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Why is the repulsion between lone pairs and bonded pairs stronger than the repulsion between the bonding pairs and the bonding pairs?

My book says "The reason for this difference is that the electron pair in a bond is further from the nucleus of the central atom than the electron pair in a lone pair."

I don't get it , how does the distance make any difference ?
I'm , taking AS level chemistry

From Coulomb's law   Wikipedia
$$\pu{ F \propto \frac{1}{r^2} }$$
If your electrons are closer to the center then the distance between them is small.
A small distance would lead to a greater repulsive force as the dependence of force is inverse square to distance between charges.

I dont understand exactly the quote either, maybe because it is isolated from the rest of the text but one way to see this is by observing that an electron pair in a bond is "trapped" between the two nuclei. Hence, it wont be able to expand too much in space. In contrast a lone pair is constrained only by one nucleus. Hence, it is more likely to be found further away in space compared to a bonded pair. Hence a lone pair which is nearby a bonded pair will push that bonded pair further away compared to another bonded pair. This is the reason why the H-N-H angles in NH3 are smaller than the H-C-H ones in CH4.


text:
Enrichment of D2O
The Wikipedia page for deuterium oxide (heavy water) says that its boiling point is 374.5 K and the one for ordinary water is 373.13 K.
Is there a way of distillation which we can use to enrich deuterium oxide?

Related but perhaps not a duplicate: https://chemistry.stackexchange.com/questions/47820/deuterium-depleted-water-through-fractional-distilation

You might be interested in the heavy water refinery process by Cody's lab on YouTube

Fractional distillation can work, but the separation per round of distillation is very low so you would need a very large multi stage process to achieve significant separation.
Nobody uses direct distillation to produce heavy water as there are better chemical processes that achieve higher enrichment per stage, for example the Girdler sulphide process.

The wikipedia page on the GS process actually mentions distillation as a common process to get from 15–20% D2O (the product from the GS process) to > 99% D2O.

For a comparison of different methods, including distillation and the Girdler processes, have a look at
Heavy Water: A Manufacturers’ Guide for the Hydrogen Century
In this 14 pages document a survey of the different available processes is given.
The distillation of water, as suggested by the OP is included.
That process is the easiest but requieres huge energy input to perform.
The distillation must be done under vacuum because the separation factor is higher than at normal pressure.
The processes that are used are based on exchange reactions.
Among these the most used is the Girdler process based on $\ce{H2S}$ but other possibilities exist.
However, this is the only one that does not require a catalyst.
Other processes are also considered briefly.
An interesting point is that the estimated cost of production is given as 300 \$/kg $\ce{D2O}$.
A table with the characteristics of the different processes is included.

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Why is hydroxide ion the strongest base only in water?
My professor mentioned that hydroxide ion is the strongest base when we only consider aqueous solutions. Then why don't the strongest bases outside of aqueous solution, like superbase, remain the strongest in water?

Possibly because they react with water to give OH-

https://en.wikipedia.org/wiki/Leveling_effect

https://chemistry.stackexchange.com/questions/63355/why-cant-the-strength-of-superacids-be-measured-in-water

If you add a superbase to water in small enough quantity that it still makes sense to think of water as the solvent, all (or essentially all) of your superbase will be converted to its conjugate acid by the reaction
$$\ce{B-}+\ce{H2O} \to \ce{HB} + \ce{OH-}$$
This will have a very large equilibrium constant (by the assumption that $\ce{B-}$ is a superbase) and will also be driven heavily to the right by the high concentration of water. If you add enough of your superbase, eventually you can have appreciable amounts of $\ce{B-}$ sticking around in the solution, but at this point it no longer really makes sense to call the solution "aqueous".

A pH electrometer measures changes of [HO-] or [H3O+] ions (from that of neutral water) due to reaction of bases or acids with water.  The reactions are expected due to solvent levelling (i.e. H2O).  On the basic side, water levels to HO-, as a result of its own auto-ionisation process.
In another solvent, a given base, may exhibit a different ability for deprotonating the solvent molecules, which will to some extent be governed by that solvent's auto-ionisation process (which will differ from that of water).  Relative to its behaviour in water, this may be expressed as an increase (stronger basicity) or decrease (weaker bacisity).  Availability of the proton changes with solvent, and the resulting BH+ or (solvent)- may be stabilised by the system, or not.
pK's are solvent dependent.  For example, in the solvent DMF, picric acid is stronger than HBr, but the reverse is true in water.
While a "super-acid" is anchored to a species having a Hammett acidity of -12 or lower, a "super-base" is not anchored to base strength, but is more of an acknowledgement that acidity changes with solvent.  As defined, the notion of a superbase is misleading, since it is only a statement concerning structure (combine known bases to form a new base) or that such a species is a stronger base than in water (which it may, or, may not be, in varieties of solvents).
In short, there is nothing "super" about a given superbase, and one should not expect it to behave as a stronger base in general, although in some solvents it MAY be stronger, while in others, weaker.

In this example we use the concentration of [H+] Ions solvated by H2O to form the Hydronium ion.   Since [H+] also happens to be how we are defining the term acid in this example... The conjugate hydroxide Base [OH]- , the  [H]+  acid, and the solvent [H2O] are in equilibrium.   Because the [H+] in pure water should be 0, it has a pH of 7.  The hydroxide ion is the most perfect species to attack and disassociate H+.
When you change the solvent to a non aqueous solution, H+ may not be present...if you’re dealing with a bronsted laury acid.  In such cases it’s better to redefine acid and base in terms off a compounds ability to transfer a  proton to another compound (acid).  Hydrogen ions or hydroxide ions don’t have to be present to have strong non aqueous acids or bases. Like n-butyl-Li or antimony pentafloride.

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Is internal energy equal to (3/2)nRT for an ideal monoatomic gas?
I came across a Khan Academy video providing the following relation for the internal energy of the system $U$:
$$U=\frac32nRT,\label{eqn:1}\tag{1}$$
where $n$ is the amount of gas, $R$ is the universal gas constant and $T$ is the temperature of the system.
I know internal energy is a state variable and only its change can be measured, not the value at a particular state. Also, isn't \eqref{eqn:1} given for the average kinetic energy
$$\bar{E_\mathrm{k}} = \frac 32nRT,\tag{2}$$
not for the internal energy $U$? I'm confused.

Internal Energy is (3/2)nRT plus some constant that cannot be measured.

@IvanNeretin That constant can be approximated to be zero is it? Because in high school physics we just use 3/2 nRT.

"Approximated" is a wrong word. Let's say it can be assumed  to be 0.

Molecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves).
By saying "ideal", energy due to intermolecular interactions is eliminated.
By saying "monoatomic", energy due to rotation and vibration is eliminated.
So only translational kinetic energy and electronic energy remains.
(3/2)nRT is the translational kinetic energy, and since almost all atoms are in the ground electronic state at low temperature, it is a good expression for internal energy as long as the temperature is low enough that essentially all atoms are in the electronic ground state.

From a statistical standpoint, the mean energy of a system is given by
$$\langle E \rangle = E \cdot P(x) = \frac{\int\limits_{-\infty}^{+\infty} E \mathrm e^{- \beta E}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{-\beta E}},\tag{1}$$
where $\beta = 1/(k_\mathrm{B}T)$ and $P(x)$ is the probability of the system being at a particular energy $E$.
Now, if your energy dependence is quadric in some variable, this is if $E=ax^2$, where $a$ is just some constant, the mean energy becomes (thanks to some pretty cool math and Gaussian integrals)
$$\langle E \rangle = \frac{\int\limits_{-\infty}^{+\infty} {ax^2\mathrm e^{- \beta ax^2}}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{- \beta ax^2}} = \frac{1}{2 \beta}= \frac{1}{2}k_\mathrm{B}T.\tag{2}$$
You should note that this result is independent of $x$ and $a$. Actually, it is easy to show that if instead of one variable, the energy depends on $n$ quadratic variables, often called the modes of the system, each mode contributes the same amount of energy $k_\mathrm{B}T/2$ to the system (all you have to do is repeat the calculations for $\left.E = \sum_{i=0}^n {a_i}x_i^2\right)$ — this is known as the equipartition theorem. Hence, the mean energy of a system with $n$ quadratic modes becomes
$$\langle E \rangle = \frac{1}{2}nk_\mathrm{B}T.\tag{3}$$
In an ideal monoatomic gas the internal energy of the system $U$ is just its kinetic energy $E_\mathrm{k}$ (there is no energy due to vibration, rotation and intermolecular interactions), and therefore
$$E=E_\mathrm{k}=\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2.\tag{4}$$
The energy is the sum of three quadratic modes, so the internal energy of the system is simply (substituting $n$ by $3$ in our expression for $U$)
$$U = \frac{3}{2}k_\mathrm{B}T = \frac{3}{2}nRT.\tag{5}$$
P.S. This might have been a bit more calculus then you'd asked for, but it is the reason this expression exists. I always find it helpful to know where things come from, so I hope this helps you as well.

There seems to be an erroneous "n" in your very last expression. This answer does not seem to address the key point of the question, which is whether $U=3/2RT$ is literally valid, or whether it is actually $U=3/2RT+constant$ and only $\Delta U=3/2R \Delta T$ is valid (without making arbitrary assumptions about the energy reference level). Even for an ideal gas, I'm under the impression that the constant might be required.

@electronpusher Think it's more like missing an N for the second expression since he's talking about internal energy of the whole system with N particles. Earlier equations were for individual particles. Can't comment on the supposed constant though; what does it even describe?

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Can bronze be electroplated with zinc?
I had to have a motorcycle part cast in bronze that was originally steel. I want to electroplate it with zinc. Will this work? Thanks

Why would you need to or want to?

Bronze is mainly copper (with a bit of tin in it) which can be electroplated with Zn and then heated to form Brass (Copper-zinc alloy): https://www.stem.org.uk/resources/elibrary/resource/33571/zinc-plating-copper-and-formation-brass-%E2%80%93turning-copper-%E2%80%98silver%E2%80%99.
Also this question has been discussed extensively here: https://www.finishing.com/126/37.shtml.


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How to filter out sulfur dioxide
Lets say I have a container that contains a mixture of air and sulfur dioxide. I want to be able to filter out the sulfur dioxide from the air, how can I do this either through the form of a filter or a chemical reaction?

