What is the dipole moment direction in the nitrosonium ion? In the recent question about the molecular orbitals of $\ce{NO^{$x$}};\ x\in\{-1,0,+1\}$ the follow-up title question arose:

  Just a random side-question: What’s the dipole moment’s negative and positive sides in $\ce{NO+}$?

At the time I was unable to answer that question sufficiently, therefore I share the addendum Q&A style. The question further deconstruct the molecular orbital components to the dipole moment.

In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With $r=107.8~\pu{pm}$ on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are attached at the bottom of the post.

I am loosely following the same procedure/ conclusions as in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.) since $\ce{NO+}$ is isoelectronic with $\ce{CO}$.

According to the definition of the dipole moment
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
positive values in the following table indicate that the negative end is at the nitrogen atom.

\begin{array}{lr}\hline
\text{Orbital} & \text{Bond Moment}           \\\hline
\text{MO 1 } (\sigma)            &   -4.8336  \\
\text{MO 2 } (\sigma)            &    5.5232  \\
\text{MO 3 } (\sigma)            &   -0.7891  \\
\text{MO 4 } (\sigma)            &   -3.0061  \\
\text{MO 5 } (\pi   )            &   -1.1642  \\
\text{MO 6 } (\pi   )            &   -1.1642  \\
\text{MO 7 } (\sigma)            &    4.7383  \\\hline
\text{MO 8 } (\pi, \text{virt.}) &   (2.3831) \\
\text{MO 9 } (\pi, \text{virt.}) &   (2.3831) \\\hline
\text{Total } \sum_i^7 \text{MO }i               &   -0.6956  \\\hline
\end{array}

While we see the same polarisation of the molecular orbitals as in carbon monoxide, the magnitude differs and the HOMO of $\ce{NO+}$ cannot compensate for the polarisation of the electron density towards the oxygen.

Appendix

Base calculation for $\ce{NO+}$ after previous optimisation, E(RHF)=-128.972090453 input file:

%chk=mp2qzvpp.chk
#p MP2/def2QZVPP
scf(verytight)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

base

+1 1
N   0.0 0.0 -0.575312
O   0.0 0.0  0.503398

Analysis calculation input file for MO 1

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

analysis

+13 1

(Both empty lines are necessary.)

Analysis calculation input file for MO 2, etc

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)

analysis

+13 1

2,1

The last line needs to contain the complete reordering procedure. The MO to be analysed needs to be in first place. Therefore for MO 3 it needs to read 3,1,2, for 4 4,1,2,3, etc.


Using sodium peroxide to scrub carbon dioxide emissions I am a student from a school and recently I came across the reaction

$\ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2}$

Can't we use this reaction to reduce $\ce{CO2}$ in the atmosphere by placing some $\ce{Na2O2}$ filters in places where there is high level $\ce{CO2}$ emission? I then realized that the filter would have to be loaded with lot of $\ce{Na2O2}$ to achieve this. So why not adsorb as much $\ce{CO2}$ as much as possible using a good adsorbent for $\ce{CO2}$. I am not sure if there are other methods better than these but I would like to know if there is any?

The problem with carbon capture is not so much that we lack the technology to do it, it's more that we lack the technology to do it in a way that is economically feasible.

Take your example - according to the US Energy Information Adminstration, it takes about 1.09 lbs of coal to generate 1 kWh of electricity. A typical coal-fired plant might produce an average of 500 MW throughout each day (this is a rough estimate), which works out to roughly 12,000 MWh, or 12,000,000 kWh. So that's something like 12 million pounds of coal each day.

If we assume that coal is 100% carbon (close but not quite), then it's about 37.8 mole C per pound of coal. That's 453.6 million moles of carbon each day being converted to $\ce{CO2}$, and since the combustion of carbon follows:

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

that means a single medium sized-coal plant produces roughly 454 million moles  of $\ce{CO2}$ each day. By now you are probably getting a sense of the scale of this problem. Since sodium peroxide and carbon dioxide have a 1:1 ratio in the absorption equation:

$\ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2}$

you will need an equal number of moles of sodium peroxide. That works out to about 78 million pounds per day.

Alibaba lists sodium peroxide at around \$400-500 per ton, so that means it would cost you (at a minimum) \$15.6 million dollars per day to scrub the $\ce{CO2}$ emissions from one medium-sized power plant. The real costs would be much higher because of the additional equipment and operating expenses.

Electricity currently sells for about \$0.10 to \$0.20 per kWh in the US, and around half of that is usually for production, with the rest being delivery. That means our 500 MW coal plant is going to be bringing in something like (at the high side) \$1.2 M each day. As you can see, the costs for scrubbing with sodium peroxide far exceed the potential revenue for the plant.

Now, there are other options:

One is to reuse the sodium peroxide, although the costs of regenerating it might exceed just buying more, and you would still need to find a place to store the $\ce{CO2}$.

Another is to use a cheaper/more effective absorbent - according to Wikipedia, amine-gas scrubbing is the most promising technology so far, although I am sure it faces the same kinds of cost scaling problems.

Another more popular approach is carbon capture, where carbon dioxide is basically just pumped into underground reservoirs in the hopes that it will stay there. While this is probably going to turn out to be the most economical approach, it has the risk of releasing large amounts of $\ce{CO2}$ accidentally, as well as having unknown effects on the local geological formations.

Yet another option is to use photosynthesis to scrub carbon dioxide. This has the advantage of being very cheap in comparison to other methods, but the downside is that the rate at which plants and algae can consume $\ce{CO2}$ is much slower than the rate at which we produce it.

It's a big problem, and all of the proposed solutions so far have been very expensive to implement - which is the main reason the fossil fuels industry opposes them.


Double anode in simple cell What would happen if I kept two copper plates and one zinc plate in a solution of dilute sulfuric acid, and joined two copper plates with wires?

(The wires would connect to a single wire and be joined from that wire to the single zinc plate.)

What would be the voltage? Will there be something wrong?

Neglecting any resistance in the connecting wires, this situation is no different than having one large copper plate. If everything is connected together, you can't measure any voltage because the fermi levels of all the metals will be equal (at equilibrium). This is essentially shorting the terminals of a battery and the zinc will start to dissolve in the electrolyte and hydrogen will be produced at the copper electrodes. You can measure a potential difference between the copper plates and the zinc plate, if they're not shorted, but it won't be any different with one copper plate or two for an open circuit.

If you're actually using the cell to do work, you may get more work out of the cell with more copper area, as the larger surface area will result in less solution resistance and better diffusion.


Why don't atoms with expanded valence shells have a formal charge? I'm working through Chemistry - Principles and Practice, 3rd Edition and I have few complaints so I don't think it's the text. At one point, I was asked to draw the Lewis structure, with formal charge, of SO3. The central sulfur atom has six covalent bonds with the surrounding oxygen atoms. Based on the calculation for formal charge, you'd expect one of +3, but there is none. How am I to know when there is or isn't a formal charge, and why?

Formal charge is the difference between the number of valence electrons that a neutral unbonded atom has and the number of valence electrons assigned to that atom when after it forms bonds in a molecule. It looks like you are making your mistake in assigning the proper number of electrons to the sulfur atom.

The electrons assigned to an atom are:

All nonbonding (lone pair) electrons on that atom in the Lewis Structure.
One half of all bonding electrons (one electron for each single bond, two electrons for each double bond, etc.).

Consider water, $\ce{H2O}$. The oxygen atom has two bonds and two lone pairs. Oxygen originally had 6 valence electrons. We assign 6 electrons to it (4 nonbonding and 2 bonding). The formal charge is $6-6=0$.

Now consider the hydroxide ion, $\ce{OH-}$. The oxygen in hydroxide has 7 electron assigned to it: 6 nonbonding and 1 bonding. The formal charge is $6-7=-1$.

Now consider the nitrogen atom in the nitrate ion, $\ce{NO3-}$. Nitrogen originally has 5 electrons. In the nitrate ion we assign 4 electrons to that nitrogen (1 for each single bond and 2 for the double bond. There are no lone pairs). The formal charge is $5-4=+1$.

As a final example, let's look at the P in phosphate, $\ce{PO4^3-} $. Phosphorous original had 5 electrons, like nitrogen. We assign 5 electrons to phosphorous in phosphate (1 for each P-O bond and two for the P=O bond). The formal charge is $5-5-=0$

How many electrons are assigned to the sulfur atom in $\ce{SO3}$? From the question it sounds like you have the correct Lewis Structure. It looks like you are making your mistake in assigning the proper number of electrons based on the number of bonds.


Would molybdenyl chloride or tungstenyl chloride work as an alternative to chromyl chloride test? (This is a follow up of this question)

Chromyl chloride test is by far the best test to determine the presence of chloride ions in a salt because no analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides. This test is considered selective for $\ce{Cl−}$.

But can molybdenyl chloride work similar to that of chromyl chloride?.  I couldn't find enough information in the internet about the feasibility of the compounds as a replacement to chromyl chloride. What about tungstenyl chloride?

In this link, I found a preparation process of molybdenyl chloride but that doesn't follow the same reaction used to make chromyl chloride. It uses common salt, sulfuric acid and molydenum trioxide. This made me curious. Common salt contains chloride that ultimately reacts with other reagent to form molydenum chloride. If I use another chloride containing compound and react with the other reactants, will molydenyl chloride form?. I am aware of thermodynamic or kinetic aspect of the reaction for its feasibility but still will the reaction count as a form of 'molybdenyl chloride test'?

In the chromyl chloride test, chromyl chloride is not a reagent that can be added to the sample to detect any chloride present. It is made in situ through a reaction of chloride containing sample, potassium dichromate and sulfuric acid.

$$\ce{NaCl + K2Cr2O7 + 3H2SO4 -> 2Na2SO4 + K2SO4 + 2CrO2Cl2 ^ (red fumes) + 3H2O}$$

Now in order to make molybdenum chloride work like chromyl chloride, you have to actually make molybdenyl chloride. For this, you need potassium dimolybdate. But it is quite unstable and is prone to decomposition. From here:

  Potassium Dimolybdate, $\ce{K2Mo2O7}$, is formed by fusion of suitable quantities of $\ce{K2CO3}$ and $\ce{MoO3}$, but is unstable; the aqueous extract of the fused mass yields large crystals which immediately begin to decompose.

Making tungstyl chloride is not feasible because in order to make it, you need potassium ditungstate which is costly, rarely used reagent and can be hardly found in chemical vendor or supplies. So, this method is out of scope and reach.

Potassium ditungstate or paratungstate has a unique formula that does not follow the general formula of potassium dichromate or potassium molybdate. Its formula is given as $\ce{5K2O.12WO3.11H2O}$ or $\ce{3K2O.7WO3.6H2O}$


Cleaning stains with hydrogen peroxide and bleach A while back, I discovered that if you have a stain on a white shirt, you can remove it by putting hydrogen peroxide on the stain, then adding some bleach to it.  I have used this for getting out wine, sweat, chocolate, etc.  The reaction produces gases and heat.  What is happening in this reaction and why does it clean stubborn stains?

