Conversion between anisotropic temperature factors and geometric ellipsoids I am using the ANISOU of the PDB standard to represent ellipsoid shaped colloids.  I am writing a Monte Carlo simulation to transform the position and orientation of the ellipsoids and want to use a PDB file to represent the result.

I am not sure how to convert between a geometric representation of an ellipsoid to the anisotropic temperature factor tensor, $\mathbf{U}$.  I am able to define an ellipsoid aligned to the coordinate axes using the diagonal elements of $\mathbf{U}$: $\mathbf{U}(1,1)$, $\mathbf{U}(2,2)$, $\mathbf{U}(3,3)$, but I am not sure how to represent a rotated ellipsoid, i.e., I am not sure how to set $\mathbf{U}(1,2)$, $\mathbf{U}(1,3)$, and $\mathbf{U}(2,3)$.  I came across this resource which introduces the topic.  This second link mentions the similarity between tensor elements and geometric definitions and says the conversion is "rather complicated" but does not provide a link for further study.

Any guidance or direction to a good resource would be appreciated.

I found the publication titled Atomic Displacement Parameter Nomenclature which seems to have the details.

When you need to deal with the anisotropic things, I think that you cannot transform or re-define an ellipsoid's matrix into U(m,n), where m is not equal to n. However, you can average the any anisotropic quantities that you want using the direct geometric transformation as long as you have the coordinate file (xyz) in your hand.


Active transport of Ca2+ ions against chemical and electrostatic gradients, assisted by ATP hydrolisis I have a homework question that I am not able to solve. No solution is available, so I was hoping someone here could help. The question is this:

  Calculate the maximal concentration ratio that can be obtained for
  transport of Ca$^{2+}$ ions if two ions are transported per ATP hydrolyzed
  at 310 K, and they are transported to a potential that is 70 mV higher
  at the receiving side. Assume that the $\Delta G$ for hydrolysis of ATP = -45 > kJ\mol.

Relevant formulas:

\begin{align}
\Delta G_{\text{electrostatic gradient}} &= ZF\Delta \Psi\\
\Delta G_{\text{chemical gradient}} &= RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}}\\
\Delta G_{\text{hydrolysis of ATP}} &= -45000~\mathrm{J/mol}\\
\end{align}

Moving two positive $\ce{Ca^2+}$ ions against both a concentration gradient and an electrostatic gradient implies that $G_{\text{chemical gradient}}$ and $\Delta G_{\text{electrostatic gradient}} > 1$. The negative value for $\Delta G_{\text{hydrolysis of ATP}}$ means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, $\Delta G_{\text{chemical gradient}}$ is multiplied by 2. Also (I presume) the concentration of $\ce{Ca^2+}$ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.

The combined expression for $\Delta G$ for the unassisted transport is

$$\Delta G = RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi$$

Rearranging this yields

$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$

I am not sure what to plug in for $\Delta G$, but it needs to account for the push from ATP. I believe this is done by summing the two free energies affecting the transport, ergo the sum of $\Delta G_{\text{electrostatic gradient}}$ and $\Delta G_{\text{hydrolysis of ATP}}$. However, this gives me a concentration ratio much smaller than 1.

Can anyone see where I am incorrect?

At equilibrium, the entire process (transport of $2~\ce{Ca^2+}$ ions and hydrolysis of $1~\text{ATP}$) involves no change in G.

\begin{align}
0 &=\Delta{G_\mathrm{total}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+},\text{ concentration gradient}}} + 2\Delta{G_{\ce{Ca^2+},\text{ electrostatic}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi\\
\end{align}

So basically I'm saying you made two mistakes, one a sign error and one a factor of 2 error.

ΔGelectrostatic gradientΔGchemical gradientΔGhydrolysis of ATP=ZFΔΨ=RTlnCinCout=−45000 J/mol

Moving two positive Ca2+ ions against both a concentration gradient and an electrostatic gradient implies that Gchemical gradient and ΔGelectrostatic gradient>1. The negative value for ΔGhydrolysis of ATP means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, ΔGchemical gradient is multiplied by 2. Also (I presume) the concentration of Ca2+ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.

The combined expression for ΔG for the unassisted transport is
Rearranging this yields

$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$


What's the difference between a nucleophile and a base? Obviously it depends on the context whether you would call a particular species a nucleophile or a base but are the two terms largely synonymous or is there a difference?

The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior?

  Nucleophilicity is a kinetic phenomenon.

Nucleophilicity is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent.

For example, a standard reaction might look like:

$$\ce{CH3I ->[Nu-][H2O] CH3Nu}$$

The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$).

  Basicity is a thermodynamic phenomenon.

Basicity is based on the position of equilibrium:

$$\ce{B + HSol <=> BH+ + Sol-}$$

When the size of a negatively charged molecule is small then it behaves as a nucleophile, but when size of a negatively charged molecule is big then it behaves as a base.  This is because of crowding around the central atom of other reactant species.


Why are tertiary amines less basic than secondary and primary amines in water than when vapourised? In gaseous state:

$$\ce{(CH3)3N} > \ce{(CH3)2NH} > \ce{CH3NH2} > \ce{NH3}$$

However, when dissolved in water:

$$\ce{(CH3)2NH} > \ce{CH3NH2} > \ce{(CH3)3N} > \ce{NH3}$$

My notes mentioned the reason for this is because of the increased steric hindrance from the greater number of alkyl group which makes the lone pair of electrons on the N atom less available. But, if this is the case, why doesn't steric hindrance affect the basicity of tertiary amines when in a gaseous state?

After some research, I found that the conjugate acid of an amine can be stabilised by solvation. But, shouldn't the stabilisation be the same, because of the attraction of electronegative O in a water molecule to the positive charge on $\ce{N}$ in an ammonium salt? (can ammonium salts still be stabilised by hydrogen bonding? I thought ions don't form hydrogen bonds with water, they form ion-dipole attractions)

Unless the effect of steric hindrance comes in here, where the alkyl groups reduce the strength of the ion-dipole attraction between the ammonium ion and thus the stability?

I'm really confused how everything links up; appreciate it if someone can kind of clear this up.

A possible answer from what I've searched on the internet so far..

The strength of basicity of amines depends on two factors: 1) the availability of the lone pair for donation to a proton, 2) the stability of the conjugate acid of the amine.

In methylamine, the conjugate acid, $CH_3NH_3^+$ will attract water molecules and ion-dipole interactions will form between the cation (the conjugate acid of methylamine) and the partial negative charge on O in the water molecule. This ion-dipole interaction or attractive force helps to stabilize the positive charge on the cation.

So, the more water molecules you have solvating your cation, the more stable the conjugate base. However, the more alkyl groups you have, the less stable the conjugate base, because these alkyl groups are hindering water molecules from coming and stabilizing the conjugate base.

[The N atom is "shielded" by the bulky alkyl groups which take up a lot of space and prevent any water molecules from stabilizing the conjugate acid. Water molecules can only come in from where the H atom is where the shape of the conjugate acid is that of a pyramid. Note that there is a positive charge on N.]

As methylamine has the least number of alkyl groups as compared to dimethylamine and trimethylamine, at this point, it would be the strongest base. However, basicity also depends on the availability of the lone pair of electrons on the N atom of the amine itself. As methylamine has the least number of alkyl groups, the inductive effect by the alkyl group is the least out of methylamine, dimethylamine, and trimethylamine.

