text:
Aluminium foil on windows: what is the residue, and how to remove it?
I put aluminium foil on some windows' glass pane wetted with plain tap water, as a typical trick to stop sunlight. After 5 days, I removed the foil to find some kind of whitish pattern on the glass. The pattern follows the imperfections in the aluminium foil: bubbles, wrinkles, even a very slight embossed pattern in the foil.
To remove the pattern, I tried window cleaner, rubbing alcohol, vinegar, washing soda. Nothing worked.
Finally I tried caustic soda, at a solution stronger than that recommended for drain opening, and it mostly worked; but still there are patches where you can see the pattern if light reflects in the glass in just the right angle.
My question is: what caused this pattern? And, how can I get rid of its remains?
My next ideas are to try bleach or a stronger solution of caustic soda, but I'm afraid of etching the glass - what would be the maximum concentration advisable?
Also - if aluminium foil reacted to tap water like this, what happens when wrapping food and even cooking with it?
(I asked basically this same question in the DIY SE, but it's not getting actual answers, so I am hoping to get better insights here)

Finally I used a glass scraper. Bleach didn't help, and I didn't dare to use even stronger caustic soda.
Fuller details at the answer I wrote at the DIY SE site: https://diy.stackexchange.com/questions/151160/how-can-i-remove-residue-left-after-removing-aluminum-foil-from-my-windows#answer-152162

I did a little bit of research on this online, and I really could not find much useful information on the topic; however, I will try to give my best educated guess as to what is happening. My guess is that because the window was wetted, aluminum oxide was likely formed; aluminum oxide is present on aluminum foil in a very thin layer; however, it is a very tough material, so it usually stays on the foil. My guess (remember this is just a guess) is that the glass (silicon dioxide) is somehow more more adhesive in respect to the aluminum oxide than the actual aluminum. If this was the case, as the glass pulls aluminum oxide off the aluminum and onto itself, it would expose a fresh layer of aluminum that could be oxidized again and undergo the same process (form aluminum oxide then adhere to the more adhesive silicon dioxide). This would likely explain the reason for the white residue. The reason why this doesn't form on food, however, is because food is not likely as adhesive towards aluminum oxide as aluminum itself is, so no aluminum oxide will leech onto your food. Now to your question on how to remove the aluminum oxide: My guess is that using an acid (possible dilute HCl) would be more beneficial to removing the Al2O3 than using a caustic base. It probably doesn't matter which is used but to me an acid seemed slightly more preferable due to the following reasons. The reason for this can be seen the reactions between the two chemicals and aluminum oxide:
$$\ce{Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O}$$
$$\ce{Al2O3 + 2 NaOH + 3 H2O -> 2 NaAl(OH)4}$$
As you can see, the reaction between the caustic base forms sodium aluminate, a soluble substance that can readily combine with other available cations like calcium and precipitate out, if you were using tap water with this, there might be a very small residue left from this reaction if the aluminate ions were able to undergo a double replacement reaction with a cation such as calcium; however, the reaction with an acid made a completely soluble substance that will not likely leave any residue. Again, I am not an expert in this field, but I hope this guess as to what is happening is helpful.

"no aluminum oxide will leech onto your food"-> Well, actually there seems to be peer reviewed info on that, and it does leach. https://www.researchgate.net/publication/221971690_Risk_Assessment_of_Using_Aluminum_Foil_in_Food_Preparation

Regarding the suggestion of using an acid - "double strength vinegar" (which should mean 10% acetic acid) didn't have any effect.

This is very interesting, in a too-complex sort of way. Aluminum is an active metal; glass has a complex surface; and water is reactive by itself, and also from minerals and chlorine that could be in it.
The suggestions for cleaning might be workable, but I'm going to suggest an entirely different track: elbow grease. With toothpaste. There are some cooktop cleaners that involve very gentle abrasion and might be better than toothpaste, but I don't think a uniform application of some mild liquid is going to clean the entire surface equally clean (so that you can't see the original pattern).

My hope is that using some liquid will remove the residue without altering the glass; I assume that any abrasion is more dangerous to the glass. But I might try in a corner and see what happens... once the liquid possibilities are discarded.

Aluminum foil on glass to stop sunlight... I wonder if UV is part of the complexity. I wonder if aluminum foil on wetted window glass in the dark will adhere and leave a residue. Experiment, anyone?


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Sign of enthalpy for exothermic and endothermic reactions
In Cambridge Chemistry Coursebook [1, p. 94] it’s written that

A rise in temperature is given a positive sign. So the value of $\Delta H$ is negative for an exothermic reaction. A fall in temperature is given a negative sign. So the value of $\Delta H$ is positive for an endothermic reaction.

Why is the sign of enthalpy for exothermic reaction negative? Doesn’t exothermic reaction rise the temperature of surrounding?
Why is the sign of enthalpy for endothermic reaction positive? Doesn’t endothermic reaction take in energy and cool down the surrounding?
Reference

Ryan, L.; Norris, R. Cambridge International AS and A Level Chemistry Coursebook, 2nd ed.; Cambridge International Examinations; Cambridge University Press: Cambridge, 2014. ISBN 978-1-107-63845-7.

Is the rising or falling of the T that is not required here. This is often see at introductory level in spire of being unnecessary. See answer

The reaction is said to be exothermic or endothermic according to the sign of the Enthalpy change. To obtain Enthalpy you subtract the enthalpy of formation for reactants from that of products. So, if the final enthalpy of the system is less than the initial one the result is negative and this difference manifests it self in the system as heat and temperature increases. The opposite is true for endothermic reactions.

This answer is incorrect. Whether a reaction is exothermic or endothermic is determined by the sign of the enthalpy, not that of the Gibbs free energy.  Indeed, for many reactions, depending upon the conditions, the enthalpy and the Gibbs free energy  have can have the opposite sign.  Thus, unless the entropy change is negligible, they cannot even be approximately equated to each other.

@theorist, You're right. the right term for free energy is endergonic or exergonic. Thank you.

In your correction, you have the signs reversed.  You wrote: "To obtain Enthalpy you subtract the heat of formation for products (enthalpy) from that of reactants. So, if the final enthalpy (heat) of the system is more than the initial one the result is negative".  In fact, enthalpy change = (enthalpy products) – (enthalpy reactants).  [This is generally true for all state functions, not just enthalpy: change = (final state) – (initial state).]  Thus if the final enthalpy is higher, the enthalpy change is positive.

Also: enthalpy is not heat. enthalpy is a state function, while heat is a path function.  I don't mean to be discouraging, but I would strongly recommend you learn more about thermodynamics before attempting to answer questions on the subject.

Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.

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Why are reactions with higher activation energy more sensitive to temperature?

In the above picture it is stated clearly, with an example, that the more the Ea of a rxn the more it's sensitive to temperature, and hence even a small change in temperature in a reaction with high Ea, would have a drastic effect on the rate of rxn, as compared to a rxn with low Ea (about the probability factor and collision frequency, it's stated in the book as - "it must be understood that we are justified in doing this only when the reaction being compared are so closely related that differences in collision frequency and probability factor are comparatively insignificant" ).
My doubt is - If the energy-barrier (Ea) is high, doesn't it mean, (kind of) that, it's difficult to reach the intermediate stage, and hence a little change in temperature doesn't do good to the reaction as the barrier has still not been overcome and the reactants have a long way to go; and hence making the reaction more "temperature resistant" ? And on the other hand, if the energy-barrier is low, doesn't it mean that even a small change in temperature would be enough to overcome it and hence would increase the rate of reaction ?
After reading a bit more, I found that the statement is repeated again as - "the larger the activation energy of a reaction the larger the increase in the rate brought about by a given rise in temperature".    WHY ?
My appeal to the answerers is that - Can you plz answer in what-is-actually-happening-on-the-backstage way and without ,if necessary, equations ?
Thank you very much and sorry for poor English.

The fact will be quite obvious from the Arrhenius relationship applied at the rate constants at two different temperatures, which states that ,$$ln(\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$

The first thing to note is that the rate constant with a large activation barrier will be far smaller than one with a small barrier at a given temperature. If the barrier is very small (e.g. 5 kJ/mol) at, say, room temperature, almost all collisions between the reactive molecules have sufficient energy to react and the barrier is crossed and this leads to a fast reaction. This means that increasing the temperature cannot lead to much of an increase, because the reaction is going almost as fast as it can at the lower temperature.
Now consider the case with a large barrier. In this case the rate constant is far smaller than when the barrier is small, and very few collisions have enough energy  to cross the barrier. Now increase the temperature and the number of filled energy levels the molecule has increased greatly, exponentially so by the Boltzmann distribution.  Now very many more collisions have sufficient energy to react and so the rate constant increases greatly, although it may still be small in value.
If you plot the Arrhenius equation (given in your text) vs temperature the slope of this gives the rate of change with temperature. With larger activation energies the slope gets larger. This is the effect also described in your text.
The figure shows rate constants calculated with the activation energies $E_a$ shown (in kJ/mol). You can see how rapidly the rate constant increases with temperature when the activation energy is large, i.e. the slope is greater when the activation energy is larger, than when the activation energy is small. (The rate constant $\displaystyle k=A_0e^{-E_a/(RT))}$ are with $A_0 =10,\; (E_a=10);\; A_0= 10^3,\;(E_a=20)$ and $A_0=10^6,\;(E_a=40)$ with $A_0$ chosen only to get them on the same scale).
The gradient of the rate constant is $\displaystyle \frac{dk}{dT}=E_a\frac{A_0e^{-E_a/(RT)}}{RT^2}$ which shows that the slope depends on the activation energy.

So as not to confuse the students, the authors have taken high temperatures (250-300 °C) and not temperatures like room temperatures, and then they have shown the variations of rates in THAT high temperature range. But you have written the first part of your answer about low range of temperatures (which kind of seems like an exception). Can you plz explain what would happen if the temp. were high enough as shown in the text ? Plz note that the authors have varied the Ea in the same range of  temperatures. Thanks.

The explanation is just the same whatever the temperature range:  the physics does not change. If you re-calculate my figure and expand it in the 200-400 range you will see the same sort of figure as above.