Peroxide already is a "bleach", and you don't say what the other one is. The gas you see evolving is most likely oxygen (from the peroxide).

Generally (most importantly) bleach oxidises double bonds (which make the colour in many organic compounds). The remaining molecule is also more polar afterwards, because an oxygen-containing group was inserted, and therefore easier to wash out with water.


Rotten meat odor in plastic I've been struggling for a few days with the cleaning of a fridge that was left off for a few weeks with raw chiken meat in the freezer compartment. When I found it, the meat had turned liquid with the smell you'd expect, spread in the entire fridge. I started cleaning thoroughly with bleach with little effect and then vinegar after stumbling on this piece of advice. But there is still a noticeable smell lingering.

My question is then: Is it possible that the plastic of the fridge has absorbed the cadaverine and/or putrecine that I assume are responsible for the smell? And if so will any amount of vinegar (or some other household acid) be able to remove the remaining smell?

I recommend you to put a large amount of activated charcoal into the fridge and left it several days. If you've still sensed the smell, replace the old activated charcoal with the new one. I hope it help you get rid of the smell.


What is the right quantum numbers of Copper? Copper have $[Ar] 3d^{10} 4s^1$ electron configuration. And Zinc have $[Ar] 3d^{10} 4s^2$ electron configuration.

My question is, what is the right quantum numbers of last electron of copper? Is it $$a. n=3,l=2,m=+1,s=-1/2$$ or $$b. n=3,l=2,m=+2,s=-1/2$$

If I use A, the electron configuration of copper is $[Ar] 3d^{9} 4s^2$ and it isn't stable. But if I use B quantum numbers, it is the same with the last electron of zinc.

The correct quantum numbers for last electron of Copper ($\ce{Cu}$) would be $B. n=3,l=2,m=+2,s=-1/2$.

  if I use B quantum numbers, it is the same with the last electron of zinc.

Yes, last electrons of Both can have same quantum numbers because Pauli Exclusion law says that no two electrons in an atom can have all four quantum numbers same but it doesn't say that two electrons of different atoms can't have all four quantum number same.

Refer this for more.


What does K3Fe(CN)6 become? What does the salt $\ce{K3Fe(CN)6}$ change to when it is dissolved in water? I mean, I think its going to ionize, but in what way? Is there a way to know this? I tried to search the web but wasn't able to find the answer.

In water it should dissociate into $\ce{K+}$ and $\ce{[Fe(CN)6]^3-}$. This is because the $\ce{Fe}$ forms an octahedral complex with $\ce{Fe}$ at the centre and six $\ce{CN-}$ ions distributed around this each at 90 degrees to one another. This ion is sufficiently stable that it will not dissociate in water. The bonding is due to $\mathrm d$ orbitals on the $\ce{Fe}$. All undergraduate inorganic text books will give details of this. The Wikipedia page on ferricyanide gives pictures of the complex's geometry.


Selective reduction of nitro group to amine, in benzene ring containing nitrile? I would like to know if there is a possible route of synthesis that may reduce the shown compound A to the following product B.

I would normally reduce a nitro to an amine using $\ce{Pd/C}$ and $\ce{H2}$, but since the nitrile is sensitive to this method, I need an alternative.

Thanks, in advance.

My answer is intended to support Waylander's suggestion elsewhere. I thought it'd be better to give OP some insight of this reduction mentioned (Ref.1). To my knowledge, after doing thorough literature search, the aromatic nitro group reduction by $\ce{SnCl2.2H2O}$ is the best method so far in the presence of other sensitive  groups such as nitrile group on the aromatic nucleus. The abstract of the reference 1 stated that:

  Aromatic nitro compounds are readily reduced by $\ce{SnCl2.2H2O}$ in alcohol or ethyl acetate or by anhydrous $\ce{SnCl2}$ in alcohol where other reducible or acid sensitive groups such as aldehyde, ketone, ester, cyano, halogen, and O-benzyl remain unaffected.

This point is depicted in following diagram as two examples given in Ref.1:

However, as shown in the separate box in the diagram (under the title, Stephen reduction of nitriles), if you use $\ce{HCl}$ (hydrogen chloride gas) with $\ce{SnCl2}$ as other suggestions, the result would be different (Ref.2). $\ce{HCl}$ is added to the nitrile group to give the imidoyl chloride salt first, which is then reduced by the tin(II) chloride. The initial product is the complex with $\ce{SnCl4}$ (which is often crystalline) that is then hydrolyzed to the corresponding aldehyde (see the insert in the diagram).

For the convenience of OP and other readers, I'd include a typical reduction procedure in Ref.1:

  The reduction of p-nitrobenzoic acid: A mixture of $\pu{1.67 g}$ ($\pu{0.01 mol}$) of p-nitrobenzoic acid and $\pu{11.275 g}$ ($\pu{0.05 mol}$) of $\ce{SnCl2.2H2O}$ in $\pu{20 mL}$ of absolute ethanol is heated at $\pu{70 ^\circ C}$ under nitrogen. After $\pu{30 min}$ the starting material has disappeared and the solution is allowed to cool down and then poured into ice. The $\mathrm{pH}$ is made slightly basic ($\mathrm{pH}\ 7-8$) by addition of 5% aqueous sodium bicarbonate before being extracted with ethyl acetate. The organic phase is thoroughly washed with brine, treated with charcoal and dried over sodium sulfate. Evaporation of the solvent leaves $\pu{1.5 g}$ ($94.5\%$) of p-aminobenzoic acid, which gives one spot on TLC and melts over $\pu{300 ^\circ C}$.

References:

F. D. Bellamy, K. Ou, “Selective reduction of aromatic nitro compounds with stannous chloride in non acidic and non aqueous medium,” Tetrahedron Letters 1984, 25(8), 839-842 (https://doi.org/10.1016/S0040-4039(01)80041-1).
James A. Knight, Harry D. Zook, “Reduction of Aliphatic Nitriles by the Stephen Reaction,” J. Am. Chem. Soc. 1952, 74(18), 4560-4562 (https://doi.org/10.1021/ja01138a031).


Why does the ink on my whiteboard move? ...or at least I thought so, when I saw that the ink that was on my whiteboard last night became really messed up the next day.

The image below shows my whiteboard before (on the left) and after (on the right). The time span was one night.

As you can see, the words became really squiggly, and they mostly travelled downwards.

When I woke up, a few hours before I noticed my whiteboard mess, I noticed lots of condensation water on my whiteboard due to my air-conditioner (my whiteboard is made of glass, if it matters).

My theory is that the water pushed the oil surfactant* of the marker ink mostly downwards. The oil surfactant in the ink coats the board to ensure easy erasing, so the oil should be the closest chemical to the board. Since oil and water are immiscible, the hydrophobic oil ended up being pushed away by the sliding condensed water (sliding due to gravity) and the ink travelled down.

However, my theory does not explain why the ink travelled upwards in some parts of the board. I have not shown it here due to sensitive information.

Is my theory lacking? What do you guys think about this phenomenon? And no, it wasn't a prank.
Can the ink travelling in directions other than downwards be explained?

  *Regarding ingredients, all alcohol-based dry-erase marker inks contain 4
  basic ingredients:

  Release agent- an oily (can be simply a mineral oil) surfactant or co-
  solvent which will coat the board to provide easy erase barrier (such that
  the colorant slides off of the board surface and does not penetrate or
  stain over time)

This is a wild guess, so I am open to criticism.

Brownian motion could explain this phenomena.

Water was on the board, so it is possible a film of water condensed on the board some time in the night.  This means the humidity was high while the board was still relatively cool. This film may have allowed the oil particles to suspend.
The ink bound tighter to itself than the non-porous glass board, and the ink itself may be buoyant in water.
The ink binding to itself explains why the letters stayed semi-intact instead of diffusing into a mess.
The water molecules are moving constantly which would allow the bigger particles, which in this case would be the letters, to do the random walk characteristic of Brownian motion.
(As a note the random walk characteristic of Brownian motion tends in the direction of gravity.)
The diffusion coefficient of the letters would be temperature dependent, so the temperature in the room may have briefly spiked just warm enough for the oil to diffuse some then it came back down so the ink did not run completely off the board.


How do I prepare 2-(3-azido-3-oxopropoxy)ethyl methacrylate from methacrylic acid? I am having some trouble finding a proper synthesis route for the following product.

Originally, I was thinking of reacting methacrylic acid with a large excess of ethylene glycol (along with an acid catalyst), in accordance with Fischer esterification, to obtain the following structure.

I subsequently thought of reacting a base such as sodium hydride, NaH, with the alcohol group (removing its 'H') in the hopes of connecting it to the other fragment, which I thought would contain an alkyl halide. However, I was unsure how realistic this step might be due to the following concern: couldn't the hydride ion, which is a very strong base, attack the ester group instead of the alcohol (despite the alcohol having the lower pKa of the two groups)?
To get around this, I attempted a synthesis, pictured below, starting from ethylene glycol. My main concern with this, however, is the potential for intramolecular Fischer esterification instead of, as depicted, the esterification taking place through methacrylic acid and the acid catalyst (e.g., HCl). How valid is this concern? Additionally, please identify any mistakes I may have made in my synthesis and suggest improvements, even if you have a different synthesis route in mind.

You're going need a protection strategy to make this work. Here is one possible route:

Start with 2-Chloro/bromo ethanol and protect the OH with a silyl
ether, $\ce{SiMe3}$ will do, but requires anhydrous workup.
React the protected haloethanol with the anion of
3-hydroxypropionate-t-butyl ester (commercially available) ($\ce{NaH,
NaI, DMF}$) then deprotect the product (mild acidic work up for $\ce{Me3SiO}$
or $\ce{F-}$ for others).
React this product with the acid chloride of methacrylic acid
(commercially available). Isolate and purify the product.
Remove the t-butyl ester with $\ce{TFA/CH2Cl2}$, concentrate to dry and
form the acyl chloride ($\ce{SOCl2}$, cat DMF).  React this with sodium
azide to give the product following the first part of this prep
(Org. Syn.).


Can a molecule be achiral whilst it's resonance form chiral? I'm trying to work out whether methylphenylsulfoxide is a chiral molecule.

It is my understanding that the sulfur uses sp2 hybridised orbitals. Why is this? The central sulfur atom is bound to 3 atoms and then has a lone pair, so surely it has to use sp3 hybridised orbitals?

Or does it use sp2 hybrid orbitals so the empty p orbital allows for extended delocalisation of the benzene ring through the sulfur and onto the oxygen?

Apparently there are resonance forms of this molecule which are chiral.

This molecule looks odd to me, would it make more sense to have electrons resonate out of the benzene ring?

Unfortunately the sloppy depiction of sulfoxides will cause such misconceptions. Your analysis is quite good:

  The central sulfur atom is bound to 3 atoms and then has a lone pair, so surely it has to use sp3 hybridised orbitals?

The only thing one should keep in mind is that the model of hybridisation starts to break down for these compounds (third period and below).
In this case it also is not strictly necessary to employ hybridisation, but as a first (very rough) approximation it can give you an educated guess.
In any case, the sulfur is approximately tetragonal coordinated.

