text:
Why does sp2 hybridization result in a greater chemical shift in carbon NMR?
Carbon NMR has a very wide chemical shift range. This is attributed to the relative proximity of electron-withdrawing groups (1-2 bonds separation) compared to proton NMR (2-3 bonds separation) along with the hybridization state of the carbon atom (plus other factors such as diamagnetic anisotropy). The reason for the chemical shift difference between proton and carbon NMR due to electron-withdrawing groups is conceptually easy; however, I am having a difficult time understanding why sp2 hybridization results in a greater chemical shift. Hybridization results in smaller, more compact orbitals  (closer to the carbon atom). This is what throws me off. I would expect greater electron density near the carbon atom due to sp2 hybridization to result in a shift upfield (not downfield). Is there any simple explanation for this?

Because sp2 hybridized carbons have $p$ orbitals that form $\pi$ systems. $\pi$ systems create electronic systems that increase effective shielding. Your statement isn't wrong, but it's just not the dominant factor.

For 13C vs 1H chemical shift ranges see: https://chemistry.stackexchange.com/q/62546/16683 (also https://chemistry.stackexchange.com/q/81127/16683), it’s not actually what you said. Non-specialised textbooks (especially those targeted at organic chemists) often don’t tell the truth about NMR.


text:
1.0 g of AlCl3 is added to 100.0 ml of water. What would be the resulting pH?

$1.0\ \mathrm g$ of $\ce{AlCl3}$ is added to $100.0\ \mathrm{ml}$ of water. What would be the resulting pH?

I think the answer may be $4.022$ but I'm not certain.
The reaction wasn't provided but I think its this:
$$\ce{Al(H2O6)6^3+ <=> H^+ + Al(H2O)5(OH)^2+}$$
Assuming all $\ce{Al}$ dissolved in water:
Initial $[\ce{Al}]=7.49\times10^{-4}\ \mathrm M$
$$x^2/(7.49\times10^{-4}-x)=1.4\times10^{-5}$$
$$x=9.56\times10^{-5}$$
$$\mathrm{pH}=-\log(9.56\times10^{-5})=4.022$$

What happened to the Cl attached to the Al?!?

I have no idea, I'm so confused with this question

I think you have the right idea, but in an excess of water the overall reaction would be $$\ce{AlCl3 + 6H2O ->[aqueous] Al(H2O)5(OH)^{2+} + H+ + 3Cl-(aq)}$$

Oh okay. Do you know how to solve this

Assume volume of solution is the same as the volume of water and that the reaction I gave goes to 100% completion.

So assume all Al dissolved in water

Yes there will be an excess of water. 100ml of water is about 5.5 moles of water.

can you confirm if 4.022 is right

No. Add your work as an answer. I'll comment on it to help you get it right, then you can accept your answer and gain some site reputation. // Start with the chemical equation. You can edit your post to see the markup I did.

okay I've shown my work

Your first mistake is "$\pu{[inital] Al} = \pu{7.49\cdot10^{-4}}$. Please show how you got that value.

Not being given any other information assume the reaction goes to completion and that it is:
$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$
molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:
$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$
$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$
pH = - log(0.07496) = 1.12

Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion.
$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)6^{3+}  +3Cl-}$$
The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied.
$$\ce{Al(H2O)6^{3+} ->[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$
$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$

oh ok? I'm very confused at this point

@nanxii - Sorry I fouled that up badly.

i'm kinda desperate at this point. I just need a simple explanation and answer

@nanxii - Since I messed up, I'll just give an answer. But this isn't a homework site. You've gotten a free ride on two questions. Don't expect more. See the site's homework policy.

Actually the OP has stated an equilibrium constant. The only problem i see is the miscalculation of the molarity. If we use the correct value you can calculate the pH assuming the first step goes to completition. (Wich is 3.0 considering the rounding for given significant digits.)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Tautomer of acrolein

Are $\ce{CH2=CH-CH=O}$ and $\ce{CH2=C=CH-O-H}$ tautomers?

There is only a shift of a hydrogen atom and a pi bond. So why are these two not tautomers?

The carbons in H2C=C=CH(OH) have to be linear. In CH2=CH−C(=O)(H) they don't.

@MaxW That's irrelevant, it's an issue only in resonance.

The molecular geometry argument is pretty common-sensical, and is true to a certain extent. But I don't think it's the cleanest argument. If you look at a typical ketone and its enol, you will notice that the enol has a more constrained geometry than the ketone:

In the keto tautomer, the highlighted carbon is free to go wherever it likes, since it's part of an sp3 chain. But in the enol tautomer, by virtue of the 120° bond angles in the alkene, the highlighted carbon must be in the same plane as the double bond.
Furthermore, it's not possible to claim that the ketone and the enol have the same geometries in the carbon chain. Many of the bond lengths in the two tautomers will be different: after all one has a C–C (and C=O) bond and the other has C=C (and C–O). Unlike resonance structures, both of these tautomers are things which exist independently, and they have different bond lengths.
The safest way is to go back to the IUPAC definition. That's a bit boring, but when you read these, you can see the effort that has gone into coming up with a definition that is (as far as possible) consistent with edge cases such as these. Tautomerism is defined as (https://doi.org/10.1351/goldbook.T06252):

Isomerism of the general form $\ce{G-X-Y=Z <=> X=Y-Z-G}$ where the isomers (called tautomers) are readily interconvertible.

Acrolein and propadienol do match the bonding pattern required for tautomerism, but don't pass the second criterion of being "readily interconvertible".
The reason is fairly simple. The process of tautomerisation involves protonating the oxygen, and deprotonating an α C–H (which way round this happens doesn't matter). To deprotonate a proton that is α to C=O, the C–H bond must be oriented parallel to the C=O π system, because the new C=C π bond is going to be made from that C=O π bond and the C-H σ bond. This cannot be the case in acrolein, because of the extended conjugation between the C=C and C=O bonds. Indeed, the α C–H bond in acrolein is perpendicular to the C=O π system.

One of the requirements of tautomerism is that the carbon skeleton of the molecule remains unchanged. Only the proton must change it's location with reference to the molecule. Acrolein and propadienol have very different molecular geometries and hence are not tautomers of each other. Here's a comparision of their structures. The first one is acrolein and second one is propadienol.

You have a link to the definition of tautomerism, which is great, but it doesn't actually stipulate in there that the carbon skeleton geometry must be unchanged.

That said, I don't entirely disagree with it. Just a little bit. There is probably a correlation between your geometry argument and my reactivity argument in that, in general, drastic changes in molecular geometry make reactions less likely to happen.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Bromination on the aromatic ring vs aliphatic chain

I thought that the answer would be a simple electrophillic aromatic substitution and as the carbonyl is a meta director, the bromine would be present meta to the group.
But the given answer was the α-brominated compound and is also found out an article which says that this happens in case a catalytic amount of $\ce{AlCl3}$ is used while the bromine is substituted in the ring if a larger amount is used. How does this happen?  Why does the amount of $\ce{AlCl3}$ matter?

Hmmm... Seems to me that something is wrong. The alpha-halogenation of the ketone is a reaction that takes place in aqueous medium with acid/base catalysis. This is because the reaction depends on generation of the enol/enolate form of the ketone. It is unlikely that alpha-halogenation would take place in an ether solvent with only $\ce {AlCl3}$ as a catalyst.

Catalytic AlCl3 enolizes the ketone. Bromination ensues liberating HBr which now acts as a catalyst while its concentration increases. Mix excess ketone with bromine in ether. Nothing happens and then the solution suddenly turns colorless. The enol content of the ketone is small but slowly bromination occurs forming more and more HBr.

The key in this reaction is ketone in propiophenone is acting as Lewis base to $\ce{AlCl3}$ Lewis acid. If $\ce{AlCl3}$ is in catalytic amount, it activate $\alpha$-$\ce{H}$ in side chain sufficiently enough (similar to acid would do to a carbonyl compound) and progress to produce $\alpha$-substituted product. If $\ce{AlCl3}$ is in excess, it seemingly deactivates aliphatic side chain completely (Ref.1), and ring substitution predominates to give $meta$-substituted product exclusively as a result (Ref.2 and 3):

The reaction with excess $\ce{AlCl3}$ should necessarily be done in solvent free conditions, and excess $\ce{AlCl3}$ is needed such that one equivalent is used to complex with carbonyl oxygen and another equivalent is used to activate $\ce{Br2}$ molecule. As a result, benzene nucleus would be substituted and 3-bromoacetophenone would be obtained exclusively (Ref.1). The crude liquid product is purified as an oil after fractional distillation in 70-75% yield (Ref.2). The reaction with trace amount of $\ce{AlCl3}$ has done in dry ether and solid crude yield has been 88-96%. When recrystallized by methanol, the yield has been reduced to 64-66% (Ref.3). Another stark different between these two reactions are addition order. In the excess reaction, acetophenone has been added to $\ce{AlCl3}$ while in the catalytic reaction, $\ce{AlCl3}$ has been added to etherial acetophenone solution.
References:

D. E. Pearson, H. W. Pope, W. W. Hargrove, W. E. Stamper, "The Swamping Catalyst Effect. II. Nuclear Halogenation of Aromatic Aldehydes and Ketones," J. Org. Chem. 1958, 23(10), 1412–1419 (https://doi.org/10.1021/jo01104a003).
D. E. Pearson, H. W. Pope, W. W. Hargrove (Checked by B. C. McKusick, D. W. Wiley), "3-Bromoacetophenone," Org. Synth. 1960, 40, 7 (DOI: 10.15227/orgsyn.040.0007) (Organic Syntheses, Coll. Vol. 5, p.117 (1973)).
R. M. Cowper, L. H. Davidson (Checked by Lee Irvin Smith, E. W. Kaiser), "Phenacyl Bromide ($\alpha$-Bromoacetophenone)," Org. Synth. 1939, 19, 24 (DOI: 10.15227/orgsyn.019.0024)(Organic Syntheses, Coll. Vol. 2, p.480 (1944)).