To understand this let's approach this way:
Pre-requisite knowledge:

At a given temperature, only a fraction of the reactant molecules possess high energies. As we try to find molecules with even higher energies, the fraction drops even more.

If the temperature increases, this fraction of molecules increases, and if the temperature decreases, the same fraction drops.

Now the question:
Reactions with higher activation energies require higher energies to carry on. Now if for such a reaction, the temperature drops, the average no. of molecules that contain the threshold energy (activation energy) becomes lesser according to pre-requisite 1.
For a low activation energy reaction, the fraction of molecules with that amount of energy is already higher compared to a high activation energy reaction.
If we drop temperatures for reactions with lower activation energies, the fraction goes down but not at the same rate as it does for higher activation energy reactions.

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Reactions on alpha tetralone to synthesize naphthalene
Alpha-Tetralone is reacted by $1.\ce{Zn(Hg),HCl}\ \ 2.\ce{ Pd/C}$ What is the product and mechanism that is followed for the 2nd reaction?
A book by SN Sanyal says that Naphthalene will be produced . I know that in the 1st reaction, clemenson reduction is followed thereby reducing the Carbonyl group to $\ce{CH2}$ group .Then how the reaction proceeds is not clear.
I may be wrong of the 1st step also. Please correct me.

Well many reactions can go both ways... and that might help you solving this

Can you help by writing the exact steps

The tetralin loses 2 molecules of hydrogen on the active Palladium surface.

Moinak Banerjee: Your concern about the mechanism of the conversion of tetrahydronaphthalene (tetralin) 1 to naphthalene 5 has been addressed in a comment by Waylander. It is important to recognize that the process is a 4-electron oxidation of tetralin 1 and a four electron reduction of the hydrogen that is liberated as 2 equivalents of hydrogen gas. Ideally, the reaction is catalytic in palladium (Pdo). Mechanistically, the reaction likely proceeds by oxidative addition of Pdo into the labile benzylic C-H bond of tetralin 1 to afford Pd(II) species 2, which, in turn, isomerizes to π-complex 3. These steps are shown as reversible because complex 3 is the likely intermediate in the hydrogenation of 1,2-dihydronaphthalene 4 to tetralin 1. Collapse of complex 3 regenerates the catalyst Pdo and an equivalent of hydrogen gas. A repetition of these steps at either the benzylic or allylic C-H bonds of 1,2-dihydronaphthalene 4 leads to naphthalene 5 and a second equivalent of hydrogen gas. These reactions are driven to the right by the loss of hydrogen gas in an open vessel. --continued--

This reaction played a critical role in early attempts to synthesize estrone since it served to reduce the number of asymmetric carbons, thereby facilitating stereochemical analysis. An example of this process is illustrated in the sequence 6 --> 8 from the work of Bachmann and Chemerda1. The double bond in tetracycle 6 is isomerized by Pdo to the tetrasubstituted position in isomer 72. The conversion to the naphthalene 8 follows the same steps as in the case of 1,2-dihydronaphthalene 4.3
1) W. E. Bachmann and J. M. Chemerda, J. Am. Chem. Soc., 1948, 70, 1468.
2) The double bond migration involves a similar C-H insertion and subsequent π-allyl complex. But that is another story.
3) See also: J. Heer and K. Miescher, Helv. Chim. Acta, 1949, 32, 1572.


text:
Thermodynamic stability and Inertness
The following is from my teacher's notes:

The reaction (where the reaction mixture is thermodynamically unstable but kinetically inert) tends to go as indicated by a positive total entropy change, yet nothing happens because the rate of the reaction is too slow to be noticeable due to a high $E_\mathrm a$.

Does that mean that the reaction is taking place but we don't see it because of how slow it is, or that the reaction will not take place in the first place?
This is A2 chemistry.

I guess that the question is down voted because of the poor editing. don't use short forms and symbols from the first time.

The reaction can be thermodynamically unstable (favorable) not only due to positive entropy change but also negative enthalpy change, both result in negative free energy. The thermodynamic stability is determined by the difference between the potential energy of products and reactants. If the result is negative the reaction is said to be favorable and vice versa.
Yet this doesn't say anything about the rate of the reaction. The rate of the reaction or reaction kinetics is determined by taking into account an intermediate step in the way from reactants to products, the transition state. If the difference between the potential energy of the transition state and the reactants is too large, the reaction will be too slow. conversely if the difference is small the reaction takes place so fast.

Here arises a question: From where will reactants get the energy to reach and pass the transition state ?
The answer is that indeed a number of molecules has this energy but their population may be too low. This population is determined from the Boltzmann distribution. Heating may be used to supply some energy that increase this population of molecules.

Returning to your question: "Does that mean that the rxn is taking place but we don't see it because of how slow it is , or that the reaction will not take place in the first place?"
Answer: Although the term "kinetically inert" is only used for complexes, the same rules hold as in the other cases. The reaction does take place but we don't notice it. An example for this is turning from diamond to graphite or burning of paper (oxidation of a cellulose).
please, Don't hesitate to ask any question in the comments if something is unclear to you.
This is a useful link:
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Kinetically_vs_Thermodynamically_Stable
Hope this helps.


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Can hydrocarbon help to split molecular hydrogen into atoms?
I am a physicist and have the following chemistry problem.
When a sample is exposed to H2 (molecular hydrogen), a tiny amount of atomic hydrogen or, maybe, protons is generated. Extremely small amount but exceeding the thermally generated amount by many orders of magnitude. The origin can be traced to the ever existing surface contamination by hydrocarbons and water (monolayer). They somehow help to generate protons/atomic hydrogen on the surface.
Any tips?
Thanks,
Hengan
PS It can be H+ from adsorbed water, too, but why molecular hydrogen is then needed?
Sample: Graphite in a hydrogen atmosphere at room temperature. The graphite has probably some remnants of PMMA from processing plus it was exposed long enough to ambient atmosphere and, hence, adsorbed water and other ambient polymers (standard contamination layer of nanometer thickness).

What sample and parameters of process?

I am sorry but your question makes no sense at all. Could you rephrase it and give us some useful data about it?

OK. Let me formulate the question like the following. How to explain a lower energy barrier for dissociation of H2 into atomic hydrogen by the presence of adsorbed hydrocarbons or water?


text:
Would atomic oxygen and atomic hydrogen result in O2 and H2 only, or some H2O too?
Let's assume we have atomic oxygen and atomic hydrogen in a closed chamber, the temperature and pressure of which we're able to monitor (and control). Number of hydrogen atoms are twice that of oxygen.
Would this result in spontaneous formation of:

$\ce{O2}$ and $\ce{H2}$ only?
$\ce{O2}$ and $\ce{H2}$ with some $\ce{H2O}$?
and how would the process vary with temperature and pressure?
And I assume creating mostly $\ce{H2O}$ is the hardest (though I would be interested in a comment on this as well).

and how could one carry out analysis on paper to figure out what would happen? (I'm from a physics background, and my chemistry knowledge is a bit rusty. Though I do have a good background on thermodynamics, enthalpy, Helmholtz free, Gibbs free energies, etc, etc).
For the sake of brevity and focus of this question, we can ignore the question of how atomic oxygen and hydrogen came about in the first place. (This assumption is even more acceptable in the special case of carrying out molecular dynamics computer simulation, something I'm interested in, because we don't have to worry about how to form atomic $\ce O$ and $\ce H$)
edit: I guess I should clarify that I'm interested in a first principles analysis of this problem (since I'm mainly interested in MD). Also what if we had two other species of atomic gases (Ar and Ne, Ar and O, Ar and H, etc, etc). I think I'm more interested in it from a physical chemistry point of view (though without quantum mechanics if possible).

Thing is, for a guy interested in MO, you're missing crucial thing - energy of your atoms. You could very well have such mixture, but especially in temperature so high that there would be no reaction at all.

@Mithoron I guess energy of atoms can be decided based on temperature, resulting in a distribution of kinetic energies with a mean. Hence, why I'm also interested in knowing what might happen at various temperatures and pressures.

It would ignite and form water vapor. It's not complicated to figure out. The conditions you have described with a mixture of monoatomic hydrogen and oxygen is what is formed when a spark passes through a mixture of hydrogen and oxygen gases.

Thanks. I guess by 'analysis on paper' I meant being able to figure out from first principles, as opposed to prior empirical knowledge. Since I'm interested in MD (molecular dynamics), I'm interested in creating a simulation with O and H, and using the interaction potentials to see what happens in the run. And first principles analysis would greatly help with that.


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Why doesn't Silicon act as the basis of a class of compounds like Carbon is for organic compounds?
Carbon is the basis of organic materials in large part due to it's four valence electrons.  Silicon also happens to have four valence electrons as well but does not serve as a basis for a large set of compounds on the level of Organic compounds.
Given the four valence electrons nature of both elements,
why can't Silicon serve as a base for a numerous swath of compounds like Carbon does for Organic compounds?

OK, we've come one level up. Now, silicon surely can and does serve as a base for a numerous swath of compounds. What was your question again?

The key issue is that carbon-carbon bonds are stable which allows a very rich chemistry of complex compounds we call organic chemicals. silicon-silicon bonds are not and the variety of silicates (containing a range of structures built on Si-O-Si bonds), while large, is nothing like as varied as organic chemicals and not apparently sufficient to support enough complexity to support life.

Silicon has a weaker affinity for electrons than Carbon does because its valence electrons are farther away from its nucleus, so it tends to act a little bit more like a metal - meaning Silicon is more likely to oxidize (give away its valence electrons, like a metal) than to reduce (fill its valence shell, like nitrogen, oxygen, etc.) Carbon just happens to be perfectly kind of "ambivalent" on whether it oxidizes or reduces, allowing it to form essentially limitless chains of single, double, and triple-bonds to itself and to nearly any other element, allowing the absurdly complex structures necessary to support life.

Silicon cannot be an effective substitute for Carbon.
This article by Raymond Dessy explains why.  Four key points are summarized below. The reason life forms are mentioned is to address the comment from @matt_black.