Both depictions are indeed missing the lone pair at sulfur.

As such the first one would has 10 electrons around sulfur, which used to be regarded as correct, but is now a deprecated viewpoint. Sulfur would need to use d orbitals, which has been proved wrong.
The charge separated structure is superior to the above, because it does not suggest the use of d-orbitals.

The issue whether or not these compounds are chiral has been answered by Jan (Can heteroatoms with lone pairs be chiral centres?):

  The sulfoxide case. Sulfoxides $\ce{R1−SO−R2}$ have a free electron pair and a trigonal pyramidal configuration on the sulphur atom. The lone pair is in an orbital with a high s-contribution meaning that the nitrogen inversion highlighted above is rather complicated. Thus, sulfoxides can be assumed to be chiral on sulphur (although in practice they are often generated racemicly).

You have asked more general though:

  Can a molecule be achiral whilst it's resonance form chiral?

It is very important to realise that resonance only affects the electronic structure. Therefore in all resonance contributors the position of the nuclei is the same. Since chirality is determined by the spatial arrangement of the atoms, the question itself is void. There cannot be a set of resonance contributors which has chiral and achiral structures. Please also see What is resonance, and are resonance structures real?


Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof) Why is the C-S bond in $\ce{CH_2SH^-}$ longer than in $\ce{CH_3SH}$? Wouldn't there be resonance in the former, forcing the C-S bond to contract?

http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431

Also while we're at it are there any non ab initio calculations/empirical studies on this interesting thiol ion?

Quick Summary

The linked paper (from 1975) is probably wrong. A slightly more recent paper (J. Am. Chem. Soc., 1989, 111 (21), pp 8112–8115) suggests the bond lengths are about equal. Some quick calculations (below) using multiple modern methods suggest that, as expected, the C-S bond length in $\ce{CH2SH-}$ is shorter than in $\ce{CH3SH}$.

For both older papers, the computational results (using basic Hartree Fock) would be considered highly suspect under modern peer review, since Hartree Fock is known to have problems with anions. Certainly the 1975 paper doesn't even mention the basis set.

Seen below, it's clear that the results are method dependent, but the best current methods, with reasonable basis sets indicate the bond length is at least slightly shorter due to some partial double-bond character.

More Details

The first thing to consider in this is that there are two forms of $\ce{CH2SH-}$, syn and anti, based on the orientation of the formally C-bound anion and the SH group:

syn form

anti form

These will have different bond lengths, since the anti form will have contribution of resonance structures II and III from Ron's answer. This gives some double-bond character and should likely have a shorter bond, although electrostatic repulsion could also be invoked.

The syn form won't have a clear effect, since the C anion lone pair and the S lone pairs are on the opposite side of the bond. My guess is that the bond length is about the same as $\ce{CH3SH}$.

I think the clear message is that we can explain a lot of things, but it's better to know if our results are accurate before trying to explain them.

What I see with more modern computational methods is that the syn form has similar C-S bond length to $\ce{CH3SH}$ and the anti form is slightly shorter. I suspect the previous result is simply wrong. Calculations involving anions are tricky, since the extra electron is usually highly diffuse. I expect the basis set was simply insufficient.

The difference in energy between the two forms is ~1 kcal/mol, so experimentally, these will interconvert readily, and an average bond length would be observed. As mentioned in the comments, this species would probably be hard to observe, even in the gas phase, since $\ce{CH3S-}$ is more stable.

Computed C-S bond lengths (in Å):

For all methods, full geometry optimization was performed using the methods indicated. I would consider the CCSD calculations to be high quality.


Is there a way to efficiently estimate the enthalpy of formation for an organic molecule in computational chemistry? I'm exploring the platonic hydrocarbons and one of the methods I'd like to use is to compare their enthalpies of formation.

The issue is that carbon's standard form is a covalent lattice and I can't think of a way to efficiently obtain its electronic and thermal enthalpies without resorting to optimising a giant PAH.

FWIW I'm using GaussView/Gaussian.

I would say yes, you could estimate standard enthalpies of formation for molecules more or less accurately by application of quantum and statistical mechanics, although there is a 'but': the usual procedure works for gas-phase heats of formation only.

And thus, if the standard state of the material of interest corresponds to the condensed phase, you have to resort to some procedure which allows you to calculate a condensed phase heat of formation given the corresponding gas-phase one. Such procedures are based on a trivial application of the infamous Hess's law, which for the case of solids can be written as follows
$$
   \Delta H_{\mathrm{solid}} = \Delta H_{\mathrm{gas}} - \Delta H_{\mathrm{subl}} \, ,
$$
where $\Delta H_{\mathrm{subl}}$, the heat of sublimation, is estimated usually from experiementally available data on simillar compounds by a correlation scheme. However, as with any empirically fitted scheme the reliability and the accuracy of any such approach are doubtful.

Alternatively, in many cases, it is enough to compare the gas-phase heat of formations, even if the standard state of material is solid or liquid. To calculate the gas-phase heat of formation for a substance in Gaussian all you need is to run frequencies calculations on the previously optimised geometry of a single molecule.

If you would like to give the gas-phase heats of formations a try, I suggest running G4 calculations first. In the route section just write

# G4

and that would do it. Alternatively, in GaussView's Gaussian Calculation Setup window first choose "Energy" at the "Job Type" tab and then at the "Method" tab choose "Compound..." in the second combobox and "G4" in the third one as shown below.


Experimenting with electrochromic prussian blue I am attempting to preform this experiment. Chemistry isn't really my cup of tea, so I have a couple questions.

The instructions say to provide 40µA per cm2, where does that number come from?

The TIO material I'm using has a different resistance than the glass listed, will that change anything?

All the pt wire I could find was very pricy so I though I would use a graphite electrode instead, but all I can find is carbon electrodes. Is that close enough?

Here is the list of materials I have come up with, let me know if any of them are wrong.

HCI
Potassium Ferricyanide
Iron III
KCI
ITO Plastic
Carbon Flat Electrode This was much larger than I thought. Will the size be fine, or should I find a smaller electrode?

Thank you! I'm really excited about attempting this, hopefully I can pull it off.

You pose a series of questions. Let me attempt to answer them all:

  The instructions say to provide 40µA per cm2, where does that number
  come from?

This comes from an article of Itaya and coworkers where they studied the electrochemical synthesis of Prussian Blue. (http://pubs.acs.org/doi/abs/10.1021/ja00382a006) In fact; the website in your question refers to an article in the Journal of Chemical Education (http://pubs.acs.org/doi/abs/10.1021/ed075p881) which in turn refers to the JACS article above.

From my understanding, this number should be seen as an experimental condition. It is given so that the experiment is reproducible.

  The TIO material I'm using has a different resistance than the glass
  listed, will that change anything?

The material you are using has roughly twice the resistivity as compared to the one used in the experiment. In principle, this would not pose too much of a problem. You can tune your electric system using the variable resistor. Just make sure you stick to 40µA per cm2.

  All the pt wire I could find was very pricy so I though I would use a
  graphite electrode instead, but all I can find is carbon electrodes.
  Is that close enough?

This has already been answered by A.K. The carbon flat electrode should work fine. Though I am not sure whether you can use it repetitively (in multiple experiments).

As a boost of confidence. The procedure you are using is both reported in JACS as well as in J. Chem. Ed.; it should work. If you are conducting this experiment at your university or high school, perhaps someone can give you a copy of the articles to which the website refers. The experimental procedures in these articles can be really helpful in preparing your experiment. Best of luck!


Modifying a TLC mobile phase in order to increase the RF values You are running a TLC using 50% ethyl acetate in hexane and find the resulting $R_f$ values are too low. How would you modify the TLC mobile phase in order to increase the $R_f$ values?
Would the hexane and ethyl acetate be considered the mobile phase/solvent?

Question: Would the hexane and ethyl acetate be considered the mobile phase/solvent?

Answer is yes: Hexane/ethyl acetate systems ($0$ to $100\%$ ethyl acetate in hexane) are frequently used mobile phases for TLC technique. However, as Waylander pointed out in his comment:

The type of mobile phase really depends on what the substrate and product are. It is worth running half a dozen trial mobile phases to see which one gives a decent separation.

Also, it is  true, as Buttonwood pointed out, a high $R_f$ value is not very useful if $\Delta R_f$ between the two compounds (or more in the mixture) of interest is small unless it increase upon change of mobile phase. For example, TLC of the mixture of benzoic acid and benzocaine (ethyl 4-aminobenzoate) shows no separation in $50\%$ ethyl acetate in hexanes. However, when mobile phase is change to $5\%$ methanol in dichloromethane, it has shown $\gt 33\%$ separation in $R_f$. So it is a trial and error method, but worth trying.


Convert Concentration (particles/ml) to (mg/ml) I am struggling with converting concentration from (particles/ml) to (mg/ml). I have the concentration for latex particle (particles/ml) and I want to convert it to (mg/ml).

I worked  out the number of moles since I know the the pressure and volume. But I am not sure what to do after that.

First lets see how to get from particles $N$ per volume to mass $m$ per volume.

Dividing the particle number by the Avogadro constant gives moles.

$$
\frac{N}{V}\cdot \frac{1}{N_\mathrm A}=\frac{n}{V}
$$

Multiplying moles with molar mass gives mass.
$$
\frac{n}{V}\cdot M=\frac{m}{V}
$$
Concluding:
$$
\frac{N}{V}\cdot \frac{M}{N_\mathrm A}=\frac{m}{V}
$$

Now your problem arises because you do not know the molar mass. Because you probably have some distribution function of particle size and not just one size. I just assume now that you made an emulsion polymerization in order to get the latex particles or bought them commercially. Then this distribution is very sharp and you can easily take the averaged molar mass as the molar mass of the most probable particle size.


why HI isn't fully miscible with cold water when HF is? HF is more soluble in water than HI , due to greater hydrogen bonding in HF , accepted
but, on the other hand HF is a very weak acid so it's dissociation will be meagre when compared to HI's dissociation , so why is HF and not HI more soluble , am I misinterpreting the concept of solubility ?

Hydrogen iodide is not listed as miscible with water because it is a gas. Miscibility is a property generally ascribed to pairs of liquids. Hydrogen fluoride is a liquid, and so it can be described as miscible with water. HF is a weak acid, but its high solubility can be explained by the high polarity of the H-F bond and the ability of HF to be both a hydrogen bond donor and accepter.

Hydrogen iodide is a gas at room temperature. It's boiling point is $-35\ ^\circ\text{C}$. Just because it is not miscible does not mean that it is not highly soluble. The solubility of HI in water is 245 g per 100 mL. In other words, a saturated solution has more HI by mass than water. HI is a strong acid, and the solubility of the gas is due to the nearly complete ionization of HI. Since HI is a gas, I presume that the solubility limit occurs when the vapor pressure of HI above the solution is high enough that vaporization begins to compete with ionization.