So, it really is a measure of which factor is more dominant or which factor outweighs the other. In this case, although trimethylamine has the most number of alkyl groups, the three bulky alkyl groups prevent a lot of water molecules from stabilizing the conjugate acid. And, for dimethylamine, although dimethylamine has a fewer number of alkyl groups than trimethylamine, the inductive effect of the alkyl groups outweighs the steric hindrance from the 2 alkyl groups as compared to the three in trimethylamine.

Hence, in an aqueous solution, $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$

However, in the gaseous state, there are no water molecules to stabilize the conjugate acid and basicity depends on the inductive effect of alkyl groups; there are no water molecules to hinder. As trimethylamine is trigonal bipyramidal in shape, the factor of steric hindrance does not apply here. Hence, as trimethylamine has the most number of alkyl groups, the lone pair of electrons on trimethylamine is most available for donation.

In the gaseous state the basicity can simply be compared on the basis of availability of lone pair.
In tertiary amine the nitrogen atom receives +I effect from all the three methyl groups which increases the electron density on the nitrogen atom ,thus its lone pair is readily available. Whereas in secondary and primary amines the nitrogen atom receives+I effect from two and one methyl group respectively. So there lone pair are not as readily available as of tertiary amines.
When in aqueous phase  the amines interact with water. This interaction also includes hydrogen bonding. This interaction causes steric repulsion. Thus it affects the basicity and causes the order to be so.
Note : the shape of amines is Tetrahedral with three surrounding atoms and one lone pair. So there is minimal steric repulsion in gaseous state. The main reason for steric repulsion in aqueous phase is the interaction with water.


Would it be possible to destroy gold? I'm a writer. I have a scenario in which a sizable amount of gold needs to be rendered unusable, preferably completely destroyed. I know an acid like aqua regia is able to dissolve gold, but would there be a way to make it impossible (or at least not worth the cost) to chemically salvage the gold from the acid, or another way to permanently destroy gold?

Other than a nuclear reactor, the only other chance is to dump it into a volcano. Having a much higher density than magma, it will just fall through until it hits earth's mantle. Then it's really gone. OK, dissolved and diluted in the Atlantic is also quite safe.

P.S. Warning: If the bullion gets stuck in the volcano, which later explodes Krakatau-style, your gold brick might land in someone's front lawn. Making a hole five meters deep, but it may be salvaged easily.

Use something that can shot waves of protons with great dynamic to make protons and neutrons separate in gold atoms so it not gonna be gold again! (but if you can decomposite it so you can reform it in some way)

( Just like you use your hand's force to cut a cheese off)

Like an explosion of a neutron star.

Or in an imaginary scenario when you use a rocket to carry all the gold amount and let it detonate itself with a nuclear bomb so there will only be gold dust in the vacuum.


Which is anode and which is cathode? A maybe (hopefully) simple question about the denotations of "anode" and "cathode". The below image is a schematic of a polymer solar cell (Source (WBM)). (The figure text is quoted as well for the sake of context).

My question is only about the naming of electrodes. Isn't it correct that

the cathode is the reducing electrode, which supplies electrons, while
the anode is the oxidizing electrode, which accepts electrons?

Why would electrons (black spheres) move towards the negatively charged cathode, while positive holes (white spheres) move towards the positively charged electron-absorbing anode? Wouldn't the opposite make sense?

Is it a mistake on the drawing or should I re-think my understanding of the terms "anode" and "cathode"?

  Figure 1. The working principle of the solar cell. Light enters the cell through the transparent anode, and is absorbed in the bulk heterojunction layer through generation of excitons (1). The excitons diffuse in the bulk heterojunction until they either recombine or reach a donor-acceptor interface, where they separate into electrons (black) and holes (white) (2). The electrons and holes will then move to the respective anode and cathode, through the donor and acceptor material phase (3).

Edit

If the figure is correct and I am the one being confused, then there are several other cases that seem to teach false. E.g. the following "PANIC"-remembering-rule to remember, which is cathode and which is anode (from this source, slide 8):

"*Negative is cathode", it says, and we are back to my question in the top: why do the electrons (black spheres) move towards the negative cathode in that schematic?

Many more diagrams agreeing with this latter one are found from a simple google search.

Regardless of the polarity, the electrode where oxidation takes place is the called the anode and therefore reduction must take place at the cathode. The electron flow from the anode to cathode is as shown in your top picture.

By way of an example, in an electrochemical cell, suppose that two beakers are connected by a salt bridge. In one beaker is a strip of zinc metal immersed in a $\ce{Zn(NO3)2}$ solution and in the other a strip of silver in $\ce{AgNO3}$ solution. The two metals are then connected by a wire and a current will flow. (The salt bridge supplies a return path so that the solutions remain electrically neutral). The redox of the Zn electrode is $-0.763$ V and that of the Ag $+0.799$ V. This means that electrons will flow from the zinc to the silver electrode. The zinc is oxidised to $\ce{Zn^{2+}}$ and the electrons are released into the metal and flow to the silver electrode through the wire. The zinc electrode is the anode and the silver the cathode.

The Cathode is the positive electrode; the anode is the negative electrode.
During discharge, positive ions flow from anode to cathode.  This makes the cathode positively charged.  Free electrons at the anode flow externally through the load to the cathode.   During charge, an external voltage source has its positive electrode connected to the cathode and its negative electrode connected to the anode.  The negative electrode of the external voltage source supplies electrons to the anode.  Excess positive ions at the cathode are repelled to the anode but move more slowly than electrons move through wires.


Why does using NaI as a nucleophile and acetone as solvent lead to an SN2 reaction mechanism instead of SN1? Is I- not considered a weak nucleophile? In a reaction with a secondary leaving group, why does using $\ce{NaI}$ with acetone as a solvent lead to an $\mathrm{S_N2}$ reaction mechanism instead of $\mathrm{S_N1}$? I thought that halogens were weak nucleophiles, and only really led to $\mathrm{S_N1}$ reactions?

As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:

$$\ce{NaI + RX -> NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$

This is in acetone solvent (I don't know how to format it along the arrow).

If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.

While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$.

This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.

Note :
When does SN2 attack occur. Let's analyse this and you will get your answer in end.

First of all the steric hindrance around the carbon atom should be least i.e if RX is 1° (primary) then SN2 attack always (mostly) occurs.

If RX is 3° (tertiary) then SN1 attack always (mostly) occurs, as steric hindrance is very high.

Now if RX is 2° (secondary), the mechanism can be SN1 or SN2, depending on the Nucleophile and the solvent.
If solvent is aprotic  and nucleophile is fairly good (strength): like DMF, acetone etc., then SN2 attack is favoured (I am not explaining why it occurs, but you can refer to the mechanism)
If solvent is protic and nucleophile is not so strong: then SN1 mechanism is favoured.

In your question, $\ce{I-}$ (iodide ion) is mild nucleophile and acetone is aprotic solvent. Thus SN2 mechanism is favoured more.


Is nifedipine a weak base or weak acid? I was wondering whether someone could explain me if nifedipine is a weak base or a weak acid.
I barely can find any information about nifedipine's physicochemical properties in studies, however some suggested that it is a weak acid. If it is a weak acid, which hydrogen atom is dissociable and what would the pKa of that group be?

Thanks in advanced.

Nifedipine's structure, as shown on wikipedia:

Nifedipine is a weak acid.

$pK_b = 10.07$
$pK_a = 3.93$
$pK_a = 4.3$

My speculation is a bit related to the answer provided by @aventurin.  In absence of experimental evidence to support this, I think in a first step the proton in (then) para-postion to the nitrogen is abstracted:

leading to a protonated, pyridinium specis, then in equilibrium to the non-protonated form.  This may be favourable by thecreation of an aromatic system.  Again, provided as a speculation including the notice that the initial deprotonation might be irreversible.