The carbonyl group attached to the benzene ring is an electron-withdrawing group and deactivates the ring towards electrophilic substitution. Thus, $\ce{AlCl3}$ in excess is required to brominate the ring in this case. If a catalytic amount is used, then the $\alpha-\ce{H}$ atom is substituted by a standard Ketone halogenation mechanism.


text:
Metamerism in amines
I came across a question where $\ce{R-CH2-NH2}$ and $\ce{R-NH-CH3}$ are considered as functional group isomers but not as metamers.
If I'm not wrong, metamers compounds which have same molecular formula but different number of carbon atoms on a functional group (for example, $\ce{CH3-CH2-O-CH2-CH3}$ and $\ce{CH3-O-CH2-CH2-CH3}).$
So, shouldn't the compounds given in start of the question be metamers considering the fact that the '$\ce{NH}$' has different number of carbon atoms on either side in the two compounds, and if not, why so?

Then shouldn’t ethanol and dimethyl ether be metamers as well...but they are not. [Compounds  having the same molecular formula but different number of carbon atoms (alkyl groups) on either side of functional group (i.e., -O-,-S-, -NH-, -C(=O)-) are called metamers and the phenomenon is called metamerism.]- Wikipedia.

Does this answer your question? Which is the correct definition for metamers (or metamerism)?


text:
Can I turn urine to odour free urine overnight?
My son leaves pee in the toilet overnight but I am a light sleeper so he doesn’t flush.
This leaves a unique smell in the toilet in the morning.
Is there any safe chemical/tablet that could be put in the toilet after he pees so the smell/colour isn’t there in the morning?

No way to neutralize the smell without doing something drastic. The only thing that you might be able to do is mask the smell with some sort of automatic mechanical room deodorizer. So you have to decide what you want - stinky bathroom or good night's sleep?

The smell is caused by bacteria which are feeding on urine. Regular cleaning of the toilet helps.

Related posts that might be somewhat helpful: Storage of Urine; Can caustic soda (sodium hydroxide) be used to dissolve urine odour?. Long story short, you might want to prevent bacteria growth; asking your son to spray a disinfectant over the toilet bowl (once he's done) might do, but do make sure he understands the danger of chemicals often used as such (e.g. chlorates/peroxides).

@imalipusram - ? She seems to be asking about the stink from a toilet bowl of urine left overnight. I assume that the son isn't spraying the whole toilet area with urine.

You could try using an automatic toilet bowl cleaner tablet.  These are placed in the tank (not the bowl), and release a cleaning agent into the water.  I don't know if this would reduce the odor, but they're cheap enough (a few dollars) that it might be worth experimenting with a couple of different formulations to see if they help.  Be careful when placing it into the tank -- the small amount of chemical released into the bowl with with each flush is relatively safe, but you want to avoid direct skin contact with the tablet itself.

....I'll add that these can have their own odor (some smell like cleaning agents), though, which you may not like (I tried using them once, and stopped for that reason).   I can't vouch for the reviews here, but this gives you an idea of the range of the products: https://www.allabouttoilet.com/best-toilet-bowl-tablets/

@MaxW - When the whole surface of the bowl is clean, less bacteria will be present. Also bacteria which are growing in the water covered drain section do matter.

People incorrectly treating their toilet has been a long occurring source of home accidents. Perhaps time for sound advice?

With respect to available home remedies, do NOT use H2O2 or Oxiclean or a generic substitute containing Sodium percarbonate ($\ce{2 Na2CO3·3 H2O2}$).
The H2O2 may react with the urea forming even more problematic Urea Hydrogen Peroxide.
Also, avoid chlorine bleach (Sodium hypochlorite, NaOCl), as this creates problematic chloramine (NH2Cl) and potentially even more unstable (as in energetic) compounds.
My advice, use powdered charcoal dust from your outdoor grill, a safe and odor removing agent. The obvious downside is now your toilet will likely display black rings and require more cleaning (which is a positive in removing germs).
I will also make a statement with respect to exposure from interaction products between urea and carbon, no expected long term negative health consequences (as in possibly carcinogenic).

I think everything derailed


text:
What is the order of adding chemicals in Hummers method for graphite oxide?
I am trying to make graphene oxide but failing miserably. The XRD peak is at 24 (2θ) instead of being near 10. Raman shows correct g and d peaks but are not very sharp. The colour of the end product is not yellow but grayish.
I have followed all the steps of Tour's and Hummers's methods.
But I am confused about should I add Graphite powder/flakes to $\ce{H2SO4}$ or $\ce{H2SO4}$ to the powder. And when should I add NaNO3 and how?
What I have tried is adding the $\ce{NaNO3}$ and graphite powder and mixing them and then added this mixture to the $\ce{H2SO4}$ and followed the rest of the process. Dried it at 60°C for over two days.
Also it will be helpful to know if the stirring rpm matters.

There have been dozens of other papers on the subject of refining Hummers´ method. Read more! ;)  And make sure you wash the end product free of alkali before drying, especially if you want to produce larger amounts!!

@Karl  I used HCl and ethanol for the washing process. Should have mentioned it (^--^) (Forgot!!)..About the papers, I have gone through a few of them and they all usually use an oxidizing agent like K2FeSO4 instead of NaNO3 or even pre-oxidized graphite. Still I am curious about the Tour's and Hummers's method. There was a correction to Tour's method about the order of adding chemicals. So I wanted to know if I am doing something wrong in the order of adding the chemicals. So far I could not get my hands on a paper that dealt with it. Any advice please... Does it affect the final product?

10.1021/acsnano.8b00128 This is the correction paper. But it did not discuss the effects of the order of other chemicals.

There shoud be a number of papers from Tölle and Mülhaupt, early 2010s.


text:
MolView is unable to generate correct 3D structure for hydroxypropadiene
MolView is unable to generate the correct structure of hydroxypropadiene. It instead generates the structure of acrolein (which has the same molecular formula). Why is this so? Is it because hydroxypropadiene is so unstable that the calculations done while generating the structure degenerate it to acrolein?

MolView is using the default GLMol renderer. Switching to JMol or ChemDoodle does not change the structure.
EDIT: The exact workflow I used was open MolView, delete the caffiene molecule, draw this molecule and click the 2D to 3D button. The generated URL is http://molview.org/?cid=7847, which, when pasted in a browser, interestingly redirects to Acrolein and not the compound that was drawn.

I suspect MolView simply fails to parse the compound's name you are trying to input. "Hydroxypropadiene" isn't the proper name, at least not by IUPAC standards. According to Google services, this name doesn't appear in any textbooks and is mentioned only in three research articles from 1980s.
The PIN propa‐1,2‐dien‐1‐ol appears to work correctly with MolView:

When you are trying to get a 3D image from the 2D drawing, there is the following issue which MolView usually reports (not sure why you haven't received the message):

Failed to load structure from sketcher. This structure may contain allenes, which cannot be represented in the SMILES notation.

MolView relies on SMILES in order to fetch the 3D structure.
The problem is likely that Ketcher generates wrong SMILES because PubChem indeed receives C=CC=O instead of the proper one, C=C=CO.

Interesting. I thought molview worked on the structure and not the name of the compound?

@AniruddhaDeb Ah, you were trying to draw the compound, not generate the structure by name. I added some info with respect to that.

Thanks. That clears it up. I received a message "the generated structure may be incorrect". I did not receive a message that specifically stated that Allenes could not be generated.

@AniruddhaDeb Yeah, the message might pop up inconsistently depending on how you draw the structure. Maybe try to add H atoms explicitly, this might help to get the error message. Also, note that allenes can normally be represented with SMILES, so the error message is likely to be interpreted more like "we cannot reliably generate SMILES in your particular case with our algorithm".


text:
What happens to the NMR of an alcohol when D2O is added?

What difference you expect in the 1H NMR spectrum of phenol in $\ce{CDCl3}$ after a drop of $\ce{D2O}$ is added to the test tube?

1) Since the $\ce{-OH}$ of the phenol is quite acidic, it undergoes rapid $\ce{H+/D+}$ exchange with the $\ce{D2O}$. My expectation is that the $\ce{-OH}$ signal from the original phenol will get reduced and broadened (maybe even disappear, but I am not sure) because it is replaced with an $\ce{-OD}$ group.
2) Aside from this, will the aromatic regions of the NMR spectrum change at all? I expect that the ortho and para $H$ can slowly turn into $D$ through EAS, but I am not sure if that is possible when only "one drop of $\ce{D2O}$ is added", especially without acid and base present.
My book does not contain an answer key, but I would like hear if my thought process is correct.

Your thought process is correct (not for aromatic protons though). The phenolic $\ce{OH}$ resonance would completely disappear due to rapid exchange with excess $\ce{D2O}$: $$\ce{Ph-OH + D2O <=> Ph-OD + DOH}$$

We would need to know the molar ratio of your sample and the D2O to say how much of the proton would be exchanged. And we would have to know the "frequency" of the NMR and the temperature to figure out whether this would be fast, slow or intermediate exchange.

In order to understand the probability of displacement of a H from any X-H bond (X is any atom), you need to know its pKa. The pKa values are as follows (ref.):

pKa of water = 14
pKa of OH (phenol) = 9.95
pKa of CH (benzene) = 43

If the pKa of a bond is $x$, it means that for one $\ce{X^- + H^+}$, there has to be $10^x$ no of $\ce{X-H}$ compound. Thus, it clear that for one dissociation of OH bond of phenol, ~$10^{10}$ phenol molecules are required, while for one dissociated CH bond of benzene, $10^{43}$ molecules of benzene is reuqired. To put it in perspective, in 1 mole of phenol, there will be ~$10^{13}$ dissociated OH bonds but only ~$10^{-20}$ dissociated CH bonds.
This hereby implies that dissociation of CH bond and exchanging with $\ce{D_2O}$ is highly improbable, but dissociation of OH bond and its exchange with $\ce{D_2O}$ is highly probable. That's why you get the OH peak to be lowered due to exchange but no similar phenomenon occurs for CH bonds in the system.


text:
Can electrodes be made from mercury?
'Mercury' was included in one of the possible answers for this exam question:

Name a metal that an inert electrode could be made from.

My answer was 'platinum' which was also a possible answer, but I’m curious how mercury, which is a liquid metal, could be used as an electrode during electrolysis?

You might like to have a look at Castner-Kellner cell

Polarography! Please look up the field of polarography in wikipedia! Look up the “dropping mercury electrode”.

Electrochemistry labs were well know to have mercury spilled in every corner so definitively look at the suggestion of Ed V.