Physical data on HI from Wikipedia


Why is benzene called "aromatic"? Why not "circular" instead? I wish to know why aromatic hydrocarbons like benzene are called aromatic when they can be just called circular to avoid the confusion?

Aromaticity is a electronic property of a molecule, not a description of its shape. An aromatic molecule contains a circular system of conjugated π bonds and p orbitals that contains 4n+2 electrons, where n is an integer.

Benzene is aromatic: it has 6 electrons in a circular conjugated system (4n+2=6, n=1).

Cyclooctatetraene is not aromatic: it appears to have a circular conjugated system, but it contains 8 electrons (4n+2=8 would require n=3/2). This molecule actually adopts a non-planar geometry to avoid antiaromaticity.

Cyclohexane is not aromatic, even though it is circular: it does not have a conjugated electron system.


How are crystals formed from a gas? In crystallization, crystals are formed from liquids by cooling.

How can crystals be formed from a gas by cooling?

It's called deposition or desublimation. It's for example how snow is created in clouds or how frost forms on leaf. When leaf is very cold, then vapour surrounding leaf looses so much energy that it changes into solid state.


Order of reaction / Rate-determining step in enzyme catalysed reactions According to my chemistry book, there are 3 simplified steps within an enzyme catalysed reaction:

$\displaystyle\ce{S + E -> ES}$
$\displaystyle\ce{ES -> EP}$
$\displaystyle\ce{EP -> E + P}$

Where S is substrate, E is enzyme, ES is enzyme substrate complex, EP is enzyme product complex and P is product.

It is my understanding that at first, when substrate concentration is low the rate determining step is step 1, as this is the slowest step in the mechanism.

When you increase substrate concentration to a certain level, where there are no free enzymes remaining to bond with them, adding more substrate will not make the reaction faster, so the order has become 0 with respect to substrate conc.

Is the rate determining step now 2 or 3? I thought it would be 3 as that is the step that frees up more enzymes which could go on to catalyse more reactions. My book says that it is step 2, as "it depends on how fast the ES can convert into EP". But surely this step won't free up more enzymes to catalyse reaction further?

Is the book correct? Can anyone help me understand this?

The rate determining step doesn't have anything to do with any of the factors you mentioned. It's not even true that the rate determining step can be determined directly from the rate of $$\ce{\mathrm{ES}->\mathrm{EP}}$$
The correct definition is that it is the biggest barrier from a stable species (reactant or intermediate) to any subsequent transition state along the reaction coordinate.

Your comment about the apparent rate of the reaction is unrelated. The rate-determining step is an intrinsic property of the reaction and its mechanism. It is unaffected by concentration, though concentration effects may change which mechanism out of several is predominant.


Helium empirical wavefunctions I know that analytic forms of the Helium wavefunctions are not known.  However, are there empirical expressions for the low-lying states of Helium?  I'd like to use them to calculate some transition matrix elements for fun.

If you want a quick and dirty expressions in terms of hydrogen atom wave functions, you can try this:

For ground state you can get a quick approximation by applying effective nuclear charge.  Then you get a hydrogen-like wave function with a modified nuclear charge between 1 and 2;  solve that and get orbital.  It will be doubly occupied, spin singlet.
The lowest excited states of He are all Rydberg states.  The roughest approximation would be to consider it He+ plus a Rydberg electron, and you can even use hydrogen-like atom formula to get the Rydberg orbital.  The total wave function would be antisymmetrized product of these two orbitals, coupled to be either singlet or triplet.

I am not sure you want transition density matrix or transition dipole matrix.  Either way, its just quite straightforward integration from there.  I think there might be ways to do this analytically, but frankly I do not remember and you can probably try that on Mathematica.


Does a molecule need to be placed symmetrically in the unit cell? I was watching an OCW by MIT (3.091X) on solid state. My previous notion of lattice point and unit cell got shattered after this example.
Where are the unit cell and lattice points in this picture?

Unit Cell is the smallest group of atoms which has the overall symmetry of a crystal, and from which the entire lattice can be built up by repetition in three dimensions.
Lattice is the arrangement of points in space such that any point us identical to any other.

So using these definitions, the answer is

Green is the unit cell, and red ones are the lattice points.
Previously I used to think that a molecule should be symmetrically placed in a unit cell such that its center of mass lies on the lattice point. But in this example, the center of mass (shown by the yellow-green circle) is way off the lattice point.
My question is, Are there any other similar examples where the center of mass of molecule doesn't coincide with the lattice point or at least doesn't placed symmetrically in a unit cell (like on the face center, body center)?

Don't think of it so much as putting the center mass of an atom on a vertex but as matching the unit cell symmetry to the crystal symmetry. Take the definition from google below, I've bolded key parts:

  The smallest group of atoms of a substance that has the overall symmetry of a crystal of that substance, and from which the entire lattice can be built up by repetition in three dimensions.

While the red boxes satisfy the first and last bolded criteria, they fail the second. The red boxes have 2-fold rotational symmetry and inversion symmetry, but the green boxes have 2 mirror planes that the red boxes do not. This lack of mirror symmetry with the red boxes does not have the same overall symmetry of the lattice and therefore the green box is the unit cell.

Think of Sodium Chloride; its unit cell face is represented as:

$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$
$\ce{Cl\ \  \color{red}{Na}\ \ Cl}$
$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$

instead of

$\ce{\color{red}{Na}\  Cl}$
$\ce{Cl\ \  \color{red}{Na}}$

Because the first example has 4-fold rotational symmetry where as the second has 2-fold and the first one has 2 mirror planes that the second example does not. Both have inversion symmetry. That said the atoms at edges/corners are centered because that is what allows for the symmetry otherwise the unit cell would be unsymmetrical which does not reflect the symmetry of the lattice.

As an example of a sort of off centered lattice there is perovskite. You can't see it in the image but the black atom is $\ce{Ti^4+}$ and is too large to actually center in the lattice so it will reside in an off-centered configuration (the titanium ion is shifted off-center towards a face). This property of off-centering makes perovskites excellent as dielectric materials for capacitors as the ion can change is off-centering by an electric field.


Absorbtion of light by a coordination compound A $\ce{[M(H2O)6]^2+}$ complex typically absorbs at around $600\ \mathrm{nm}$. It is allowed to form a new complex $\ce{[M(NH3)6]^2+}$ that should have absorbtion at?

The absorbtion is because of transition of $\mathrm{e_2g}$ electron to $\mathrm{t_{2g}}$ and vice versa which depend on crystal field stabilization energy as:
$$\Delta = \frac{hc}{\lambda=600\ \mathrm{nm}}\approx2.0\ \mathrm{eV}$$

But I don't know how to convert it in terms of $\Delta_\text o$ so that I can know the actual configuration.

In $\ce{H2O}$ there will be no pairing of electrons as it is weak and reverse with $\ce{NH3}$.

This looks useful but isn't without the metal configuration:
$$\frac{\Delta_{\text o,\ce{H2O}}}{\Delta_{\text o,\ce{NH3}}}=\frac{\lambda_{\ce{NH3}}}{\lambda_{\ce{H2O}}}$$

The given wavelength of $\lambda=600\ \mathrm{nm}$ may be directly converted to various other quantities, namely wavenumber $\tilde\nu$, frequency $\nu$, photon energy $E_\text{p}$, and molar energy $E_\text{m}$:

$$\tilde\nu=\frac{1}{\lambda}=16\,700\ \mathrm{cm^{-1}}$$
$$\nu=\tilde\nu c=\frac{c}{\lambda}=5.00\times 10^{14}\ \mathrm{s^{-1}}=500\ \mathrm{THz}$$
$$E_\text{p}=h\nu=\frac{hc}{\lambda}=3.31\times 10^{-19}\ \mathrm{J}=2.07\ \mathrm{eV}$$
$$E_\text{m}=N_\text{A}h\nu=\frac{N_\text{A}hc}{\lambda}=199\ \mathrm{kJ\ mol^{-1}}$$

With regard to electronic spectra of coordination compounds, usually wavenumbers $\tilde\nu$ expressed in $\mathrm{cm^{-1}}$ are used.

The electronic configuration of $\ce{M^2+}$ in the given complex $\ce{[M(H2O)6]^2+}$ is unknown. Nevertheless, we may assume that the complex has octahedral geometry and that the observed absorption maximum probably corresponds to promoting one electron from the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$) to an $\mathrm{e_g}$ orbital ($\mathrm{d}_{x^2{-}y^2}$ or $\mathrm{d}_{z^2}$). Thus, it directly gives an estimate for $\Delta_\text{o}=10\ \mathrm{Dq}$ as $16\,700\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{NH3}$, the value of $\Delta_\text{o}=10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are increased.

An empirical equation for estimating the value of $\Delta_\text{o}$ for any pair of metal and ligands is $$\Delta_\text{o}=f \cdot g$$ where $f$ is the ligand parameter and $g$ the contribution from the central atom. The parameter $f$ describes the ligand strength relative to water; i.e. the value for $\ce{H2O}$ is $f=1.00$. The empirical value for $\ce{NH3}$ is $f=1.25$.

Since the ligand field splitting for $\ce{[M(H2O)6]^2+}$ is $\Delta_\text{o} = 16\,700\ \mathrm{cm^{-1}}$, the corresponding ligand field splitting for $\ce{[M(NH3)6]^2+}$ may be estimated as $\Delta_\text{o} = 1.25 \times 16\,700\ \mathrm{cm^{-1}}= 20\,900\ \mathrm{cm^{-1}}$.

The resulting wavenumber $\tilde\nu=20\,900\ \mathrm{cm^{-1}}$ may be converted to a wavelength of approximately $\lambda=480\ \mathrm{nm}$.

Since $\lambda=\frac{1}{\tilde\nu}$, the new wavelength could have been directly calculated from the given original value of $600\ \mathrm{nm}$ using the parameter $f=1.25$ for $\ce{NH3}$:
$$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$


Why is acetate a monodentate ligand? Why is the acetate ion considered a monodentate ligand as opposed to a bidentate ligand considering both oxygen atoms can donate their respective lone pair of electrons?

This depends a lot on the compound in question.

In general, a bidentate acetate would lead to a four-membered ring with the central metal; since the $\ce{O-C-O}$ angle is relatively fixed at approximately $120^\circ$ and since we expect the resulting compound to be at least somewhat symmetric, it would typically lead to unusually small bond angles or a $\ce{C\bond{...}M}$ distance that is too short. If, for example, we imagine $\angle(\ce{O-M-O})=90^\circ$, that means each $\ce{M-O-C}$ angle would be $75^\circ$ which is not what we usually observe in such compounds without a significant force holding it together. Coordinate bonds are usually not very strong, do not liberate enough energy and it would thus be more favourable for one of the two $\ce{O-M}$ bonds to be broken leading to acetate as a monodentate ligand.

There are, however, exceptions. My favourite is probably $\ce{[Cr2(OAc)4(H2O)2]}$ or chromium(II) acetate monohydrate. In this compound, four acetate ions arrange around two chromium atoms in a bidentate fashion; however, each individual acetate bonds to two different chromium ions via its two oxygens. In addition, a water molecule sits on each end of the $\ce{Cr-Cr}$ axis and it turns out the bond between the two chromium ions is, in fact, a quadruple bond. The quadruple bond permits the chromium atoms to come close enough to each other so that the acetates can arrange in a preferable configuration.