Minerals in water solidify when heated? Is it possible for minerals in drinking water to solidify when high temperatures are applied? The minerals in mind are: iron, copper, aluminum, magnesium.

Background: I have small, brittle, darkish flakes appearing in my hot water system. The manufacturer of my tankless hot water heater said it's possible for these elements to solidify in this manner over time.

Looking at their website, I found this brief explanation:
http://support.noritz.com/article/faq-what-is-water-hardness-46.html

  "Water hardness is the measure of dissolved calcium, magnesium, and other minerals in your drinking water. It is typically measured in grains of hardness which equates to 17.1 mg/L of dissolved calcium carbonate. If not treated hard water can cause mineral build-up in your heater. Mineral build-up reduces flow and produces flakes at your fixtures."

We got out water hardness measured, and it came back as 68 mg/L.

The temperature of the water heater is set to 130F. I believe that the heater has a higher internal temperature.

I'm not sure why they are dark/black in color, all the pictures I'm seeing for calcium carbonate are white.

Thank you

Limescale (calcium and magnesium carbonate) precipitates in the heater, the brown colour is most likely some iron compound that co-precipitates. A fresh surface of limescale likes to adsorb other ions, which then get trapped in it as the scale grows. If you had a water kettle, you'd see the same stuff in there after using it a few times. (Not exactly, because carbonate precipitation depends on the pressure, but iron (hydroxide?) not. I wouldn't want to predict.)

The limescale first appears on the surface of the heater cell, and then breaks off during temperature changes, i.e. when you open the faucet. As you have no tank, the flow rate of the water is high in the heater and carries the scale out of there.

Very convenient, that. In a tank, the scale builds up and can fill the whole thing after a few years, depending on your water source.

Your heater is set way to hot.
Set it at 108° then turn your shower as hot as it goes.
You can't handle any hotter.
Your adding cold water to cool it down to 108.
Calcium starts dropping out of solution around 125°
Two birds with one stone!


Is [PdCl2(PMe3)2] diamagnetic or paramagnetic? Please tell me where I am wrong. The complex will have $\ce{Pd^{2+}}$ ion, which has a $\mathrm{d^8}$ configuration. So, it will have $2$ unpaired electrons. Unpaired electrons will mean that it is paramagnetic.

My reference book has this line:

  $\ce{[PdCl2(PMe3)2]}$ is a diamagnetic complex of Pd(II)

Shouldn't it say "paramagnetic"?

You have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\ce{Pd^2+}$ ion adopts the $\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic.

Image source: Wikimedia Commons

Image source:Wikimedia Commons

this is a special case of jahn teller distortion....in pd(II), weak ligands can remove the degeneracy of eg orbitals & as the energy difference between two eg orbitals are high, the ekectrons pair up in the dz2 orbitals. Thus form a diamagnetic complex


Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution? Why is phenol more reactive towards electrophilic substitution than methoxybenzene? Isn't the lone pair on the methoxy group oxygen more available for participating in resonance into the ring than the lone pair on hydroxyl oxygen due to the +I effect of the methyl group? Or is it because of the steric hindrance of the methoxy group that will favour lesser substitution at the ortho- position of the benzene ring?

Does the hydrogen bonding of phenol have anything to do with this?

First of all we must have to note that the difference in reactivity between anisole (methoxybenzene) and phenol is small. Small enough for the undergraduate texts I own to not mention anisole separately from phenol, just referring to it as having "phenol-like reactivity" (i.e. strong ortho-para directing). We can more or less measure the difference in ortho-para directing strength by looking at the proton NMR spectra of anisole and phenol.

Source.

Source.

As you can see they are almost identical to the point where you can ask yourself if there is any practical difference in reactivity at all. The A protons nicely reflect the ortho-para direction since they haven't shifted from benzene's 7.26 ppm signal and the B and C protons are all upfield as expected.

Also I don't think steric hindrance is a big factor either since aromatic electrophilic substitution goes first trough a $\pi$-intermediate and a single methyl-group is not that bulky.

This is because phenolic oh group is having a negative charge ( phenol is acidic) extra which is absent in methoxy oxygen. So oxygen in phenol can get into resonance easily than oxygen in methoxy benzene..


Acidic strength of para substituted benzoic acids Comparing the acidic strength, the nitro-substituted group is definitely more acidic. This can be reasoned by fact that a higher delta positive charge which develops on the H due to the strong -M nature of the nitro group.
Another way to check the acidity is to check the stability of the conjugate base.

The conjugate base would be the carboxylate ion which being a -M group creates a lesser electron density at the ortho and para positions. This positive charge is however stabilised by the methoxy group, which is a +M group, and destabilised by the nitro group. So from this viewpoint, the methoxy derivative should have been more acidic.

Why is this not so? What is the fault in my reasoning?

You are somewhat on the right track. However, consider that acidity depends on the relative stabilization of the acid vs. the conjugate base, since the expression for $\mathrm{p}K_\mathrm{a}$ is as follows: $$\mathrm{p}K_\mathrm{a} = -\log\left(\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\right)$$

Consider this figure for a visualization:

Electron-withdrawing groups like p-$\ce{NO2}$ (nitro group in the para position) increase acidity by making the carboxylate ion stable relative to the protonated form. However, electron-donating groups like p-methoxy decrease acidity relative to standard benzoic acid by destabilizing the (already negatively charged) carboxylate anion!

TL; DR: An electron-withdrawing group like p-$\ce{NO2}$ will better stabilize the negatively charged $\ce{COO-}$ group than an electron-donating group like p-$\ce{OCH3}$.

I found a neat picture from UC Davis that summarizes what I have said:

In Methoxy Benzoic Acid,we have Resonance or mesomeric effect which stabilizes the carboxylate anion.

In Nitro Benzoic Acid, we have Intermolecular Attractions (Some may call it as Hydrogen Bonding) between the Nitrogen(of Nitro group) in one molecule and the Carboxylate Anion in another molecule.

So, since the Intermolecular Attraction is more stronger than the Resonance effect, the Nitro benzoate is more stabilized than Methoxy Benzoate. So, Nitro Benzoic Acid is more acidic.


What exactly is "B-strain"? On factors that influence the stability of carbocations, I came across this peculiar one:

  Bulky groups attached to the positively charged $C$-atom (of the carbocation), stabilize it by B-strain, which prevents the C-atom for returning to the $sp^{3}$ state.

   Source: Pearson's Organic Chemistry (7e), Morrison and Boyd

The issue? I've never heard of "B-strain" before. I mean, it's possible that bulky groups stabilize the positive $C$-atom via inductive (and other) effects...but what's with "B-strain"?

I ran a cursory Google search, but I couldn't find any reliable Chemistry-literature that deals with this. In fact, it raised more questions than answers, most notably, the presence of another effect called the "F-strain".

So I must ask:

What is the "B-strain", and how does it contribute to stabilization of carbocations?

Additionally, is there such a thing as the "F-strain"? If so, what's it all about?

B-strain

Both B- and F-strains are the terms I found were originally used to describe steric hindrance in complexes [1, pp. 66-67]:

  The concept that steric hindrance in a monodentate ligand can affect the affinity of that ligand for a metal was first introduced by Brown and Bartholomay (1944) who suggested that co-ordination of a tertiary amine with bulky substituents would create strain within the amine. This strain, termed B-strain on account of it occurring behind the ligand atom with respect to the metal, would arise because although in the unco-ordinated amine any steric interaction between the bulky substituents could be relieved to some extent by an increase in the $\ce{C-N-C}$ angles.