Silicon oxidizes to the solid Silica, thus this would post waste disposal problems for life forms.
Silicon does not form many compounds that have handedness.  In other words, it cannot process or store energy in a stable process like Carbon based enzymes can. Thus the energy could not sustain a living organism.
The Silicon analogs of carbon compounds cannot be formed as they are too unstable or too reactive despite the reagants available to Chemists. This is supported by @matt_black's comment.
The narrow range of temperature and pH levels where most of life processes occur are suitable for Carbon compounds, but not Silicon.

OP wasn't asking why silicon cannot be a substitute for carbon though. Edit: Oh... hi OP.

Silicon very commonly forms at least one chiral compound.    Quoting Wikipedia: "Although optical activity is normally thought of as a property of fluids, particularly aqueous solutions, it has also been observed in crystals such as quartz (SiO2) ... Rotation of the plane of polarization is observed due to the relative rotation between crystal planes, thus making the crystal formally chiral as we have defined it."


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Effect of nitro group basicity order of o,m and p nitroanilines?
Would someone please help me with this? The order of basicity of o,m and p nitroanilines is m>o>p. My book states that ortho isomer has -I effect, -M effect as well as ortho effect, and meta isomer has only -I effect while para isomer has -I and -M effect. These effects destabilize the conjugate acid.
My actual doubt is why exactly the nitro group has a negative mesomeric effect on ortho and para isomer, when actually it is a meta deactivator. Shouldn't it be having no effect at ortho and para positions, and destabilize the conjugate acid at meta position? I'm confused.

First we have to understand what affects basic nature of a compound.

Tendency to donate lone pair
Electronic effects— Inductive, resonance etc
Factors stabilizing conjugate acid (or acidity of conjugate acid
Nature of solvent

Tendency to donate lone pair is affected by the acidity of conjugate acid and electronic effects.
Grab a pen and paper and draw conjugate acids of

$$2-nitroaniline$$
$$3-nitroanliline$$
$$4-nitroaniline$$

And you know that electron withdrawing inductive (–I) effect increases acidity and electron withdrawing resonance (–M or –R) also increases acidity.
In conjugate acid of

$$2-nitroaniline$$
You will find there is only –I inductive effect and there is no –M effect of $-NO_2$ group because of SIR (In SIR - STERIC INHIBITION OF RESONANCE - the nitro $-NO_2$ group shows steric hindrance to $-NH_3^+$ group and the p–orbital of Nitrogen (in $-NH_3^+$ group) moves out of plane of p–orbital of Carbon (in benzene ring) which inhibits resonance or does not allow resonance to happen). So in conjugate acid of $2-nitroanliline$ there is only –I effect which increases acidity.
$$3-nitroanliline$$
You will find that there is no –M effect obviously!! But there is –I effect of nitro ($-NO_2$) group which increase the acidity of conjugate acid of $3-nitroanliline$ but it is less than what we have seen in $2-nitroanliline$ (due to distance factors). Therefore conjugate acid of $2-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$
$$4-nitroanliline$$
You will find there are both –M and –I effect which increase acidity of conjugate acid of $4-nitroanliline$. So conjugate acid of $4-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$ or $2-nitroanliline$

Conclusion
Order of acidity of conjugate acids
$$conj.\ acid\ of\ 3-nitroanliline<conj.\ acid\ of\ 2-nitroanliline<conj.\ acid\ of\ 4-nitroanliline$$
Since the order of basicity of 'orginal' compounds is just opposite of order of acidity of conjugate acids, so answer to your question is as follows
$$3-nitroanliline>2-nitroanliline>4-nitroanliline$$
Or m>o>p

You should be more careful with formatting. Maybe check out https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

But small groups such as $ NH_2 $ are not affected by SIR. Do they?

@user541396 you can apply SIR in that case too. But the main cause of Ortho substituted aniline to be weaker base is the instability of acid formed on protonation due to steric factors.

First thing we have to know about a nitro group (NO2) is this " NO2 is a meta director". But, NH2 is a Ortho, Para director. Secondly, the basicity is the tendency of donating a pair of electron. How spontaneously a compound donates a pair of electron that will determine it's basicity.
Among the o,p&m nitro aniline, the meta nitro anilines is the most unstable isomer. The para isomer is more stable than Ortho isomer and hence the p-nitro aniline has no need to donate a pair of electron. The m-nitro aniline is unstable and so it can donate a pair of electron easily than the other two. Thus, the order of basicity of the nitro substituted anilines follows m > o > p.


text:
Does carbon monoiodide (CI) exist?
Many of my students get chlorine $\ce{Cl}$, confused with carbon monoiodide $\ce{CI}$, despite my promises to use serif fonts whenever possible and to never give them carbon monoiodide in a question.
Everything I know about chemical bonding tells me carbon monoiodide should not exist ($\ce{CI4}$, sure), but weird things happen sometimes. My only results from internet searches for carbon monoiodide bring up a set of Quizlet flash cards.
Will my students prove me wrong if I tell them that carbon monoiodide does not exist (except possibly in extreme laboratory conditions).

Impossible to prove a negative so for the moment a comment rather than a answer as I can't provide a reference. No, a simple diatomic CI molecule does not exist as something you can put in a bottle and use later. Compounds with the stoichiometry CI do exist, such as C2I2 (http://www.sciencemadness.org/member_publications/diiodoacetylene.pdf) but that's as close as you can get. But personally I think the kids are taking the proverbial.

Define "exist."

If carbon monofluoride it seems likely that carbon can bond singly with any of the halogens, even if the molecule is highly reactive or unstable.

Provided that mellitic anhydride ($\ce{C12O9}$, empirical formula $\ce{C4O3}$) is considered one of carbon oxides, then e.g. hexaiodobenzene ($\ce{C6I6}$, empirical formula $\ce{CI}$) could be considered a form of carbon(I) iodide, formally. But please rather don't teach that in regular lectures.

Usually "oxocarbon" is used to describe carbon-oxygen compounds when speaking more broadly than CO and CO2. And yes, quite a few are known.

Sure
Formula:    CI
Element System: C-I
Element Names:  Carbon, Iodine
Molecular Weight:   138.915 g/mol
Name(s):    CI <g>
Carbon Monoiodide gas

See:
http://lb.chemie.uni-hamburg.de/static/ES/2_C_C-I.php?content=527/pz0cDmNo

Then there are carbon monoxide and cobalt to confuse them. There are some fonts, though, such as those designed for computer source, that do a better job of distinguishing 1  and l, O, o and 0.


text:
Why do we call O2 oxygen?
I have been taught that oxygen is a chemical element, in other words a certain type of atom that has 8 protons in its nucleus. So why is O2 called oxygen? It is not a type of atom but rather a molecule.

You can call it molecular oxygen all the time, or at least until someone tells you it's a waste of time.

Or maybe gaseous diatomic molecular oxygen, just to be really sure no one thinks you're somehow talking about ozone or red oxygen.

Note that the same thing applies not just to oxygen, but also to hydrogen, nitrogen, fluorine, bromine, chlorine, and iodine. All of these elements prefer diatomic molecular form.

I'm voting to close this question as off-topic because it's about semantics.

What's in 02 that is not oxygen?

:/ I think the original misunderstanding here was not getting that oxygen as an element is not a type of atom - an element is a collection of all atoms with same number of protons.

I think what you may find most helpful is to know a bit of the history of element discovery and atomic theory.
The first pure substance containing only the element oxygen to be isolated was dioxygen ($\ce{O2}$), in 1774, though it was called "dephlogisticated air" until 1777 when Lavoisier used the term "oxygen" for the first time. This was some 30 years before John Dalton even proposed the first empirical atomic theory. Back then, we only barely had an understanding of stoichiometry, such that Dalton famously claimed the molecular formula for water was $\ce{HO}$. The fact that dioxygen is a substance made of molecules containing two atoms of oxygen probably wasn't widespread knowledge until at least 1811, with the gas stoichiometry experiments of Amadeo Avogadro.
Basically, for a point in time, we knew that there was a substance composed of a single type of atom, which could not be broken down into anything simpler. This fit the then-prevalent definition of an element; "a pure substance that could not be decomposed into any simpler substance". We knew that Lavoisier's "oxygen" had to be $\ce{O_n}$, for some n, but we had no reason to assume $n \neq 1$ for decades. By the time we figured out $n = 2$, the name "oxygen" was already widely used to refer to dioxygen. The fact that $n = 3$ also forms a stable compound in ambient conditions (ozone) would also not be known until 1867. A similar story happened with (di)nitrogen (octa)sulfur, (tetra)phosphorus, and so on. The only elements which form stable monoatomic substances in reasonable conditions are the noble gasses.
There is an interesting aspect to consider behind all this. There are some (such as Eric Scerri) who claim we are doing Chemistry a disservice in muddling together the properties of elements and the pure substances which they make. Nowadays our definition of an element is solely dependent on the number of protons inside an atomic nucleus, with no reference to reactivity or in what form the pure substance can be found. In this sense, the elements do not have "reactivities", "melting points", etc.; these are all properties of the pure substances. The only true properties of the elements are things such as electronic distribution, ionisation energies, and so on. However, it is common to see periodic tables stating the melting and boiling points the pure substances of each chemical element, and even Wikipedia bundles the physical properties of dioxygen with the atomic properties of elemental oxygen. For better or for worse, we're stuck with this subtle ambiguity in nomenclature.

What do you define as "stable monoatomic substances" or "reasonable conditions"?  Mercury in a puddle will evaporate at a significant rate in open air at 25C, producing a vapor whose molecular weight matches mercury's atomic weight, but mercury is hardly a noble gas.

@supercat That's an excellent point. I considered metal vapours briefly, but evidently I forgot about mercury.

Technically, $\ce{O2}$ is molecular oxygen. $\ce{O}$, atomic oxygen, only exists under rare conditions; e.g. for a very short time after the molecule is dissociated by radiation, or where air pressure is so low that dissociated oxygen does not find a partner to recombine quickly.
Whenever one refers to an element, it is likely in the form of molecules, or metallically bound, except for the (usually) inert gases. The common yellow form of sulfur is as $\ce{S8}$ molecules. However, it is still considered elemental sulfur, not a compound, because all the atoms are of the same type.
BTW, later comments indicate some confusion with the number of protons, found in the nucleus, and the electron cloud, found around it. Think of an atom as a fluffy ball of cotton candy (electron cloud), with a few sugar crystals (neutrons and protons) in the center. When atoms combine to form a molecule, only the cotton candy sticks together; the nuclei are separate, with the same number of protons and neutrons (no croutons, though) as in the atomic element. Nuclei can combine, with enough pressure, and that process is nuclear fusion -- which could be discussed better in the physics section of StackExchange.
In short, an element can be found in molecules; it's still an element.

