it is $\ce{C3H7NO2Se}$, PubChem says it is $\ce{C3H6NO2Se}$.

Which one is correct? Or are both reasonable?

Selenocysteine can be either of the following: $\ce{C3H8NO2Se+, C3H7NO2Se, C3H6NO2Se-}$ or $\ce{C3H5NO2Se^2-}$. This is because the neutral form of selenocysteine is both a base and a diprotic acid: the amino group can be protonated and both the carboxylic group and the selenol can be deprotonated.[1]

The above graph taken from the chemicalize.com property calculation site[2] shows the distribution of the different forms of selenocysteine in aqueous solution. The underlying data has been calculated by chemicalize.com and is not experimental data. For the cell, neutral pH is most important. We can see that the most common form by far at neutral pH is $\ce{C3H7NO2Se}$, but a small quantity is present as $\ce{C3H6NO2Se-}$. Once selenocysteine has been incorporated into a protein, only the selenol function remains potentially acidic in aqueous solution; however, it’s $\mathrm pK_\mathrm a$ value is higher than neutral pH by three logarithmic units. Thus, we can assume that it will not be deprotonated notably.

The actual $\mathrm pK_\mathrm a$ values have been measured by Huber and Criddle.[3] These show that chemicalize.com’s predictions are rather bad. They determine the $\mathrm pK_\mathrm a$ values to be $2.01,5.24$ and $9.96$ for the carboxylic acid, the selenol and the amino group, respectively. Using these values, we expect selenocysteine to be present in the monoanionic form $\ce{C3H6NO2Se-}$ at cellular pH. A distribution graph based on these values (provided by Canadianer) can be seen below.

PubChem’s suggestion of $\ce{C3H6NO2Se}$ is wrong. If you add up valencies, you will notice that it is only possible as a radical structure. It would be acceptable if a negative charge had been included (which I don’t see on the site).

On the other hand while the chemical formula on Wikipedia or ChemSpider is correct and gives a neutral compound with all spins paired, this formula does not correspond to the species predominantly present in strongly buffered aqueous solutions.

Notes:

[1]: This way of writing it assumes the uncharged neutral form. In aqueous solution, the molecule is expected to adopt a zwitterionic neutral form with the carboxylate being deprotonated and the amino group being protonated. From this form, the carboxylate is the base while the amino group is one of the two acidic functions.

[2]: Unfortunately, I am not aware of a way to link directly to the calculation. However, this calculation is free if you register, since the molecule contains less than twelve heavy atoms.

[3]: R.E.Huber, R.S.Criddle, Arch. Biochem. Biophys. 1967, 122, 164–173. DOI: 10.1016/0003-9861(67)90136-1.


How will the degree of conjugation effect the light emitted? (glow stick reaction) So regarding the chemiluminescence in a glow stick reaction, the glow is due to the unstable compound 1,2-dioxetanedione decomposing into carbon dioxide, releasing energy which is then absorbed by the electrons in the molecules of the dye. This promotes the electron to an ‘excited state’ and when the electrons fall back to their ‘ground state’, they emit photons of light, where the exact energy of the light given off is dependent on the structure of the molecule.

Since I learned in organic chemistry that the more conjugated the molecule, the less energy required to promote the electrons to an excited state, I assume that is the reason why more conjugated systems (in the photo, Rhodamine B) emit lower frequencies of light (red), since lower energy is required to excite the electrons.

Am I correct?

In addition, if that is so, wouldn't the amount of conjugation of dye affect the efficiency of the chemiluminescence reaction? So for example, since more conjugated systems require less energy for the electrons to be excited, they will emit more light with the same amount of energy compared to less conjugated systems that need higher energy (emits higher frequency light).

I have asked my teacher multiple times, but she does not provide me with a satisfying question, I pray for help.

Thank you.

The initial chemiluminescent (CL) species transfers its excited state energy to the dye. This depends on the overlap of the CL emission spectrum with the absorption spectrum of the dye and their spatial separation, i.e. better spectral overlap and higher concentration more transfer and more dye emission. (see Forster energy transfer for mechanism).  The dyes used have 80-100% fluorescence quantum yield, so v. high chance of emission. The intensity you experience will also depend on how your eyes respond to different colours.