Interestingly, further it is pointed out that at the same time there is no other solid evidences that B strain exists. Although modern organic chemistry textbooks are widely using this term for tertiary compounds.

From [2, p. 417]:

  Steric effects of another kind become important in highly branched substrates,
  and ionization can be facilitated by relief of steric crowding in going from the tetrahedral ground state to the TS for ionization. The relative hydrolysis rates in 80% aqueous acetone of $t$-butyl $p$-nitrobenzoate and 2,3,3-trimethyl-2-butyl $p$-nitrobenzoate are 1:4.4.

  This effect has been called B-strain (back strain), and in this example only a modest rate enhancement is observed. As the size of the groups is increased, the effect on rate becomes larger. When all three of the groups in the above example are t-butyl, the solvolysis occurs 13,500 times faster than in t-butyl p-nitrobenzoate.

According to [3, p. 350]:

  Not all steric effects decrease reaction rates. In the hydrolysis of $\ce{RCl}$ by an $\mathrm{S_N1}$ mechanism ..., the ﬁrst step, which is rate determining, involves ionization of the alkyl chloride to a carbocation:

  The central carbon in the alkyl chloride is $\mathrm{sp^3}$ hybridized, with angles of $\sim 109.5^\circ$, but when it is converted to the carbocation, the hybridization becomes $\mathrm{sp^2}$ and the preferred angle is $120^\circ$. If the halide is tertiary and the three alkyl groups are large enough, they will
  be pushed together by the enforced tetrahedral angle, resulting in strain. This type of strain is called B strain (for back strain), and can be relieved by ionization to the carbocation.

  [...]

  Substitution of ethyl groups for the methyl groups of tert-butyl chloride
  does not cause B strain; the increase in rate is relatively small, and the rate smoothly rises with the increasing number of ethyl groups. The increase is caused by normal ﬁeld and resonance (hyperconjugation) effects. Substitution by one isopropyl group is not greatly different. But with the second isopropyl group the crowding is now great enough to cause B strain, and the rate is increased 10-fold. Substitution of a third isopropyl group increases the rate still more.

Further in [3, p. 419]:

  For some tertiary substrates, the rate of $S_N1$ reactions is greatly increased by the relief of B strain in the formation of the carbocation. Except where B strain is involved, $\beta$ branching has little effect on the $S_N1$ mechanism, except that carbocations with $\beta$ branching undergo rearrangements readily.

There is also a Baeyer strain aka angle strain most noticeable for three-membered rings ($60^\circ$ is far from expected tetrahedral angles), which can also probably be abbreviated alike. [3, p. 193]

It also seems like earlier sources used B-strain as a reference to H. C. Brown [4, p. 291]:

  Hammond has proposed a steric "B-strain" (see H. C. Brown, J. Chem. Soc. (London) p. 1248 (1956)) argument to explain the greater stability of primary over secondary or tertiary carbanions. From the viewpoint of hybridization, structure A with $90^\circ$ geometry is preferred in carbanions. But this conformation brings the substituents too close together, and a compromise structure with larger bond angles results. Methide ion, $\ce{^-CH3}$, with the smallest substiuent groups, should be the most stable simple carbanion, since it can more nearly adopt the small angled conformation A preferred on electronic grounds. The more alkyl groups are added, the less able sterically are
  the anions to adopt the favorable conformations with smaller angles, and
  the less stable and more reactive the intermediates become.

F-strain

[1, p. 68] illustrates the origins of the term F-strain:

  The boron trifluoride and boron trichloride complexes of diethyl ether are less stable than those of dimethyl ether, and the same order of stability is observed for the complexes of diethyl and dimethyl sulphides. As expected, the steric interaction is decreased as the distance between the metal and ligand atom is increased. Thus it is decreased when the metal atom is changed from boron to aluminium, and when the ligand atom is changed from oxygen to sulphur. It is largely eliminated, of course, if the substituents on the ligand atom are held back away from the metal, and thus very stable complexes are formed by the tertiary amine quinuclidine, while complexes of tetrahydrofuran and tetrahydropyran (and also the analogous sulphur heterocycles) are far more stable than those of their acyclic analogues.

  Strain which arises as the result of this type of steric interaction has been termed F-strain by H. C. Brown -- who has carried out extensive investigations in this area -- on the grounds that the interactions responsible for the strain occurs mainly in front of the ligand atom with respect to the metal. For a critical but now rather dated review of H. C. Brown’s work, see Hammond (1956).

References

Houghton, R. P. Metal Complexes-Organic Chemistry, 1st ed.; Cambridge University Press: Cambridge; New York, 1979.
Carey, F. A.; Sundberg, R. J. Advanced organic chemistry Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.
Hartley, H.; Karabatsos, G. J. Advances in alicyclic chemistry; 1966; Vol. 1.

Now consider an acid (Say Boron trifluoride) forms acid-base adduct with a base  with bulky substituents (Say tripropylamine), then there will be steric hinderance between the bulky substituents which results in opening bond angle (Propyl-Nitrogen-Propyl) of the alamine. This opening in bond angle causes more s-character in the N-R (Propyl-Nitrogen) bond more p-character left in the lone pair of nitrogen due to which lone pair is poorly situated for donation. This phenomenon is called B-Strain.
Note: F-strain decreases the basicity of the base.
       More s-character results larger bond angle
         For more detail Refer Inorganic Chemistry By James E. Huheey, Ellen A. Keiter, Richard L. Keiter and Okhil K. Medhi


Why calcium chloride is used to melt ice over sodium chloride I'm not quite sure I understand this. My question is:

  Calcium chloride is a salt used widely to melt ice on sidewalks and roads.  Explain why one mole of $\ce{CaCl2}$ would be more effective than one mole of $\ce{NaCl}$.

So from everything I've learned, $\ce{NaCl}$ would be more effective for the melting, because calcium chloride dissociates into three ions and sodium chloride dissociates into two, so that would make the boiling point of water with calcium chloride higher.  Therefore, it shouldn't be used to melt ice over sodium chloride, as it would make the water have a higher boiling point so wouldn't melt it as well!

That one mole of $\ce{CaCl2}$ is more effective in melting ice than one mole of $\ce{NaCl}$ is explained by the van 't Hoff factor. The freezing point depression of a solution is calculated by
$$\Delta T=K_fbi$$
where $K_f$ is a cryogenic constant which is specific to the solvent, $b$ is the molal concentration of the solute, and $i$ is the van 't Hoff factor, which indicates number of solute particles. In this case, it is the number of ions produced upon dissociation. Because $K_f$ and $b$ are constant, and $i=3$ for $\ce{CaCl2 \rightarrow Ca^2+ + 2 Cl-}$ and $i = 2$ for $\ce{NaCl \rightarrow Na+ + Cl-}$ the freezing point depression is greater for $\ce{CaCl2}$.

The reason $\ce{CaCl2}$ is widely used as melting agent may have more to do with keeping the concrete intact. Concrete is composed of many calcium-containing species. When water flows over the concrete, the calcium can be leeched out, and the concrete becomes brittle. The higher the calcium content in the water, the less leeching occurs, and the concrete remains structurally sound.

The answer is simple. One of the unique properties of any salt is that it completely breaks down when dissolved in water. By that I mean each of the atoms clings to water molecules, preventing them from using their natural ionic bonds to each other. For every NaCl molecule, you can occupy 2 water molecules. For every CaCl2 molecule, you can occupy 3 water molecules. This makes it, Mol for Mol more effective an ice melt.


Why isn't ethane used for cooking? We commonly use methane and propane for cooking (and home heating), but not ethane. I would expect ethane to be suitable for this, being in between the two, but I've never heard of anyone using it for this purpose. Why is that?