But why do we call O2 oxygen is my question? I thought only an atom with 8 protons can be called oxygen?

Each atom does have 8 protons. A molecule of two atoms with 8 protons each is not one atom with 16 protons. BTW, it's called oxygen because the discoverer thought all acids contain it (oxy means biting, sharp or acid in Greek, ὀξύς). So there is a misnomer.

Not much of a misnomer, though, because oxygen is a good oxidiser; it completely destroys heavier alkali metals, for instance, and eats very quickly through warm hydrocarbons.

@wizzwizz4 The term "oxidiser" was created later, and it's named that way because oxygen is the most common oxidiser.

@NieDzejkob But oxidisation does bite through things.

To be really specific, one should call the $\ce{O2}$ molecule dioxygen, to distinguish it from ozone ($\ce{O3}$) and other, less common allotropes of oxygen. And in some contexts one might want to further specify that one is talking about the usual triplet state, rather than the metastable singlet oxygen ($\ce{^1O2}$).

To elaborate on @DavidRicherby's point, words don't generally have unique meaning.  Soy milk doesn't come from a mammal; the town hall is neither a corridor nor a large room; and iron can denote a substance or an object, as well as an element.

@J.Smith "I thought only an atom with 8 protons can be called oxygen?" I'm not sure why you would think that. As you know, the word "oxygen" is used all the time, by everyone, including professional chemists, to refer to two of these atoms joined together into a molecule. The meaning of English words is determined entirely by usage -- there's no organization with the authority to declare "This word means X" -- so the fact that people (especially including experts) use the word "oxygen" to refer to the molecule implies that the molecule is one of the meanings.

It's simply a colloquialism for ease of reference. It is sort of like the way we call a heap of corn, corn. Instead of corn off the cob, like the comedian Mitch Hedberg jokes about. He says "corn" should mean "corn on an ear of corn", and a pile of corn kernels should be called "corn off the cob", but since we encounter "corn off the cob" much more often than a big ol ear of corn, we simply call it corn.
The more common something is, the less likely it is to be accurately named, and more likely it be colloquialized.

12.5% keyword density of the word 'corn' in an answer to a question about oxygen, that's got to be a world record

I am not sure the rest of us get these "corn" (AKA maize) references.

Words can mean more than one thing. There is nothing whatsoever inaccurate about referring to individual kernels of corn as "corn".

@DavidRicherby Duh; after all, corn originally meant "a grass with big seeds we use to make food", and still does in Traditional English. But the point mark makes is a correct one anyway - the uses that are more common get preferential treatment, because humans like to optimize their communication. Therefore, "molecular dioxygen" is called "oxygen", while "monooxygen" is called "elemental oxygen", if the distinction even needs to be made; just like you're more likely to use "car" rather than "a horse-less horse-drawn carriage". Accurate terminology helps communication, but so does efficiency.

@Luaan but is it a colloquialism?  It's just one sense of the word oxygen.  How is it a colloquialism?

It would be very odd to try to deny all allotropes of an element that element's name, or to require an additional description as in the phrases "molecular oxygen" or (to exclude non-modal allotropes) "diatomic oxygen". Why would it be odd? Because it may require some very specific descriptions that quickly become distracting, difficult to get right, and unimportant to the point you want to make. This is especially true if an element exists in multiple allotropes, whose chemical reactions of present interest might only care about the overall atom count present.
It's easy to get used to monatomic and diatomic forms of elements (metals usually treated as honorary examples of the former), and forget about such creatures as:

$\ce{As4}$ and multiple allotropes that don't admit as neat a formula (which will be a theme),
$\ce{B12}$ and other allotropes of boron
$\ce{O3}$, $\ce{O4}$ and $\ce{O8}$
$\ce{P2}$  vs $\ce{P4}$ and multiple polymeric forms of phosphorus
$\ce{Se8}$ and its crystalline and polymeric counterparts
$\ce{S_n}$ where $n$ can be any of $2$ to $9$, $15$, $18$ or $20$
and some elements having pretty much no natural $X_{\mathrm{small}\,n}$ allotropes such as antimony, carbon, germanium and tellurium.

What would you like to call them all?

When we say that certain elements “exist as molecules” what does that mean? From my understanding, an element is a species of atoms with the same number of protons, so when we say that oxygen, a species of atoms “exists” as a molecule, what does that mean precisely?

@J.Smith Well, from my understanding, an element is a collection of atoms that all have the same proton number, said atoms possibly not separate. Obviously we can define terms however we want, but given what atoms usually do it would be strange to define a lot of matter as neither an element nor a compound nor mixture.

So like a physical collection of atoms? If that’s the case, then when we talk about the reactivity of sodium why do we say that “sodium has one electron in its valence shell” as to imply that sodium is a type of atom?

@J.Smith Well, there are contexts where you can assume a particular form is intended. Never mind sodium or oxygen; you see the Swiss-Army knife nature of language when we discuss something as simple as hydrogen. It means very different things to stellar astronomers vs Jupiter experts vs spectroscopers vs quantum mechanists vs organic chemists vs fuel storage researchers vs those discussing acids.

@J.Smith Why wouldn't we? Sodium is an atom. It's also a metallic substance composed predominantly of sodium atoms. We only need to distinguish the two when it's important to what we're talking about, and the intended meaning isn't obvious. Whenever you're discussing anything, you need to be extremely mindful of context. Valence electrons are important because they affect how chemical bonds form - it should be pretty obvious why you want to discuss them when you're considering how sodium reacts with itself and other elements.

So when we say sodium is reactive, how do you know that we are talking about the atom "sodium" and not the metallic substance?

@J.Smith for the purposes of having the property of being reactive, they are the same thing.

@J.Smith if you're writing a technical paper, you define your terms more precisely.  If the distinction between atomic sodium and metallic sodium is significant, you mention it explicitly before or after making the statement that sodium is reactive.

"Oxygen" is a set which has "atomic oxygen" and "molecular oxygen" as elements(of a set).
It's like asking why is a "Ford automobile" and a "Chevy automobile", each called an "automobile". So, calling "atomic oxygen", "oxygen" is not inconsistent. Neither is calling "molecular oxygen" oxygen inconsistent.
We can add ozone to this as we can say, "ozone is the triatomic form of oxygen".
There are no inconsistencies here.
     Your question is not of science but of grammar. The singular and plural forms of the chemical elements are the same. We don't say, "oxygens", or "aluminums" unless we are talking about a property as "oxygens melting point is...". This question doesn't really belong here, but belongs in the category of English grammar.

But I have been taught that “elements” are species of atoms. If oxygen is a species of atoms with 8 protons then how can we call O2 oxygen?

Diamond and graphite are both arrangements of carbon atoms, each with very different properties and yet they are both pure carbon, as is a single carbon atom, of course. Likewise $\ce{O2}$ is as much oxygen as atomic oxygen is. The only complication is that what we habitually think of as oxygen is oxygen as a gas comprised of $\ce{O2}$ molecules. Like Humpty Dumpty in Alice in Wonderland, "a word means what [we] choose it to mean" and often we have to add modifiers or alternate terms to avoid ambiguity. Such is language...

If O2 is a type of molecule, can we refer to a collection of O2 molecules simply as O2? Because it doesn’t matter how much we have, it is still O2 right?


text:
Comparing the Lattice energy of different compounds
I am confused about some concepts related to Lattice energy.
First of all,how to compare lattice energy between $2$ compounds belonging to different groups and periods?
1.Between $\ce{AlF3}$ and $\ce{MgO}$ (In $\ce{AlF3}$ product of ionic charge is $3\times1$ and in MgO it is $2\times 2$. So,according to me $\ce{MgO}$ should have higher lattice energy but the opposite is true.
2.Between $\ce{KF}$ and $\ce{LiF}$ (According to Fajan's rule $\ce{KF}$ is more ionic than $\ce{LiF}$ and lattice energy depends directly on ionic character).But, value of lattice energy of $\ce{KF}$ is $\pu{808 kJ mol-1}$ and $\ce{LiF}$ is $\pu{1030 kJ mol-1}$. How is this possible?
So,can you please explain me how to compare lattice energies of different compounds and which factor is more dominating size of cation and anion or charge on ions.

I don't think this question is too broad - see answer.

You might want to look at
https://scilearn.sydney.edu.au/fychemistry/prelab/e12.shtml
and the related page
https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml
which has a nice lattice energy calculator that allows you to play with the parameters and see how the lattice energy varies, but I'll try to summarise those the argument below.
Here we are dealing with the ionic model - everything is totally ionic, there is total charge separation, all binding is electrostatic. Thus our system consists of a set of point charges. We can write the energy of such a system as
$$
E=N_AM \frac{z_1 z_2 e^2}{4 \pi \epsilon_0 r}
$$
where $N_A$ is the Avogardro constant, $z_1$ the charge on the first ion, $z_2$ the charge on the second, r is the closest separation of the ions and M the Madelung constant. Now note the Madelung constant only depends upon the structure of the crystal - all Sodium Chloride type crystals, whether it be NaCl, KF, or FeO, all use the same value off M. Looking at the formula we can see that FOR A GIVEN STRUCTURE (i.e. with the same M)

High charges on the ions mean high lattice energy
Small separation means high lattice energy

Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. The first point explains why MgO has a higher lattice energy than NaF. However it doesn't cover your first example, as $\ce{AlF3}$ and MgO have different structures (and actually I don't think it is a very good question).
Comparing different structures is very difficult. About the best we can do is the electrostatic term from the Kapustinskii equation (https://en.wikipedia.org/wiki/Kapustinskii_equation)
$$
U_{L}={K}\cdot {\frac {\nu \cdot |z^{+}|\cdot |z^{-}|}{r^{+}+r^{-}}}
$$
This is an approximation to the above. Here K is a constant which is INDEPENDENT of structure and $\nu$ is the number of ions in the empirical formula, 2 for MgO, 4 for $\ce{AlF_3}$. Thus assuming ${r^{+}+r^{-}}$ is the same for both of these $\ce{AlF3}$ has a bigger lattice energy because $\nu\cdot |z^{+}|\cdot |z^{-}|$ is bigger - For $\ce{AlF3}$ it is 4*3*1=12, while for MgO it is 2*2*2=8. What this is really saying is that the $\ce{AlF3}$ structure has a much larger Madelung constant than the NaCl one, enough to overcome the charge differences, and thus it has a higher lattice energy.

in AlF3  in case of each AlF bond,  charge of Al =3
=> product of charges in case of each bond=3
=>sum of products of charges =3x3=9>4
so, AlF3 has lattice energy more than MgO.


text:
Inorganic Chemistry
Which of the following is a covalent bond?
1.Al2O3                 2.AlF3
3.AlCl3                   4.Al2(SO4)3
The correct answer is AlCl3.But I have a confusion between option c and d.A/c Fajan's rule more the size of anion more is the covalent character. So,SO4^-2 have large size than Cl^-3 .So option d is correct but option c is the right answer.Please explain this question .