Just add your gas mix in a container with NaOH solution and shake well or bubble it continuously through the solution. SO2 dissolves in water to form H2SO3 (and H2SO4 in the atmosphere when it rains resulting in what we call acid rain) which will be neutalized by NaOH to give dissolved Na2SO3. Your air should be freed of most SO2. This is how it is done in the industry. There are many other ways as well but most rely in neuralizing the SO2. Google Flue-gas desulfurization for more details.


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Does Langmuir model predict evaporation rates?
Slide #16 of this presentation on amalgams says that "Langmuir's equation" provides a specific "theoretical maximum evaporation rate" for liquid Hg as a function of temperature and surface area.
The only common equation bearing Langmuir's name is his adsorption model.  I can vaguely imagine that model being applied in reverse to characterize evaporation rates, but I can't see or find how in general, much less in the specific case asserted in the paper.  Can someone provide a reference or example of how that (or any other model) provides a "theoretical maximum evaporation rate" of a particular liquid?

That most likely refers to the Hertz-Knudsen equation, also known as Knudsen-Langmuir equation. It is the most simplified model of evaporation in which no discussion is given into the diffusivity and the chemical potential gradient (let alone other driving forces like gravity and pressure gradient): one just assumes the leaving mass flux is dictated by the system's thermal energy. This model then suggests a direct relation between evaporation rate and vapor pressure, temperature and molar mass.


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Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement?
Background: Many organic reactions involve the migration of an alkyl group from one position to the adjacent one. The migration of the alkyl group is decided by its migratory aptitute i.e. electron-richness. It generally follows the priority order of hydride > phenyl > higher alkyl > methyl.
Main question: The Beckmann rearrangement also involves an alkyl migration. However, this migration is not governed by migratory aptitude. In fact, the $\ce{-R}$ which is in anti-position to the hydroxyl group in the oxime migrates, irrespective of its migratory aptitude! My question is why is this so? And are there other organic reactions which have such a rule for alkyl migration?
MasterOrganicChemistry  and Wikipedia don't even mention the word "trans" or "anti" anywhere. My textbook obviously doesn't mention anything either, hence I am driven to ask this question.

In the mechanism of Beckmann rearrangement, as  $H_2O$ leaves the nitrogen atom and adjacent carbon forms another bond with Nitrogen. At that point the migrating group forms a new bond with N leaving C. So, to form a strong covalent bond, the migrating groups orbital has to form a strong overlap with Nitrogen's orbital containing orbital. At the same time the orbital of Nitrogen which was overlaping with O before, has to now overlap with C. If the migrating group is at cis to $OH$ both the overlaps can't be done effectively. I think, this may be the reason.

@SoumikDas Thanks! Are you saying that if both the migrating alkyl group and the leaving $\ce{-OH}$ group are on the same side of the pi bond, then the overlap will be difficult to achieve, as compared to when the alkyl group migrates from one side and the $\ce{-OH}$ leaves from the other side?

Yes. I actually tried to say that.

Oximes can undergo the Beckmann rearrangement.  Oximes also exist as stable syn and anti isomers.  In the figure below, $\ce{R_1}$ is anti to the hydroxyl group; another isomer exists where it is syn the the hydroxyl group.
In the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group.  As the hydroxyl group begins to depart, the $\ce{R}$ group which is anti to the leaving group begins to migrate and form a bond with nitrogen.

(Wikipedia)
This step is said to be concerted (rather than stepwise) with the $\ce{N-O^{+}H2}$ bond being broken and the new $\ce{R-N}$ bond being formed at (more or less) the same time.  Whether you think of the $\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\ce{R}$ group as starting to bond to the available $\sigma^{*}$ orbital of the elongating $\ce{N-O}$ bond (equivalent descriptions), you see that having the $\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\ce{N-O}$ bond would be preferred. Hence, the $\ce{R}$ group anti to the $\ce{N-O}$ bond migrates preferentially.
There are many reactions showing this geometric preference, the Baeyer-Villiger oxidation being an example.


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Why we need halogens of different periods in wurtz-fittig reaction? + which is the correct mechanism?
For the mechanism of wurtz-fittig reaction I got two different ones on the net.
1. From another question   on stackexchange:

From tutorvista:

They are not really the same. In the first one the alkyl group is acting as the nucleophile and as Sn reaction do not occur at sp2 hybridized orbitals, it goes through an SnAr reaction. But in the second one the aryl group is acting as a nucleophile. So the same reasoning does not apply here, because here Sn2 is happening at a primary sp3 hybridized carbon whereas the aryl group is just the nucleophile. So is the second mechanism wrong or I am missing some point? Neither of the mechanisms really show why we used two different halogens in the alkyl and aryl group (or maybe I can't understand), in fact the first mechanism is using Cl on the alkyl group whereas the second mechanism is using iodine (which is a far better leaving group). But the aryl group has bromine on both. Really confusing if I am to search for explanations myself. Please clear me up.

"From another question of stackexchange: " Which SE question did you get it from? I am surprised how SN2 can take place on aromatic ring position. "it goes through an SnAr reaction." I was told that SN2Ar only occurs in presence of electron withdrawing groups like nitro on the ring at ortho/para to the halo group. The given benzene ring has none of it.

The asker actually asked that very question, that if an Sn2 occurs at that sp2 hybridized carbon. But the answers showed that it was an SnAr. I picked this picture, just to show the case where alkyl group acts as a nucleophile. Maybe I should have picked the picture from the answer. Anyways here's the link: https://chemistry.stackexchange.com/questions/81983/does-wurtz-fittig-reaction-involves-sn1-or-sn2


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Is it possible to form an alloy of gold and zinc?
I know you can dissolve gold in mercury, but i do not have access to any. Would it be possible to dissolve it in molten zinc instead mercury.

Of cause, if you melt both metals together you would get an alloy of both.

Yes. There are alloys and intermetallic compounds up to $\ce{44\%Au:56\%Zn}$.  This phase diagram is from here.

That said, crystal structure changes with composition, so appearance on solidification might be hard to predict.
Are you trying to emulate Hevesy?

No, my planned use for this is to process gold too fine to pan from black sand, without using expensive chemicals or mercury.

Interesting idea... it also avoids use of poisonous Hg. If practical, it might be necessary to blanket in inert gas, as Zn is reactive.

I not sure, but it will need a fair amount of sulfuric acid to disolve the zinc.   As for a inert gas CO2 should work.

@Brad not sure about CO2.  Could form CO+ZnO at elevated temp.  N2 is safer.

Gold looks pretty soluble in molten zinc. The melting temperature increases up to about 625 C with 50% gold.Charcoal on the surface of the melt will partly protect the zinc ; I have seen charcoal used on lead but not zinc.


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Why does cobalt have no negative charge?
I would like to know why cobalt cannot have a negative charge (or at least why a negative charge for cobalt isn't typical). I am not sure where I have gone wrong in my reasoning.
The electronic configuration for cobalt is:
$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7$
Clearly, you could give cobalt a $+2$ charge in order to get a more stable ion as follows
$\ce{Co^{2+}}$: $[\ce{Ar}] 3d^7$
And from there, you could go further and remove $2$ more electrons from the $d$-orbital to become even more stable like this:
$\ce{Co^{4+}}$: $[\ce{Ar}] 3d^5$
If this logic is correct, then could I apply this same thinking but for adding electrons? Such that
$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7\to$ $\ce{Co^{3-}}$: $[\ce{Ar}] 4s^2 3d^{10}$
Since the $d$-orbital has been filled, the atom should have become more stable. Is it true?
Thus, am I able to argue that Cobalt has $3$ main ion charges: $+2, +4, -3$?

Short answer: electronegativity.

+TAR86 So if you were solely to use electronic configuration (as I had done), you wouldn't be able to come to that conclusion.

Without a doubt there is at least one cobalt atom in deep space floating around with a -3 charge. So what?

So in this way, there would be $\ce{Ni^{2-}}$, $\ce{Fe^{4-}}$ etc. but that's not happening.

Cobalt can be found in various oxidation states including negative ones such as Co(-I) in NaCo(CO)4. In this instance however the excess electron density on the cobalt center is relieved by backbonding with the CO ligands and therefore such a low formal oxidation state of the Co atom is stabilized. On the other hand the free Co$^-$$^1$ ion cannot hold the extra electron first due to the low electronegativity of Co as TAR86 hinted out in his comment and second due to the fact that the extra electron is placed in a d orbital which is not very  penetrating and thus shielded by all other electrons. Hence, it is unlikely to encounter compounds such as [Na]$^+$[Co]$^-$ where there are no ligands on the cobalt to stabilize the excess electron density. The opossite applies when you remove electrons from Co (and any other transition metal): the atom becomes positively charged and the nucleus attracts the remaining electrons even more effectively (Higher Zeff). Therefore, positive oxidation states are more common.


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How to determine the pH range at which maximum hydrate is present?
Question:

Find the $\mathrm{pH}$ range at which maximum hydrate is present in a solution of oxaloacetic acid (given $\mathrm{pK_a}$ of $\ce{-COOH}$ in the left is $=2.2$ and on the right is $=3.98$):

I understand that the hydrate will be formed at the carbonyl group position, resulting in a product like this:

I know that the gem-diols are usually stablized by H-bonding, as is the case here as well (between the lower hydroxy and left acid group). I can also see that loss of the $\ce{H+}$ ion from the carboxylic acid group on the left will create an $\ce{O-}$ ion which will have an even stronger H-bonding. However, I am not sure how I can use this fact to relate the given $\mathrm{pK_a}$ values and the expected $\mathrm{pH}$ range.
I need help completing my train of thought. Thank you!

https://chemistry.stackexchange.com/a/42796/9961

@Mithoron That question doesn't have $\pu{pH, pK_a}$ nor does it have a hydrate alongside carboxylic acids. It's only slightly related, due to both question discussing formation of a stable hydrate.

Your thought about H-bonding stabilization is leading you astray. Ron's answer ^ says what's important for diol - ketone equilibrium.


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Naming tetrahalogenated benzene

ChemSketch generates the IUPAC name of the compound above as "2-bromo-1-chloro-5-fluoro-3-iodobenzene" but I have a different answer.
I know that I have asked a related question before as well. The key takeaway from orthocresol's answer was that:

first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants

I have applied the same here: the prefixes ordered in alphabetical order are "bromo, chloro, fluoro, iodo". Now, I have to choose a locant set that minimizes the locants. A possible locant set is "1-bromo-2-chloro-4-fluoro-6-iodo" which correctly locates the substituents. Also, the locant set "1246" is lower than "2153", as noted in this answer. So, according to me, the preferred IUPAC name should be "1-bromo-2-chloro-4-fluoro-6-iodobenzene".
What exactly am I doing wrong here?