On a related note, why is butane used for cigarette lighters and basically nothing else (in ordinary life, I mean)?

Probably the biggest drivers behind using methane as a fuel is that it is abundant in natural gas and is (currently) mostly useless as a chemical feedstock.
Ethane makes up a few percent of natural gas and can also be obtained as byproducts of petroleum refining, but the big difference from methane is that ethane is extremely useful in chemical synthesis (mostly to make polyethylene).
In fact, household natural gas often contains a bit of ethane, which may vary depending on the current demand for ethane as a feedstock.
In short: it burns just as well, but ethane has other important uses.

Propane has some use as a feedstock, but it is an attractive fuel because it is easily stored as a liquid without requiring huge pressures.

Methane and ethane are very hard to be liqufied with respect to propane and butane. To make methane liquid requires highee pressures and this means more cost and danger.

But especially, n- butane may be easily liqufied and liquid phase covers less volume. This makes it more suitable for lighthers.


Why does ionization energy increase as we go from left to right in a period? Why does ionization energy increase as we go from left to right in a period?

In my textbook, the explanation is as follows:

"This is consistent with the idea that electrons added in the same principal quantum level do not completely shield the increasing nuclear charge caused by the added protons. Thus electrons in the same principal quantum level are generally more strongly bound as we move to the right on the periodic table, and there is a generally more increase in ionization energy values as electrons are added to a given principal quantum level."

This, however, doesn't make any sense to me. From what I understand, in a stepwise ionization process, it is always the highest-energy electron (the one bound least tightly) that is removed first. So when we go more to the right in the periodic table, there are more electrons, and thus much more possibility for these to shield the outer electron from attraction to the nucleus. This would make it easier to remove that outer electron, and so I would say the opposite: the ionization energy should lower when we go to the right.

So can someone explain to me why this is not so? Thanks for any clarifications.

Crash Course on Ionization Energy:

As we all know, atoms prefer a full valence shell. So as we go right in a period, we are increasing $e-$. And also ADDING PROTONS. Because we are adding protons, the size of the atom gets smaller because the nuclear charge will be more powerful. Adding protons in a period trumps the addition of electrons. At the end of the day, we have a small atom with many electrons in it's valence shell that does not want to let go of them. Especially the Noble Gasses and Halogens.
Ionization energy decreases as we move down a group because:

As we move down, a new full energy level is being added. More electrons means more repulsion.
This creates the shielding effect where the addition of the shells, shields the outer electron from receiving the nucleic charge. NOTE: Here, however the addition of another energy level trumps the addition of protons.

This is just a piece of the whole picture.

Each time we move one atom across the period there is one more proton in the nucleus, this means that the nuclear charge is increasing. Also each time we move across the period we gain one electron, this electron goes into the same shell (all though the sub shells are slightly different in energies, the difference can be discarded when considering the attractive strength of the protons). Which means there is NO EXTRA SHIELDING of the valance shell electrons, as the electron is going into the valence shell. This means that the atomic radius decreases across the period as the nucleus is pulling all the shells closer to the nucleus as the nuclear charge is greater.

When considering ionisation energies we know that the first ionisation energy it is the energy required to remove the outer most electron. Regardless of the fact that there are more electrons the valance shell is closer to the nucleus therefore there is a greater attraction between the nucleus and the valance shell electrons, this greater attraction means it requires more energy to remove the outer most electron, hence the general trend of increasing first ionisation energies increasing across the period. Of course there are exceptions when you shift across the blocks and when pairing starts in the sub-shells, but that's a different question.
Hope this helps.


Which functional group is more polar in a drug: primary amide or a carboxylic acid? There is a lot of speculation as to which functional group is more polar, I've read a lot around this on the web and I can't decide which one is actually more polar? Can someone help because I need to know how to identify the most polar groups in drugs to create some drug monographs.

It depends on your exact definition of polarity
The measured dipole moments of formamide (3.7 D)1 and formic acid (1.4 D)2 in the gas phase indicate that the amide is intrinsically more polar.

However, this neglects additional factors relevant to the discussion: ionization and solvation. Under physiologic conditions, carboxylic acids are generally deprotonated resulting in a formal charge separation with a -1 charge delocalized across two oxygens. Depending on what parameters you are concerned with, this may be a more profound effect than the difference in gas phase polarity.

A carboxylic acid would be more polar.  Amide bonds are essentially how proteins are bonded together, and are the less-polar condensation product of an acid and and amine.  One way to think of it is, amino acids are more polar, and are converted to proteins, which are less polar.  Polarity may change with pH, but is the case at least for the only pH that you'd care about, which is that of the human body.


Protonation of Carbonyl oxygen in Fischer Esterification In the Fischer Esterification, why is it the acid's carbonyl oxygen getting protonated and not the hydroxyl oxygen? This protonation occurs in Step 1 of the mechanism below.

I read in some textbooks that this is due to the carbonyl oxygen being more neucleophilic and the hydroxyl oxygen. Why is this so?

There are a few reason why the carbonyl oxygen gets protonated -

1) The carbonyl oxygen is sp2 hybridized which is more electronegative than the sp3 oxygen. So, it has more tendency to give electrons since it is more negative.

2) The lone pairs of carbonyl carbon are more available since they are not in conjugation whereas the lone pairs of the other oxygen are not much available since they are in conjugation with carbonyl carbon.

3) When carbonyl oxygen donates its electron the resulting structure is more stabilized (because of conjugation) than the structure formed when the other oxygen atom donates.

So, that's why carbonyl oxygen has far more tendency of donating electrons than the -OH grp.


Strictly speaking, is air homogeneous? Why would air be considered homogeneous in its gaseous state (by Wikipedia and several other sources) even though its concentration keeps varying with time?

A homogeneous fluid as i think of it is as follows: If I take a small part of the fluid the number of particles of each kind is always the same.

Strictly speaking you might have a point there. But it is probably more difficult to judge.

I was unable to unearth an official recommendation by the IUPAC, so I unfortunately have resort to Wikipedia ([1] Kristi Lew (2009). Acids and Bases. Essential Chemistry. New York: Chelsea House Publishing.):

  Homogeneous is a term in physical chemistry and material science that refers to substances and mixtures which are in a single phase. This is in contrast to a substance that is heterogeneous.[1] The definition of homogeneous strongly depends on the context used. Generally it refers to smooth variation of properties with no discontinuities or jumps.

If we talk about all the air around the world, it is certainly not uniform. There are concentration differences across certain areas. When we consider a vector into space, the total concentration will decrease along this vector. As we know, we will pass certain areas with different chemical compositions, for example the ozone layer.
But all these changes will be gradually; there are no discontinuities. In this sense one would call air homogeneous, even though it does not fit in the definition of phase.

Additionally, if you take an arbitrary (volume) unit of air, all at the same temperature and pressure, and you allow it to establish an equilibrated state, then yes, this will not only be homogeneous, it will also form one phase.
The above linked article further considers air as a homogeneous mixture. It rambles on with considering it a gaseous solution of oxygen, carbon dioxide, etc. in nitrogen. This is technically incorrect, since the gaseous state of matter is specifically excluded from the IUPAC definition of solution. Also a solution can never be at the same time a mixture, as the former can be separated by mechanical means, while the latter cannot.

The problem here is obviously the lack of a unique and rigorous definition for homogeneity. For whatever purpose when it really matters, one should probably use a more robust definition like phase, which is essentially what your second statement describes.

Depending on the context you might even obtain a homogeneous mixture of water and oil - the question is then more or less how stable such a suspension is. After a while it will certainly separate in two phases.