Perhaps because the charge carrying oxygens in sulphate are in resonance?

But resonance also mostly happens in covalent compounds

So you have to pull the electron density towards the cation. If those electrons are busy oscillating elsewhere, then won't it be difficult to pull such a distributed electron density? That's what I feel..

Fajan's rule is correct, but the sulfate ion is complex. We need to ask what is being polarized, a large sulfate ion as a whole, or an oxygen atom specifically. As it turns out, the aluminum acts on a nearest oxygen, so that even if this oxygen is more polarizable because it has a sulfur to draw on, a chlorine is still more polarizable.
Another thing is that aluminum chloride is dimeric in the gas phase, whereas aluminum sulfate is just an ionic solid - although you might still say there is some covalent character in the aluminum-sulfate bonds.

But since charge is distributed throughout the molecule via resonance, then how can you visualize that only oxygen is being polarized rather than the entire molecule?

That's the issue that makes the question complex. If the charge is distributed freely, then the anion is large. But if a single oxygen is polarized first, and then, using its slightly deformed charge distribution, slightly polarizes the sulfur, which then slightly polarizes the other oxygens, then you might as well just concentrate on the first oxygen (small anion). So I suppose that the charge does not distribute equally over the whole molecule if only one oxygen is polarized. Degrees of resonance might be an explanation.

What are 'degrees of resonance'?

@YUSUFHASAN You don't get what "resonance" is supposed to be. It's about any oscillation. It's about mesomeric structures contributing to resonance hybrid. Also this is a place for  discussion about that, since it has negligible importance for this topic.

Ok, I get it. Just a final quick question: when the carbonate ion is polarized highly, then it can liberate carbon dioxide gas. Which gas will sulphate liberate in such a case: SO2 or SO3?

At the anode, electrons will be removed from something and oxygen will be evolved. The simplest equation doesn't even involve SO4--. The oxidation of water is: H2O - 2e -> 1/2 O2 + 2H+.  If you must put SO4-- in there, then: SO4-- - 2e-> SO4 -> SO3 + 1/2 O2. Then SO3 + H2O -> H2SO4, so you get the SO4-- back, along with 2 H+. So neither SO2 nor SO3 is liberated.


text:
Substances - A Macroscopic Concept
Does the term “substance” refer to any sample of matter that is tangible? In other words is “substance” a macroscopic concept or can we also refer to individual particles such as atoms, molecules, etc. as substances?

It depends on the context
"Substance" is not a tightly defined term so the argument is largely linguistic not about chemistry. Unless, that is, you are talking about observable chemical properties and use the word. Then, the context of use matters and you might need to be careful in how the word is used.
With gases it doesn't much matter how much stuff you have as the individual molecules don't interact much. But for solids and liquids when you talk about "substance" the apparent chemical properties may depend on how many molecules or atoms you have. So you might need to be careful how many of the components are present. The reason why I say this is that some collections of small numbers of components have different properties to larger collections.
Nanoparticles of gold, for example, have different properties to bulk metallic gold. Some show much stronger colours that metallic gold (eg small colloidal gold clusters are used to impart a red colour to some specialty glass products). So the small clusters have different properties to bulk gold and you don't see the bulk properties until the amount of gold is large enough.
So although the substance of both bulk gold and colloidal clusters of gold atoms might be said to be the same, it is probably better to be more specific and talk about colloidal clusters as being a different "substance" to bulk gold. But this isn't universally agreed as usage of substance is not precisely defined.
In short, if you are talking purely about what something is made from then the amount doesn't matter and you can say substance for any amount but if you are talkingg about the properties of the material it can be better to be specific as the properties do change with the amount of the components involved.

So if I were to say that “sodium” is a reactive substance would I be referring to sodium atoms or sodium metal?

@Interestinggg For sodium it would make no difference. Pure metals contain atoms not molecules of their element so, in reactions, act like they react one atom at a time in simple reactions.

Alright so elements are substances “made up of one type of atom”. And sodium is an element. So why can we say that sodium “has one valence electron” when sodium is a substance not a type of atom?

@Interestinggg Context. And note that just because sodium is a substance doesn't mean we can't talk about sodium atoms. Sodium is both a type of atom and a substance. If we talk valence electrons the context tests us we are talking about atoms; if we talk "soft, reactive metal" we are talking about the bulk substance.

Could you just explain what the “made up of one type of atom” thing means. This would clear a lot of confusion. Regards.

@Interestinggg It isn't complicated. A lump of pure sodium metal only contains sodium atoms and no other type of atom. An oxygen molecule only contains oxygen atoms; oxygen gas contains oxygen molecules which consist only of two oxygen atoms.

An element is more like a form of matter isn’t it? As opposed to a physical aggregation of particles?

@Interestinggg "element" is a much broader term than that. Single atoms are by definition elements. small aggregates of atoms can be mede up from a single element. Large aggregates can be mad from a single element but don't have to be. Many of the properties of an element in large aggregates are often called the properties of the element but some elements form more than one type of aggregate. Ozone and dioxygen are both molecular forms of oxygen; diamond and graphite are both aggregates of carbon.


text:
Turning hot water to cold water
Is there a way to turn hot water instantly to cold water? Is there some sort of chemical reaction or element that allows this process to happen?

Instantly? No!!

Liquid nitrogen would cool it off pretty fast, though not instantly. This technique is used to make "instant" ice cream.

Makes a pretty good explosion too!

If you put a strongly endothermic reaction in a water bath, or put something extremely cool near or through it (dry ice, liquid nitrogen, etc. like @iad22agp said), then it would cool down.
One thing to note is that water has an extremely high heat capacity compared to most substances, so it takes a considerable amount of energy to turn "hot" water to "cold" water.


text:
Side chain oxidation with KMnO4/H+

I know that in side chain oxidation with the given reagent follows oxidation of benzylic carbon (if atleast one benzylic hydrogen is present). So my answer would be

But in my answer key only it is correct for nitro substituted anthracene. But why? What is the reason?
Thanks for clearing my doubt!

In each case the permanganate oxidises the most electron rich ring.
In the nitro-substituted case, that is the unsubstituted ring, probably first at the position alpha to the nitro group, forming a phenol, then para to that to a quinone but permanganate keeps on chewing until the phthalic acid is formed.
In the aniline case the most electron rich ring is the one with the NH2 attached. Oxidation of anilines by acid permanganate give a polymer called "aniline black" Wiki link. Though I can't find a reaction specific to 1-Naphthylamine other than one that documents the destruction of it by sulfuric/permanaganate without identifying the products here. The site of first oxidation is p to the aniline. This creates a reactive intermediate which is attacked by a second aniline as the first step of polymer formation.


text:
Can nuclear shielding affect relaxation time?
Can nuclear shielding affect the transverse relaxation time, i.e extend the relaxation time the more the nuclei is shielded?

I'd say yes, but not by any detectable amount. In some relaxation mechanisms (e.g. dipolar relaxation, a major relaxation pathway for $\ce{^1H}$), the rate of relaxation varies with $\langle B_\mathrm{loc}^2\rangle$, i.e. the magnitude of the random fields that the spin experiences due to the magnetic momenta of all the other spins. The magnitude of any field that a nucleus experiences will be attenuated by the induced magnetic field arising from its own electrons, so it stands to reason that – all else being equal – a more shielded nucleus will have a smaller $\langle B_\mathrm{loc}^2\rangle$, and a longer relaxation time, than a less shielded nucleus. This applies to both $T_1$ and $T_2$.
Bear in mind, however, the most obvious effect of nuclear shielding, which is to change the resonance frequency of the nucleus. The spread of frequencies is on the order of $10^{-6}$ relative to the strength of the external magnetic field. If we crudely estimate that the spread in $T_1$ and $T_2$ arising from sheilding differences is also on the ppm level, that corresponds to ~µs levels of difference, which I strongly doubt is detectable.
In any case, for two protons that have different chemical environments, there would be other factors which affect the relaxation times to a much greater extent.


text:
What determines the coefficients in the wavefunction of a hybrid orbital?
I was studying the hybridization of carbon atoms and I came to figure out that the $sp$ hybridization wavefunction is given as
$$|sp_{+}\rangle =\frac{1}{\sqrt{2}}\left(\psi_{2s}+\psi_{2p_x}\right)$$
$$|sp_{-}\rangle =\frac{1}{\sqrt{2}}\left(\psi_{2s}-\psi_{2p_x}\right)$$
From a simple mathematical point of view, Why don't I have states that look like the ones below?
$$\frac{1}{\sqrt{2}}\left(-\psi_{2s}+\psi_{2p_x}\right)$$
$$\frac{1}{\sqrt{2}}\left(-\psi_{2s}-\psi_{2p_x}\right)$$
Even for the $sp^2$ hybridization, the states are
$$\frac{1}{\sqrt{3}}\psi_{2s}-\sqrt{\frac{2}{3}}\psi_{2p_y}$$
$$\frac{1}{\sqrt{3}}\psi_{2s}+\sqrt{\frac{2}{3}}\left(\frac{\sqrt{3}}{2}\psi_{2p_x}+\frac{1}{2}\psi_{2p_y}\right)$$
$$-\frac{1}{\sqrt{3}}\psi_{2s}+\sqrt{\frac{2}{3}}\left(-\frac{\sqrt{3}}{2}\psi_{2p_x}+\frac{1}{2}\psi_{2p_y}\right)$$
I don't know how the construction of the hybrid orbitals are done mathematically...all that I know is that in case of the $sp^2$ hybrid orbitals, the formulation should be
$$\psi_{sp^2}=\alpha\psi_{2s}+\beta\psi_{2p_x}+\gamma\psi_{2p_y}$$
where $\alpha^2+\beta^2+\gamma^2=1$. But, mathematically there should be infinite solutions to this equation with this one constraint, where do this extremely precise values of $\alpha$, $\beta$ and $\gamma$ arise from?
P.S. :  I am not a student of chemistry, but just an engineer.