The locant set ‘1,2,4,6’ is not lower than ‘2,1,5,3’.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the lowest locant set is defined as follows:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)

Especially note that the locants are cited in order of increasing value when they are compared.
Thus, the locant set ‘2,1,5,3’ is regarded as ‘1,2,3,5’, which is lower than ‘1,2,4,6’.
Therefore, the correct name is 2-bromo-1-chloro-5-fluoro-3-iodobenzene.

Thanks that'll do! I understand OC is supposed to be systematic but I don't understand why they couldn't make rules easier to follow in a simple language.

@GaurangTandon Hm, I checked the previous definition in the 1998 IUPAC Guide Principles of Chemical Nomenclature, but it isn’t much different: “(…) Lowest locants are determined by comparing alternative sets of locants. When compared term-by-term with other locant sets, each in order of increasing magnitude, a set of lowest locants has the lowest term at the first point of difference; (…)”

I didn't mean to refer to the previous edition of nomenclature. I meant to say that these nomenclature guides are written in a complicated manner that presents a steep learning curve for beginners to master it.

@Gurang Tandon it as such because uniquely name  every single molecule in a virtually infinite set of molecules is a big task :)


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Voltaic cell and Nernst equation

It is given voltaic cell
  $\ce{Cu}|\ce{CuSO4}(\pu{10^-3 M})||\ce{CH3COONa}(10^{-2})|\ce{Pt}(\ce{H2})(\pu{1 atm})$.  Find the potential at $\pu{50^\circ C}$.
  Given $K_\mathrm{a}$ for $H= 1.8 \times 10^{-3}$.
  $E_{\ce{Cu^2+/Cu}} = \pu{0.336 V}$

When voltaic cell starts working: $\ce{H2}$ becomes $\ce{2H+}$. In the other hand $\ce{CH3COO-}$ hydrolizes forming $\ce{CH3COOH}$ and $\ce{OH-}$. We know $K_\mathrm{a}$ and we can find $K_\mathrm{b}$. And from that we can find the concentration of $\ce{OH-}$ and after that using the formula $\mathrm{pH}=14-\mathrm{pOH}$  we can find the concentration of $\ce{H+}$.
Now how we can find the concentration of $\ce{Cu^2+}$ in order to replace it into Nernst equation?
\begin{align}
  E &=E^\circ - \frac{RT}{nF} \cdot \ln\{\text{Qc}\}
  \text{, where}&
  \text{Qc} &= \frac{[\ce{H+}]}{[\ce{Cu^2+}]}.
\end{align}

A correct standard notation for the galvanic cell would also include the physical state of each species involved. As such, the correct notation for the given cell is:
$$\ce{Cu(s)}~|~\ce{CuSO4(aq)}~(10^{-3}~\mathrm{M})~||~\ce{CH3COONa~(aq)}~(10^{-2}~\mathrm{M})~|~\ce{Pt(s)}~(\ce{H2(g)})~(1~ \mathrm{atm})$$
Recall that aqueous solutions of salts are generally accompanied by their molarity in solution. Also, metals are usually solid, while gaseous species are accompanied by a reference electrode (platinum in this case) and their pressure value.
In your case, notice that the ionic salt $\ce{CuSO4}$ is assumed to be a $100\%$ dissociated ionic salt, hence, its aqueous solution will produce $[\ce{Cu^{2+}}]=[\ce{SO4^{2-}}]=10^{-3}\mathrm{M}$. That should answer your final query.
Apart from that, your expression for $\mathrm{Q_c}$ is wrong. It should be $\displaystyle\mathrm{Q}=\frac{p_{\ce{H2}}}{[\ce{Cu^{+2}}]}$. Since $\ce{H2}$ is present in gaseous form, use its pressure value in the expression for $\mathrm{Q}$  (and not molarity which doesn't exist)
PS: The concentration ($\pu{10^−3M}$) that's given to you is the concentration of copper ions after equilibrium has been achieved for the interconversion of $\ce{Cu+2(aq)}$ and $\ce{Cu(s)}$.


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Does 95.8% silver tarnish as fast 92.5% silver?
I've been trying to find out whether 95.8% (Britannia silver) would tarnish as fast as 92.5% (Sterling silver)? I haven't been able to find much on tarnish for Britannia silver.
Given that both silvers are kept in the exact same environment, would the Britannia silver tarnish at the same rate as Sterling silver? If not which of the two tarnishes quicker?

I suspect they are the same; However, I think the only way to know is to test or find a reference .  For example ; adding  0. 3 % Cu to steel makes a measurable increase in atmospheric corrosion resistance , which I doubt could have been deduced .

If you look here: https://blog.centimegift.com/does-sterling-silver-tarnish/, it says that pure silver is very tarnish resistant. Sterling silver on the other hand being mixed with some Cu to enchance the metal's mechanical properties tarnishes as a result of the Cu in it. Therefore, considering that the amount of Cu in Britannia silver is smaller, it should tarnish more slowly than sterling silver. In practice i dont know if there will be a significant difference as the increse in silver purity is not that much.


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Intensity of “surface plasmon resonance “ absorption peaks for a nano-rod
The SPR absorption spectrum of a nano-rod is seen to have two peaks with the longer wavelength peak corresponding to the absorption by the longitudinal plasmons and the other to the transverse plasmons. When the intensities are compared.. longitudinal plasmon peaks are almost always more intense than transverse ones. Why?

this clearly belongs to physics SE

The topic is taught as the part of a course that addresses nano sciences. Nano sciences being inter disciplinary and my being a chemistry graduate sure does make this relevant to the chemistry SE too.

@permeakra I disagree, I think this belongs equally well on chemistry; the language of physicist in nano is different from that of chemists.

The intensity of a plasmon resonance is related to the magnitude of the transition dipole moment. Longitudinal resonances have a larger transition dipole moment compared to transverse resonances, and so have more intense absorption peaks. You can rationalize this by imagining the change in dipole moment from a longitudinal versus transverse electron density oscillation.

I did think about that. But the catch is , the ratio of the intensities remain unchanged, even when the lengths of the nano rods are increased , keeping their diameters the same. The intensity seems to be dependent on something more in addition to the magnitude of dipole moment change..


text:
Chemical tests given by glucose (biomolecule)
We know that glucose gives Tollen's test because it's a reducing sugar, and having a hemiacetal group, on hydrolysis can be converted into an aldehyde group, giving the positive test.
But, when we come to tests like the Schiff's test or with sodium bisulphite, it does not show any reaction which are shown by aldehydes. Why? Is the answer that hemiacetal group cannot be converted into aldehyde or anything else? Probably there shouldn't be any water used in these reactions, so no hydrolysis, no conversion from hemiacetal to aldehyde. Is there any water present in Tollen's reagent?

Stronger oxidising agents will open the ring and oxidise the glucose but the mild oxidising agents won't do it. As it has hemiacetal group Schiff's base and sodium bisulphite doesn't react with glucose as they are mild oxidizing agents. I guess stearic repulsions may also be the reason.

This are two different types of reactions.
Tollens is a redox reaction in which silver is reduced and glucose is oxidized.
This reaction depends on the redox potential of glucose only.
Schiff and bisulphite tests are addition reactions. The reagents condense with the aldehyde group.
This occurs also in the case of glucose but it is not sufficient to give a positive result.
In the case of bisulphite, the adduct must precipitate, but it is too soluble.
In the case of the Schiff test, the adduct must dehydrate, but it forms the glucosamine derivative instead.


text:
Basicity of nitrogen heterocyclic compounds

Question :
The order of basicity for the following compounds is :

The logic I've applied is that since 4 has a conjugated system, the lone pair of N will be delocalised and hence would be the least basic. Conjugate systems/Resonance is not seen in the other three compounds
I'm not entirely sure about how to differentiate between the other three. For morpholine 2, If we see the stability of the conjugate acid formed, the O would exert (very less) -I effect, which should destabilize the positive charge on N (on formation of conjugate acid). Now between 1 and 3, 1 has a five other C atoms attached to the N, while 3 has four. Now CH3 is a +I group, hence more CH3 in 1 should stabilize the positive charge on N more than 3.
I don't think Inductive Effect should matter much, considering Inductive Effect becomes very ineffective with increase in chain length. But this is the only reasoning that strikes me. So ultimately, order of basicity becomes 4 < 2 < 3 < 1. Is my reasoning and the final answer correct?

See nitrogen inversion that's a PDF. Inversion will be less in a five membered ring.

Thanks for the link, but how is nitrogen inversion related to the question? Could you elaborate more on how inversion or less inversion affects basicity?

Related (but not a duplicate): https://chemistry.stackexchange.com/q/89880/5026

These are piperidine and pyrrolidine. Look at their pKa values. Do you really expect to explain a difference that small on the qualitative level?

@skb You didn't see my wonderful PDF if you had then you wouldn't be asking me to elaborate :(

@IvanNeretin That's weird, I wasn't aware that. I'm extremely sorry if this was a meaningless question.I asked this question here because I got it in one of my tests (I'm in grade 11).

@skb That's not your fault at all.

@AvatarShiny Hey I just read your "awesome" PDF. Correct me if I'm wrong, but essentially more inversion means more delocalisation of lone pair electrons of N, right? So due to more ring strain in the five membered ring, inversion would be less; delocalisation of l.p electrons would be less (as compared to (I)) and hence (III) should have been  more basic than (I), right?

@AvatarShiny But then how does (II) fit into the inversion model? How would we differentiate between (I) and (II)?

Inductive effects $$\ce{3>1>2>4}$$

Since on bases of conjugate acid 4 is more acidic and 2 is less so we get that 4>1>3>2

Nitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.

The PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.
Inductive effects explain why morpholine 2 is less basic, and resonance effects explain why pyrrolidone 4 is even less basic. However, since the pKaH's of piperidine 1 and pyrrolidine 3 are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.
Simply looking at the data, though, the order you want is 4 < 2 < 1 < 3.