Air is heterogeneous if you think practically. It is a mixture of nitrogen, oxygen, $\ce{CO2}$ etc. But the concentration of oxygen and $\ce{CO2}$ is not uniform everywhere. Polluted regions have more $\ce{CO2}$ while $\ce{H2O}$(steam) concentration is more in tropical areas or sea sides. Unpolluted air, which is rare to find, however is homogeneous.

Most text books don't mention the difference between polluted and unpolluted air....so there this is might be a good explanation!


Why does ZnO turn yellow when it is hot? Why does ZnO's color become yellow when it is hot? I have read that it is because of physical properties, but I need a more elaborate explanation.

The thermochromism of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed.

Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.

Upon heating zno anionic vacancies are created and when an electron occupies that vacancy it imparts yellow colour(metal excess defect/f-centre).


What happens when butane and methane both absorb 100 J of heat? Two cylinders, one containing 1 mole of $\ce{C4H10}$ gas at $\pu{1 atm}$ and the other containing 1 mole of $\ce{CH4}$ gas at $\pu{1 atm}$, are at $\pu{288 K}$.  If each gas absorbs $\pu{100 J}$ of heat under conditions of constant volume, which of the following is true?
  (a) The temperature of the $\ce{CH4}$ increases more than the temperature of the temperature of the $\ce{C4H10}$.
  (b) The internal energy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (c) The heat capacity of the $\ce{C4H10}$ is less than the heat capacity of the $\ce{CH4}$
  (d) The entropy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (e) The heat transferred to the $\ce{C4H10}$ is greater than the heat transferred to the $\ce{CH4}$.

The correct answer is (a). I am trying to eliminate answers, but I seem to get stuck.

We know that (b) is clearly false as we've added heat and no work is done, therefore the internal energy for both gases must increase. Likewise, (d) is false as the entropy should increase as the molecules have more heat and become more disordered. Finally, (e) is incorrect as we transferred $\pu{100 J}$ of heat to both systems.

But here is where I get stuck. I can't seem to eliminate (c) from just the information given.

First, consider this from a nice write-up on heat capacity:

  The amount of energy needed to raise the temperature of a substance by one degree in temperature will depend upon how many different ways the atom or molecule can store the energy in the form of these thermal motions and how much energy each kind of thermal motion involves.In general, for molecules with similar bonding the bigger the molecule the more heat has to be added to raise its temperature by one degree.

Now, examine the degrees of freedom for each molecule: there are more ways to store energy in the larger butane molecule (with 36 vibrational modes in addition to the translational and rotational modes) than there are in methane (9 vibrational modes plus translational and rotational ones).

Therefore, we'd expect a greater increase in temperature for the smaller molecule than the larger one for an equivalent amount of heat added to the system.

A is the correct answer.

C is incorrect because butane is larger than methane. This means butane has more London Dispersion Forces than methane, and so needs to absorb more heat to gain the same amount of temperature.

A is correct for the same reason: given that both absorbed the same amount of energy, methane is going to have more temperature than butane because it has fewer intermolecular forces holding it together.


How to determine which gas would have the lowest pressure? If the volume of each sample of gas was reduced to one-tenth of its original size and the temperature remained the same, which gas would have the lower pressure. I'm getting confused because my options are $\ce{C6H4}$, $\ce{C3H4}$, $\ce{N2}$, $\ce{O2}$, and $\ce{Cl2}$. Our teacher told us to pick the polar compound but the problem is all of them are non polar. Would the answer be $\ce{C3H4}$? I have a feeling I might be drawing the lewis structure wrong because I'm positive the other four are non polar.

Not all of them are polar. The gases  $\rm N_2$, $\rm O_2$,  $\rm Cl_2$ are not polar.

The $\rm HC\equiv C-CH_3$ is polar, but $\rm H_2C=C=CH_2$ should not be and $\rm C_6H_4$ is also polar, if it is right chemical formula you wrote. But I'm not sure which is more polar, but I will say that it is $\rm C_6H_4$

.

In more polar compound there will more interaction between molecules and that is why pressure will be smaller, simply because they "want" to be liquid form - not in gas.

A homodiatomic gas cannot be polar, by symmetry arguments. At best it can be polarizable by dispersion forces overall.  The lowest pressure would be the most strongly attracting molecules if any (e.g., water and hydrogen bonding), the most inert molecules, the smallest molecules (van der Walls equation coefficients), and then "anomalous" behaviors.  Look up the coefficients (CRC Handbook)to get the real world answer.

One would guess nitrogen for its being very tightly bound and all electrons bonded.  Look it up, as above.

/_\PV = energy, 101.325 J/liter-atm. Adiabatically compressing a gas tenfold will heat it.  A diesel cylinder compresses 15 - 22 to one, minimum 550 C at the top of the stroke.


How is sulphur trioxide formed and why is it stable? Structure of $\ce{SO3}$ (sulfur trioxide):

In the molecule, if each oxygen atom shares two electrons with sulfur atom then how does the sulfur atom remain stable? It already has 6 valence electrons and needs only 2 more to become stable but the oxygen atoms share total of 6 electrons with sulfur atom. So how is the molecule stable?

Synthesis

Synthesis of $\ce{SO3}$ can be done by heating $\ce{CaSO4}$ with $\ce{SiO2}$, which produces $\ce{CaSiO3}$ and $\ce{SO3}$. Another way of synthesis is the heating of $\ce{NaHSO4}$ to $\ce{Na2SO4}$, $\ce{SO3}$, and water. (source)

As to why $\ce{SO3}$ is stable...

Sulfur forms an expanded octet. That means that it doesn't really obey the octet rule, allowing it to take on extra electrons. Sulfur is a 3rd-period element; hence it can use its 3d orbitals to make more than 4 bonds. Wikipedia has a page on hypervalency, and from an answer here:

  Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: C,N,O, and F.

EDIT:

Users such as @Mithoron have pointed out that this analysis is not correct. Papers seem to show that there is no d-orbital utilization. According to this comment:

  ...the doubly bonded picture is a very crude oversimplification, though. -@Martin

This enlightening image

shows that sulfur trioxide is actually a resonance of several different structures, and should be more accurately expressed as several single bonds with a dashed line to show that the bond order is not actually two. This also solves the question as to why it is stable with apparently six bonds, as in reality it only has four bonds.

The sulfur element has D-shell available for bonding. That is why it can form six bonds and have an expanded octet. Oxygen on other hand uses p shell for bonding which is completely filled by sharing 2 electrons. As to why sulfur expands, sulfur has vacant d orbitals lying with comparable energy to the bonded ones hence it is energetically favorable to form more bonds. Also, note that typically atoms with large sizes form expanded octet with small surrounding atoms as there is less inter-electronic repulsion.


Help understanding the pressure's effect on equilibrium and Le Chatelier's principle So let's say we have a closed system in the following reaction:

$\ce{N2O4 -> 2 NO2}$

The only way to increase the system's pressure without adding any mole of gas and keeping temperature constant is reducing the system's volume. Once we increase the pressure by reducing the total volume, formation of $\ce{N2O4}$ is observed.

In many text books, the explanation for this is that 2 moles of $\ce{NO2}$ occupy a bigger volume than 1 mole of N2O4, thus the equilibrium is restablished when more $\ce{N2O4}$ is formed. What doesn't make sense for me is that both the substances are contained in the same volume, so how 1 mole of $\ce{N2O4}$ would occupy less volume than 2 moles of $\ce{NO2}$? It only makes sense to think about partial pressures, not partial volumes, right?