For the $sp$ case, the other wavefunctions you listed would be equally valid. However, these are just $sp_+$ and $sp_-$ with the signs flipped. Since we square wavefunctions to get the probability density, these would give the exact same physical observables.
As for $sp^2$, the reason for the particular choice of $\alpha,\beta,\text{ and }\gamma$ comes precisely from the name of this hybridization: we want an orbital that is 2 parts p orbital for every 1 part s orbital. In other words, we want a mixture that is 1/3 s and 2/3 p which happens with exactly the normalization coefficients you listed. This is a second constraint. Now this still will lead to an infinite number of different solutions, as we can consider rotating these orbitals by any amount in the xy plane, which would give another valid solution. One of the nice things about the choice you listed is that one of the MOs is aligned with the y-axis and the other two are mirrored across the y-axis, so it is easy to consider the effect of axis symmetry operations on this choice of orbitals. If we include this third constraint of having one of the MOs aligned with an axis, we get a unique solution (besides possible sign flips and that we could align the orbital along the x axis instead) for $sp^2$.


text:
Identifying food, quick and cheap?
I’m asking this question in the most technical perspective of identifying substances, so I’m not considering the use of chemical tests like Benedict’s solution, biuret solution, Sudan III stain, or indicator dichlorophenolindophenol, as valid answers since these methods rely on identifying the color by observation. The answers I’m really looking for are based less on physical and more on chemical properties without the need of human observation or intervention.
(At this point, if you have an answer please reply… the rest is clarification to my question)
With that being said spectroscopy, chromatography, electrophoresis, and biochemical analysis are more along the lines of the answers I’m expecting. However, I’m really asking which of these methods or others in connection with other techniques, are the most convenient, smallest in size (in terms of equipment), and cost-effective in identifying food sources.
So in attempt to sum up my question, here's a scenario: I have a yellow onion. I extract juice from the onion and I give it to a skilled chemist/engineer labeled unknown. Using an analytical technique, in about a few seconds, the chemist supplies a prediction based on the components found within the unknown food (amino acids, peptides, proteins, enzymes, lipids, phospholipids, carbohydrates, alcohols, fat-soluble vitamins, water-soluble vitamins, organic acids, organic bases, phenolic compounds, bittering substances, pigments, aroma compounds, and dietary fiber)
What is/are the analytical techniques used? Is there a device that does such a thing?

Unfortunately, your confidence in what a chemist can, is quite exaggerated. Though flattering.
Analyzing the entire range of parameters that you list, requires several analytical techniques, including sample preparations like homogenization, extraction etc.
A technique like Near-infrared spectroscopy can give an easy and fairly good estimate of some major components (protein, fat etc) in selected types of food. FOSS Analytical (and probably several others) purchases equipment for this.
Routine methods exist for analyzing components like vitamins, pigments or metals, but they require a laboratory setup and often very expensive equipment.


text:
When there is more than one amino group in a molecule, which group is more likely to react when forming an imine?

Why does the third group react in this example and not the first?

Perhaps, you can also elaborate on your thoughts as well?

Well the 1 group is an amide and the 2-3 group is a hydrazide.

Imine formation essentially involves an nucleophilic amine and an electrophilic carbonyl group. To decide which is the nucleophilic $\ce {N}$ atom which performs the nucleophilic attack on the electron-deficient $\ce {C}$, we would have to first consider each $\ce {N}$ atoms nucleophilicity, which we can interpret here to be the ability of the $\ce {N}$ atom to donate its lone pair of electrons.
For 1 and 2, the lone pair is delocalised into the carbonyl group and they can be seen as behaving as amides. This delocalisation significantly reduces their nucleophilicity. As discussed by Ron here, the delocalisation is so strong that restricted rotation about the $\ce {C-N}$ bond is observed. For 3, there is no such delocalisation present, thus the lone pair of the $\ce {N}$ is readily donated to electrophiles, making it the likely nucleophilic atom for the imine formation.
Another possible effect that raises the nucleophilicity of that $\ce {N}$ atom is the alpha effect, also discussed here. The following diagram (p. 513) from Clayden, Warren, & Greeves (2012) illustrates the molecular orbital theory basis of the effect. Although it is used to illustrate the effect for peroxide, the same concept can be extended to our case of nitrogen atoms. The adjacent $\ce {N}$ atom with the lone pair can increase the nucleophilicity of the $\ce {N}$ atom. However, the alpha effect may be less prominent for our case since the lone pair of the adjacent $\ce {N}$ is also being strongly delocalised into the carbonyl group. Nonetheless, it is an interesting consideration, further justifying why 3 should be most nucleophilic among the three.

Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

This is semicarbazide semicarbazone reaction which is a nucleophilic addition reaction with the elimination of water. This addition reaction, like any other, involves a nucleophile (lone pair on N here, which one - we will see) on the nucleophilic center (carbonyl group). Thus the reaction will go more forward if the reacting nucleophile is more electron rich and thus more nucleophilic.
Now, 1st and 2nd groups are participating in conjugation (I presume you know about resonance, etc.) with -C=O group and hence their lone pair electrons (cause for nucleophilicity for them)are delocalised and thus less available making them weak nucleophiles. Thus, the 3rd group is the most nucleophilic and proceeds with the reaction.
Hope this helps!


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Reaction of grignard with beta Keto ester
A reaction takes place between $\ce {CH3COCH2COOC2H5}$ with 4-chlorophenyl magnesium bromide, followed by acidic work-up. What will be major product of the reaction? Would the product be the same as the original $\beta$-keto ester due to the active methylene group, at which the acidic doubly-$\alpha$ $\ce {H}$ gets deprotonated?

It would be helpful if you could provide illustration of these molecules. Also, could you also provide your thoughts on the matter?

You will get the ethyl acetoacetate back. It is sufficiently acidic (pKa ~11) to quench the Grignard and form the Mg salt of acetoacetate.

@Waylander What counts as sufficiently acidic? For example, why isn't the alpha H of a ketone deprotonated by grignard?

I don't know precisely where to draw the line, somewhere around 18/19 perhaps.  EtOH (pKa 15.9)  and tBuOH (17.9) quench Grignards. Acetone (pKa 19.2) certainly undergoes Grignard addition readily enough.

@Tan Yong Boon: Plenty of examples exist where enolization of ketones competes with addition upon treatment with Grignard reagents and organolithium reagents.


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Sugar: A molecule or crystal
Is sugar a molecule or a large crystalline structure composed of many molecules? What do we mean when we say sugar?

Sugar is white powder used for sweetening our tea; the moment you realize it consists of molecules, it ceases to be called "sugar". The related molecule, and also the crystalline structure composed of these molecules are both called "sucrose".

You're asking another time about the same misunderstanding! Sugar is compound which can be in multiple states, just like water, oxygen etc. Compound or element doesn't equal atom/molecule/crystal /gas/liquid but can be in one of these forms. Is it clear now?

So basically, “element” is a way of referring to a certain type of material made of one type of atom regardless of how much you have? So one atom of sodium is “sodium” and a collection of sodium atoms can still be called “sodium”?

yes, but generally we use extra words to help remove ambiguity, for example sodium metal, or sodium vapour or sodium ions, or sugar solution or sugar crystals.

Terminology can be confusing especially when some terms are only vaguely defined.
"Sugar" is one of those vague terms. There is a whole family of sugars recognised by chemists. In casual use what people usually mean is sucrose, a specific sugar molecule made from two other sugars, fructose and glucose, joined together. this is produced from sugarcane or sugar beet and is the dominant form of sugar seen by non-chemists.
But is sugar the molecule or the crystal? Both. Many pure chemicals form crystals, indeed crystallisation is a very common way of producing purer chemicals. Those crystals consist of a regular 3D array of a single molecule (for a pure chemical). So "sugar" describes both the molecule and the crystals of that molecule which consist of a large number of them arranged in a regular structure.

To answer your question I will separate it into two parts, crystal and sugar.
I wills start with the concept of a crystal, for that I will cite wikipedia since I couldn't find the IUPAC definition of a crystal:

A crystal or crystalline solid is a solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

So basically stating a crystal is a structure that has a unit-cell that repeats in all direction. Every unit-cell is exactly the same as the other ones.
Sugar is a carbonhydrate. It may constitute of one (example glucose), two(example sucrose), or even a chain of sub-units (example starch).
The white sugar that one buys at a store is crystalline sucrose. It is not a big crystal as a slab of Aluminum is. But it is still a crystalline structure made of a repeating unit-cell which is sucrose molecules.

A slab of aluminum is not one big crystal either.

"carbonhydrate" is not a correct spelling


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Formulae distinguished only by case
A recent question (on $\ce{CI}$ v $\ce{Cl}$) reminded of this question: are there any real (as opposed to contrived) formulae which would be the same if case was ignored?
One certain example is $\ce{Co}$, Cobalt, and $\ce{CO}$, Carbon Monoxide.  Of course, confusing Cobalt and Carbon Monoxide is rather unlikely.
In my search for others, the best example I could find was $\ce{BaS}$, Barium Sulphide, and $\ce{BAs}$, Boron Arsenide.  Are these real substances and is the stoichiometry just $1:1$?  I had trouble verifying that.
Are there any more interesting examples?
One real example that I saw once was a poster in a university chemistry lab showing some complex organic molecules.  $\ce{Ac}$ appeared at many points but it seemed very unlikely that these compounds contained Actinium.  I then noticed $\ce{Me}$ and $\ce{Et}$ and guessed that they were organic units e.g. methyl and ethyl so $\ce{Ac}$ may have been acetyl.