Also look up “Hanging mercury drop electrode” in wikipedia! And @Alchimista I “inherited” one of those mercury contaminated labs when the department’ electrochemist retired. There was even mercury droplets in a cabinet drawer.

See saturated calomel electrode

By the way, mercury is not inert, relative to platinum or glassy carbon electrodes, for example. But the over-voltage for hydrogen ion reduction is important for aqueous solution work, and the mercury surface is harder to foul than Pt or Au, etc. With the dropping mercury electrode (DME), you get a fresh growing drop of Hg every few seconds or so, so the surface is relatively clean and you do not have to clean and polish the electrodes.

Trick question. There are noble metals that are "resistant to corrosion and oxidation", but the only metal that is chemically inert at STP (other than, "with sulfur at atmospheric pressure") is iridium.

There is nothing in the question that states the assembly needs to be operated at room temperature — mercury could operate as a solid electrode at lower temperatures.

The dropping mercury electrode (DME) is a working electrode made of mercury and used in polarography. Experiments run with mercury electrodes are referred to as forms of polarography even if the experiments are identical or very similar to a corresponding voltammetry experiment which uses solid working electrodes. Like other working electrodes these electrodes are used in electrochemical studies using three electrode systems when investigating reaction mechanisms related to redox chemistry among other chemical phenomena.

Source: dropping mercury electrode
Massive thanks to Ed V for advice in the comments prompting my research

(+1) Good work! Glad to see you answer your question!

@EdV thank you very much for your help

If I understand it correctly, the mercury in polarography is not an inert electrode - it directly participates in the redox chemistry. On the other hand, the platinum electrode acts as a catalyst for gases to undergo redox reactions (and as a conductor).

I disagree about the mercury participating (via redox itself) in the typical applications. This P.A.R. application note is quite good: https://www.ameteksi.com/-/media/ameteksi/download_links/documentations/library/princetonappliedresearch/application_note_p-2.pdf?la=en

Related only - but posted as an answer as, while not 100% what you asked about,  it's related enough to be worth noting.
In the past some angle detection or motion detection switches used rolling mercury as a bridge between two metal contacts.
The term "Mercury switch" was commonly used. The Mercury is necessarily inert relative to the associated metallic contacts on the environment used. Substantial currents can be handled in some designs. See references below for details. Thase switches are much less common now due to the hazards of Mercury contamination and the increased availability of electronic based alternatives.
Mercury Jet contacts have also been used in devices where minimising contact force is important.
Wikipedia Mercury Switch
Many images here each linked to a webpage.
Search What metals are used as contacts in a mercury switch?

Often Platinum, but also Iron, Molybdenum and more.
Never Gold! :-)

_____________________________________
Related / interesting:
Material for liquid metal contacts
Mercury metal oscillator !!! & inverter driven by one.
Links suggested by uhoh:
NIST NanoFab Tool: Mercury Probe

The Four Dimensions CV92A mercury probe system provides non-destructive and rapid measurement of current-voltage and capacitance-voltage curves to derive resistivity, breakdown voltage/field and film thickness of thin dielectric films without having to metallize the wafer and fabricate capacitors, as is required by traditional probe stations. The capacitor top contact (”gate”) is simply formed by a mercury contact. The accuracy of the measurements is guaranteed by the precisely defined contact area.

MERCURY C-V PROFILING

MCV systems allow to eliminate the need for costly metal and poly deposition processes by using a pneumatically controlled, non-damaging probe design and a top-side mercury contact. The system features an extremely stable contact area and uses only a small quantity of mercury to make highly repeatable C-V and I-V measurements for process development and process monitoring applications.

________________________________
Mercury switch - source Wikipedia

@EdV Thanks. You checked up on me :-).

+1 also https://www.nist.gov/laboratories/tools-instruments/nanofab-tool-mercury-probe and https://semilab.com/category/products/mercury-c-v-profiling-0

@uhoh Thanks - I added your references to my answer

In a "previous lifetime" I'd used one to characterize thin oxide layers on silicon wafers.

@uhoh I did wonder how you were aware of them. | I too have been known to use "In a previous lifetime" :-). 23 years in corporate employ and for the last 27 years having fun under my own steam. (That makes me old, right ? :-) ).

@RussellMcMahon oh I'm not sure if you can qualify as old yet, I think you'd better keep working on it! I hadn't heard that about the "reconstruction" at Borobadur, and now I can't remember if I was there before or after that, will have to (go find and) check the snapshots. I like living in this hemisphere because so many destinations are close and accessible (except for right now with the pandemic thing).

@uhoh Borobadur - several reconstructions - earliest early 20th century. Latest 1975-1982 -wikipedia - they essentially dismantled the whole edifice, all stones marked for location, laid it all out on the surrounding plain, rebuilt the foundation with drainage and strengthening and then rebuilt it. [!!!]. Have you met the "Borobadur Boat". - imge in the stone of the craft(2?) used to settle  Madagascar from Borneo before Magellan sailed by? [!!! again].  The majority people of Madagascar are thus ethnically 'Indonesian'.

An old method of electrolytic production of sodium from water solution uses a mercury cathode. It is "inert" in regard to not taking part in the electrolysis itself. It dissolves sodium and prevents it from reacting back with water.

@NilayGhosh Just for interest - kathode is the German language version. (Possibly also Dutch?)

Yep. But I speak neither German nor Dutch, nor any other language that has "th" in "cathode". I made a honest and genuine spelling mistake.


text:
Electrophilic Aromatic Substitution of phenols
What happens if 2,4,6-tribromophenol is forced to undergo further bromination?
In phenol, the benzene ring is activated towards electrophilic substitution. Therefore, phenol on bromination gives 2,4,6-tribromophenol. If now, 2, 4, 6-tribromophenol is forced to undergo bromination, the group attached to the benzene nucleus makes phenol not suitable to undergo Friedel–Crafts reaction due to adduct formation with Lewis acid.

Please make your last statement If now, 2, 4, 6-tribromophenol is forced to undergo bromination the group attached to benzene nucleus makes phenol not suitable to undergo Friedel Craft reaction due to adduct formation with Lewis acid. a bit more clear.

2,4,6-Tribromophenol may be converted to the tetrabromide, 2,4,4,6-tetrabromo-hexa-2,5-dien-1-one. It is an effective, crystalline brominating agent of enols. https://pubs.acs.org/doi/suppl/10.1021/jo061522l/suppl_file/jo061522lsi20060905_042700.pdf

Way overthinking this.  Kohn reports that 2,4,6-tribromophenol can be further brominated to pentabromophenol by adding more bromine and an iron catalyst.  Subsequently the pentabromophenol is oxidized by fuming nitric acid to give a fully brominated quinone.


text:
Is it possible to oxidize beryllium metal around NTP or not?
Perhaps this seems like a really simplistic question to many of you, but I’ve read too many conflicting answers at this point.
Simply put, most articles I’ve read have insisted that it isn’t possible to ignite pure beryllium below temperatures of about 1000 to 1200 Celsius. However, the explanation behind this usually has something to do with an oxide layer forming on the outside of the metal that prevents oxidation of the metal atoms that aren’t on the surface anywhere near NTP (20 C, 1 atm). This logic doesn’t track for me, as it clearly suggests that oxidation at NTP can occur in an attempt to explain why it can’t.
I guess my question is, if you were to take a sizable chunk of beryllium metal, grind it up to an extremely fine powder — or rather, since we’re thinking theoretically, pound it out into a single-atom-wide sheet — and then expose it to oxygen gas at a pressure of 1 atm and a temperature of 20 C, how violently would said metal powder/sheet react?
To me, all literature on the subject seems to imply that it reacts at an imperceptibly slow rate like an iron fence rusting, although even iron oxidation becomes noticeable with fine powders (e.g. in single-use handwarmers). Is this true, or would something more noteworthy occur?

Bulk metals and pyrophoric powdered metals behave differently. Bulk aluminum and zirconium, for example, are not easily ignited. But their ultra-fine powders are pyrophoric. Surface area and oxide coatings make a big difference. Beryllium, by the way, is a dangerously toxic element, but BeO insulators used to be available for semiconductor heat sinking usage.

A metal forming a passivating oxide layer is well known.

Another point - Pure oxygen greatly accelerates any oxidation reaction compared to air. // Hydrocarbon grease is used on all sorts of things and normally thought of as being relatively resistant to oxidation. However is is absolutely forbidden to use hydrocarbon grease on pure oxygen fittings because of the danger of spontaneous combustion.

Mind you the same things happens for both silicon and aluminum and neither one is going to readily ignite. Self-passivating oxides can be really nice things.

@EdV - note that the problem is actually with the oxide leading to berylliosis.

@JonCuster Yeah, I saw that after I posted the comment and read the wiki article. Amazing that the next generation space telescopes will have Be mirrors or gold-plated Be mirrors. I had read somewhere that other beryllium compounds were toxic, so I sent my bottle of beryllium sulfate off to our university environmental health and safety folks.

@EdV I’m aware of the fact that bulk materials and powders have different properties. Even flour is a good example of this. The surface area to volume ratio is obviously the driving factor in these differences. However, I’ve seen several videos now of people attempting to ignite even powdered beryllium to no avail. My guess is that the grains of the powder, obviously not being single atoms, developed their own oxide layer before the attempt was made. Is this correct? And if so, how much would you expect the results to differ if an instant change from vacuum to pure oxygen was performed?

Presumably very fine Be powder could by made by grinding Be in vacuo (Be is brittle, which helps), effectively eliminating the oxide coating that might exist on the starting bulk Be. My gut feeling is that exposing that powder to pure oxygen would result in formation of the oxide layer. But I don’t actually know. Of course, experiments could be done to detect any resulting oxide, but I do not know what specific experiment might be done. Perhaps x-ray diffraction or something of that ilk.


text:
Conversion of methylcyclopentanol to cyclopentanone
How to convert methylcyclopentanol to cyclopentanone?
I tried eliminating the alcohol group as $\ce{H2O},$ but that gives Saytzeff product. If only I can get the Hofmann product (methylene cyclopentane), I would subject it to ozonolysis and get cyclopentanone.
I don't know exactly what to use as the reagent to get that Hofmann product. Could there be other methods for the conversion?