I'm trying to use entropy and energy microstates to solve this problem, because it seems more fundamental than this mantra in the general chemistry textbooks. From quantum mechanics, I know (can't explain, just know; need further studying on this subject) that translational energy states become closer when a gas expands in a bigger volume, so I'm assuming that in the reaction above, with a bigger pressure (and thus a smaller volume) these energy states will be more spaced between each other. With energy states more far apart, why are there more possible energy microstates when $\ce{N2O4}$ is formed?

I can't get past this! Why does this implies in the formation of the substances with less gaseous coeficients? How are the energy states of reactants and products and their changes on pressure related with that?

Edit: to make myself clearer, I'll use this example from this textbook I'm reading:

Chem1 Virtual Textbook, Stephen Lower, link

In the endothermic reaction H–H (g) ⟶ 2 H (g), you only get hydrogen atoms at very high temperatures, because you'll have a higher distribution of populated states on the 2 H side, thus you favor the products at higher temperatures for endothermic reactions. I can completely understand the effects of themperature in Le Chatelier's principle, and I'm trying to understand an analogous thought for changes in pressure in the microscopic point of view.

What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.

In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG < 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.

A molecule's rotational states do not become closer as the pressure is increased, neither do the vibrational levels, they are a property of the molecules themselves. The population in these levels is determined by the temperature which in your case is constant.   More collisions per second as the pressure is increased means more possibilities of reaction. This means that there are more chances that at a given total energy (comprised of translational, rotational & vibrational energy) the molecules will have enough energy to cross the energy barrier (transition state) between reactants and products and vice versa, i.e the number of barrier crossings both ways is increased.

Note that this works both ways, a collision between two $\ce{NO2}$ can form $\ce{N2O4}$ as well as a collision destroying it and forming $2\ce{NO2}$, i.e. a collision between $\ce{N2O4}$ and $\ce{NO2}$ (or between two $\ce{N2O4}$ ) can sometimes convert $\ce{N2O4}$ into $2\ce{NO2}$.

The species are in equilibrium $\ce{N2O4 <=> 2NO2}$ and the equilibrium constant has a fixed value at constant temperature and is unaffected by changes in total pressure, because, as noted above, the population of energy levels depends only on temperature.

[Notes: The equilibrium constant can be expressed as the ratio of partial pressures. At equilibrium $\Delta G^\mathrm{o}=-RT\ln(K_p)$ where for a reaction $\ce{xA + yB<=>rC + sD}$ the equilibrium constant is $\displaystyle K_p= \frac{P_C^rP_D^s}{P_A^xP_B^y}$. The change in $K_p$ with temperature is given by the Van't Hoff isochore, $\displaystyle \frac{d\ln(K_p)}{dT}=\frac{\Delta H^\mathrm{o}}{RT^2} $ ]


Why does fluorine stabilise a carbocation? I have read that halogens have a stonger -I (inductive effect) than +M (mesomeric effect),then why do they stabilise a carbocation instead of destabilising it?

I asked this doubt to my teacher and he said that halogens like fluorine do not exert -I on a carbocation and hence they stabilise a carbocation through +M,is this true?

Fluorine belongs to the same period as carbon. Carbocation is electron deficient which has its two $\mathrm{2p}$ orbitals vacant. One electron of carbon in $\mathrm{2p}$ orbital goes into the formation of $\ce{C-F}$ σ bond. $\ce{F}$ has $\mathrm{2p}$ orbital containing a lone pair which can be donated to the $\mathrm{2p}$ vacant orbitals of carbocation. Same size $\mathrm{2p}$ orbitals of carbon and fluorine overlap effectively to form a π bond.

Carbocations are stabilized by neighboring atoms with lone pairs. The key stabilizing influence is that the neighboring atom can donate a pair of electrons to the electron-poor carbocation. In this way halogenes, nitrogen or oxygen can actually be an electron-donor group and stabilize the carbocation.


Why cannot HCN react with ethene? I expected ethene ($\ce{C2H4}$) to undergo electrophilic substitution due to $\ce{HCN}$ but it doesn't happen, as I am told by my textbook.

I couldn't figure out the reason, as I found out that $\ce{HCN}$ does react with $\ce{CH3CH2Br}$ by nucleophilic substitution.  Upon ionisation, the proton of $\ce{HCN}$ is an electrophile,while the cyanide group is a nucleophile.  Correct?

The "classical" way

Direct addition of $\ce{H-X}$ to alkenes proceeds via the addition of a proton to the $\ce{C=C}$ double bond in the initial step, i.e via a carbocation. This works nicely for strong mineral acids, but not for the weak acid $\ce{HCN}$ with the rather weak nucleophile $\ce{CN-}$.
Synchronous addition?
Q: Can't $\ce{H-CN}$ add synchronously add to $\ce{C=C}$?
A: NO! Such a reaction would involve the interaction of the HOMO (highest occupied molecular orbital) of one reaction partner with the LUMO (lowest unoccupied molecular orbital) of the other. The LUMO of an alkene has a nodal plane orthogonal to the $\ce{C-C}$ bond, while there is - obviously - no nodal plane but a $\sigma$ bond between hydrogen and carbon in $\ce{HCN}$.
The synchronous addition of $\ce{H-CN}$ to $\ce{C=C}$ is symmetry-forbidden!
Catalysis?
Yes! Since the first catalytic hydrocyanation using $\ce{Co2(CO)8}$ (see Philipp's excellent comment above), other catalysts have been developed, namely $\ce{Ni(0)}$ based species using a trypticene-derived chelating phosphine ligand (see image).

Summarizing Phillip and Siddahrtha's comments.

Under normal conditions ethene normally undergoes addition since the double bond is nucleophilic.  In order to have a substitution reaction on ethene H $^{-}$, would need to be the leaving group.

$C_2H_4 + HBr \rightarrow CHCHBr + H^{-} $

This is not energetically favorable under normal conditions.  Under high pressure and with a catalyst the reaction would be possible.


Why do unboiled eggs last longer than boiled eggs? Just like the title says, I watched this video https://youtu.be/CHMY4G9gTPA and then became intrigued.

An eggshell is a complex structure.  From Wikipedia:

Boiling the egg removes the waxy cuticle from the outside of the egg, dissolves a small but not insignificant amount of calcium carbonate from the shell, damages the protein matrix that holds calcium carbonate crystals in place in the shell, and can disrupt or destroy the two shell membranes.  All of these contribute to allowing bacteria a means to enter the egg and spoil it.

Note: Though calcium carbonate is actually less soluble in hot water vs cold, the rate of dissolution is substantially greater.  Unless the egg is boiled in a saturated solution of calcium carbonate, it will lose material to the solution when boiled, and any pores in the shell become enlarged as a result.

In addition to Jason answer: egg is a live organism with repair and immunity capabilities. Small punctures in egg membrane may be healed, and bacteria are most likely to get killed by white blood cells using antibodies or other techniques. Boiling kills everything, including egg immunity system - existing antibodies are denaturated (they are protein too) and there are no white cells to produce fresh ones.


Why can't alcohols form hydrogen-bonded dimers like carboxylic acids? Carboxylic acids such as acetic acid are capable of forming dimers:

I'm wondering why alcohols like ethanol don't generally form dimers. In the diagram below, the oxygen atom on the left ethanol molecule should be capable of accepting a hydrogen bond from the partially positive $\ce{H}$ atom on the right ethanol molecule. Does it have something to do with the $\ce{-OH}$ bond angle?