(Image taken from Wikimedia, where a full list of authors is available)


Finding concentrations in a voltaic cell A voltaic cell with $\ce{Ni/Ni^2+}$ and $\ce{Co/Co^2+}$ half cells has the initial concentrations of $\ce{Ni^2+}$ $\pu{0.80 M}$ and $\ce{Co^2+}$ $\pu{0.20 M}$.
  (a) Find initial cell potential;
  (b) Find concentration of $\ce{Ni^2+}$ when $E_\mathrm{cell}$ reaches $\pu{0.03 V}$

In part (a), I correctly found the initial $E_\mathrm{cell}$ to be $\pu{0.05 V}$.

In part (b), the book doesn't have any other examples that are explained showing how to get one of the concentrations, it's supposed to be $\ce{0.50 M}$, but I'm not sure how they get to that. From my tables, I found $E$ half cell for $\ce{Ni}$ to be $\pu{-0.25 V}$ and $E$ half cell for $\ce{Co}$ to be $\pu{-0.28 V}$, and the overall then to be $\pu{0.03 V}$.

Tried plugging in everything I have into the Nernst equation, but that still leaves me coming up short.

I think you were on the right track!

You were right that you need the Nernst equation, which must be how you determined the initial $E_\text{cell}$.

$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$

You correctly identified that your $\ce{Co|Co^2+||Ni|Ni^2+}$ cell has the cell standard potential

$$E_\text{cell}=E_\text{cathode}^\circ-E_\text{anode}^\circ=-0.25\ \mathrm V--0.28\ \mathrm V=0.03\ \mathrm V$$

So the question is essentially asking what concentrations are needed for your cell to have the standard potential. Again looking at the Nernst equation, you can solve for the reaction quotient $Q$ when $E_\text{cell}=E_\text{cell}^\circ$

$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$

$$E_\text{cell}-E_\text{cell}^\circ=0=\frac{RT}{nF}\ln Q$$

$$\ln Q=0$$

$$Q=1$$

At this point, you should be able to solve for the concentration of $\ce{Ni^2+}$ with simple algebra.


In Crystal Field Theory what does Δo=10Dq mean? While studying the Crystal Field Theory I was told $\mathrm{Dq}$ is a unit, related to the unit $\Delta_{\mathrm{O}}$ by the relation $\Delta_{\mathrm{O}} = 10\ \mathrm{Dq}$. But aren't $\Delta_{\mathrm{O}}$ and $\mathrm{Dq}$ variables, not units? The unit is of energy, such as $\mathrm{eV}$ or Joules, these are symbols to represent a particular value of energy, which could be anything. So what do we mean when we say $\Delta_{\mathrm{O}}= 10\ \mathrm{Dq}$? What exactly is $\mathrm{Dq}$ (I haven't been able to find this in any of my textbooks, they just either state this relation, or not at all) and are they actually units of energy?

Though I don't think I can provide a full answer, I did see this in a lecture series. $\Delta_{\mathrm{O}}$ is the ligand field splitting and it is a measure for the strong or weak field that the ligands of a complex create. Depending on the ligands the energy levels of, for example octahedral iron complexes (below), are closer to each other in a low field and farther apart for a higher field. This is dependant on $\Delta_{\mathrm{O}}$. This is also an energy difference and the units are in joules or electron volts. See the picture below. You might also want to look up Tanabe-Sugano Diagrams.

I was told (sorry, no proper reference) that the expression $\Delta_{\mathrm{O}}= \pu{10Dq}$ comes from quantum mechanics. The reason that this term is simply stated instead of defined, lies in the origin of CFT. The theory was largely empirical and  thus $\Delta_{\mathrm{O}}$ was used. This difference in energy levels could also be described in quantum systems and is called $\pu{Dq}$ after its quantum number. The difference between the levels appeared to be $\pu{10Dq}$.

I hope this is not a gross oversimplification, but a rather helpful one!


In my homemade electrolysis setup, only the negative end bubbles? I've created an electrolysis setup by connecting a $6~\mathrm{V}$ battery to a cup filled with saline water via pencils; I am confused as to why only the negative pencil bubbles though. After running the setup for longer than five minutes, the water turns brown but so far that information has not been useful. Why is the negative pole the only one bubbling and if possible, why is the water turning brown?

Your setup, using table salt ($\ce{NaCl}$ plus additives) will initially electrolyse according to the following two reactions, separated by cathodic reduction and aniodic oxidation:

$$\ce{2 H+ + 2 e- -> H2 ^}\tag{CatRed1}$$

$$\ce{2 Cl- -> Cl2 + 2 e-}\tag{AnOx1}$$

These reactions assume no counterions. However, you don’t have dissolved protons in your solution because you aren’t electrolysing hydrochloric acid. Instead, the proton is taken from water molecules and if we include them (and sodium counterions) we get the following improved cathodic reduction equation:

$$\ce{2 Na+ + 2 H2O + 2 e- -> H2 ^ + 2 NaOH}\tag{CatRed2}$$

This still displays the hydrogen gas bubbling away at the cathode but chlorine gas at the anode should also bubble after some time (or at least evolve noticeably). However, with the increase of $\mathrm{pH}$ due to the generated hydroxide, the chlorine produced at the anode will react further:

$$\ce{Cl2 + 2 OH- -> Cl- + OCl- + H2O}\tag{AnOx2}$$

And both of these are soluble ions. The overall reaction of your electrolysis setup is:

$$\ce{NaCl + H2O -> H2 ^ + NaOCl}\tag{RedOx}$$

There will be hydrogen evolution at the cathode but no gas evolution at the anode. The browning will likely be due to a number of side reactions induced by the oxidant hypochlorite anion.

In case you’re wondering why water’s oxygen is not oxidised, this is due to its higher standard electrode potential.


Equilibrium constant expression for heterogeneous equilibrium Consider this reaction:

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$

for this, $K_c=\left[\ce{CO2(g)}\right]$

For a given temperature, $K_c$ will remain fixed. That means, the  amount of $\ce{CO2}$ released will be same, irrespective of amount of $\ce{CaCO3}$ taken.

But, that means, at the same temperature, irrespective of  whether we start with 1 g of $\ce{CaCO3}$, or 10 kg of $\ce{CaCO3}$, the amount of $\ce{CO2}$ at equilibrium will be same(!) because, according to the $K_c$ expression, it should be constant.

This is against my intuition, because, it seems that if we start with 10 kg of $\ce{CaCO3}$, we should end up with a much larger amount of $\ce{CO2}$ at equilibrium than if we started with just 1 g of the carbonate.

Am I being mistaken somewhere, or is this is what will actually happen? I am confused.

First, as kaliaden commented, the true equilibrium constant

$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx
p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$

where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar.
Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain

$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$

Now your question is: how is it that this constant does not depend on the amount of matter that we have?

The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium.


How do chemical reactions taking place in an alkaline battery produce electrons? I'd like to understand the chemical reaction in a typical alkaline battery a little better and how that produces a flow of electrons.  I read wiki, how stuff works, and some more advanced articles.  They all seemed to either stop short of explaining the nitty gritty details, or went straight to chemical formulas.

I understand that a chemical reaction occurs between the cathode and the anode and that the electrolyte helps make that happen.  I don't really understand the flow of electrons though, and how the anode/cathode are consumed.  I read a bit about ions but not sure I grasped how that was moving electrons about.

Also, does this reaction start right away or only after a circuit is completed and electrons can flow?  Or has the reaction already started, and there is an excess of electrons on one plate that only start flowing after the circuit is completed?

What forces the potential to stop at 1.5V?

An alkaline battery is a specific type of galvanic cell.  As I am sure you found in your research, it involves an oxidation-reduction reaction between zinc($\ce{Zn}$), and manganese(IV) oxide ($\ce{MnO2}$.)  They are called alkaline batteries because these chemicals are surrounded by an alkaline (basic) solution of potassium hydroxide, $\ce{KOH}$.

Your question is really one that is more about galvanic cells in general.  When an oxidation-reduction (a.k.a. redox) reaction occurs, electrons are transferred between species.  This sort of reaction is exceedingly common.  That's all that happens in a battery, it's just that people have figured out how to catch the electrons on their way from species A to species B and use their energy to do work.

If I took a piece of zinc and placed it in a slurry of $\ce{MnO2}$ and $\ce{KOH}$, it would react according to the following balanced chemical equation and produce a lot of heat.

$\ce{Zn + 2MnO2 <-> ZnO + Mn2O3}$

Suppose that I separated the two pieces of the reaction into their own jars.  A piece of zinc placed in a solution of $\ce{KOH}$ in one jar and some $\ce{MnO2}$ mixed with $\ce{KOH}$ in another.  The reactants are all present, but they have no way of interacting, so nothing would happen - there's no way for electrons to move from the zinc to the $\ce{MnO2}$.

Now I attach a piece of wire to the zinc on one end and into the $\ce{MnO2 - KOH}$ slurry at the other. Now electrons can move from one jar to the other.  If I have a voltmeter that can measure a very brief signal, I can actually see a quick pulse of electrons flowing through the wire, from the zinc to the $\ce{MnO2}$.  The pulse ends almost immediately though, because the movement of electrons away from the zinc and toward the $\ce{MnO2}$ creates a charge separation.  The former becomes positively charged and the latter negatively charged.  Continuing to move electrons against this charge separation is progressively more difficult, and it doesn't take long for the electrostatic force of the charge separation to balance out the electrochemical potential of the reaction and the entire thing just stops.

To fix this problem, I need a way for the charges to balance back out without allowing the reactants to mix directly.  This is done with some sort of salt bridge.  A salt bridge allows positive and negative ions to move between the jars, but it still keeps the jars separate from one another.  This could be as fancy as an actual salt bridge (a glass tube filled with a gel containing some sort of salt) or as simple as a piece of paper soaked in the $\ce{KOH}$ solution.  Now when I connect the wire, electrons flow from the zinc to the $\ce{MnO2}$, negatively charged $\ce{OH-}$ ions move through the salt bridge toward the zinc, and positively charged $\ce{K+}$ ions move through the salt bridge toward the $\ce{MnO2}$.  The charge stays balanced in both jars, so there is no electrostatic force preventing electrons from moving, and they flow freely through the wire.  If I put a motor in the middle of the wire, it turns.

Specifics of Alkaline Batteries

That all was regarding galvanic cells in general.  A battery has additional requirements.  It needs to be compact and safe, in particular.  The construction of an alkaline battery involves a zinc core with a paste of manganese(IV) oxide and potassium hydroxide wrapped around it.  The paste and zinc are separated by a paper sheet (the salt bridge in this galvanic cell.)  Because all of the reactants and products are solid, there is no concern that they will diffuse through the cardboard; it just keeps the reactants in their places.