Just to make things clear, do you mean 1-methylcyclopentanol is the precursor and methylidenecyclopentane is the Hofmann product?

If you leave it to E1, you will not get the selectivity you need.
The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction.
In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is typically accomplished by first converting to a tosylate or a mesylate.

Thank u so much... So that means I will have to treat it first with TsCl and then with tertiary butoxide (a bulky base) to get the Hoffman product and carry out the further reaction.

If you are dealing with 1-methylcyclopentanol, this reactive tosylate may form the chloride in situ. Why not treat the alcohol with HCl to form the chloride initially and then proceed?

@user55119 I strongly doubt that that tosylate will react as is without heating (Please provide a reference if you have reason to believe my intuition is wrong. It might be.). Sn1 reactions can be messy, and I have a strong preference not doing that when possible.

From my own experience. Used to put reaction in a freezer to form tosylate and avoid formation of chloride in more complex tertiary alcohols. It doesn't matter much in this case if it is a mixture of tosylate and chloride.

Ah OK. Still good to know though. Thanks for the feedback. @user55119


text:
Why is it necessary to bubble hydrogen through the electrolyte in a hydrogen electrode?
I searched from the internet but i got confused. I was not sure about my answer.

What exactly do you find confusing?

What exactly bubble hydrogen does in a hydrogen electorde

Ehhm, assures its functionality? Look, you probably want to give Wikipedia article a good read and refine your question.

Since it has not been mentioned yet, you appear to be referring to the Standard Hydrogen Electrode, called, for short, SHE. So your internet search was not so effective.

The presence of $\ce{H_2}$ on the electrode is necessary, in order to satisfy Nernst' law. The potential of the $\ce{H_2}$ electrode is defined by the following expression : $$\ce{E = E° + (0.0592/2)·log{[H^+]^2/p(H_2)]}}$$ In this equation, $\ce{E°}$ is equal to Zero by definition. If we want the measured $\ce{E}$ value of the Hydrogen electrode to be also equal to Zero, the platinum electrode must be in contact with $\ce{H^+}$ in a solution where the activity is equal to $1$ (nearly $1$ M). But the platinum electrode must also be in contact with $\ce{H_2}$. The trouble is that $\ce{H_2}$ is not a metal. It is a gas, and it is not easy to have a gas in contact with a platinum electrode. Furthermore, the gas should be at $1$ atmosphere pressure. The simplest way of carrying out this contact is to have some pure $\ce{H_2}$ bubbles crossing the solution, with the hope that at least one or two droplets gets attached or adsorbed at the surface of the platinum. The yield is not remarkable, as nearly all bubbles are simply crossing the liquid to be lost at the surface of the liquid. But the small amount which stays fixed on the platinum electrode is sufficient to get the whole electrode at Zero potential.

Not an expert in electrochemisty, but I would think the gas doesn't have to be in contact with the electrode. Rather the solution has to be saturated with the gas. This is why you bubble the gas. Contact between bubbles adhering and releasing from the electrode would probably generate unwanted noise.


text:
Diluting 4 % NaOCl to get a 1 % solution
Assuming that one has a 1 litre bottle of liquid bleach containing $4~\%$ sodium hypochlorite. If I wish to dilute it down to $1~\%$ sodium hypochlorite (to use as disinfectant), would adding 3 litres of water to it, be sufficient? This should yield 4 litres of $1~\%$ sodium hypochlorite.
And that should be same as taking $\pu{250ml}$ of the liquid bleach, and then adding $\pu{750ml}$ of water, to yield 1 litre of $1~\%$ sodium hypochlorite.
Now the liquid bleach also contains a stabiliser to reduce the rate of decomposition of $\ce{NaOCl}$. How will diluting the solution affect the action of the stabiliser?

Comments are not for extended discussion; this conversation has been moved to chat.

There is a simple dilution formula: $\mathrm{C_iV_i=C_fV_f}$, which is valid for all concentration units. The "i" and "f" indicate initial and final concentrations or volumes.
Using this formula, you can see that you don't need to prepare buckets full of 1% bleach. All you need to decide is what should be your final volume and final concentration.
As you highlighted, the issue, you cannot prepare a 1% solution and expect it to last for a day. It is better to prepare on need basis. The stabilizer in bleach is sodium hydroxide itself. Upon dilution, of course the base is also diluted and the pH is reduced.
One can look at the distribution diagram of "bleach" as a function of pH (yes this is from a Handbook of Surfactant. It is a 4000 paged volume- books are still useful)

Perfect explanation and completely understandable answer. Thanks, especially for extracting this diagram from a 4000pg book :-)

As the dilution part of the answer has been given, and I will provide an answer on actually the very important point on maintaining the integrity of the diluted composition, so as to answer the specific question:

Now the liquid bleach also contains a stabiliser to reduce the rate of decomposition of NaOCl. How will diluting the solution affect the action of the stabiliser?

The most common stabilizer per this patent source in agreement with my observations is, to quote:

Small amounts of free caustic (NaOH) or soda ash (Na2CO3) may be used to buffer the finished product to a desired pH level.

So, simple dilution should be accompanied by addition of pure NaOH or Washing Soda (Na2CO3) to preserve the starting target pH of the chlorine bleach product. Failure to do so, could result in a rapid deterioration of the aqueous Sodium hypochlorite.
Additionally, as the chlorine is highly sensitive to transition metal impurities, to cite the source again:

Chorine bleach degrades rapidly due to many factors including the presence of trace amounts of di- and trivalent transition metals ions such as Ni, Co, Cu, and Fe etc.

The water source employed for dilution should be transition metal-free. Also, as dust particles can contain trace amounts of said metal contaminants, take precautions on cleanliness and air flow exposure.


text:
The particle in a two-dimensional box model and graphene
I have been having trouble with understanding and solving the following problem.

For two-dimensional conjugated systems, we may use the particle in a two-dimensional box model. In this case the total energy can be written as follows:
$$E_{n_1,n_2} = \frac{h^2}{8m} \left( \frac{n^2_1}{L^2_1}+\frac{n^2_2}{L^2_2} \right)$$
where $L_1$ and $L_2$ are the lengths and $n_1$ and $n_2$ are the quantum numbers of the first and second dimensions, respectively, $m$ is the mass of the particle and $h$ is the Planck constant.
Graphene is a sheet of carbon atoms in the form of a two-dimensional hexagonal lattice in which one atom forms each vertex.

The distance between two adjacent carbons in the hexagonal 6-carbon unit is approximately $1.4$ Å. For a square shaped graphene sheet with $L_1$ = $L_2$ = $11$ Å:
1) Calculate the number of electrons in a ($11$ Å $×$ $11$ Å) sheet of graphene. For this problem you may ignore edge electrons.
2) Calculate the energy of the HOMO.
3) Calculate the energy of the LUMO.

For the first question, I simply divided the area of the graphene sheet by the area of a hexagon shaped graphene unit. This resulted in approximately 24 units, which means  there is a total of 48 carbon atoms, since each unit contributes 2 carbon atoms. Therefore there would be a total of 288 electrons in the sheet (from multiplying 48 by 6, since each carbon atom has 6 electrons).
In case this is right, I don't know how to go about calculating the HOMO and LUMO energies. I'm having a bit trouble understanding the quantum numbers in the formula given above: I don't know what the values of the quantum numbers should be when calculating the HOMO or LUMO energy. I do understand how the quantum numbers combinations and energy are related, but I don't know, which specific combinations are used when calculating the HOMO and LUMO energies.

You can use the number of occupied MOs to get to the HOMO. Then it'll straightforward to get to the LUMO. Part of the exercise is to relate the quantum numbers to the MOs.