To compare between 1 and 3, let's take into account the interaction between the non-bonding electron pair on nitrogen and the anti-bonding MO of the C-H bond (σ*C-H).
In order for the interaction to be effective, the lone pair on nitrogen and the C-H bond must be planar. In the six-membered ring, there is always a C-H bond next to the nitrogen lays in the parallel position against nitrogen's lone pair, which is the axial one. In the five-membered ring, the C-H bond does not totally lie on the same plane with the lone pair of nitrogen, so the interaction is less effective.
This interaction decreases the energy of the non-bonding MO, which consequently decreases the basicity of the nitrogen. As this interaction is more stressed in the six-membered ring, its basicity is less then that of the five-membered ring.
Basicity: 4<2<1<3

I would like to answer it,
KEY POINT: More electron density on N, more availability of the lone pair, more good base.
$(1)$ The nitrogen is directly connected inside the ring, surrounded by lots of carbon atoms. N being more electronegative than C, snatches the electrons towards itself as the nearest carbon atoms shows +I effect, it drops down as we move away from N. So, electron density on N increases, which means the lone pair is more available for donation. Hence, it's a good base!
$(2)$ The ring is the same but this time, a highly electronegative O came now. As it is more electronegative than N, it will snatches the electrons towards itself from the downside ring. Thus, decreasing the electron density on N. Hence, it's now no more willing to donate its lone pair. It is more tightly bound to the N as the repulsions from the electrons around it are decreased. Hence, not so good base.
$(3)$ Everything still the same, but the number of carbon atoms reduces to $4$ around it. Hence, +I effect reduces compared to $(1)$. So, electron denisty reduces comapred to $(1)$. Basic character reduces compared to $(1)$.
$(4)$ Less basic than $(3)$ as N got a $\ce{sp^2}$ hybridized carbon around it which is more electronegative than the previous. Hence, that +I effect reduces. So, a weak base than $(3)$ . But, the O here isn't connected directly to the ring. So, it will not effect any other C atom to perform their +I effect. Hence, the density on N reduces a bit, whereas in $(2)$, it reduces a lot as the O here will effect every C to not to give their best at +I effect, it will also snatches electron density around N a bit. Hence, it's the least Basic. So, correct order:
$\ce{(1)>(3)>(4)>(2)}$
Actually I do not at all like chemistry, but i don't know why i posted this as my ever first unexpected answer on this site. This is probably because this topic is my favourite by mistake from my class $12$ chapter: Amines.

Thanks @Selena, really appreciate the effort :) I do have a few doubts, though. Firstly, isn't the effect of Resonance more dominant than Inductive effect? There would be resonance in (4) which I believe should be the more dominant factor as it literally delocalises the lone pair of N. Which is why (4) should have been the least basic. Now coming to differentiating between (1) and (2) and (3), the Inductive Effect logic seems pretty rational but the thing is Inductive Effect is a very weak effect and its effect decreases with increase in chain length.

Yeah, that is also the case, i forget to add in answer. The lone pair on N delocalizes to give a + charge on N, hence the lone pair is less available. So, it's the least basic.  And......, The basic strength of first three compounds can be understood only due to inductive effects...that's a pretty good reason....

"So, it's the least basic" please edit to your final ordering of bases in the answer to reflect that.


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Reason behind conversion of acid to acid bromide in Hell-Volhard-Zelinsky reaction
Why in Hell-Volhard-Zelinsky (HVZ) reaction the acid is first converted to acid bromide by using $\ce{PBr3}$ and then alpha halogenation is carried out?

How do you think the alpha-bromine is incorporated into the acyl bromide? Homework???

Google is your friend!

The $\alpha$ halogenation occurs through the enol form.
The formation of the acyl halide is key because carboxylic acids do not form enols readily since carboxylic acid proton is removed before the $\alpha$ hydrogen. Acyl halides lack the carboxylic acid hydrogen.
The phosphorus halides form the acyl halides readily because phosphorus has a huge affinity for oxygen.
Sometimes, phosphorus tribromide is not enough to form the acyl halide but, since there is also bromine in the media, phosphorus pentabromide is formed which can form acyl halides even of solid acids.
The halogenation is performed by the $\ce{X2}$, either bromine or chlorine.
When $\ce{PBr3}$ and $\ce{Cl2}$ is used, the alpha-chloro acid is obtained and not the bromo acid.

This was a question where AJ needed to make some effort in offering a solution.

@user55119 Understood. However, I hope it will serve others as well.


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How to explain the products of the nucleophilic aromatic substitution of 1-chloro-3-methylbenzene with NaOH?
I guess this reaction is a nucleophilic aromatic substitution where $\ce{ONa-}$ is the nucleophile $\ce{Cl-}$ the leaving group, is that right?

Methyl group is a ortho-, para- director. So why should I have a meta product here? Don't we need an electron withdrawing group ($\ce{NO2}$ , carbonyl) in a NAS-reaction which is at ortho or para positions to the leaving group?
I am a bit confused about the shown picture.

Could be because of ipso substitution. This link might help.

thank you for the link. the ipso subst leaving group seems to be at ortho or para positions as ı mentioned. but in the reaction ı provided it is at meta. that is what is confusing me

What is going on here is a benzyne process. Under the conditions of intense heat the hydroxide removes one of the protons either side of the Cl. Cl- then leaves to create a triple bond, giving a benzyne. As you might imagine this is a highly reactive species and reacts with the OH- with no regioselectivity; either end can get hydroxylated and you have two possible benzyne intermediates. Thus you get the mixture of products.


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Are certain logP values eluted when 100% methanol is used as the solvent in a C18 column?
I have used a C18 sep-pak to concentrate metabolites in a chemically complex solution. The column was first washed with pure water, 75% methanol:water mix, and finally the compounds eluted with 100% methanol.
What I am wondering is what chemical properties that I can infer from this information? I know I would have washed the very polar compounds away, and in pure methanol, I would expect compounds of a semi-polar nature to be eluted. Can I assume that these compounds would have a specific LogP range? I cannot find an answer online anywhere, so I guess the answer must be no, and if so, why not?
There is one (or more) compound of a biologically active nature in this complex mix. I am wondering if I can use the elution conditions to understand some of the chemical properties of the compounds eluted. Further, I did FT-ICR-MS on this mix, and I would like to look at a list of compound hits from the mass spec databases and understand how feasible a particular hit is based on its polarity.

HPLC is often used to determine lipophilicity in industry, as the original methods to determine logP by partitioning between octanol and water was not particularly amenable to high-throughput assay.
Lipophilicity values measured by HPLC are often reported as cLogP / cLogD (where c is relating to the fact that the value has been determined by some form of chromatography).
Experimentally, a C18 column is used with an acetonitrile/water solvent system, buffered appropriately. Unlike with the octanol/water procedure, different companies run these HPLC assays differently, and as such they aren't as comparable (i.e. if you ran 20 compounds through a GSK assay, and the same 20 compounds through a Pfizer assay, the precise values for cLogD obtained in each case would vary, even if the general trend remained the same).
In terms of you doing this yourself, you'd first need to find a training set of compounds with known and reliable values. If you know paracetamol comes of at x mins and has a  literature logP of y, and ibuprofen comes off at z mins and has a literature logP of w, you can interpolate. When you then run an unknown sample, its retention time using your method can be used to predict the cLogP/D (keep in mind that you'd need a far greater training set than 2 compounds).

Thanks so much for your reply. So because I am just using the disposable C18 sep pak, and not HPLC, it sounds like I am out of luck when it comes  to drawing conclusions regarding the nature of the molecules that were eluted.

Oh sorry. I never even made the connection that C18 might be anything other than a HPLC column!  But yes, you'd be unwise to draw too many conclusions from that.


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Biomolecules : Enzymes
My class $12$ board exams are approaching and i'm having a doubt in this question:
This is from my class $12$ NCERT exemplar book

Activation energy for the acid catalysed hydrolysis of sucrose is $\pu{6.22 kJmol^-1}$, while the activation  is only $\pu{2.15kJmol^-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain

I've found nothing so good from the web. All I know is this:
Enzymes, the biocatalysts, work even under mild conditions, they do not require elevated temperatures or other physical conditions. While with ordinary chemical catalyst, this might be the case. Hence, activation energy for chemical catalyst reaction is little more than enzyme catalyst.
Is this reason valid? Or is there any other reason?

I'll just quote the official answer given in the NCERT Exemplar book. It reads:

Enzymes, the biocatalysts, reduce the magnitude of activation energy by
  providing alternative path (my emphasis). In the hydrolysis of sucrose the enzyme sucrase
  reduces the activation energy from $\pu{6.22 kJ mol^–1}$ to $\pu{2.15 kJ mol^–1}$.

The key thing to note here is an alternate path is provided to the reaction. The values given in the question ($6.22$ and $2.15$) are only to distract you from the systematic answer. You are not supposed to rote learn them. Focus on the lowering of activation energy due to an alternate pathway. Here is a graph depicting such a situation:
 (source)
The mechanism how the catalyst exactly achieved this lowering of activation energy, that too specifically for sucrose, would be difficult to describe unless this is a research paper, and is surely not required from a 12th grader. All that you are required to answer is what I put in the yellow blockquote above.
I hope it helps!

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

Well, in general, no matter if biocatalyst or homogeneous or heterogeneous catalyst in organic chemistry or the industry the role of the catalyst itself is to lower the activation energy of the reaction. Often people have two mistakes when they think about that. One is that they think, that the catalyst is not changed in the reaction and the other one is that the catalyst can change any thermodynamic energy level or so.
About the first point, the catalyst takes part in the reaction. It is true that in general, it leaves the reaction as it entered (there is decomposition but we ignore that for a second) but inbetween the catalyst will take part in the reaction itself. And therefore we are speaking of a completely different reaction at all, which leads to the second point. The energy level of starting material, transition state (so activation energy or reaction barrier) and product is set. The catalyst does usually not influence this. Therefore you have to cross another reaction path.
For example, let's take the hydrogenation of ethylene, you add $H_2$ to ethylene to create ethane. If you do this with a metal catalyst on the metal's surface you will end up in an activation of the hydrogen itself fist. So the hydrogen molecule adsorbs on the surface of the metal and the bond is then cleaved. And those two separated hydrogen atoms can now hydrogenate the ethylene, which also adsorbs to the surface. So metal can help here and lower the overall energy of this reaction. The new path consists of several steps like coordination, cleavage, reaction and desorption from the catalyst. And if you choose this correctly then those energies are lower than for the uncatalyzed one.
And in biocatalysis, it's basically the same. You can check for example the catalytic triad to see how many amino acids and the different sites in the catalytic center of the enzyme help to perform the reaction.
About your specific reaction if have no idea, I am no biochemist, this is the only thing I could find about that.