Well, this is a legitimate concern indeed, which is why we feel irate to no end when somebody messes up cases in chemical formulae.

BaS is indeed a known compound, with the NaCl prototype. BAs is also a known compound with the ZnS prototype.

Well I never put it together that actinium acetyl have the same abbreviation.

Letter case in chemical names sometimes matters too! E.g. when o-methylphenol (o-cresol, ortho-cresol, 2-methylphenol, I) improperly capitalized to O-methylphenol (instead of o-Methylphenol), it can be interpreted as a different compound, phenol methylated at oxygen (anisole, methoxybenzene, II).


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Do beta hydroxy aldehydes respond to Fehling's test?
Do beta hydroxy aldehydes respond to Fehling's test?
I was told that it readily loses a molecule of water forming alpha beta unsaturated compound. But it still has an alpha hydrogen and I presume that it meets all requirements for a compound to respond to Fehling's test.
Please correct me if I am wrong

If it's an aromatic aldehyde, Fehling's test anyways not gonna work. But if it's an aldehyde not having a bulky side group, I think it should respond. It is not so spontaneous for any $\beta$-hydroxy aldehyde to lose a water molecule to form the unsaturated aldehyde.


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Does the Nernst equation suggest voltaic cell voltage remains constant over time?
Say the following reactions are occurring in my voltaic cell:
$$\ce{Mg->Mg^{2+} + 2e-     +2.36V}$$
$$\ce{Cu^{2+} +2e-->Cu     -0.34V}$$
Then, the Nernst equation would become:
$$E_{cell}=2.70-\frac{RT}{nF}ln\frac{[Mg^{2+}]}{[Cu^{2+}]}$$
However, if the initial volumes and concentrations of the copper and magnesium solutions were equal, the fraction within the natural logarithm would constantly equal 1.
Does this mean there is no change in voltage as the concentrations decrease, or did I understand something wrong?

You got it wrong. As the reaction progresses, [Mg2+] changes; so does [Cu2+], and also their ratio, and also ln thereof.

@IvanNeretin thank you, I just realised I was thinking of the Mg2+ and Cu ratio


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Displacement of leaving groups in organic reactions
When I initially learned about substitution reactions, there would be times where a poor leaving group had to be changed into a better one in order to proceed with a substitution reaction. For instance, changing a poor leaving group (-OH) to a better one, such as water:

After the "good" leaving group is made, it is displaced via an SN2 reaction. The displacement of the water required a nucleophile. However, I recently came across this mechanism (dehydration):

In this case, the water leaves by itself. My question is:
If I ever come across an instance where water (a good leaving group), must leave, how should I assess whether it can leave BY ITSELF or not?

Generally, you look at the cationic intermediate to see if it is stabilized or not. For example, creating a cation from the primary alcohol is too disfavored. Secondary cations on the other hand, are reasonably attainable though some forcing conditions may be required.

That does make sense since a primary carbocation is far too unstable. So would it be alright to just displace water if it results in a secondary or tertiary carbocation?

Sn1 reactions of secondary and tertiary alcohols are standard fare in a beginning organic chemistry course.

In that kind of reactions, the full explanation would require advanced quantic calculation to determine the energy of all the present entities.
In chemistry, we usually prefer giving a half-quantitative explanation to explain the formation of the observed products.
In your example A —> B + C + D, A is a secondary alcohol and it CAN be transformed to a carbocation leading to B... but you can see it is far from perfect since B represents only 50% of the yield. The result would be much more clear-cut with a primary alcohol (unstable carbocation) or a tertiary one (very stable carbocation). I think you are expected to explain all that.

Dehydration represents 100% of the yield as shown in the figure. 3 possible isomers result from the secondary cation; this is not an indication of how stable or otherwise the secondary cation is.

@Waylander Yes you are right. I think I was too focused on the questions I had at the time and I will probably change my answer


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What is the major product of the nitration of N,N-dimethylaniline?
What is the major product when N,N-dimethylaniline, $\ce{C6H5N(CH3)2}$, reacts with concentrated nitric acid (conc. $\ce{HNO3}$) in presence of concentrated sulphuric acid (conc. $\ce{H2SO4}$)?
Here, nitrogen  gets protonated. So, it becomes deactivating, but it can't be stabilised by resonance because it would exceed valency of nitrogen. So, is meta or para the major product?


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Confusion about electron arrangement in elements
My textbook says, "The first shell can hold only 2 electrons. The second and subsequent shells can hold 8 electrons to give a stable noble gas arrangement." Can someone explain me if it is MAXIMUM 8 electrons in the second and subsequent shells or is it possible to have more than that. I searched on the internet but it keeps showing this equation: 2n2. As I am a high school student, I have no idea about "quantum" stuff. Can someone explain? Thank you.

Why aren't you satisfied with $2n^2$ (which is true, BTW)?

You should know that you are been taught a "simplified" version and it becomes less and less simplified when you can understand the whole version. It is the same everywhere in the world

@IvanNeretin You see it all started when I tried making up an electron arrangement for elements in the Periodic Table. Uptil Calcium it was ok. COLUMNS: Group I, II, III, V, VI, VII, VII/O (between II and III are "transition metals"). ROWS: Period 1, 2, 3, 4, 5, 6, 7. Group # = # of electrons in outer shell. Period # = # of shells. Calcium has 20 electrons and occurs in Group II and Period 4. So logically, first shell has a maximum of 2, second shell = 8, third shell = 8 and fourth shell (outer shell) = 2 (2 corresponds to Group#2). But after Ca, I can't make sense of groups and periods.

Basically, it depends on subshell. If you google something like "subshells", you'll see that s-subshell can hold 2 electrons, p - 6, d - 10, f - 14, g - 18. It's very unlikely that you'll need subshells f and g, since you're school student.
Look at this picture (about Aufbau principle).
It shows you the order of orbitals. Each s orbital holds up to 2 electrons, p - 6, so on. Fill them with all your electrons in given order.
For example, you have 12 electrons. The order (using Aufbau principle) is 1s-2s-2p-3s-3p-4s-3d...
1s holds 2 electrons (12-2 = 10), 2s holds 2 (10-2=8), 2p - 6 (8-6=2), 3 s - 2 (that's all). We see that 1st shell has 2, 2nd shell has 8, 3rd shell has 2, but 3rd is not full (since you still have empty 3p and 3d subshells).
So let's check. n is for number of shell. $2*n^2$ is for maximum number of electrons that shell can hold. For 1st, $2*1^2=2$, as we've seen. For 2nd, $2*2^2=8$, as we've seen. For 10th shell, the maximum is 200 electrons.
[edited]
Use Aufbau principle as well as info about subshells. Let's look at Y(39 electrons), which is in G3 P5. So we expect 5 shell to be outermost, and 3 valence electrons. Not "electrons in outer shell", but valence. For school course of chemistry, outer shell electrons are usually referred to as "valence", but these concepts are not always interchangeable.
Back to our example, Y has 1s-2electrons, 2s-2,2p-6,3s-2,3p-6,4s-2,3d-10,4p-6,5s-2,4d-1 (sum - 39). So yes, "maximum shell" is 5, as expected. But why do we have 1 electron on the outermost subshell, when we need to have 3? It's because valence electrons are those on 5s and 4d (2+1=3). Usually d subshell electrons are also considered "valence", when you have 1-10 of them (if total number of electrons, starting with 4d electrons is 18 or less). I highly recommend you to use periodic table with 18 groups, it's much more convenient for these purposes.

@KellySheppard You see it all started when I tried making up an electron arrangement for elements in the Periodic Table. Uptil Calcium it was ok. COLUMNS: Group I, II, III, V, VI, VII, VII/O (between II and III are "transition metals"). ROWS: Period 1, 2, 3, 4, 5, 6, 7. Group # = # of electrons in outer shell. Period # = # of shells. Calcium has 20 electrons and occurs in Group II and Period 4. So logically, first shell has a maximum of 2, second shell = 8, third shell = 8 and fourth shell (outer shell) = 2 (2 corresponds to Group#2). But after Ca, I can't make sense of groups and periods.

Check [edited] part of my answer, please. Maybe it'll be helpful.


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Chemical Equilibrium in case of copper(II) ions

$\pu{Cu^{2+}}$ ions react with $\pu{Fe^{2+}}$ ions according to the following reaction:
  $$\ce{Cu^+2 + 2Fe^2+ <=> Cu + 2Fe^3+}$$
  At equilibrium, the concentration of $\pu{Cu^{2+}}$ ions is not changed by the addition of which ion $\pu{(Cu/Cu^{2+}/Fe^{2+}/Fe^{3+})}$.
  Explain the result.

This appears to be a homework question, please share your thoughts and attempts towards the solution. If you receive useful answers, consider accepting one. The homework tag is deprecated, please don't use it.

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Obviously $\ce{Cu}$, as it is a solid and has no effect on Equilibrium constant and thus no effect on altering equilibrium concentration of $\ce{Cu^2+}$.

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.

Cu is the answer. As the concentration of any solid component is fixed and does not effect the Kc value at all.
let concentration = $C$$ , mass = $$m$$ , molar mass = $$M$$ , volume= $$V$$ , density=$$d$$ and no. of moles = $$n$;
then$$C=\frac{n}{V}$$ and $$n=\frac{m}{M}$$ and by the 2 equations we get
$$C=\frac{m}{MV}$$ which equals $$C=\frac{d}{M}$$ and as $$d,V$$ are constant for a solid as Cu, adding more substance would not change its concentration.

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


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I need some help on this Chemistry question can any one help please!! I need to find the cost of a single atom of aluminum
A 75 ft2 roll of aluminum foil costs $3.79.  A 12"x12" pieces has a mass of 1.0 g.  How much does a single atom of aluminum cost, in dollars?
i need to show all my work i am so lost.