The key to answering this is to remember the following principles:
1) electrons are fermions and therefore two electrons cannot share all quantum numbers (degeneracy of all QM numbers is forbidden)
2) electrons have spin quantum number, which means an energy state corresponding to a given set of $n_1$ and $n_2$ quantum numbers can be doubly occupied (doubly degenerate)
3) the Aufbau principle, which means lowest energy states are occupied first
Once you know the total number of electrons, you have to assign each a pair of quantum numbers $n_1$ and $n_2$ (ignoring $n_s$, the spin quantum #). The first two principles listed above mean that to each $(n_1,n_2)$ state you can assign two and only two electrons (with opposite spin quantum #). The third principle means that, since the energy increases with $n_i$, and the sides of the box are identical (making states that permute the $n_i$ degenerate), you should assign the $n_i$ in increasing order of energy as follows: the lowest E state has $n_1=1,n_2=1$, or (1,1); followed by (2,1) or (1,2) - these are degenerate in energy; then (2,2); then (3,1) or (1,3) -again degenerate; then (2,3) and so on.
In tabular form the energies increase as follows:
$$\begin{array}{c|c|c|c}\hline i & n_1 & n_2 & n_s  \\ \hline 1  & 1 & 1 & \alpha \\   & 1 & 1 & \beta \\ 2  & 2 & 1 & \alpha \\   & 2 & 1 & \beta  \\   & 1 & 2 & \alpha \\   & 1 & 2 & \beta  \\ 3  & 2 & 2 & \alpha  \\   & 2 & 2 & \beta\\ 4  & 1 & 3 & \alpha \\   ...&...&...& \\ \hline\end{array} $$
If you counted 288 electrons you need 144 states that are doubly degenerate (permutations of $n_i$, differing only in $n_s$) to populate the box. The HOMO is the occupied state with highest energy.
Using a script to crack the problem (rather than searching for a mathematical formula that encodes the pattern) produced the following list of states (each can be populated by two electrons, one with $n_s=\alpha$, the other $n_s=\beta$). The HOMO is triply degenerate (ignoring spin) and the LUMO doubly degenerate.
$$\begin{array}{c|c|c|c}\hline i & n_1^2 +  n_2^2 & n_1 & n_2  \\ \hline 1 & 2 & 1 & 1 \\
     2 & 5 & 1 & 2 \\
     3 & 5 & 2 & 1 \\
     4 & 8 & 2 & 2 \\
     5 & 10 & 1 & 3 \\
     6 & 10 & 3 & 1 \\
     7 & 13 & 2 & 3 \\
     8 & 13 & 3 & 2 \\
     9 & 17 & 1 & 4 \\
     10 & 17 & 4 & 1 \\
     11 & 18 & 3 & 3 \\
     12 & 20 & 2 & 4 \\
     13 & 20 & 4 & 2 \\
     14 & 25 & 3 & 4 \\
     15 & 25 & 4 & 3 \\
     16 & 26 & 1 & 5 \\
     17 & 26 & 5 & 1 \\
     18 & 29 & 2 & 5 \\
     19 & 29 & 5 & 2 \\
     20 & 32 & 4 & 4 \\
     21 & 34 & 3 & 5 \\
     22 & 34 & 5 & 3 \\
     23 & 37 & 1 & 6 \\
     24 & 37 & 6 & 1 \\
     25 & 40 & 2 & 6 \\
     26 & 40 & 6 & 2 \\
     27 & 41 & 4 & 5 \\
     28 & 41 & 5 & 4 \\
     29 & 45 & 3 & 6 \\
     30 & 45 & 6 & 3 \\
     31 & 50 & 1 & 7 \\
     32 & 50 & 5 & 5 \\
     33 & 50 & 7 & 1 \\
     34 & 52 & 4 & 6 \\
     35 & 52 & 6 & 4 \\
     36 & 53 & 2 & 7 \\
     37 & 53 & 7 & 2 \\
     38 & 58 & 3 & 7 \\
     39 & 58 & 7 & 3 \\
     40 & 61 & 5 & 6 \\
     41 & 61 & 6 & 5 \\
     42 & 65 & 1 & 8 \\
     43 & 65 & 4 & 7 \\
     44 & 65 & 7 & 4 \\
     45 & 65 & 8 & 1 \\
     46 & 68 & 2 & 8 \\
     47 & 68 & 8 & 2 \\
     48 & 72 & 6 & 6 \\
     49 & 73 & 3 & 8 \\
     50 & 73 & 8 & 3 \\
     51 & 74 & 5 & 7 \\
     52 & 74 & 7 & 5 \\
     53 & 80 & 4 & 8 \\
     54 & 80 & 8 & 4 \\
     55 & 82 & 1 & 9 \\
     56 & 82 & 9 & 1 \\
     57 & 85 & 2 & 9 \\
     58 & 85 & 6 & 7 \\
     59 & 85 & 7 & 6 \\
     60 & 85 & 9 & 2 \\
     61 & 89 & 5 & 8 \\
     62 & 89 & 8 & 5 \\
     63 & 90 & 3 & 9 \\
     64 & 90 & 9 & 3 \\
     65 & 97 & 4 & 9 \\
     66 & 97 & 9 & 4 \\
     67 & 98 & 7 & 7 \\
     68 & 100 & 6 & 8 \\
     69 & 100 & 8 & 6 \\
     70 & 101 & 1 & 10 \\
     71 & 101 & 10 & 1 \\
     72 & 104 & 2 & 10 \\
     73 & 104 & 10 & 2 \\
     74 & 106 & 5 & 9 \\
     75 & 106 & 9 & 5 \\
     76 & 109 & 3 & 10 \\
     77 & 109 & 10 & 3 \\
     78 & 113 & 7 & 8 \\
     79 & 113 & 8 & 7 \\
     80 & 116 & 4 & 10 \\
     81 & 116 & 10 & 4 \\
     82 & 117 & 6 & 9 \\
     83 & 117 & 9 & 6 \\
     84 & 122 & 1 & 11 \\
     85 & 122 & 11 & 1 \\
     86 & 125 & 2 & 11 \\
     87 & 125 & 5 & 10 \\
     88 & 125 & 10 & 5 \\
     89 & 125 & 11 & 2 \\
     90 & 128 & 8 & 8 \\
     91 & 130 & 3 & 11 \\
     92 & 130 & 7 & 9 \\
     93 & 130 & 9 & 7 \\
     94 & 130 & 11 & 3 \\
     95 & 136 & 6 & 10 \\
     96 & 136 & 10 & 6 \\
     97 & 137 & 4 & 11 \\
     98 & 137 & 11 & 4 \\
     99 & 145 & 1 & 12 \\
     100 & 145 & 8 & 9 \\
     101 & 145 & 9 & 8 \\
     102 & 145 & 12 & 1 \\
     103 & 146 & 5 & 11 \\
     104 & 146 & 11 & 5 \\
     105 & 148 & 2 & 12 \\
     106 & 148 & 12 & 2 \\
     107 & 149 & 7 & 10 \\
     108 & 149 & 10 & 7 \\
     109 & 153 & 3 & 12 \\
     110 & 153 & 12 & 3 \\
     111 & 157 & 6 & 11 \\
     112 & 157 & 11 & 6 \\
     113 & 160 & 4 & 12 \\
     114 & 160 & 12 & 4 \\
     115 & 162 & 9 & 9 \\
     116 & 164 & 8 & 10 \\
     117 & 164 & 10 & 8 \\
     118 & 169 & 5 & 12 \\
     119 & 169 & 12 & 5 \\
     120 & 170 & 1 & 13 \\
     121 & 170 & 7 & 11 \\
     122 & 170 & 11 & 7 \\
     123 & 170 & 13 & 1 \\
     124 & 173 & 2 & 13 \\
     125 & 173 & 13 & 2 \\
     126 & 178 & 3 & 13 \\
     127 & 178 & 13 & 3 \\
     128 & 180 & 6 & 12 \\
     129 & 180 & 12 & 6 \\
     130 & 181 & 9 & 10 \\
     131 & 181 & 10 & 9 \\
     132 & 185 & 4 & 13 \\
     133 & 185 & 8 & 11 \\
     134 & 185 & 11 & 8 \\
     135 & 185 & 13 & 4 \\
     136 & 193 & 7 & 12 \\
     137 & 193 & 12 & 7 \\
     138 & 194 & 5 & 13 \\
     139 & 194 & 13 & 5 \\
     140 & 197 & 1 & 14 \\
     141 & 197 & 14 & 1 \\
     \text{HOMO}\,142 & 200 & 2 & 14 \\
     \text{HOMO}\,143 & 200 & 10 & 10 \\
    \text{HOMO}\, 144 & 200 & 14 & 2  \\
     \text{LUMO}\, 145 & 202 & 9 & 11 \\
    \text{LUMO}\, 146 & 202 & 11 & 9 \\
     147 & 205 & 3 & 14 \\
     148 & 205 & 6 & 13 \\
     149 & 205 & 13 & 6 \\
     150 & 205 & 14 & 3 \\
     151 & 208 & 8 & 12 \\
     152 & 208 & 12 & 8 \\
     153 & 212 & 4 & 14 \\
     154 & 212 & 14 & 4 \\
     155 & 218 & 7 & 13 \\
     156 & 218 & 13 & 7 \\
     157 & 221 & 5 & 14 \\
     158 & 221 & 10 & 11 \\
     159 & 221 & 11 & 10 \\
     160 & 221 & 14 & 5 \\ \hline\end{array} $$

I'm still a bit confused how you ended up with the combinations (47,48) and (48,47). Do I need to list 144 different combinations in increasing order?

@PöytäLaatikko Well, I saw that that would involve a lot of effort, so I tried finding a way of computing the highest QM number. Each ni QM # occurs a max of 6 times, and from your question 6x48=288, so the HOMO has to have QM n=48 in one dimension. It looks like I made a mistake at this point, however. It seems the HOMO should be a combination of (48,49) and (49,48), and the LUMO is (49,49). I changed my answer to refect this and again, while the principles I state should guide you, ...

... enumerating states explicitly is not a good way to solve the problem, you should instead find some other way to count them. Look for a pattern in the way the states are filled.

Have you figured the right combinations out yet? I'm still very unsure about them.

@PöytäLaatikko No, unfortunately not yet.I could write a script that would crack it but that would be a totally crude and unsatisfying solution. My guess is that the answer lurks in some math textbook or perhaps over at math se. I can give it a little thought over the next hour...

@PöytäLaatikko I gave it another shot with a script. Not pretty and no guarantees but it looks right. A solution involving a closed math formula would have been nicer but I am not a mathematician.

Thanks for the effort! After some inspection of my own, I think I ended up with the same result!


text:
What are the major differences between different reagents used for hydrolysis?
While doing questions on hydrolysis, i often get confused about what will happen if different reagents are used in hydrolysis... Such as H+ ; H+/H2O ; H+/H2O/heat ; etc.
What exactly can we concur when these different reagents are given?

As a noobie to ChemSE you are expected to show your thoughts lest your post be closed.

Usually, H+ and H+/H2O mean the same thing

Those are the same reagents ( so does H+ and H3O+ ), but expressed differently. Heat is just different condition, speeding up reactions.


text:
Is there a mistake in diagramatic representation of RS perturbation theory described in Szabo and Ostlund textbook?
I'm trying to understand rules of the diagrammatic representation of the many body perturbation theory that described in Szabo and Ostlund "Modern Quantum Chemistry" textbook. One of the first rules they introduced is the following:
"Each pair of adjacent dots contributes the denominator $$\sum E_{hole}^{(0)} - \sum E_{particle}^{(0)}$$ where sums run over the labels of all hole and particle lines crossing an imaginary horizontal line separating the two adjacent dots."
According to that rule and other rules described in this book, the sum of the following diagrams:

can be translated as follows:
$$-\sum_{n,m\neq i}\frac{2|V_{ni}|^2|V_{mi}|^2}{(E^0_{i}-E^0_{m})(2E^0_{i}-E^0_{n}-E^0_{m})(E^0_{i}-E^0_{n})}$$
The idea of authors was that if we sum all possible n-dots diagrams we will get the exact expression for nth-order correction to energy. However, for the fourth-order perturbation theory there are no terms with such a denominator factor:
$$(2E^0_{i}-E^0_{n}-E^0_{m})$$
Am I right that in the case of 4th-order perturbation theory there is no direct mapping between those diagrams and strict algebraic formulations?


text:
Can a substituted cyclohexane progress through SN1?
I have read that when the leaving group substituent is stuck in the equatorial position for some reason (like from a t-butyl group or a trans decalin), SN2 is either very slow or just doesn't happen at all because the rear attack is hindered by axial hydrogens/other substituents. I haven't been able to get a straight answer as to whether it can instead progress through SN1 at an acceptable rate, since the attached carbon will at least be secondary.
The only thing I can think of which would prevent this happening is the difficulty of forming a planar sp2 carbocation in a such a rigid system.
Any help is appreciated, thanks.