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Most activated position on para-terphenyl for EAS
Para-terphenyl: it doesn't look pretty with all those math-y numbers, but those are going to come helpful in answering my question!

A question asked me to tell the expected product when this reacts with $\ce{Br2/FeBr3}$. Now, I have done such questions with all types of fancy organic molecules ranging from benzene to picric acid to parantiroanisole etc. I know that I am supposed to find the most activated position1 for EAS. But I have never done more than one phenyl ring at once.
So, in this question, I have drawn the eight resonating structures (without charge separation) of paraterphenyl manually. Here they are:

But, as you can see, every position seems to be activated identically. Each position has their own share of electrons through a pi bond, but it ends up being completely symmetric.
So, how do I find the most activated position in this compound?

1:Clarification: By "most activated position" I mean the position with maximum electron density. For example, in aniline, those are the ortho and para positions.

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

You need to draw the resonance structures with charge separation. They contribute to the electron distribution as well.

So, how do I find the most activated position in this compound?

Start by drawing resonance structures of the various possible intermediates (sigma complexes) formed when $\ce{Br^{+}}$ attacks the different ring positions in p-terphenyl. Whichever intermediate has the most resonance structures is likely to be the most stable (lowest energy) intermediate. The lowest energy intermediate will have the smallest activation energy and consequently be the kinetically favored product (barring steric effects).
I've drawn just a few of these resonance structures in the following diagram.

In the top row I've drawn a few of the possible resonance structures for electrophilic attack at the para position on a terminal phenyl ring.  You can draw many more resonance structures delocalizing the charge around the various rings, but notice that it is possible to delocalize the positive charge over all 3 aromatic rings. In the second row I've drawn a few resonance structures showing charge delocalization when electrophilic attack occurs in the center ring.  Importantly, in this case charge can only be delocalized over 2 of the aromatic rings.
This suggests that based on resonance effects, electrophilic attack on a terminal benzene ring would be preferred over attack at the central benzene ring.
Further, we would expect ortho and para attack on the terminal rings to be preferred since the phenyl substituent acts as an activating o-p director in electrophilic aromatic substitution. The rate of attack at the ortho position might also be decreased somewhat due to the steric effect of the adjacent phenyl (actually biphenyl) substituent.
In a paper by Shafig and Taylor (J. Chem. Soc., Perkin Trans. 2, 1978, 0, 1263-1267. DOI: 10.1039/P29780001263), they explore your question by studying the electrophlic protonation of tritiated p-terphenyl.  After correcting the rates for statistical effects (e.g. there are twice as many ortho positions on the terminal rings as there are para positions) they find that:

the para position (your positions $10, 16$) reacts at a relative rate of $273$
the meta position (your positions $9, 11, 15, 17$) reacts at a relative rate of $1.54$
the ortho position (your positions $8, 12, 14, 18$) reacts at a relative rate of $176$, and finally,
the $4$ identical positions on the middle ring (your positions $2, 3, 5, 6$) react with a relative rate of $59.1$

These results support our predictions that

attack on the terminal rings will be preferred over attack on the central ring and
that attack at the ortho and para positions is preferred over attack at the meta position on the terminal rings.

Does this mean that all the three rings would be planar in the intermediate?

@ApoorvPotnis  Yes (or at least close to coplanar, remember that p-orbitals can twist +- 20° while still maintaining significant overlap), at least when substitution occurs on the terminal benzene ring.  When substitution occurs in the middle ring, then only 2 rings need to be close to coplanarity.

Electronic structure, it is indeed almost planar at room temperature http://www.tandfonline.com/doi/abs/10.1080/00268977900101021


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Directionality of ring current-induced magnetic fields in purines

What would be the directions of ring current-induced magnetic fields in purine-like heterocycles in a strong magnetic field?
Would be the antiparallel state be a default state for these substances in the absence of strong magnetic field at warm temperatures due to heat motion?

This is an interesting problem, I think it is best for a moment before I get stuck in that we considered a slightly different compound. I will consider it with an external magnetic field applied to it. Napthalene, now napthalene has two aromatic hydrogen environments in its NMR spectrum. These are very similar in chloroform to benzene in terms of chemical shift.
This makes me think that the ring currents go in the same direction in this compound.

It may be better to consider the ring current as flowing along the perimeter of the aromatic system rather than the loops formed by the six membered rings.

It is important when considering aromatic rings to escape being trapped by six membered rings. A lot of aromatic molecules have six membered rings but there are plenty of five and seven membered aromatic rings. It is more important to understand that a pi system which is cyclic and has 4n+2 pi electrons is the defining thing which makes a ring system aromatic.
If napthalene had ring currents going in different directions I would expect that if the interconversion of the sites (by ring current reversal or flipping of the molecule) was slow then I would expect to see four different chemical shifts in the proton spectrum.
If this exchange of sites was fast then I would expect to see two sites which are each the average of the two sites. I would also expect the chemical shifts of the napthelene to be very different to that of benzene.
If we allow ourselves a little diversion for a moment, the cyclopropanes have a rather rare proton chemical shift, this is becuase the C-C bonding in cyclopropane has some pi character and the hydrogen nuclei are above the ring in such a place that the ring current has the opposite effect on the NMR shift to the magnetic field due to the ring current in benzene.

Now if we move onto purine, in case anyone is not sure what the shape of it is then here it is.

If we look at the literature (S2-75) in the supporting literature of Thomas H. Graham, Wensheng Liu, and Dong-Ming Shen, Org. Lett., 2011, 13 (23), pp 6232–6235
DOI: 10.1021/ol2026813. This compound in DMSO has three very similar proton NMR shifts. This suggests that the ring current goes the same way around in both rings. These NMR shifts are all in the aromatic area and are further left (higher ppm) than benzene.


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Finding required drug concentration given Kdiss

The enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9 mol/l). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7 mol/l).
Compute the total concentration of [D1]tot that is needed to bind 90% of the HIV-RT at the given concentration [E]tot. You do NOT have to consider competition betwwen the drugs D1 and D2! They are administered separately.

I understand how to calculate the concentration when [E]eq = [ED]eq when 50% of [D]eq is bound to HIV-RT because Kdiss = [D]eq and it makes for a simple calculation because [E]eq and [ED]eq cancel out when calculating Kdiss, but what changes mathematically need to be made to account for the change in ratio (90% rather than 50%)? Any help would be appreciated.

Chemical reaction is
$$\ce{D1 + E <=> ED1}$$
Initial concentrations: $[\ce{D1}]_\textrm{tot}; \quad [\ce{E}]_\textrm{tot}; \quad 0$.
If $X$ percent of $E$ "reacts", then stoichiometry tells us that:
Concentrations at equilibrium: $[\ce{D1}]_\textrm{tot} (1- X) ;\quad [\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X;  \quad [\ce{D1}]_\textrm{tot} X$.
The equilibrium expression is
$$K_\textrm{diss} = \frac{ [E][D_1]}{[ED_1]}$$
Therefore
$$K_\textrm{diss} = \frac{ [\ce{D1}]_\textrm{tot} (1- X) ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)}{[\ce{D1}]_\textrm{tot} X}$$
Which we can simplify to
$$K_\textrm{diss} = \frac{(1- X)}{X} ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)$$
So you can solve this for $X = 0.90$, $[\ce{E}]_\textrm{tot} = 2.00 \ \mu M$ and $K = 1.00 \ nM$.


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How can Phosphate be charged at -3?
I am studying chemistry in High School and I am studying my ions(Because I failed too in the beginning, oops) and I have a question on this reasoning: In Phosphate(PO4), it has a charge of -3. The only way for this to make sense is if half the atoms of oxygen are positive and half are negative bring a positive charge of 4 and negative charge of -3 along with phosphorous being -3. Meaning that there is a total of Negative charge of 7 and positive charge of 4 leaving the total charge of the ion of -3. How come this happens?

The Wikipedia page about phosphate describes all 4 protonation states of phospahe: 0, -1, -2 and -3, you may want to take a look: https://en.wikipedia.org/wiki/Phosphate

Cryptic your words are, young padawan.

Sorry? What if the oxidation state of P was +5 and each oxygen had -2 like we usually expect it? I bet you'd end up with a total charge of -3 on the phosphate as well.

All the oxygen atoms have the same oxidation number $-2$. The charge balance then tells that the phosphorous atom has be at a state $+5$. That's the entire story.


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How to find chemical market value
I would like to retrieve pricing information for specific chemicals. Is there a centralized resource for this?
Edit: I want to clarify that this type of data is to conduct a techno-economic study. I know that for certain markets there are centralized databases which collect prices, and report price statistics, I was wondering if there is something similar for the chemical industry.

https://www.sigmaaldrich.com/united-states.html

No, you should pick a supplier. How much something costs can be widely different depending on the supplier.

You need to be specific when you look for a vendor.  For example ,reagent grade Na Cl has a very different price from a box of salt from the grocery store, very different from a truck load of food grade salt. very different from a rail car of road salt ( Na Cl ).

I am thinking about the bulk market. If someone is to develop a new chemical synthesis process, which products should they target? The market value is an important consideration when answering that question.

Regarding the opinion based hold: I think my edit should clarify that this is not an opinion based question. Perhaps such market tracking system does not publicly exist (I am certain all the major chemical companies dedicate plenty of resources to study the market).

Until 2005 or 2006 the best source was Chemical Market Reporter.
It had the form of a newspaper and was edited once a week.
The last pages listed market prices for commodities, some speciality chemicals and also agricultural products.

Hundreds of products were listed and you had some idea of what were the options you had for designing an industrial process.
You could easily decide which solvent to use for an extraction, which metal for a reduction or which raw material for a synthesis.
The publication was on subscription base but it was available in some University libraries or you could get an older copy from a contact in the big chemical industry.
Now this has been converted on a pay per product information scheme.
You can enter in www.icis.com and ask a report for one product.
If you want to compare prices you have to buy the information for every product.
It is almost impossible to have a complete picture of the economic trend in the Chemical Industry.