We are given that 75 square feet costs \$3.79, so 1 square foot costs \$3.79 divided by 75, or just over 5 cents. We are also given the mass of one square foot as 1.0g, so from the atomic weight of aluminium, one mole costs approximately 27 times this, or \$1.36. One mole is $6.022\times10^{23}$ items (in this case atoms), so how much does an atom cost?
Adjust significant figures to taste.

could you show me how you got the 27 times and $1.36. im sorry im so  lost

The atomic mass of aluminum is $26.98 \text{ g/mole }\approx 27$

thanks for fixing the formatting, MaxW - I tried the "/$" but that kept messing up for me.


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Does a portion of a solution have the same molarity as the entire solution?
For example the solution contains $\pu{0.5 mol}$ of a solute and the total volume is $\pu{0.5 L}$, the molarity would be $\pu{1M}$. But if I take a portion of that solution, let's say $\pu{0.1 L}$, would that molarity be still $\pu{1M}$? If it were that way, there would be  $\pu{0.1 mol}$ of the solute in the portion.
Or, do the moles of the solute stay the same, $\pu{0.5 moles}$, and the molarity would change to $\pu{5M}$?
So my question is, does the amount of solute decrease when considering only a portion of the solution?

If the concentration of the whole soln is 1M, then a portion taken out remains 1M unless you do something to it.

So the solute would be 0,1 moles in 0,1L of the the dissolution, right? Because the concentration would be 1M?

Correct. If you concentrate it down to dry you would expect to get 0.1 moles of the solute

Or if I take out a portion of it to mix it with another dissolution, right?

Exactly so. If you take out 50ml to react with something else, that 50ml is at a concentration of 1M

Great!! All cleared up, thanks for the fast reply :) appreciate it

Molarity is simply a ratio between the number of mols of a solute within a liter of solvent. So, a 1M solution tells us that there is one mol of solute per liter of solvent. If I take half a liter, 0.5 L, I will also take along with that half-liter 0.5 mols of solute. However, the molarity hasn't changed, as it is still 1M.
If your question is: are there less moles in a solution of lesser volume, compared to a solution with greater volume, but with equal concentrations of the same solute? The answer is yes, the ratio will not change assuming we haven't added or removed any volume of the solvent. A solution of 1M and 5L has 5 mols. A solution of 1M and 2L has 2 mols, and so on.
If your question is: if I have the same amount of mols, but I decrease volume of solvent, does my concentration go down? The answer is no. Concentration would increase. If I concentrate a 1M solution that has 1 mol and 1 L by fixing the amount of mols and removing solvent, say to 0.5 L, my molarity would increase to 2 M, thereby increasing the concentration. We can get rid of solvent by boiling it off, which allows us to increase the concentrations of solutions without removing any mols of solute.
Though this question is old, this answer may help others. My advice is to come up with some arbitrary concentrations, mols, volumes, etc, and see what happens as each variable changes.

Yes , the solute particles are homogeneously spread throughout solution so when we take a spoonful of solution , the number of solute particles in the spoon is less than those which were present in solution.
Molarity is an intensive property of a solution so it remains same no matter you calculate it by taking a solution filled beaker or by taking only a spoonful amount.

No, solute will not decrease if volume will decrease.But molarity will increase for sure, cause molarity is the amount of solute in litre of solution.
Consider container A in which you took 1 litre of water and added two spatulas of NaCl and in another container B you took 0.5 litre of water and added two spatulas of NaCl. Now which container has greater amount of solute , the second one that is container B ,hence molarity of container B solution will be greater than container A.
Hope it will work to clear concept.

Yeah I get that, but I'm talking about only ONE glass of a dissolution.

Well, no. Unless you are doing a different experiment. If you have a homogeneous solution the molarity remains the same unless you do something to it. If you dissolve the same amount of solid in a smaller amount of liquid then you are right, but that wasn't the question.


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Proper explanation for depression of freezing point
Why does depression of freezing point takes place if a non-volatile solute is added to the solution?
Depression of freezing point

A simple "intuitive" oversimplification: intrusive foreign molecules get caught between solvent molecules, getting in the way of crystallization. That, however, would contradict the similar boiling point elevation.
Another oversimplification: the solute molecules pull the solvent away from the incipient crystals.
Yet another simplistic answer: osmosis exerts a force on the solvent, pulling it back into solution, away from the pure crystal of solvent.
Perhaps the last explanation is closest to a statistical analysis of freezing point depression.
Hopefully, chemistry will not be a depressing subject.

osmosis is also a colligative property, but then both are not very well intuitively understandable. ;-)


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Why is syn conformer of methyl formate more stable than trans conformer
As stated, I've read this in a textbook and it seems not explainable solely based on steric clash or Coulombic interactions. I know for such a simple molecule, it can be predicted ab initio using quantum mechanical models (A paper on this), yet, I'm seeking a simple explanation for this conformer preference. Would anybody kindly explain to me the logic behind this? Thanks.

https://chemistry.stackexchange.com/questions/20185/why-are-lactones-more-acidic-than-esters

https://chemistry.stackexchange.com/questions/59458/which-electronic-effects-are-responsible-for-the-reduction-of-nucleophilicity-of


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Maximum metal-metal distance for antiferromagnetic coupling
Imagine a transition metal complex with multiple open-shell transition metal cations. If the metals are relatively close in distance, it is possible that antiferromagnetic coupling will occur between the metals. However, at approximately what distance does ferromagnetic vs. antiferromagnetic alignment of spins become isoenergetic due to negligible coupling between the metals? To provide some context, I'm asking about this in the context of density functional theory calculations. To make sure we're on the same page, let's call two magnetic configurations "isoenergetic" if the energies are, say, within ~1 kJ/mol or so.

The annoying answer is that this will be system dependent. I'll try my best to explain with a quick recap of magnetism.

Unpaired electrons produce a magnetic moment
One of the factors that determines the strength of the magnetic moment is the number of unpaired electrons.

If you have a specific system in mind, I would suggest performing a series of single-point (no ionic relaxations) calculations where you increase the distance between the ions. Contrast the energetic between FM and AFM configurations. If you have a license, VASP can provide information on the magnetic moments.


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Organic chemistry( basics)

I have a confusion here .Why option (a) have minimum enol content.According to me enol content is related to stability ,so only compound a is undergoing resonance so it must have a maximum resonating structure ,but the reverse happens here.Please explain me if stability is not the factor which determine enol content or I am wrong??

Keto-enol enol is considered after sufficient time has passed for the equilibrium to be attained, and thus is a form of thermodynamic control regulated by Gibbs free energy

In this question, enol of (d) will be stabilized by a 6 member hydrogen bonded ring. (C) is forced to undergo tautomerism to minimize dipolar repulsion between the oxygens. (B) on tautomerizing forms an aromatic compound while (a) forms an anti aromatic one. So (A) is least stable

@YUSUFHASAN option d contains only 5 carbon. So how it form 6member ring.If c have repulsion than option b have more repulsion than c.And,how a form anti aeromatic ring can you explain this question to me with the help of diagram please.

The last option is forming a six members ring via intramolecular H-bonding, and the said ring consists of 3 carbons, 2 oxygens and 1 hydrogen. For option B, you are right, dipolar repulsions will be present, and on top of that, the 2π electrons(denoted by e-s in my diagram) will ensure that the tautomerism takes place. Although dipolar repulsions still exist after this step, but since we are out of acidic hydrogens, so this is the best configuration the molecule can acquire. Finally, option A will form an enolate possessing 4π electrons, and by Huckel's rule, it is antiaromatic in nature
Diagrammatic representation


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What is/are the product(s) of the reaction of (5,5-²H₂)cyclopenta-1,3 diene with NaH, and quenched with H₂O?

$\ce{NaH}$ breaks as $\ce{Na+}$ and $\ce{H-}$. Here $\ce{H-}$ acts as a nucleophile and attacks at a place where the electron density is less. But the $\ce{C-D}$ bond is stronger than the $\ce{C-H}$ bond, so the reaction can't be feasible in forward direction. The correct option given is '(d) all of these'. Where is the other $\ce{D}$ gone and how is $\ce{D}$ now attached at all the places?

C-D bond is the same as C-H. The difference is really, really small, much smaller than you might think after reading these words, and even smaller than that.

H- does not act as a nucleophile, it is a strong base.

I think that the deuterium isotope effect may have something to do with this

Nothing to do with the deuterium isotope effect. Consider cyclopentadienyl anion - is it aromatic?

@Waylander Yes it is. But what does it mean?

An interesting thing to note here that prima facie, it seems that the deuterium is being substituted at different position in the molecule, but looking at the reaction mechanism we see that in fact the second deuterium is not moving at all after abstraction of the first, rather it is the resonance of the molecule and protonation at different sites which is responsible for the varied products

I can't access that link.Can you please explain the mechanism of this reaction in detail?

@kaushki  The key thing to realise is that H- deprotonates the deuterocyclpentadiene to give cyclopentadienyl anion. This is a stable and aromatic species so when it is quenched there is an equal probability of each carbon being protonated.


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Methylation of nitroalkanes with diazomethane
It is known that diazomethane can methylate various compounds with acidic hydrogens. Diazomethane is most well-known for the efficient synthesis of methyl esters from carboxylic acids. Clayden, Greeves, & Warren (2012) mention that it can also methylate phenols ($\ce {pK_a = 10}$), but not straight-chain alcohols ($\ce {pK_a = 16}$). The latter not being acidic enough. I was just wondering if nitroalkanes ($\ce {pK_a = 10}$) would be able to be methylated by diazomethane?
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Never seen this reaction - which doesn't mean there are not examples though

That pKa value looks wrong to me, for simple nitroalkanes. I'm sure it is substantially higher for nitromethane, eg. ca. 17.

@OrganicChemistryExplained I am referring to the pka in water.

@OrganicChemistryExplained According to Clayden's, pka of nitromethane is indeed 10.

OK, it looks like the ~10 value may be in water, although I haven't been able to find the original ref, whereas the pKa is 17.2 in DMSO - JACS 1975, 97(24), 7006. Like @Waylander, I haven't seen this reaction. I suspect that if it is acidic enough in the right solvent to treat with diazomethane, it give O-methylation rather than C-methylation, and that O-methyl product may readily hydrolyse on workup or attempts at purification. Interesting question!