I think it's difficult to say for sure without actually making experiments. It should depend a lot on the conformation of the reactant, the substituents, the leaving group and also on the solvent. Nevertheless, I would imagine that it should be possible in some cases if you have a good leaving group.
In the case of a substituted cis-decalin, for example (I suppose you actually meant to refer to cis because in the trans isomer both hydrogens are stuck in axial positions), the elimination of the leaving group must indeed induce some strain, but maybe not that much. The structure could adapt with small changes in angles all over the molecule and both cycles would keep chair-like conformations. A tertiary carbocation would also be quite favorable.
So yes, in my opinion it should be possible in some cases, specially if heating is employed or you have a good leaving group. However, I should reinforce the fact that it really depends on many factors. This type of system can get quite complex if you consider the influence of the solvent, the influence of the substituents and the possibility of side reactions, including eliminations reactions, which commonly compete with substitution (on the presence of bases).


text:
What is the electron count in this nickel complex?
Our class has started learning about electron counting using the ionic method. I was having a little difficulty, especially when there are two metals in one complex, so I looked at Wikipedia for help.
The Wikipedia page for organonickel complexes says:

In $\ce{(allyl)2Ni2Br2}$ and $\ce{(allyl)Ni(C5H5)}$, nickel is
  assigned to oxidation number +2, and the electron counts are 16 and
  18, respectively.

I understand that in $\ce{(allyl)Ni(C5H5)}$, the allyl and cyclopentadienyl ligands both have a $-1$ "charge", so the nickel has a +2 "charge". Thus, the electron count is: $4 + 8 + 6 = 18$.
However, I am struggling to calculate the electron count in $\ce{(allyl)2Ni2Br2}$. Using the same logic, $\ce{Ni}$ is in the +2 oxidation state. Since the structure (see below) does not have a metal-metal bond, we do not have to add $1$ when counting electrons. So, I think the electron count should be $\frac{4*2 + 8*2 + 2*2}{2} = 14$, not $16$. How does Wikipedia get that $\ce{Ni}$ has an electron count of $16$ in $\ce{(allyl)2Ni2Br2}$?

The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $4×2$ in the numerator where you have $2×2$.  Then it will come out to $16$.

"bromide ions contribute an electron pair on each side of the square they form with the nickel centers". What do you mean by each side of the square? There are 2 sides of the square that connect to a nickel, and if each side counts as an electron pair, doesn't that mean 2*2 = 4 electrons?

Yes, bromine can contribute four electrons.  In fact if you follow my link you see it doing that routinely in organic chemical reactions.


text:
Carboxylic acid bond lengths in 3-carboxycyclobutane-1-carboxylate

Select correct order of bond length of the above bond P, Q, R & S:
(A) P > R > S > Q
(B) Q > R = S > P
(C) S > Q > R > P
(D) None of these

I think the answer should be (D) because in this molecule, both carboxylic acid groups are equally acidic and the $\ce{H+}$ may migrate from one acid moiety to the other. This would be somewhat similar to the idea of resonance, although it isn't actually the same idea.
However, the given answer is (B) which we will get if this structure remains static. Because of the proton transfer, I don't think this can happen. Am I right or not?
Moreover, if we were to use some spectroscopic method to study the molecule, would we get an image in which the molecule has four equivalent bond lengths (due to proton transfer), or not?

The $\ce{H+}$ is not so loosely bound as to migrate from one acid group to another.

Over time, yes, the bond lengths will all average out to be equal. But you said it yourself: this isn't resonance. At any one instant, this molecule has three different equilibrium bond lengths.

@orthocresol I am a school student. I don't know much about what I am asking about imagining a molecule in this comment. If we were to take IMG. Of molecule from any meth. May be raman spectroscopy (not at all sure) than would we get IMG. Of molecule with all bonds equal or different.

That's a really good question to ponder about. The general answer is beyond the scope of this question, but it comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is an average over those $x$ seconds. Bear in mind that everything has a finite time, including the interaction of photons with the system. There is no such thing as an instant method. So it comes down to how fast the system is moving, vs how fast your method is.

Of course, for photons you need to phrase that in a more quantum-y way, using e.g. energy-time uncertainty principle. But that's the general idea, at least. If you use a method that is "slow"er than proton transfer between the acids (e.g. NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR??), then you may well find that they are inequivalent. But the question is probably asking in the context of a (hypothetical) infinitely fast method.

@orthocresol thank you for your such a detalied  explanation. I appreciate your help

No problem. You've touched on a very important idea for all analytical methods, so kudos. I will try to gather my thoughts properly into an answer at some point in time.

It's most certainly true that proton transfer between the two acid groups can occur and is facile (particularly in an intermolecular manner). However, that only means that over a sufficiently long period of time, the four C–O equilibrium bond lengths will all average out to be the same. It does not mean that at any one instant in time, they are all the same.*
As you said, this is not the same thing as resonance. Resonance in a carboxylate group is a phenomenon whereby at all points in time, the two bonds are equal. There is no flip-flopping between two different structures.

Before we get to the answer for the actual question, I'll talk a little bit about your last question.
You bring up the idea of using spectroscopy to measure bond lengths, which is a very good point. The general answer is beyond the scope of this question, but it basically comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is a time average over those $x$ seconds. Bear in mind that every method of spectroscopy takes a finite time, including the interaction of photons with the system. There is no such thing as an instant method; there are merely faster and slower ones. So it comes down to how fast the system is moving versus how fast your method is.
It's exactly analogous to taking a picture of a moving object, like a fast car.† If you try that with an phone camera with a long exposure time you will probably just get a blur. But a high-speed camera is good enough to take clear snapshots of Formula 1 cars during the race.
Of course, for photons you need to phrase that in a more quantum mechanical manner, using e.g. the energy–time uncertainty principle. But that's the general idea, at least. If you use a method that is "slower" than proton transfer between the acids (perhaps NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR?), then you may well find that they are inequivalent.

Back to your question.
Generally, it is the instantaneous (static) picture which chemists deal with. That means that, at least for the purposes of this question, the dynamic motion of the system is not relevant, even though you are very correct to bring up the proton transfer. Equivalently, we could say that these questions are usually meant in the context of a (hypothetical) infinitely fast measurement method.
An infinitely fast method would be able to tell the difference between the $\ce{CO2H}$ and the $\ce{CO2-}$, because at any one point in time, one of them is protonated and one isn't. They can't be in a state where they're both half-protonated.
But an instantaneous method can't differentiate between the two oxygens in the carboxylate group. Those are entirely equivalent by resonance. And therefore you get three different bond lengths: Q > (R = S) > P.

* Of course, we are neglecting vibrational motion of the bonds, which technically mean that at any point in time it's far more likely that all the bonds are different. For example, even in the carboxylate group in which both oxygen atoms are equivalent (due to resonance), if an asymmetric stretch occurs then the instantaneous bond lengths will be different. Here, when we talk about "bond lengths", we are only referring to equilibrium bond lengths. We are also neglecting single bond rotations...
† I am not an expert on photography, so you may want to check out Photography Stack Exchange.


text:
Benzene to phenol in a single step
My teacher told me we can convert benzene to phenol within a single step using $\ce{V2O5}$/ $\ce{O2}$ at 300 ⁰C (not 500 ⁰C). Later I searched for that reaction but I couldn't find it in all over Google except in Allen's Chemistry handbook.
Does that reaction really occur and if it occurs then what is the mechanism of that reaction?

I don’t think that we can get phenol from benzene just by $\ce{V_2O_5}$. As you know, $\ce{V_2O_5}$ is an oxidising agent, it will oxidise our benzene.

Related(for the maleic anhydride case): https://chemistry.stackexchange.com/questions/8209/what-is-the-mechanism-for-the-oxidation-of-benzene-to-maleic-anhydride-using-van

Also related: https://pubs.acs.org/doi/abs/10.1021/i200015a003

TL; DR: Not exactly ratta maar, a mechanisms is there.. Perhaps it maybe beyond your current level of understanding,or tour teacher is unaware of the same

The terminology given to that reaction by your instructors is very odd. But the more important thing is that $\ce{V_2O_5}$ isn't a reagent in converting benzene directly to phenol, rather it is a catalyst. The first few pages of the book linked by @M.Farooq writes

[...] Various sources have reported the use of nitrous oxide to supply the active oxygen [1-8]. Iwamoto et al. appear to be the pioneers in the early 1980s, where they employed $\ce{N_2O}$ over a $\ce{V_2O_5 - SiO_2}$ catalyst and reported a benzene conversion of $11\%$ and a phenol selectivity of $45\%$.

And even in this article it is written about vanadium oxides that

These materials have been tested in the hydroxylation of benzene to phenol in liquid-phase with molecular oxygen in the absence of reductant. The catalyst exhibits high selectivity for phenol (61%) at benzene conversion of 4.6%, which is a relatively good result in comparison with other studies employing molecular oxygen as the oxidant.

So, you notice the oxygen for phenol comes from either from $\ce{N_2O}$ or molecular oxygen itself, while $\ce{V_2O_5}$ is just a catalyst. But in your original post you have written

My teacher told me we can convert benzene to phenol within a single step via Ratta Maar Reaction using $\ce{V_2O_5}$.

There is no mention of nitrous oxide or molecular oxygen (although in images you do have molecular oxygen). So, the gist is that your instructors told you something but didn't care to explain the intricacies and complexity of conversion.

The reaction of benzene over $\ce{V2O5/ PtAu}$ catalyst at lower temperatures, can convert benzene  to phenol with some success.
See this book here Direct hydroxylation of benzene
[The original question before editing was: What is the Ratta Maar Reaction?] Someone has played a prank with you with the named reaction. "Ratta maar" is a slang for "rote memorization" in Hindi/Urdu, which implies keep on memorizing without understanding and perhaps this is what your teacher wanted too. I assume this Allen's Chemistry Handbook is being taught or used in India.

Allen is a coaching institute for the JEE exams... You'll find many such "unexplained" phenomena in the coaching manuals. Half of the traffic on SE is due to these questions caused by loose ends which were not explained fully, because they won't ask the reason in the exam.