Sigma Aldrich was mentioned, and is a very good source of relative pricing; it can help eliminate certain possibilities when you see compounds listed in 1 gram quantities vs 1 kg or more.
Another item is the intended size of purchase and perhaps continued purchases. Prices in bags, pallets (2000 lb.) or pails, drums, totes or truckload (20 tons) will vary widely due to bulk discounts, distributor markup and freight.
Another item to consider is purity: technical grades may have different degrees of purity or different contaminants depending on suppliers.
Another item is availability: overseas suppliers may have warehouses in the country, and may keep an item stocked. Or not, which requires careful planning and internal stocking, which costs money.


text:
What exactly is the mechanism of "nascent hydrogen"?
I put the quotes because Wikipedia says it's an outdated theory.
What exactly happens when nitrobenzene is converted to aniline using $\ce{Sn/conc HCl}$?
According to a few books which haven't been updated, even $\ce{Na}$/ethanol, $\ce{Zn/dil H2SO4}$ produce this. I guess a metal may have an important role in the actual mechanism.
Is there any specific reason why  $\ce{Na}$/ethanol is used to reduce cyanides to amines, and $\ce{Sn/conc HCl}$ to reduce nitrobenzene? (and not the other way around)
Even $\ce{KMnO4}$ produces "nascent oxygen"? What exactly is going on?

Which question are you asking? Currently there's three ? marks in your post (excluding the one in your title) :( As for the middle one, I was taught that tin+conc HCl will reduce cyanide to amine, but then further reduce it to aldehydes because it is in an aqueous medium. A similar thing won't happen with sodium/ethanol (alcoholic medium). I am not sure of the exact mechanism though.

https://chemistry.stackexchange.com/questions/42549/acids-generating-nascent-gas-atoms https://chemistry.stackexchange.com/questions/24726/aqua-regia-and-nascent-chlorine?s=2|45.1855


text:
How is it that I could obtain endothermic Temperature ~ Time plots for a calculated Exothermic Enthalpy of Solution?
I recently conducted a coffee-cup calorimetry experiment in which we attempted to measure if the dissolution of inorganic crystals such as LiCl, NaCl, and SrCl were endothermic or exothermic based on a change in temperature over time plot, including measure the enthapy of solutions, but I'm not sure how to do that.
Methods:
We measured 7.9204 grams of SrCl2, 4.2143 grams of LiCl, and 2.9290 grams of NaCl and decanted these into separate coffee cups filled with 50 mL of deionized of water. (These were about 1 - 2 mole/liter solutions.). A temperature probe measured the temperature change over 600 seconds, recording a measurement twice a second. I've only displayed about 140 seconds of the measurement, because the temperature was just continually oscillating 0.1 degree Celsius after that point as it reached equilibrium.
We made plots for each of the chemicals, where an increasing curve shows an exothermic reaction, and an endothermic reaction displays a decreasing curve. Over time, these solutions (Temperature ~ Time) curves should equilibriate to room temperature.
Theoretically, we predicted we could determine if the enthalpy of solution like so ...
Enthalpy of solution = Lattice Enthalpy + (Hydration Enthalpy of cation + Hydration Enthalpy of Anion),

where the Lattice Enthalpy is a measure of the stability of the compound crystal which is at least a function of the charges of the cation and anion, as well as their size, and the structure of the crystal body centered cubic vs face centered cubic, and the hydration enthalpy is the energy it takes to dissolve the gaseous ions after breaking the crystal lattice.
Schematically, the Lattice Enthalpies ( Δ H):
LiCl(s) ----> Li+ (g) + Cl- (g); ΔH = 852 kJ / mole
NaCl(s) ----> Na+ (g) + Cl- (g); Δ H = 787 kJ / mole
SrCl2(s) ---> Sr2+ (g) + 2 Cl- (g); ΔH = 2153

Source:
These Lattice Enthalpies were obtained from Table 18B.4 of my textbook:
Physical Chemistry Thermodynamics, Structure and Change, 10th Edition, VOlume 1, by Peter Atkins & Julio de Paula
Then the Enthalpies of Hydration are these:
Li^(+) (g) ---> Li^(+) (aq); ΔH = -519 kJ / mole
Cl^(-1) (g) ----> Cl^(-1) (aq) ; ΔH = -381 kJ / mole
Na^(+) (g) -------> Na^(+) (aq); ΔH = -409 kJ / mole
Sr^(2+) (g) -------> Sr^(2+) (aq); ΔH = -1443 kJ / mole

Source of Enthalpies of Hydration (Tables 1 - 2):
The Enthalpies of Hydration that I've used are experimental.
Smith, Derek W. "Ionic hydration enthalpies." Journal of Chemical Education 54.9 (1977): 540.
Therefore, the enthalpy of solution for LiCl, NaCl, SrCl2 is this:
For LiCl,
ΔHsolution, LiCl = (852 kJ / mole) + [(-519 kJ / mole) + (-381 kJ / mole)] = -48 kJ / mole < 0, so this reaction should be exothermic and it is as predicted.
For NaCl,
ΔHsolution, NaCl = (787 kJ / mole) + [(-409 kJ / mole) + (-381 kJ / mole)] = -3 kJ / mole < 0, so this reaction should be exothermic.
I've been searching the web about this reaction and most sources say it's endothermic which I my graph shows, but my calculations do not show that. What?
Doing the same kind of mathematics, the Enthalpy of solution for SrCl2 is this:
ΔHsolution, SrCl2 = [2153 + (-1443 - 2 * 381)](kJ / mole)
= -52 kJ / mole < 0, so this reaction should be significantly exothermic.
BUT MY PLOT SHOWS that It's significantly endothermic!
What's going on? Is my mathematics faulty. Am I using too concentrated solutions? Is the SrCl2 that we are using impure or not really SrCl2? I suppose that's possible, but we didn't do any validation because this lab took quite some time. How is it possible that my experiments don't match theory mostly?
Thank you for you for your help.


text:
Does RedP+HI reduce all carbon functional groups to alkane?
I've seen the related questions 1 and 2, they're not duplicates
So. when introducing Red P+HI reduction, our teacher said that it is the strongest reducing agent, and it can reduce any functional group i.e. all of these: carboxylic acid,  acid anhydride, ether, amide, acid halide, carbonyl, and nitrile
However, today I stumbled upon a question in my book involving a step in the synthesis of ibuprofen, where they did not reduce the carboxylic acid group even after using RedP+HI. I searched a lot on the internet, and finally came across this paper which confirms this:

paper - notice they've only reduced the alcohol in step (ii)
This now makes me wonder if what I was taught is actually correct. Which brings me to the question:

Does RedP+HI really reduce all carbon functional groups to alkane?

Of course, this is  the very same methodology employed by Kiliani in the 1880's to prove, via P/HI reduction of the cyanohydrins of glucose and fructose to form heptanoic acid and 2-methylhexanoic acid, respectively, that glucose was an aldose and fructose a ketose. Look here.

Somewhat but not completely related question: https://chemistry.stackexchange.com/questions/46573/reduction-of-glucose-to-hexane-with-hydroiodic-acid

Nope, in general it doesn't. You are correct. See also for example:

A. Aramini, M. R. Sablone, G. Bianchini, A. Amore, M. Fanì, P. Perrone, A. Dolce, M. Allegretti, Tetrahedron 2009, 65 2015–2021. DOI: 10.1016/j.tet.2009.01.005:

M. Dobmeier, J. M. Herrmann, D. Lenoir, B. König, Beilstein J. Org. Chem. 2012, 8, 330–336. DOI: 10.3762/bjoc.8.36 Open access PDF. (This also includes your one as a reference, too.):

If it did, it would be a pretty poor reagent since it would have no functional group tolerance, so not very useful. However, it is not unlikely that under special conditions it might be able to reduce carboxylic groups as well.

Instead of just linking the entire pdf, if you could cite it as a reference, and pick out the specific extracts, or page numbers from it, that'd be better. In fact, if you could also list the carbon functional groups that are actually reduced by red P +HI, that would make your answer much more complete. Thank you! ^_^

@GaurangTandon From the looks of it, HI + P is mainly aimed at reducing the OH to H. If you read the second paper they also have a table where they report decomposition and lack of selectivity using the reagent for some compounds.

Thanks for the details! Yes, you're correct. This does seem to be primarily aimed at alcohols. From the table, it's clear that they did not (1) reduce either at step 5 (2) ketone at step 4 (3) amide at step 6 (4) ester at step 12. Though, I'll wait for some conclusive evidence before the green checkmark. Thanks for answering though! :)


text:
Structure of Dolomite
This the structure of Dolomite.
Can someone explain this structure to me?( like what are the blue, yellow etc atoms are? why are they arranged this way?)
Any help would be appreciated.

There are quite a few levels of abstraction between this picture and the real compound. It would be good to know which level you don't understand. To begin with, do you know which spheres corresponds to carbon atoms?

First, the tag [tag:molecular-structure] is out of place. Second, strictly speaking, crystal structure is an infinite periodic three-dimensional lattice and cannot be depicted with a single image. What you show here is a unit cell.

What do you want explaining?

I don't really like the picture but let's try to do this with that.
So, you probably know about the NaCl lattice? In there sodium is surrounded by six chlorides, which forms an octahedron. Chloride has six sodium around it to form another octahedron. And you have probably heard that also other compounds crystallize in this lattice, let's say sodium bromide (so increasing the anion in size) or potassium chloride. It does not change the crystal structure.
Now, what about more complicated units. Let's deform that spherical anion to an elongated one. What about calcium carbide? The C-C unit with its triple bond is unlikely cleaved during the crystallization, so we can count the $C_2^2-$ unit as one part. This forms an elongated cell but still, it is a NaCl-type lattice.
If we make it flat instead we end up with something like carbonate for example. As you may no the carbonate unit is flat, it's basically a triangle with the carbon in the center. If we see this unit as one part then we can just form another NaCl-lattice. And as the jump from sodium chloride to potassium chloride was possible, we can assume that if calcium carbonate crystallizes in a deformed NaCl-lattice, then magnesium carbonate will do the same.
And the final step is to combine them now, instead of separating the two phases, 1/2 of the calcium-ions are just exchanged by magnesium to form the dolomite-structure.
So you could say that carbonate forms a cubic packing where 50% of the octahedral sites are occupied by calcium and the 50% by magnesium for example. All carbonate-units are parallel to each other therefore creating one direction in which they do not require that much space causing the deformation of the cell.
The metals are coordinated by (and I had to load this in my solid state software as well to see it) six oxygen on the carbonates, or if you take the center carbon of each carbonate unit also by six carbonates, which forms an octahedron again. On the other hand, each of these center carbons is coordinated by three oxygen in a triangle or if you count the counter ions by six metals in the shape of an octahedron.
In your structure you can start off by looking at the dark atoms, which are the carbons, and connect them to the red ones, so can see it will always form triangles. For all the others it's hard to see in this picture. If you want I can also highlight you the different polyhedra in my software so you can see it better. Because this is just a unit cell so it is cut at some positions creating the feeling that some of the connections are wrong. You'd have to watch several unit cells aligned to see the full structure of it. But if you imagined the yellow and the blue ones to be the same you'd end up with the Calcium carbonate or magnesium carbonate structure.
EDIT: I also added the crystal structure now:
EDIT II: Because there were still some problems I tried to further explain the structure a bit. I don't know how others see this structure but I like this layer explanation and it's also the one you can find in many books. So please just look through the new attachment and tell me if this helps you more.