The reaction does not start until the battery is put into a device and the device is turned on.  Flipping that switch gives the electrons from the zinc a pathway to reach the manganese(IV) oxide.  Until the circuit is complete, there is no buildup of electrons at one end of the battery.

Electrical potential (voltage) is the ratio of the energy carried by an object to that object's charge.  One electron in this system has a potential of 1.43V, two electrons carry twice as much energy, but they also have twice the charge, so they also have a potential of 1.43V.  Electrical potential is sometimes called electromotive force, and that idea, that we're measuring how hard the electrons are being pushed through the wire, may help you understand the battery's behavior better.

Why 1.43V though?  It's because of the properties of the chemical reactions the battery is based on.  You probably know that if you burn some amount of propane it will produce a given amount of heat, and if you burn an equal amount of acetylene, it will produce a different amount of heat.  The same thing is true here.  The energy associated with the half reactions in each part of the battery depends on the chemicals involved.  Rather than list a total amount of energy available, batteries list the energy per charge, voltage.  This battery's half reactions are:

$\ce{Zn + 2OH- -> ZnO + H2O + 2e-}$  E°=1.28V
and
$\ce{2MnO2 + H2O + 2e- -> Mn2O3 + 2OH-}$ E°=0.15V

Each half reaction gives a specific amount of energy to the electrons involved in it.  Since the energy:charge ratio (E° above, called a reduction or oxidation potential) is constant, we can use that to find the energy:charge ratio of the entire battery just by adding them together.  Why doesn't it go above 1.43V?  For the same reason that a fire doesn't burn hotter.  The chemicals involved only release so much energy per reaction.  Note: The potential can be changed in some systems by playing around with the concentrations of the chemicals in the cell.

As the battery is used up, the zinc slowly but surely turns into zinc oxide and the manganese(IV) oxide turns into manganese(III) oxide.  As their amounts fall to zero, the battery's potential falls and it quits working.


How to derive the total number of molecular orbitals based on the given number of atomic orbitals? If four orbitals on one atom overlap with four orbitals on another atom, how many molecular orbitals will form?

Why are there eight molecular orbitals and not four? Is not, in molecular orbital theory, each atom's orbital overlaps with another atom's orbital to form a bond?

The question itself deals with molecular orbital theory and by extension with the approximation of combining atomic orbitals to form these molecular orbitals; in quantum chemistry this approach is called LCAO (Linear combination of atomic orbitals).

Let us look at one of the simple cases, the dihydrogen molecule. You have one 1s orbital at the first hydrogen, i.e. $\chi_A$, and one 1s orbital at the second hydrogen, i.e. $\chi_B$. You can now combine these orbitals to form a bonding $\eqref{eq:bonding}$ and an antibonding $\eqref{eq:antibonding}$ combination to create the molecular orbitals $\phi_{A+B}$ and $\phi_{A-B}$.
\begin{align}
\phi_{A+B} \equiv 1\sigma_\mathrm{g} &= \chi_A + \chi_B
  \tag1\label{eq:bonding}\\
\phi_{A-B} \equiv 1\sigma_\mathrm{u} &= \chi_A - \chi_B
  \tag2\label{eq:antibonding}
\end{align}
A scheme of this can be found on Wikipedia created by the user CCoil.

This does in principle mean that for every bonding configuration of a molecular orbital you will always have an antibonding configuration. In other words the number of atomic orbitals is the same as the number of molecular orbitals.

In your case, when there are four orbitals on one atom and four orbitals on the other atom, you will end up with eight molecular orbitals. A quite complicated case of bonding that shows this is the $\ce{C2}$ molecule, which has previously been discussed here: Bonding in diatomic C2, a carbon-carbon quadruple bond?


How to simulate phase separation in a discrete system using LAMMPS? I would like to simulate phase separation in LAMMPS and need advice what kind of two particle systems and interactions will yield phase separation.

For example, a meta stable phase separation is achieved with a continuous phase flow model governed by Cahn–Hilliard equation, where $c$ is the concentration, $\mu$ is the chemical potential, and $\gamma$ is a constant:
$$\begin{align}
\frac{\partial c}{\partial t} &= D\nabla^2\mu&
\mu &= \left(c^3-c-\gamma\nabla^2 c\right)\\
\end{align}
$$

I tried to use a discrete two atom system in LAMMPS. The first type of the particles has an inter-LJ interaction and the second type has an inter-soft-potential interaction. Between the distinct two types there is also a soft potential:

Initial configuration (interleaved lattice of the two particles):

Steady state configuration at $T=0.1$ (reduced LJ units):

Configuration file:

units lj

timestep 0.001

dimension 2
boundary p p p
atom_style atomic
neighbor 0.3 bin
neigh_modify every 20 delay 0 check no

variable temp equal 0.1

lattice hex 0.5
region simbox block 0 50 0 25 -0.1 0.1
create_box 2 simbox

lattice hex 0.5
create_atoms 1 region simbox

lattice hex 0.5 origin 0.5 0.5 0
create_atoms 2 region simbox

mass 1 1
mass 2 1.0

velocity all create ${temp} 1234567 dist gaussian
pair_style hybrid lj/cut 2.5 soft 0.5
pair_coeff 1 1 lj/cut 2.0 0.8 4.0

pair_coeff 1 2 soft 0.5
pair_coeff 2 2 soft 0.5

fix 1 all nvt temp ${temp} ${temp} 0.1

dump 1 all atom 500 out/dump_temp_${temp}.lammpstrj
write_data data.all
thermo 500
run 500000

How do I make my system to look more similar to the continuous model suggested by Cahn and Hiliard? I want to get a more prominent phase separation like in the continuous model. I am not sure how to calibrate my system/interactions.
Is there a different simple system I could use to simulate phase separation?
If you think Matlab could be more suitable to simulate phase separation, please elaborate.

Firstly, a couple of comments.

The Cahn-Hilliard model, for which you gave a snapshot, represents the two fluid phases on a much larger length scale than the atomic scale. It's a continuum model. So, whatever particle-based model you use, you would need an enormous system to get close to similar-looking pictures. The interfaces will always look more irregular, in a practical particle-based simulation. Don't expect to see nice circular droplets.

Secondly, even that Cahn-Hilliard simulation is not yet at equilibrium. The droplets are in the process of coalescing, and the final state will be a single large droplet of one phase in the other (or possibly the two phases will form slabs, with planar interfaces between them, it depends somewhat on the boundary conditions and dimensions of the surrounding box). The driving force is the reduction of the total interfacial area. This whole process takes time. Something similar should happen in the particle-based simulations, but because the length scales are so different, the time scales will also be different.

Given all that, your first choice of models may be suitable. However, you need to choose a temperature which is high enough for both the phases to be liquid. You haven't done this! The triple point temperature of 2D Lennard-Jones is $k_BT/\epsilon\approx 0.4$. At your temperature of $0.1$, the Lennard-Jones atoms will be trying to crystallize; in practice they will get "stuck" in irregular semi-crystalline semi-amorphous clumps. Nonetheless, if you raise the temperature above $0.4$, you might see liquid-liquid coexistence for this system. I can't be sure, and you may need to experiment.

I have some suggestions for a simpler approach, if all you want to do is see liquid-liquid phase separation using LAMMPS. Firstly, it is OK for both the fluids to be identical. For example, calling the two species A and B, the A-A and B-B interactions can both be Lennard-Jones. If you choose the A-B interaction to be different, and unfavourable compared to Lennard-Jones (for example, the completely repulsive Weeks-Chandler-Andersen version of Lennard-Jones), you should see phase separation.
The advantage of this approach is that the Lennard-Jones phase diagram is well known, making it fairly easy to choose a density and temperature at which both the phases will be liquid.

Even simpler, you can choose A and B to interact through a coarse-grained potential such as the dissipative particle dynamics DPD potential
$$
v(r) = \begin{cases} \frac{1}{2}a(1-r)^2
& r<1 \\ 0 & r>1
\end{cases}
$$
where the cutoff distance is taken as the unit of length.
LAMMPS has a DPD option, and you can find a perspective article on DPD
by Español and Warren J Chem Phys, 146, 150901 (2017)
also available as a preprint. There are also tutorials on DPD readily available on the internet.
DPD has the advantage of being very fast;
also it is sort-of intermediate between an atomic model and a continuum one,
since the particles in DPD really represent regions of fluid,
not individual molecules.

Once again, the A-A and B-B interactions can be made identical,
for example $a=25$, with a typical temperature $k_BT=1$
and a density $\rho=3$ in 3D; you would need to experiment for suitable parameters in 2D, but the precise values may not be crucial. The A-B interaction could be the same, but with a larger value of $a$, for instance $a=50$.
This approach has already been used to model fluid-fluid separation,
for instance in papers by Novik and Coveney
Phys Rev E, 54, 5134 (1996)
also available as a preprint,
and Phys Rev E, 61, 435 (2000)
also available as preprint.
Those papers are a bit involved,
and there are many complicated technical issues with
the DPD method.
It is not essential to go into those in depth.
If you are not essentially interested in the hydrodynamics,
it is quite satisfactory to use the DPD potential
in a standard molecular dynamics simulation with a thermostat of your choice.


pKa value of an indicator Can we deduce the actual value of $\mathrm{p}K_\mathrm{a}$ of an indicator if the $\mathrm{pH}$ range has been given?

I know that $\mathrm{pH}$ range is physically the range of values over which an indicator changes its colour from its acid form to its base form, but can we judge the $\mathrm{p}K_\mathrm{a}$ value from the data?

For example: if $\mathrm{pH}$ range of $3.4$ to $4.6$ has been provided, then I know for certain that the $\mathrm{p}K_\mathrm{a}$ value of the indicator lies in between, but can we deduce the actual value?

Visually, we can only estimate the  $\mathrm{p}K_\mathrm{a}$ value to be near the middle of the indicator range, shifted to more intense colour.

The $\mathrm{p}K_\mathrm{a}$ of an indicator could be determined more precisely by  combining $\mathrm{pH}$ meter with photometry, where we would get the well known " round step function" of absorbance versus $\mathrm{pH}$.

E.g phenolphthalein has indicating range $\mathrm{pH}=8.2 - 10.0$ for reaction
$$\ce{HInd(clear) <=> H+ + Ind^-(violet)}$$

(I will not fight over the colour name. It may be magenta, carmine, purpur, dark pink -  you know the 16 colours of Windows jokes. )

Let suppose the phenolphthalein $\mathrm{p}K_\mathrm{a}$ is in the middle of the range, i.e. $\mathrm{p}K_\mathrm{a}=9.1$.
Let suppose the ratio of concentrations of clear and violet forms at $\mathrm{pH}=8.2$ is $\ce{[Ind-]/[HInd]}=X$.