@Micelle Yeah,I know about Allen and such related cram schools(once prepared for such exams myself). And sadly,your point is very true; this reaction does have a mechanism as I have higlighted in the comments above,but if the teacher doesn't know it or isn't willing to teach it,he proclaims it as ratta maar. This creates a false impression about the nature of chemistry in high school students, and hence there questions are highly downvoted on SE. It's painful to watch as there questions receive negative ratings and are closed ultimately :(

@Mr. Yusuf ; on my online video lecture my organic chemistry teacher told that we can convert benzene to phenol by using O2/V2O5 at 300⁰C which called ' RMR ' and he told its full name as Ratta Maar Reaction. Whatsever, we had not asked him anything he was just teaching the theory . So there is no cause to use any slang . You should not abuse any educator or any coaching centre like that.

@zombie, I am sorry but the coaching centers, the way they make the students cram for exams are curse in the education system. These systems can never create original thinkers. The same college teachers who are unable to teach in a govt. college become perfect instructors in coaching centers. Not all coaching centers are like that but a majority it.

@YusufHasan, If you are interested in why chemistry never attracts the best and the brightest, you should read "Chemistry: Fact or Fiction" by Gillespie from 1970s-yes the same person who came up with VSEPR theory. Despite being a 50 year article, the facts remain true today.

@zombie You didn't get my point. I haven't used any slang; he did as he gave the reaction a made-up name RMR. Moreover,it is the duty of the teacher to teach the mechanism of the reaction along with the reaction equation(if a mechanism exists). If he doesn't do so, atleast he or she should state outright that a mechanistic pathway exists for this reaction,but we won't be covering it here. Stating a reaction outright as "Ratta Maar" (meaning rote memorization) when studies on mechanisms have been done is both incorrect as well as undermining to the scientists who have worked on the research....

.... I am not abusing anyone,just stating what should have been the way to handle this situation

@M. Farooq Ok,will definitely give it a whirl :)

I agree with @M.Farooq, IIT JEE is a kind of smudge in the education system that needs to be stained off. No exam will ever want a student to memorise 50-100 without even knowing “where can we find vanadium for making $\ce{V_2O_5}$?”.

@M.Farooq Can't find a link to the aforementioned work; can you provide me a source for "Chemistry: Fact or Fiction"?

@YusufHasan, You are right, I cannot locate an online copy of his article. I had a scanned copy shared by a professor. The full citation is Gillespie, Ronald. "Chemistry–fact or fiction? Some reflections on the teaching of chemistry." Chemistry Canada 28.11 (1976): 23-8.

Note that the question has been edited to remove the reference to "Ratta Maar Reaction", just in case some readers are wondering why this answer mentioned it.


text:
What is the mechanism for the reaction of alkenes with diborane and chloramine?

$$\ce{CH3-CH=CH2 ->[B2H6/THF][NH2Cl/H2O, Δ] Product (P)}$$
What is $\ce{Product (P)}?$

First of all, in this reaction I performed hydroboration to form an alkylborane. But I don't have any ideas what to do next. I tried a lone pair attack of nitrogen on the boron atom of the alkylborane, but what should I do next? What is the product and by what mechanism is it formed?

Since others already answered this question, I will not reinvent this wheel, but the OP may want to compare the result with a similar question. In both cases a carbon atom made nucleophilic by attachment to a more electropositive atom is acting as a nucleophile, displacing a chloride ion from the chloramine.

Organoboranes react with chloramine to give the corresponding amine as reported by Brown et al. [1], so P is n-propylamine (propan‐1‐amine).
The mechanism is similar to the oxidation of boranes using $\ce{H2O2}/\ce{NaOH}$, but instead of migration to an oxygen atom, the alkyl chain migrates to a nitrogen atom. The $\ce{N-B}$ bond is cleaved in the same way.

Reference

Brown, H. C.; Heydkamp, W. R.; Breuer, Eli.; Murphy, W. S. The Reaction of Organoboranes with Chloramine and with Hydroxylamine-O-Sulfonic Acid. A Convenient Synthesis of Amines from Olefins via Hydroboration. J. Am. Chem. Soc. 1964, 86 (17), 3565–3566. DOI: 10.1021/ja01071a036.

For a more elaborate variation of this reaction, see: https://www.sciencedirect.com/search?authors=mueller&cid=&issue=&offset=0&page=2925&pub=tetrahedron%20letters&qs=&show=25&sortBy=relevance&volume=

I like Waylander's answer, which is 100% correct. However, I have a little concern about the question itself. The conditions given in OP's question are not exactly correct. It may mislead the viewers and people seeking answers for similar type of questions. H. C. Brown had specifically noted (Ref.1) the following condition for his groundbreaking discovery (I used OP's example):
$$\ce{CH3-CH=CH2 ->[\text{i.} B2H6/THF][\text{ii.} NH2Cl/H2O/NaOH, Room Temp] CH3-CH2-CH2-NH2}$$
After $\pu{1 h}$ stirring of substrate with $\ce{B2H6}$ under $\ce{N2}$, water was added to the reaction mixture to destroy residual $\ce{B2H6}$ and then $\pu{3 M}$ aq. $\ce{NaOH}$ has been added before adding $\ce{NH2Cl}$ in THF. The mixture was then stirred another hour at room temperature before acid/base workup to isolate pure amine.
The heating to reflux was done for the reaction of substrate with the other reagent, hydroxylamine-O-sulfonic acid:
$$\ce{CH3-CH=CH2 ->[\text{i.} B2H6/THF][\text{ii.} NH2OSO3H/\Delta/3 h; \text{iii. aq.} HCl] CH3-CH2-CH2-NH3+Cl-}$$
Here though, solid $\ce{NH2OSO3H}$ was added without any mention of water addition. Acid/base workup follows to isolate pure amine.
Reference:

Herbert C. Brown, Wolfgang R. Heydkamp, Eli. Breuer, William S. Murphy, “The Reaction of Organoboranes with Chloramine and with Hydroxylamine-O-sulfonic Acid. A Convenient Synthesis of Amines from Olefins via Hydroboration,” J. Am. Chem. Soc. 1964, 86(17), 3565–3566 (https://doi.org/10.1021/ja01071a036).


text:
How to draw the 2D representation of ammonia molecule
In the 2D representation of the ammonia molecule, we see one solid line, one solid triangle and one dashed triangle

A plane can be made to pass through any three given points. In other words, we can always find a plane which will pass through any three given points in space. So the central N atom and two H atoms will lie in a plane. Only one H will be out of plane. So I think, in the 2D representation, we need to show two solid lines and one solid triangle. Please give your opinion about this. Thanks

The image is fine. In the image the H with the straight line is in the plane with the N. The H with the dashed triangle is behind the plane, and the H with the solid triangle is in front of the plane.

You can draw the way you proposed too if you omit the lone pair (l.p.). Currently, the plane is defined by $\ce{\text{l.p.}-N-H}$ and this way the graphical depiction is also consistent with the one of tetrahedral environment of a carbon atom with $\mathrm{sp^3}$ hybridized orbitals.

According to the rules, yes, Ammonia can be drawn with three of it's atoms in one plane and the other outside the plane. Here's an example, along with a 3D projection:

However, what do you comprehend from this structure? This structure does not convey the trigonal pyramidal geometry of the molecule very convincingly. It might be mistaken for trigonal planar at a glance, and this defeats the purpose of using the wedge/dash notation. Ammonia (and several other tetrahedral molecules) are represented in that manner since it is easiest to interpret their 3d structure from a 2d image.

The choice is between trigonal pyramidal and trigonal planar.


text:
Citric acid cycle original paper
According to wikipedia, it was published by Krebs&Johnson to the Enzymologia journal in 1937 [source, "citric acid cycle" section]. However, I'm really struggling to find a digital copy. I thought maybe someone has come across it? It's a pity that there's no copy of a paper that won a Nobel prize.

It is reprinted here

Interesting to see "close votes" for a genuine research query and marking it as a homework question!  So much negativity. Enzymologia is not online but journal citation is here https://www.cabdirect.org/cabdirect/abstract/19371405235. Many multilingual journals which published in French, German and English vanished with the passage of time. Such journals had Latin names to make them appear neutral. Some of them still do well like Talanta, Analytica Chimica Acta, Helvetica, Chromatographia. Older scientists often had a reading knowledge of German/French/Russian.

Awesooome, thanks @MaxW!
@M.Farooq I think for the research community as a whole it's good to have a common language, but there's definitely some nostalgic charm in having to learn a foreign language in order to read a publication of high interest:)

@M.Farooq - understand foreign languages?!? I think you expect too much from us poor Americans. We can't even speak real English. ;-)

@MaxW, Actually, I have just submitted a sequel to my previous article for people like you and me...how to best use machine translation for scientific purposes and my focus was French and German. The theme addresses exactly your concerns. Many fundamental articles or articles of historical interest remain in other non-English languages. I took help from native speakers.


text:
Use of Me2SOI/NaOH

In V triazole ($\ce{N^1}$) will substitute -Cl group. However from V to W, I don't get what $\ce{Me2SO+I-}$ in presence of base will do. Any hint? I also wish to know how to guess the reactions, in case one don't know them.

I think there is a small but crucial error in the question. The reagent should be Me3SOI. The protons on the Me group are acidic due to the adjacent S + centre.

See: Johnson–Corey–Chaykovsky reaction

Trimethyl sulfoxonium iodide (and the closely related trimethylsulfonium iodide) form a methylene epoxide from a carbonyl. The protons on the Me groups are activated due to the proximity of the S+ centre. Deprotonation, more often with NaH, gives a nucleophilic species that attacks the carbonyl group. The resulting O- attacks the incoming methylene-S centre displacing DMSO to give the epoxide.