I am unable to understand how are we combining both of them? And from where are 1/2 calcium atoms are being replaced by magnesium ions?

Well if you read my picture from the left to the right I started off with all metals being the same and being blue, then I seperated them into Mg and Ca by adding the green color as well.  In a simplified view you can say that you get a layer-structure, where you have a layer Mg, then a layer carbonate, a layer Ca then carbonate, Mg, carbonate, Ca, ....

In the second last diagram, in the upper right carbonate ion, the oxygen atom has formed 3 bonds( with carbon, magnesium and calcium) how is that possible?

Well in solid compounds this is like dissolving something in water. NaCl would dissociate and you'd expect the Chloride to be encapsulated by water and not just one water. It's the same here but just a solid type of matrix or solvent if you prefer to imagine it like that. There are layers of positive charge above and below the oxide, so it will certainly 'feel' something and that connection to carbon is a covalent 'real' bond anyways.

To draw in bonds helps to see why for example some things are not well centered by kind of move to the opposite charge. And it also helps to organize everything in your mind. It's the same as for complex compounds when we talk about ligands and polyhedra forming around an ion.

Here is my understanding of it:
1)  The black (carbon) and red (oxygen) atoms form the triangular carbonate ($\ce{CO_3}$) units.
2)  Each oxygen atom is bonded to a gold magnesium atom and a blue calcium atom.
3)  Look carefully at the bonds to each calcium or magnesium atom.  The bonds shown fit onto the vertices of a regular octahedron, we see a pair of oppositely directed metal-oxygen bonds at each metal atom.  The metal atoms are in fact octahedrally coordinated to oxygen.
So why do we not see all six metal-oxygen bonds at each metal atom?  The answer is that the oxygen ends of the "missing" bonds are in adjacent unit cells.  I would have drawn those bonds through the faces of the cell, even if I have to leave off the oxygen atoms so that the bonds appear to end in "thin air".  But having the "thin air" bonds makes it easier to piece multiple cells together so that you see the octahedral coordination come together.

Well that's basically it. For the coordinations you can simply look at my picture, I just took the published data on dolomite and drew in the octahedra. And yes, you cut of at the borders. What you see is the unit cell, the repeating unit. You'd have to look at a supercell to see more connections, I could add that but the structure is already quite big and it becomes quite hard to see anything.


text:
HF co ordinate covalent bond?
Is it a co ordinate covalent bond,I think that due to high electronegativity of the fluorine atom it will attract the hydrogen electron towards itself.Am I right in thinking so?

The bond between $\ce{H}$ and $\ce{F}$ atom in $\ce{HF}$ molecule is a polar covalent bond.
Though the electronegativity of $\ce{F}$ is sufficiently high, but $\ce{H^+}$ is a tiny ion with a positive charge on it, which creates high electronic charge density on the atom and due to its much smaller radius,The polarising power ($\phi$) becomes very high. So, according to Fajan's rule, there is a significant covalent character.
If you want to think in terms of valence bond theory, then $1s$ orbital of $\ce{H}$ containing a single electron pairs with one of the $2p$ orbital of $\ce{F}$ containing a lone electron to form a covalent bond.Similarly, if you consider molecular orbital theory, the molecular orbital diagram will be something like this,

So, the bonding electron resides in a $\sigma$ bonding orbital, which indicates there is a covalent bond.


text:
Type of the shown reaction

Could you please let me know what the type of this shown reaction is?

Homework or homework like questions need to be accompanied by the asker's approach/reasoning to the question or according the the homework policy they are closed.

Alright, this is very obviously not a homework. I am a student and this was a question in my past papers where ı was expected to fill in the red molecules. and I am struggling to get to know what type of reaction this is. is there something wrong here?

@Taylan This question is so specific that it does not add to the mission of SE. If you had a question like 'Why is NaOEt required for this class of reaction.' it might work, as it gets at mechanisms and explaining this one, or  ' Given the product of this reaction, how can the reactants be predicted.' , leading to an answer that can explain that the addition occurs at the -CHO site.

alright thank you, ı am new to the website so wasnt sure about how to formulate the question in a suitable way for the website. ı removed the pic after being told that ı should not have asked it but someone edited it and put it back. thank you for the clarification!

This is a classic example of the Knoevenagel condensation
https://www.name-reaction.com/knoevenagel-condensation
https://chemistry.tutorvista.com/organic-chemistry/knoevenagel-condensation.html

You should try to make more elaborate answers and don't answer questions which may violate homework policy.


text:
How long can solid DPPH powder last for in terms of shelf-life before its free-radical status is quenched by the atmosphere?
I am writing in my research on the free-radical scavenging activity of various flavylium compounds. I may need to have DPPH shipped as a result of a lack of supply; I am just wondering what sort of precautions I might have to take and how viable it is to do so.
I understand that DPPH solutions need to be stored very well, but I'm assuming DPPH is most likely shipped in its powder form. What I'm wondering is what its stability as a whole is in solid form. Should it be constantly kept in darkness so as to prevent photooxidation and kept in a cold place, i.e. a freezer?
Even so, how long would its shelf life be by then, and how long would it be without taking precautions and leaving it at room temperature and normal lighting?
I'm asking so as to prevent or minimise the loss of its free-radical status.


text:
Which hydrogen does the nucleophile attack in E2 reaction?

This reaction will be both an E2 and SN2, where the E2 will be major and the SN2 minor (If I have done it correctly). I was drawing out the mechanism as I noticed that I was making my nucleophile attack the wrong hydrogen, the one attached to the alpha-carbon. I know that the nucleophile is supposed to attack the hydrogen on the beta-carbons however how do I determine if the $MeO^-$ attacks the hydrogen on the second carbon to the left or to the right of the alpha-carbon? Or does it attack both and since the sturcture of the compound formed is the same either way, it doesn´t matter which hydrogen it removes?

Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2.
As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options.
In future similar examples, you may come across reactants with multiple possible reactive positions that are not equivalent. For example, consider 3-chlorohexane (instead of 3-chloropentane, as in the original problem). In this case, again there are two possible sites to be deprotonated (Ha and Hb). Looking at the two possible products, both are disubstituted alkenes, so they will be close in stability. There isn't much difference in the groups nearby Ha and Hb, so the kinetics of the deprotonation are probably similar. So which is the "correct" answer? In reality, both are correct. They may not be formed in exactly equal amounts, but they will probably both be formed in substantial amounts.

There are a couple of key lessons here for those studying organic chemistry:
1) Draw out all of the possible products that come from mechanisms that you think make sense. This might be lots of drawing! Nevertheless, it is much easier to realize the differences if everything is drawn out in front of you.
2) Organic chemistry is messy. Often times multiple mechanistic pathways will be viable and multiple products will form. Much of the creativity in organic chemistry is finding ways to ensure that only one mechanism can occur or one (desired) compound is formed as the major product.

Re: "In reality, both are correct." I was taught small nucleophiles favor Satyzeff alkene, while bulky nucleophiles (like tert-butoxide) favor Hoffmann alkenes. So, how do you justify this statement? Thank you!

@GaurangTandon in this case, which is which? They are both di-substituted. And to answer a question with a question, what does "favor" mean? Is the product distribution 100:0, 75:25, 51:49? Those all have a "major product" but they vary greatly in how much each is favored.

"in this case, which is which?" i'd say the one on the right is the Saytzeff alkene, since it is more stable by an extra hyperconjugative structure; "what does "favor" mean?" Well I was just asking a question based on the knowledge I got at high school ^_^ I haven't done the experiments myself, so I really don't know what ratio it is exactly; but in this case the alkenes are so close in stability I guess you are right in that "both be formed in substantial amounts."


text:
What does it mean if I am unable to write RS configuration for a compound?

We had to write all possible optically active monochloro derivatives of 2-methylbutane.
I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?
P.S. I’m still a beginner at this...

Hi Aura, welcome to Chem.SE! A well posed question! After having read your question properly, I think your query simply is: "Are all molecules without any chiral carbon optically inactive?" Please confirm if I interpreted your question correctly.

Yes I think so.

Yes, I did. Correct me if I am wrong, these compounds don’t seem to have planes or centres of symmetry. Atleast in the way their fischer projections are drawn...

Actually the answer is overly simple.  No achiral carbon no R or S configuration around it. It is like asking for why I cannot determine if ethane is cis or trans. Get it? Plus you might have other issues, so look at the others comments and answer.  My comment answer to your question as it reads in the title.

Fischer projections are not the best way to figure out R or S...

I think I get what your question is and will try explain the things to you.

I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?

The crux of your question lies apparently in two problems:

you assume that molecules 1 and 3 are optically inactive, but are unable to visualize this with a center/plane of symmetry
you seem to assume that molecules without chiral carbons are optically inactive

So, let's tackle this:

Correct me if I am wrong, these compounds don’t seem to have planes or centres of symmetry.

Well, you're wrong. Note that in simple saturated hydrocarbons (with or without halogen substituents), it is usually safe to assume that if no chiral carbons are present, then the molecule will be optically inactive (without the need to validate this assumption by manually drawing out their planes/centers of symmetry). Here are two examples:

 (source)
notice how the plane of symmetry cuts-through through the blue and green atoms, using the remaining two grey identical atoms to obtain a plane of symmetry, to prove the molecule is optically inactive. This was expected, since the molecule had no chiral carbon.

it is possible to visualize the plane of symmetry in 2-chloro-2-methylbutane from its eclipsed Newmann projection, as seen above. The plane of symmetry cuts-through the chlorine atom and the methyl radical in front of it, rendering the molecule optically inactive, as expected.