The energy gap in the dyes does not greatly affect their fluorescence efficiency, this is primarily governed by interactions within the dye to produce triplet states which drain energy out of the fluorescent singlet state.


How do noble gases bond with themselves? Noble gases have full electron shells, which virtually blocks any other element from bonding with it. However, I've heard about cases where they bond to each other  - for example, helium can apparently form a dimer $\ce{He2}$.

How is this possible?

Noble gases usually do not form strong bonds between their atoms - it takes fair amount of energy to dimerise them into excimers, but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms. On the other hand there are many stable molecules created by heavier noble gases (mainly xenon) with other elements.

As mentioned in comment, you may heard about detection of so called van der Waals molecules of helium, which aren't "true" molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it.


If the temperature of a solution decreases, what is the sign of the enthalpy change? Here is what seems to be a straightforward question: you dissolve some ammonium chloride in water at $25\ \mathrm{^\circ C}$, lowering the temperature of the solution. Is the change in enthalpy for the dissolving process positive or negative?

My reasoning: if the temperature of the solution decreases, it means heat was released into the surroundings, so the enthalpy change is negative.

Apparently the answer is the enthalpy change is positive. What am I doing wrong? Whatever it is, it must be stupid, but please forgive me for its late here.

When I was doing enthalpy, this was one of the tricky concepts that I still sometimes get wrong. Here is how you are supposed to remember it.

  Identify the System

  The system is where the reaction happens. You cannot physically monitor the change in heat here. The change you always measure happens in the surroundings.

  Identify the Surroundings

  This is the surroundings where usually the change is expressed as.

Now that we got the basics cleared away, let us look at your problem.

  My reasoning: if the temperature of the solution decreases, it means
  heat was released into the surroundings, so the enthalpy change is
  negative.

This is what I usually think of, but now you must understand that as I've said before the change always happens in the surroundings and is associated with the surroundings. So the temperature of the solution decreases, which means that the surroundings temperature decreased. The solution is part of the surroundings. This means that the temperature of the system increases. Thus making the $\Delta H$ positive and the reaction endothermic.

Good Question, always remember that you cannot, in most situations, physically measure the change in the system.

Thanks.


Transiesta Electrode Run I am using a (1,1,1) gold electrode for electron transport experiments using Transiesta. I have never used it before. I am wondering if when I do the electrode run, do I build a system where it is the left and right electrodes connected to each other? or do I use the entire device (molecule AND electrodes) and just set a flag in the file?

(In response to andselisk's request for more information about Transiesta: it is a computational chemistry program that is a part of the Siesta Suite. It is used mainly for electron transport calculations.)

There is a manual available that states (page 111).

  TranSiesta
  defines the Left Electrode to be the first atoms specified in the SR
  .fdf
  file,
  and the Right Electrode to be the last ones.  The transport direction has to be considered
  to be the the third cartesian axis, the
  z
  axis.  The Left Electrode atoms must have smaller
  z
  components than the Right Electrode atoms.  It is also crucial that the atomic positions
  specified at the left (right) EL calculation must be equivalent to the left (right) electrode
  part of the SR setup.  Here, equivalent means that they can be made equal by a simple
  translation  in  space.   It  is  also  possible  to  use  buffer  atoms.

There is more information that I think may address further doubts in the manual as well. I am not very familiar with the program. So, this is as much as I can help you. You could try asking the mailing list, though SIESTA's mailing list does not get many responses to be honest.


How are there three possible tautomers of 2,2,4-trimethylheptane-3,5-dione? The answer for the above question is three. But I am able to draw four structures:

Where am I getting it wrong?

In your attempt the 4th compound is not reported to have significant percentage in solution (acidic or basic). This is because once the tautomerisation occurs for example let's say the 1st compound, there is a tautomer having a six-membered conjugated ring involving H-bonding. This thereby increases the stability of the compound and consequently increasing the activation energy for the tautomerisation you proposed in compound 4. This is why compound 4 is not formed.

Also keep in mind that stereo-chemistry is also important while finding number of possible compounds.