You showed me the path, how to think that. That was the kind of answer I was expecting.


text:
How do I display a potential energy surface with molden?
I calculated a gaussian scan job for propylcyclohexane with two dihedral angles. Now I want to display a potential energy surface for that molecule with molden. I am new to molden and only found the geom. convergence button so far. This creates an two dimensional plot, but I need a three dimensional surface, with the energy plotted against the two dihedral angle changes. Is there a way to do this in molden?
This in my input file for the calculation:
%cpu=16-31
%mem=32GB
%chk=propylcyclohexan_scan_10.chk
#p PM6 Opt=modredundant

 propylcyclohexan_scan_10

0 1
C         -2.32729        0.36923       -0.26826
C         -1.86086        1.79592        0.04793
C         -1.42371       -0.68619        0.35867
H         -2.35296        0.22494       -1.35567
H         -3.35311        0.22391        0.09146
C         -0.38557        2.05033       -0.34666
H         -1.99132        1.97106        1.12477
H         -2.51564        2.50378       -0.47169
C          0.03534       -0.45199       -0.00958
H         -1.73512       -1.68573        0.02974
H         -1.53451       -0.66689        1.44998
C          0.49676        0.97326        0.33008
H          0.17109       -0.63513       -1.08350
H          0.66387       -1.18248        0.51357
H          0.48096        1.10355        1.41947
H          1.53881        1.06369        0.01138
C          0.03241        3.54672       -0.05839
H         -0.30194        1.90196       -1.43267
C          1.54273        3.82216       -0.15556
H         -0.26571        3.76488        0.97694
H         -0.55219        4.21744       -0.69950
C          2.23970        5.10589       -0.65389
H          2.00099        3.01727       -0.72714
H          1.90136        3.75452        0.88334
H          2.45381        5.07336       -1.75369
H          3.23327        4.96387       -0.25958
H          2.02974        6.11114       -0.31960

D 6 17 19 22 S 35 10.0
D 12 6 17 19 S 35 10.0

I am not aware that molden is capable of plotting potential energy surfaces (PES). This is not what molden is primarily made for. Instead you can extract the PES as a table from the gaussian output file and then use any general plotting software (e.g. gnuplot or Python/matplotlib) to display it.
The "geom. convergence" feature is intended to show how a geometry optimization converges over its iterations. This is something completely different from a PES.

Chemcraft might be able to do it out of the box, but it's not free. I'm not sure if Avogadro can do it; it's worth checking though. Otherwise plotting it should work quite well, obviously you lose the ability to view the structure interactively.

This question might fit better on the new Materials Modeling SE, as it is more of a software issue than a chemical problem.


text:
Why is chromate stable in basic medium and dichromate stable in acidic medium?
When I was studying the equilibrium existing between chromate and dichromate ions in the solution, I came to know that

Chromate ion is more stable in basic medium while dichromate ion is stable in acidic medium.

However, I am not able to understand what is the governing factor to decide which species will exist to a greater extent and how the equilibrium shifts the direction on changing the nature of the medium.
Why is chromate stable under basic conditions and dichromate stable under acidic conditions?

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
The preface of the book says:

The approach: These reactions represent important facets of the elements involved but are typically presented as no more than facts. (Why does boiling white phosphorus in alkali lead to hypophosphite and not phosphate?—Current texts make no attempt to address such questions.) Arrow pushing demystifies them and places them on a larger logical scaffolding. The transformative impact of this approach cannot be overstated. Almost to a person, students who have gone through our introductory course say that they cannot imagine how someone today could remain satisfied with a purely descriptive, nonmechanistic exposition of inorganic main-group chemistry.

Potential concerns: Given the plethora of advantages of a mechanistic approach, it’s worth reflecting why it has never been adopted for introductory inorganic chemistry. A plausible reason is that, in contrast to common organic functional groups, simple p-block compounds such as hydrides, oxides, halides, and so forth, tend to be much more reactive and their vigorous and even violent reactions have been much less thoroughly studied. As good scientists, inorganic chemists may have felt a certain inhibition about emphasizing an approach that has little grounding in experimental fact. This is a legitimate objection, but hardly a dealbreaker, in our opinion, for the following reasons.

Our ideas on main-group element reactivity are not taken out of the blue but are based on parallels with well-studied processes in organic and organoelement chemistry.
Second, it no longer necessarily takes a prohibitive amount of resources to test amechanistic proposal, at least in a preliminary way. Quantum chemical calculations, particularly based on density functional theory (DFT), very often provide an efficient and economical way of evaluating reaction mechanisms.
Third, and perhaps most important, it’s vastly better to be able to formulate a hypothesis on how a reaction might happen than to have no inkling whatsoever about the mechanism

What all of this basically means is that in this approach, you use the classical organic chemistry concepts of electrophiles and nucleophiles to write an arrow-pushing mechanism for the reaction. It will only be a probable mechanism, but often, it's not much far away from the reality, and this methodology can serve to be a pretty great teaching tool for inorganic chemistry, so long as you accept the fact that the mechanism you have made can only be proved by a research paper. But having an idea is also good.
Having got the introduction and disclaimer of this method out of the way, we can now start to explain your phenomenon by arrow-pushing diagrams.
The dichromate anion $\ce{Cr2O7^2-}$ has got a structure consisting of a $\ce{Cr-O-Cr}$ bridge, four $\ce{Cr=O}$ bonds, and two $\ce{Cr-O^-}$ bonds.
Here, the $\ce{Cr=O}$ bonds can be said to be structurally similar to the carbonyl group.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Cr}$ atoms due to the electron-withdrawing effect of two $\ce{Cr=O}$ groups per $\ce{Cr}$. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage as follows:

The $\ce{HCrO4^-}$ formed can  again be deprotonated in the basic medium to form $\ce{CrO4^2-}$
In acidic medium, the $\ce{CrO4^2-}$ anion can be protonated to form $\ce{HCrO4-}$ , which can react with another $\ce{CrO4^2-}$ anion in the following manner to give $\ce{Cr2O7^2-}$ and expelling water as a leaving group:

Any alternate protonation pathways, for example, protonating $\ce{CrO4^2-}$ twice to form $\ce{H2CrO4}$ and then reacting it with $\ce{CrO4^2-}$ might also give the same product with a few extra proton transfers here and there, but the way I have presented thing would be the most probable way, as the reaction is progressing stepwise with each protonation.
Note: Remember that these are just probable pathways we have drawn to reach our product, but there's a good chance they resemble the reality( which can only be determined by kinetic experiments) and as the authors have stated, it is a good learning tool, and not too far off the mark.

Hehe! Your pics are full of smilies and poker faces, not the song ;)

@Zenix Oh right, the way I draw my negative charges. Yeah my professor used to draw the minus charge like that,and I have got into the same habit :)

I’m not sure if those are actually double bonds but yes, nice mechanism.

@Jan Well,I just picked up these structures from the Wikipedia page of Chromate and dichromate. The liberty I took was that they have depicted a resonance hybrid while I took canonical structures. Although I don't claim wherher these salts exist in such a form

The chromate and dichromate ions exist in equilibrium as follows:
$$\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$$
In a basic medium, $\ce{[H+]}$ is less, which favours the formation of reactant $\ce{CrO4^2-}$ whereas in an acidic medium, $\ce{[H+]}$ is high, which favours the formation of product $\ce{Cr2O7^2-}$
Sources: Wikipedia

Yes,that was my question.Could you give the reason for that?

That is the reason: varying concentrations of hydrogen change the position of equilibrium. If you want an explanation of the reaction mechanism, @YusufHasan has posted an answer explaining that.


text:
What is the meaning of the J value in term symbols
I understand how to work out the J values of term symbols.
e.g. a d1 configuration gives J values of 5/2 and 3/2. resulting in 2D 3/2
However I dont understand the meaning of it, and why it arises.

For multi-electron atoms where the spin-orbit coupling is weak, it can be presumed that the orbital angular momenta of the individual electrons add to form a resultant orbital angular momentum $\bf{L}$.
Likewise, the individual spin angular momenta are presumed to couple to produce a resultant spin angular momentum $\bf{S}$ . Then $\bf{L}$ and $\bf{S}$ combine to form the total angular momentum $\bf{J} = \bf{L}+ \bf{S}$.


text:
Need help finding equilibrium constant for equilibrium system

The reaction between iron (III) ions and thiocyanate ions, SCN$^-$, is an equilibrium reaction:
Fe$^{3+}$ (aq) $+$ SCN$^-$ (aq) $\ce{<=>}$ FeSCN$^{2+}$ (aq)
When $20$cm$^3$ of $1.0\times10^{-3}$ mol dm$^{-3}$ Fe(NO$_3$)$_3$ (aq) is mixed with $20$cm$^3$ of $1.0\times10^{-3}$ mol dm$^{-3}$ KSCN (aq), the concentration of FeSCN$^{2+}$ (aq) at equilibrium is $8.0 \times 10^{-3}$ mol dm$^{-3}$. What is $K_{c}$ for the equilibrium system?

The question also includes a hint, saying that,

After mixing, the concentration of each reactant is half, because the
volume of solution is doubled.

My attempts to solve this question are as follows:
The initial number of moles of Fe$^{3+}$ and SCN$^- $$= 1\times 10^{-3}\times20\times10^{-3}=2\times10^{-5}$mol
At equilibrium, the concentration of Fe$^{3+}$ and SCN$^-$$ = \displaystyle\frac{2\times10^{-5}}{20\times10^{-3}} - (8\times10^{-3})=1\times10^{-3} - (8\times10^{-3}) $mol dm$^{-3}$
I have obtained a negative value for the concentration Fe$^{3+}$ and SCN$^-$, so I think I have made some wrong calculations somewhere. I am confused about the volume and concentrations of Fe$^{3+}$ and SCN$^-$.
Could you tell me where I made a  mistake?

You don't not to write that cumbersome codes, your first equation is \ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq)}

I didn't see your calculations, but, concentration of $\ce{FeSCN^{2+}}$ is too high to be true.

I think there's a typo: $\ce{[FeSCN^2+]}$ can't be higher than $\ce{[Fe^3+]}$ or $\ce{[SCN^-]}$. I think it should be $8 \times 10^{-5}$

I find following your logic pretty hard since you are randomly omitting units. The measured quantity consists of a number and a unit, and you are not allowed to omit either of them. Also, your markup is indeed pretty cumbersome; please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

The $\ce{[FeSCN^2-]}$ at equilibrium should be $\lt 5.0 \times 10^{-4}$.

if $x$ is the equilibrium amount of the product in the reaction $ \ce{A + B = C}$ and initially A and B are present at $a_0,b_0$ then $(a_0-x)$ of A and $(b_0-x)$ of B are present at equilibrium, next write down the equilibrium const and  find its value. As concentrations must be positive, you can see why the equilibrium amount of FeSCN$^{2+}$ must be < Fe$^{3+}/2$  and SCN$^{-}/2$