Are all molecules without any chiral carbon optically inactive?

No, this is wrong. Common examples like ortho-disubstituted-biphenyls (tl;dr: they are propeller shaped rings), allenes (tl;dr: their terminal groups are in perpendicular planes), etc. have no chiral carbons, yet are optically active. I won't detail these as you're still a "beginner" in this, but you'll definitely come across these examples in the very end of your course on optical activity.

In regards to your edited question, it is unreasonable to draw Fischer projection for achiral carbon atoms. For example, when drawing the Fischer projection of carbohydates, the terminal achiral carbon is written as a $\ce{-CH2OH}$ substituent to the penultimate chiral carbon, and not expanded out into its own Fisher projection.
I hope it helps!


text:
What are the degrees of freedom (from Gibbs phase rule) for an azeotropic mixture?
For an azeotropic binary mixture, the number of phases is 2 and number of components are 2. So, according to Gibbs phase rule, the number of degrees of freedom should be 2. But, the right answer is 1.
Can someone explain this?

Gibbs' phase rule simply counts the number of variables required to specify the system and the number of relationships between these variables (constraints). Any failure of Gibbs' phase rule is symptomatic of additional variables or constraints that have not been taken into account.
This system is generally specified by four variables, the mole fraction of one species in the liquid phase, $x$, the same in the gas phase, $y$, the pressure $P$, and the temperature $T$. Equilibrium of the liquid and gas phases in both species provides two constraints. The specification of an azeotrope, for which $x = y$, provides a third constraint, as has been mentioned in the comments. One degree of freedom remains.

The Gibbs Phase rule can be applied to a P-T thermodynamic diagrams for pure substances. On any such of diagram, each point represents a state of the material. Once the point has been located, all other thermodynamic properties can be read off.
For a single phase, the required point can fall anywhere on the region of the plane representing this phase. Thus, two coordinates are needed to specify a point within that part of the plane. For example the horizontal and vertical axis values, T and P, can be given. (F=2)
A two phase region on a P-T is represented by a curved line. Thus, to specify a point on this curve only one additional coordinate, either T or P is needed. (F=1)
The triple point is represented by a single point. No additional information, neither T nor P is needed. (F=0)

An azeotropic mixture is not a single component system. Your answer is perfectly valid for a single component system with multiple phases. For a multi-component system also the Gibbs phase rule predicts the degrees of freedom except in few cases. An azeotrope is one such case.

@SriramKrishnamurthy You are correct. However, looking at the rationale behind the rule, it is easy to use still. An azeotrope for instance, represents a single constraint on the system, and this reduces the DoF by 1. The real challenges to using the rule starts when you get immiscibility.


text:
Why does benzoate show Friedel Crafts alkylation?
Why does benzoate show Friedel Crafts alkylation?
I know that benzoic acid does not show Friedel Crafts alkylation because the Lewis acid forms a bond with the lone pair of oxygen itself. So, shouldn't the same happen with benzoate as well? The oxygen in benzoate has a negative charge, which should make bond formation with the Lewis acid even easier. But my book states that it will show Friedel Crafts alkylation.

Related question on Chem.SE but it didn't get answered conclusively.

@GaurangTandon Thank you for the link but it increased my confusion even more. All the reasons due to which benzoic acid does not show Friedal Crafts should be valid with benzoate also. But they aren't.


text:
Is a ring intermediate formed in hydroboration?

  This was the mechanism taught to us for hydro-boration

Something really strange in that mechanism is that in step 2  hydrogen has two bonds.
Also , the  the doubly bonded $\ce{H}$ , has one bond $\ce{C-H}$ which contains no electrons

Is this mechanism plausible?

In clayden they have shown the cleavge of $\ce{H-B}$ bond  and formation of the $\ce{H-C}$ bond in the same step.
But it doesn't mention the formation of a ring anywhere.

Section 1
Let's assume that the ring does form but after cleaning the $\ce{H-B}$ the electron pair goes into the bonding orbital of H not the anti bonding molecular orbital between $\ce{C-H}$ .

Primarily my doubts  are

The bond with no electron in it (does it actually form)
The reasoning in section 1 (the previous section).

"Something really strange in that mechanism is that in step 2 hydrogen has two bonds." I'll have to check, but that's probably a transition state (not an intermediate), and one of those two solid lines should be a dashed line, indicating a bond being formed/broken.

That should be a transition state, not an intermediate.

Also, there should not be a positive charge on your boron in the final compound you drew.


text:
Why does white phosphorous exhibit chemiluminescence?
My work - I only know that chemiluminescence means glowing in dark.

Long story short, nobody really knows.

Oh! I thought that there was a reason to explain it.

You might go with "white phosphorus oxidizes in the air and emits light in the process". This is true, but does not really explain anything.

Actually, I got something from a source .

Chemiluminescence is the emission of cold light as a result of a chemical reaction. In a chemiluminescent reaction, reactive intermediates are formed which enter electronically excited states. Subsequent transition back to ground state is accompanied by a release of energy in form of light.

Sounds good. Same as I said plus a few smart words and hand-waving. Yes, I guess this is what they want from you. Anyway, nobody really knows what is that intermediate.

Among the excimers that undergo emission the most likely are PO2 and HPO, the role of the latter indicated by deuterium effects. In all cases the luminescence of white phosphorus does requires oxygen, and is a consequence of a reaction.  This should already be useful to you, as it is not evident if it was clear what chemiluminescence is (emission of light as follow up of a chemical reaction, which is not the case of fluorescence and phosphorescence).

From here:

The best-known and first well-documented example of deliberately
  induced, man-made chemiluminescence was the discovery of elemental
  phosphorus by a German alchemist H. Brand. The waxy white phosphorus
  is slowly oxidized at room temperature giving off faint greenish glow.
  The formation of excited reactive intermediate $\ce{(PO)2}$ and
  $\ce{HPO}$ is responsible for the green glow.

That's where the term phosphorescence come from.


text:
Which of the following compounds do not undego electrophilic aromatic substitution?
Which of the following does not undergo electrophilic aromatic substitution?

Answer is option (D). But I am not able to understand why. All of them have lone pairs. I can notice only two differences. Functional group in Option (d) has the largest size and lowest electronegativity.
I think that may be because of a large difference in size of Selenium and Carbon, pie bond formation in between them will be a bit difficult, but that bond is not playing any role here except stabilizing the compound.
Low electronegativity seems like a good thing to me as it does not deactivates the compound.
So what am I missing here?

@AvatarShiny Couldn't you say the same for sulphur?

Se has bigger d orbitals. It's softer. @Gaurang

@AvatarShiny I still can't understand why will d- orbitals effect the reaction.

@AvatarShiny Oh so you're saying that it will be easier to take electrons from Se because of d orbitals? That must be it. Thank you.

@VJ Your thinking is correct.  Selenium is the largest atom of the series, it is in Period 4 while the others are in Periods 2 and 3.  It's larger size precludes effective overlap with the carbon pi orbitals.  Whereas molecules A-C are "aromatic" because the heteroatom lone pair is involved in overlap with the dienyl system to make a 6 pi electron ring, the selenium atom can't participate like this so molecule D is not "aromatic" like the others

@ron Oh. Thanks for helping. So conclusively, there is no aromaticity and no substitution either (according to Avatar Shiny) right?

@VJ  Yes, that is correct under normal reaction conditions.  Although I suspect that under "forcing" conditions, something will happen.

Wikipedia disagrees.  Selenophene does undergo EAS, and apparently faster than thiophene.
None of the proposed answers to the original question is correct.


text:
Structure of phosphate ion
I understand that the hybridization in the phosphate ion is $\mathrm{sp^3}$ since phosphorous forms $4$ sigma bonds with $4$ of the oxygens and there are no lone pairs on the phosphorous atom.
The outer electronic configuration of phosphorous is $\mathrm{3s^2 3p^3}$. In the case of $\mathrm{sp^3}$ hybridisation, the $\mathrm{3s}$ and the all of the $\mathrm{3p}$ orbitals are involved. Hence, among the $4$ equivalent orbitals now obtained, $1$ of them contains a pair of electrons, while the other $3$ are singly occupied. So shouldn't phosphorous be able to form sigma bonds with only $3$ of the oxygen atoms since it has only $3$ unpaired electrons?
I am trying to compare this case with that of carbon tetrachloride. There, by $\mathrm{sp^3}$ hybridization, we were able to get $4$ equivalent orbitals each with an unpaired electron, and therefore explain its tetravalent nature. In the case of phosphate ion, one of the $4$ orbitals obtained already has a pair of electrons. So, could anyone please tell me where my understanding is flawed?

Electrons do not know what atom they belong to. Strip one electron from P. Now it is the same as C. Now give that electron to one of the O.

wouldnt that affect the hybridisation of oxygen?

Why? Which way? And besides, how is that important at all?

Disclaimer: hybridization is a mathematical concept and fails at higher levels.  As per the linked answer, it is clear that the participation of $d$-orbitals in phosphate ion is minimum. But this is just a high-school level, layman's explanation for your question.

So, could anyone please tell me where my understanding is flawed?

Your reasoning is flawed in assuming that carbon atom has four unpaired electrons in its orbitals after hybridisaton, while phosphorus does not. Actually, both of them do.
In fact, carbon atom in its ground state does have a lone pair of electrons in its $\mathrm{2s}$ orbital. It is only when it gets into its excited state, that that lone pair is broken up, yielding four unpaired electrons. The excited carbon atom now undergoes hybridization as depicted in the figure below, yielding four equivalent covalent bonds as in carbon tetrachloride.

(source)
As you noted, phosphorus also has a lone pair of electrons in its $\mathrm{3s}$ orbital. But, that is its ground state electronic configuration. Once it gets excited, it will have five unpaired electrons, the fifth electron going into its $\mathrm{3d}$ orbital. Now, phosphorus will undergo $\mathrm{sp^3}$ hybridization, to form $4$ sigma bonds. The fifth electron helps form the $d\pi-p\pi$ bond with the neutral oxygen atom.
I hope this helps!

sigh OP needed explanation how dipolar bond works not some outdated "d-orbital" nonsense.

@Mithoron The way OP framed his question with "excitation of electrons in carbon" and "hybridization" it is his clear that he needed a simplistic explanation of the things going on as per high school level. The question you linked to involves Natural Bond Orbital analysis, and I am not sure how that would have helped the OP in any way. I've edited my answer to make it more clear