The equation for indicates are the same as for any other weak acid:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind-]}}{\ce{[HInd]}}$$

If we consider the mentioned border $\mathrm{pH}$ values,
then
$$\begin{align}
8.2 &=9.1+\log X \\
10.0 &=9.1+\log \frac 1X \\
X&=0.126 \\
\end{align}$$

The problem is the relative eye sensibility to small additions  of one form of an indicator to the other form.
 We notice much easier a small amount of violet substance in clear substance, than a small amount of clear substance in violet substance.

$$\begin{align}
\mathrm{pH_{low}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{noticeble}]}}{\ce{[HInd_{abundance}]}} \\
\mathrm{pH_{high}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{abundance}]}}{\ce{[HInd_{noticeble}]}} \\
\end{align}$$

As the consequence, the $\mathrm{pH}=10.0$ is significantly closer to $\mathrm{p}K_\mathrm{a}$ than $\mathrm{pH}=8.2$, therefore

$$\mathrm{p}K_\mathrm{a} \gt \frac {8.2+10.0}{2}$$

Let suppose for now the real $\mathrm{p}K_\mathrm{a}=9.7$.

Then $\ce{[Ind-]/[HInd]}$ is

2 for pH 10.0, i.e full violet weakened by 1/3
0.03 for pH 8.2, i.e 3% of the full violet.

The similar visual shifts happen to other indicators as well.


What do the quantum numbers actually signify? I know how to calculate them and such stuff, but I wanted to know what they actually signify. I have a vague idea that they have something to do with an electron's position in an atom but what do all of them mean? Any help would be greatly appreciated!

Quantum numbers give information about the location of an electron or set of electrons.  A full set of quantum numbers describes a unique electron for a particular atom.

Think about it as the mailing address to your house.  It allows one to pinpoint your exact location out of a set of $n$ locations you could possibly be in.  We can narrow the scope of this analogy even further.  Consider your daily routine.  You may begin your day at your home address but if you have an office job, you can be found at a different address during the work week.  Therefore we could say that you can be found in either of these locations depending on the time of day.  The same goes for electrons.  Electrons reside in atomic orbitals (which are very well defined 'locations').  When an atom is in the ground state, these electrons will reside in the lowest energy orbitals possible (e.g. 1$s^2$ 2$s^2$ and 2$p^2$ for carbon).  We can write out the physical 'address' of these electrons in a ground-state configuration using quantum numbers as well as the location(s) of these electrons when in some non-ground (i.e. excited) state.

You could describe your home location any number of ways (GPS coordinates, qualitatively describing your surroundings, etc.) but we've adapted to a particular formalism in how we describe it (at least in the case of mailing addresses).  The quantum numbers have been laid out in the same way.  We could communicate with each other that an electron is "located in the lowest energy, spherical atomic orbital" but it is much easier to say a spin-up electron in the  1$s$ orbital instead.  The four quantum numbers allows us to communicate this information numerically without any need for a wordy description.

Of course carbon is not always going to be in the ground state.  Given a  wavelength of light for example, one can excite carbon in any number of ways.  Where will the electron(s) go?  Regardless of what wavelength of light we use, we know that we can describe the final location(s) using the four quantum numbers.  You can do this by writing out all the possible permutations of the four quantum numbers.  Of course, with a little more effort, you could predict the exact location where the electron goes but in my example above, you know for a fact you could describe it using the quantum number formalism.

The quantum numbers also come with a set of restrictions which inherently gives you useful information about where electrons will NOT be. For instance, you could never have the following possible quantum numbers for an atom:

$n$=1; $l$=0; $m_l$=0; $m_s$=1/2

$n$=1; $l$=0; $m_l$=0; $m_s$=-1/2

$n$=1; $l$=0; $m_l$=0; $m_s$=1/2

This set of quantum numbers indicates that three electrons reside in the 1$s$ orbital which is impossible!

As Jan stated in his post, these quantum numbers are derived from the solutions to the Schrodinger equation for the hydrogen atom (or a 1-e$^-$ system).  There are any number of solutions to this equation that relate to the possible energy levels of they hydrogen atom.  Remember, energy is QUANTIZED (as postulated by Max Planck).  That means that an energy level may exist (arbitrarily) at 0 and 1 but NEVER in between.  There is a discrete 'jump' in energy levels and not some gradient between them.  From these solutions a formalism was constructed to communicate the solutions in a very easy, numerical way just as mailing addresses are purposefully formatted in such a way that is easy that anyone can understand with minimal effort.

In summary, the quantum numbers not only tell you where electrons will be (ground state) and can be (excited state), but also will tell you where electrons cannot be in an atom (due to the restrictions for each quantum number).

Principle quantum number ($n$) - indicates the orbital size.  Electrons in atoms reside in atomic orbitals.  These are referred to as $s,p,d,f...$ type orbitals.  A $1s$ orbital is smaller than a $2s$ orbital.  A $2p$ orbital is smaller than a $3p$ orbital.  This is because orbitals with a larger $n$ value are getting larger due to the fact that they are further away from the nucleus.  The principle quantum number is an integer value where $n$ = 1,2,3... .

Angular quantum number ($l$) - indicates the shape of the orbital. Each type of orbital ($s,p,d,f..$) has a characteristic shape associated with it.  $s$-type orbitals are spherical while $p$-type orbitals have 'dumbbell' orientations.  The orbitals described by $l$=0,1,2,3... are $s,p,d,f...$ orbitals, respectively.  The angular quantum number ranges from 0 to $n$-1.  Therefore, if $n$ = 3, then the possible values of $l$ are 0, 1, 2.

Magnetic quantum number ($m_l$) - indicates the orientation of a particular orbital in space.  Consider the $p$ orbitals.  This is a set of orbitals consisting of three $p$-orbitals that have a unique orientation in space.  In Cartesian space, each orbital would like along an axis (x, y, or z) and would be centered around the origin at 0,0.  While each orbital is indeed a $p$-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space.  Therefore, for a set of $p$-orbitals, there would be three $m_l$, each uniquely describing one of these orbitals.  The magnetic quantum number can have values of $-l$ to $l$.  Therefore, in our example above (where $l$ = 0,1,2) then $m_l$ would be -2, -1, 0, 1, 2.

Spin quantum number ($m_s$) - indicates the 'spin' of the electron residing in some atomic orbital.  Thus far we have introduced three quantum numbers that localize a position to an orbital of a particular size, shape and orientation.  We now introduce the fourth quantum number that describes the type of electron that can be in that orbital.  Recall that two electrons can reside inside one atomic orbital.  We can define each one uniquely by indicating the electron's spin.  According to the Pauli-exclusion principle, no two electrons can have the exact same four quantum numbers.  This means that two electrons in one atomic orbital cannot have the same 'spin'. We generally denote 'spin-up' as $m_s$ =1/2 and spin-down as $m_s$=-1/2.


How slow is a ester reaction without a catalyst? (in a whiskey barrel?) The typical description of a high school esterification reaction of an alcohol and an organic acid uses a strong acid catalyst like $\ce{H2SO4}$; otherwise the reaction is said to be "slow".

I would like to know, how slow? For context I am interested in the esterification reaction which happen in the time frame of distillation of whiskey and whiskey maturing in a barrel.

New spirit (what goes into a barrel to age and become whiskey) is approximately 60% EtOH, 39% water and less than 1% (I think) other alcohols, and organic acids like acetic and lactic acid.

How long would this system take to come to equilibrium, without any catalyst? (Hours, days, months, years?)

The best info I could come up with is from wikipedia (with no citations for the relevant sections)

https://en.wikipedia.org/wiki/Fischer–Speier_esterification

For reaction with an excess of one reactant and an acid catalysts, it says:

  "Typical reaction times vary from 1–10 hours at temperatures of 60-110
  °C."

Later it talks about ester reactions in wine:

  "Of course, when compared to sulfuric acid conditions, the acid
  conditions in a wine are mild, so yield is low (often in tenths or
  hundredths of a percentage point by volume) and take years for ester
  to accumulate."

Which is a bit of an Apple to Oranges comparison, and is not whiskey, but it gives an indication of 'Yes, slowwwwww'. Or something like that.


Calculate Gibbs energy of dinitrogen pentoxide using its vapor pressure and sublimation I am having trouble with the following problem:

  Dinitrogen pentoxide, $\ce{N2O5}$, is a solid with a high vapor pressure. Its vapor pressure at $\pu{7.5 °C}$ is $\pu{100 mmHg}$, and the solid sublimes at a pressure of $\pu{1.00 atm}$ at $\pu{32.4 °C}$. What is the standard Gibbs energy change for the process $\ce{N2O5(s) → N2O5(g)}$ at $\pu{25 °C}$?

Okay so my attempt is to first use the Clapeyron equation to find $\Delta H$ for usage in the reversible process entropy formula. The Clapeyron equation is $$\ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ Plugging everything in I got $\pu{50807.79 J}$.

Next I found the change in entropy in a reversible process:

$$\Delta S = \frac{\Delta H_\mathrm{vap}}{T_\mathrm{vap}}.$$

Now here is where I have a problem. I calculated the enthalpy of vapor pressure. So I would expect to use the temperature that corresponds to the enthalpy of vaporization. Problem is, the book lists this as the wrong answer. I got within a few significant figures when I used $T_\mathrm{sublimation}$.

I calculated the final answer by using

$$\Delta G = \Delta H - T\Delta S$$

The corresponding values are plugged in. I am not sure if I even solved the problem correctly, but if I am, why is that temperature used?

First, here is a way to solve it that avoids calculating entropy:

Calculate the enthalpy from the given data.
Calculate the vapor pressure at 25 $^\circ$C. This is equal to the equilibrium constant for the process.
Calculate $\Delta G^\circ$ as -RT ln(K).

This way is efficient because you already set up the Clapeyron equation in the first step and you just use it again in the second. You could even solve 1. and 2. graphically, and then go from there.

  I got within a few significant figures when I used Tsublimation.

You need to calculate $\Delta S^\circ$, i.e. the change in entropy when all species are at standard state. At the normal temperature of sublimation, $\ce{N2O5(g)}$ is at the standard state, so that is the temperature you need to use. The enthalpy is not concentration dependent, but the entropy and the Gibbs free energy are.

  I am not sure if I even solved the problem correctly[...]

You solved it correctly in that you got the correct answer. You did not solve it as the problem author intended, I suspect, because you had to look up the normal sublimation point (unless you used the Clapeyron equation to estimate it).


Most acidic compound amongs the following unsaturated ketones? In the given picture, I know that the most acidic will be (a), because of potential of ring expansion. I tried a few times but wasn't able to get a satisfactory six membered ring. I think (a) is the most acidic compound among the 4 due to ring expansion only. Or if I am wrong, please correct me with a detailed solution.

The answer given is (a) but explanation is needed

Its not because of any ring expansion but because of formation of an aromatic compound.

Lets Start with (C) and (D). Both compounds after formation of carbanion are stabilized by extended conjugation:

But (A) forms an even more stable carbanion because of aromatic nature of:

Cyclopentadiene has a $p_{ka}$ of 16. Its high acidic strength is because of the above reason.

You can analyse (B) yourself. But it wont form a carbanion as stable as (A).


The walls of a plastic vial with Mercury inside are darkening. What is causing this? I have a sample of Mercury in a plastic vial,
The vial is a BD Vacutainer (genuine), made of polyethylene terephthalate. It is used for blood samples, and the one I used was new.

When the vial was new, it was almost as transparent as glass, as seen in the image:

vacutainer http://mm.claflinequip.com/catalog/product/cache/2/image/750x750/9df78eab33525d08d6e5fb8d27136e95/i/2/i20885.jpg

However, after about a month, here is what the vial looks like:

As you can see, the vial has darkened.

The vial was sealed with a small initial layer of hot glue about 1 centimeter from the top, then the last centimeter was filled to the brim with epoxy, to ensure it won't leak.

I have been handling the vial quite a lot, but it has been stored in the padded box as seen in the image.

What is causing the plastic to darken? Once again the vial is made out of polyethylene terephthalate if that could be useful.

Thanks!

Mercury doesn't react with PET under normal conditions, but mercury vapor will diffuse into plastics.  The diffusion will change the optical density of the plastic in the area where diffusion occurred.

Here's a link to a nice study where the mercury was doped with a radioactive isotope of mercury.  This doped mercury was placed in a glass chamber which also contained plastic discs held at a known distance above the mercury.  The plastic discs were removed after various times and then microtomed.  The various layers of the microtomed plastic disc were measured for radioactivity.  Analysis of the resulting data allowed the investigators to measure the rate constant for diffusion of mercury into the plastic.

One final point, plastics are sometimes contaminated with small amounts of various impurities.  Sometimes these impurities were intentionally introduced, such as radical chain terminating materials (sulfur is an example) in order to produce polymers with a certain molecular weight distribution.  So while the plastic itself might not react with mercury, an impurity might.  The above link also provides analysis of plastic contaminated with sulfur and finds a much higher mercury diffusion constant in such materials.


Is it not safe for health to finish a bottle of wine during a week People said that wine is good for health. So I take a cup (less than 100 mL) of wine every evening. (Yes, wine is cheap where I live.) So to finish a bottle of wine (750 mL), it takes me a week.
(Of course I put it in the refrigerator.) I have to say that it doesn't taste good after 3 days (but still OK).

But I don't know if it is not safe to do so. Maybe the wine begins to produce something not good for health after a few days when it is opened. So, my question is: what chemical species are formed after opening a wine bottle, making it go bad? Are they only bad for the taste, or also bad unhealthy?

The bad taste left after drinking wine can be attributed a number of agents found in wine, particularly alcoholic wines.

Alcoholic wines typically contain 8-15% ethanol. When ethanol mixes with oxygen in air it gets oxidised, gradually producing acetic acid (vinegar).

Furthermore, the reaction of ethanol with urea (naturally present in wine) to form ethyl carbomate (urethane) can produce a saline, bitter taste, although probably not toxic.

The ethanol in wine is produced through the fermentation of carbohydrates by yeast. This process also results in the production of acetaldehydes which contribute to its complex set of flavours.

Ethanol itself however is tasteless. It is however toxic and a class 1 carinogen according to the International Agency for Research on Cancer (IARC).

When consumed, ethanol is oxidised by alcohol dehydrogenase (ADH) enzymes in the body to produce acetaldehyde, which is 10-30 times more toxic than alcohol and possibly carcinogenic in humans.

Recent scientific research on cancer seems to indicate even low levels of alcohol consumption (1 cup per day) 'increases the risk of cancer of oral cavity and pharynx, esophagus and female breast' Source: http://annonc.oxfordjournals.org/content/early/2012/08/21/annonc.mds337.full

So a bad taste may not be the worst thing alcohol leaves in one's mouth.

With regard to the health effects of alcohol, please refer to the World Health Organisation report which is based on data collected scientifically from around the world over many years.


Why does bromine react with iron(II) but not iodine? My question comes from the answer (C) to this question:

Firstly, (correct me if I'm wrong please) I thought that the answer could either be C or D since the other answers are negative ions which can't react with iodine or bromine since they (as diatomic molecules) have a full shell of outer electrons. Now, from looking at the question, I knew it had something to do with oxidation and reduction as well as displacement reactions, however, as iodine and bromine aren't on the reactivity series, we can't determine which would react with which metal. From there, I just had no more ideas. Could anyone provide me with a hint or a guide as to how to do this ?

All four of the ions are reducing agents. But usually halogen gases' tendency to undergo reduction to become halide ions decreases down the group (along with their reactivity). Out of the $4$ ions, only $\ce{Fe^{2+}}$ has a mild enough reducing power.

$$\ce{2Fe^{2+} +Br2 -> 2Br^- +2Fe^{3+}}$$

But $\ce{Fe^{2+}}$ is not strong enough to reduce $\ce{I2}$ as it is relatively stable.

In fact, $\ce{SO3^{2-}}$ has the strongest reducing power among the four options, as it can reduce both $\ce{I2}$ and $\ce{Br2}$.

$\ce{OH^{-}}$ is a mildly strong reducing agent, it can react with both $\ce{I2}$ and $\ce{Br2}$, and its typical reaction is

$$\ce{4OH^- -> 2H2O +O2 + 4e^-}$$

$\ce{Mn^{2+}}$ is the least reducing, because it is the product of a strong oxidizing agent that has been reduced:

$$\ce{MnO4^{-} + 8H+ +5e^- -> Mn^{2+} +4H2O}$$

Here, permanganate ion $\left(\ce{MnO4^-}\right)$ is a very strong oxidizing agent; the product thus would be weakly reducing.

Actually, a similar idea can be seen with $\ce{SO3^{2-}}$ (sulphite) ions. Since it is a strong reducing agent, we can expect the product formed after it is oxidized is a weak oxidizing agent.

$$\ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-}$$

Indeed, sulphate ions alone are weak oxidizers, and stable.

I suggest you refer to E.C.S. (electrochemical series) to know whether a reaction would occur. It is easy to find online; e.g., at Wikipedia.


Is there a way to quantify the 'surface area' or simply the shape of organic compounds? I wish to write an investigation into the effects of the shape of organic compounds and the location of the functional groups on the enthalpy of combustion of the organic compounds. Thus, I wonder if there is a way of quantifying these variables, as databases I have looked at do not provide much of an answer.

It is possible to compute the surface area of a molecules, and to compute the surface area of each individual atom in a molecule. These areas depend on the conformation of the molecule. An example of a software that does this is MSMS (http://mgltools.scripps.edu/packages/MSMS).

With respect to the shape, it is possible to categorize molecules in a triangular diagram describing similarity with "rods", "disks" and "spheres" (e.g. http://pubs.acs.org/doi/10.1021/ci300415d). Details about this analysis were published here: http://pubs.acs.org/doi/10.1021/ci025599w.


Acetic acid odor while testing CO2 sensor; can it be easily removed, is there a better way? A friend asked me to help test a hobby-quality $\ce{CO2}$ sensor MG-811 (updated link). It uses a solid electrolyte and should provide a sensitivity of something like -40 mV per decade. I though I'd just use a small amount of acetic acid + baking soda to make roughly 4,000 ppm in a large container, and compare to "fresh air" which should be roughly 400 ppm, or at least within the 350 to 600 ppm range, as a very rough initial test.

But then I noticed/remembered (duh!) that the acid smells quite strong. Are these simply molecules of $\ce{CH3COOH}$ that are evaporating, or are they bringing along molecules of water in some kind of cluster?

I realized that they could potentially contaminate the sensor, so I'd like to know what I'm up against in terms of removing them from the gas before immersing the sensor.

I've tried to look up the vapor pressure of acetic acid, and it seems to be quite large, about 10 Torr at 20C. See for example here and here. So I'm thinking of using some kind of ad-hoc cold-trap to try to remove it. Might there be another way to do this without using special laboratory equipment? Would passing the gas over dry $\ce{NaHCO3}$ or a solution of it help to "getter" the remaining acetic acid vapor?

Or, is there a better way to produce a controlled amount of $\ce{CO2}$ to begin with?

You could use a less volatile acid, like oxalic. But I think the reaction with NaHCO3 probably gives off a mist as the reaction bubbles, and that might be why the odor is so strong.

How about using seltzer water (or Perrier? or Diet Coke!) as the source of CO2? I estimate that the pressure of CO2 is about 45 psi in the unopened bottle, and if you attach a balloon to a freshly opened bottle and let it outgas slowly, the balloon will fill up and be essentially pure CO2. If a latex balloon is unsuitable or too hard to handle, a mylar balloon ($1 at the Dollar Store) will work just as well. The gas will be saturated with water vapor.


Can we picture metallic bonding as an equilibrium between electrons and cations? Can we picture metallic bonding as an equilibrium between electrons and cations?

Suppose:

  $$\ce{Al^3+ + 3e- <=> Al}$$

In metals, electrons are non-localized, forming a "sea" of electrons, rather than having them localized, as in the $\ce{Na+Cl-}$ lattice of crystalline salt. See Metallic bonding for a more complete description.

It is, of course, a matter of degree, as covalent, ionic and metallic bonding can "blend" from one to the other. A bond can be considered partially ionic and covalent, for example; see these helpful graphics


Use of pipette during experiments for determination of oxygen by Winkler's method We were performing estimation of $\ce{O2}$ by Winkler's method and we were told by the instructor that while adding $\ce{MnSO4}$ and $\ce{NaOH}$ / $\ce{KI}$ the pipette tip should be well below the surface of the solution.

When asked as to why this should be done, he said that if the reaction took place at the surface then there's a possibilty that the atmospheric $\ce{O2}$ may get dissolved.

Is it true?

[...] the pipette tip should be well below the surface of the solution [...]

Usually, one would indeed keep the pipette above the solution to avoid that a (minor) part of the solution forms a drop (or film) on the tip of the pipette and is carried away.

Here, when the amount of oxygen dissilved in the solution is to be determined, another effect prevails.

If you add your reagents from greater height, you might create a splash, just like throwing a rock into a lake. As a result, you will mix some oxygen from the gas phase into the solution, which might taint the measurement.


Why 4‐methylidenespiro[2.4]heptane cannot be converted to 1‐chlorobicyclo[3.2.0]heptane? According to the solution, the following reaction is not correct:

But why? I have tried solving in the following steps:

Forming a carbocation by breaking the double bond.
Expanding the ring.
Adding a $\ce{Cl-}$ anion.

This should thereby form the compound that is shown. But the solution is given that this reaction is not correct. Can you please explain to me why?

Here the concept of angular strain comes into the picture. The ideal tetrahedral and is around 109° 29'. As the angle between the C−C bond increases, it is more away from the ideal $\mathrm{sp^3}$ hybridization and hence will become more unstable.

For a cyclohexane molecule, the angle between the C−C bonds is 120°, therefore the deviation from the ideal angle is 10° 31', whereas for a cyclopentane the angle between C−C bonds is 108°, therefore the deviation from the ideal tetrahedral angle is around 1.5°. This is a very low deviation.

Since the transition is from cyclopentane which has a relatively lower deviation from the ideal angle to cyclohexane which has a relatively higher deviation from the ideal angle, the above transition does not take place.

Plus the number of carbons are not balanced in your reaction.