The strength of a hydrogen bond somewhat depends on the $\ce{X-H\bond{...}X}$ angle that the hydrogen-bonding hydrogen forms with the two electronegative elements $\ce{X}$. In our case, carboxylic acids or alcohols, $\ce{X} = \ce{O}$ so the angle is $\ce{O-H\bond{...}O}$. The ideal angle for this fragment is $180^\circ$.

As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire $\ce{(R-COOH)2}$ feature as a benzene-like ring streched in a single direction. Most importantly, the acceptors and donors line up nicely, the carboxyl function features an angle of $120^\circ$ and the $\ce{C-O-H}$ angle (and the $\ce{C=O\bond{...}H}$ one!) is also close to $120^\circ$. These arae the theoretically predicted, unstrained angles.

For two alcohol (e.g. ethanol) molecules to attempt a similar arrangement, we would end up with only four atoms that have to form a rectangle with $\ce{O, H, O, H}$ at the four corners. Thus, the $\ce{O-H\bond{...}O}$ hydrogen bond angle would be much closer to $90^\circ$ — a very bad arrangement. Furthermore, this would put the two electronegative oxygens (which thus feature a negative partial charge) closer together with nothing in-between which would cause destabilisation due to like charges approaching. The hydrogen-bonding hydrogens cannot alleviate this unfavourable interaction since they are at the corners of the square.

The situation for alcohols is much better if they create a network of a number of molecules that allow for much more favourable angles. With four molecules, you could already create a cyclic structure of approximately $100^\circ$ $\ce{H\bond{...}O-H}$ angles and $170^\circ$ $\ce{O-H\bond{...}O}$ angles — much more favourable. Of course, in the actual solution this number will fluctuate strongly as hydrogen bonds are broken and reformed constantly and different ring sizes happen all the time.

The main reason for the higher extend of H bonding in carboxylic acids is due the higher acidity of the carboxyl group (e.g. ethanol: $~\mathrm pK_\mathrm a = 16$, acetic acid: $~\mathrm pK_\mathrm a=4.5$) making it a good hydrogen bond donor, and the ability of the carbonyl oxygen to act as a good hydrogen bond acceptor.
Thus the extend of H bonding can be explained solely based on acidity/basicity arguments. Compared to that a hydroxide is just too basic to be formed out of the hydroxyl group, so hydrogen bonded dimers of alcohols make no sense. What is also important to point out, is that the resulting dimer will be completely symmetric (due to resonance) making the situation energetically even more favourable.


Best way to raise pH without alkali metal? I have a process that requires pH~14-14.5 to work. I’ve been achieving this with concentrated NaOH solutions, but the Na cations are later causing side reactions which are detrimental.  Is there any organic base/non-alkali base that has the solubility in water and pKa to reach these pH values?

Tetramethylammonium hydroxide

This is a strong base like NaOH.

See Behavior of Polyethylene Oxide Based Nonionic Surfactants
in Silicon Processing Using Alkaline Solutions J. Electrochem. Soc. 1995   volume 142, 621-627:

  alkaline cleaning solutions based on quarternary ammonium hydroxides such as choline ($pK_b = 0.1$) and tetramethyl ammonium hydroxide ($pK_b = 0$)

and Ceramic Nanomaterials and Nanotechnology II which says:

  In this research, tetramethylammonium hydroxide (TMAH) was used, instead of KOH, to adjust pH of the precursor solution to 14.0

Also, TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society (1905):

  An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.

Water solutions of amines can get to pH+ 12 or so. Piperidine, $\ce{(CH2)5NH}$, is about pH 11.

To get much higher pH+ requires somewhat exotic fluorinated organic chemicals, such as $\ce{9-COO-Me-fluorene}$, $\ce{(4-Me-C6F4)2CHCN}$ or $\ce{(4-Me-C6F4)(C6F5)CHCN}$. See Paenurk et al, A unified view to Brønsted acidity scales: do we need solvated protons? and download the PDF for an interesting explanation.


If the pH is zero, what would be the normality of a sulphuric acid solution? If pH of the solution is zero, what would be the normality of $250 \, \mathrm{ml}$ $\ce{H2SO4}$ solution?

My work:
$\mathrm{pH} = 0$, hence the concentration of $\ce{H+}$ ions is $\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\pu{1M}$ hence the normality of the solution must be $2~\mathrm{N}$.  Is this correct?

$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$

At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$

At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or  $\ce{SO4^{2-}}$.

Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.

Normality of an acid is the concentration of the acid times the number of acidic protons per molecule.  Sulfuric acid has two acidic protons per molecule.

Normality $= \pu{1M} \times 2 = \pu{2N}$

Here the $\mathrm{pH}$ is $0$, it means it has concentration of $\pu{1 M}$, i.e. $[\ce{H+}]$ is $1$.
$$[\ce{H+}] = 10^{-\mathrm{pH}} = 10^{-0} = \pu{1 M}$$
Hence the molarity will be $1$. So its normality will be,
$$\text{normality} = \text{molarity}\times\text{n-factor} = 1\times 1 = \pu{1 N}.$$
Here we have taken $\text{n-factor}$ for $\ce{H2SO4}$ will be $1$, because $\mathrm{pH}$ is given as $0$, so it will be a strong acid and in strong condition the $\text{n-factor}$ or valency factor becomes $1$ and so it forms to ions $\ce{H+}$ and $\ce{HSO4-}$, i.e. $\text{n-factor} = \text{no. of $\ce{H+}$ replaceable ions is 1}$, hence normality will be $\pu{1N}$.


Is alcohol really a solution? A friend recently sent me the following chemistry joke:

As a lover of puns, I'd like to enjoy it, but I think that it is actually false the way it is phrased.

A solution is defined as: "a homogeneous mixture composed of two or more substances" Alcohol alone would not constitute a mixture of multiple substances and would therefore not be a solution.

But in English the word "alcohol" is also commonly used to refer to many recreational drinks containing ethyl alcohol in addition to other types of molecules. Are there any common examples of alcoholic beverages which are solutions in the scientific sense that would make the pun valid, such as liquor, wine, or beer?

Technically, alcohol is the name of a class of organic compounds containing one or several hydroxyl groups.

Colloquially, the term "alcohol" is understood as you have described: A solution (of varying degree of purity) of ethanol and water. Pure ethanol is impossible to create via the traditional method of purification (atmospheric distillation) as the water-ethanol mixture is azeotropic:¹

  An azeotrope [...] is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because, when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture.

  Source: Wikipedia

This is why a chemist would never think "100% pure ethanol" when you mention "alcohol".

If you check for example this vendor page for the purest ethanol (>99.5%) I could find, you see various substances listed as impurities:

<0.2% Water
<0.1% Methanol
Other organic substances, such as isopropanol and acetone

Concerning your second question, I would argue that many distilled spirits (vodka, whisky, and cognac just to name a few) are solutions of ethanol, water, and various flavour compounds.

Some alcoholic beverages, such as unfiltered beer and wine, usually also contain solid particles and would as such be correctly termed "suspensions".

¹ I wonder now whether it would be possible to produce 100% ethanol from another mixture somehow. The problem is probably that every common starting mixture already has some amount of water. But synthesizing ethanol in, say, an ionic liquid under an inert atmosphere and then distilling it away might be an interesting approach.

I would argue that these are mixtures, not solutions. A solution requires at least two different components. A solution might look like alcohol (being the solvent) that would take water (the solute) into solution, making $\ce{H+}$ cations and $\ce{OH-}$ anions. That would be an acid and a base, that would immediately turn back to water. That would never happen. The water molecules would just stay together in the first place. The fact is that it is a mixture, not a solution.


Parallel 1st Order Reaction Rates Consider some general parallel first order process, where $\ce{A -> B}$ and $\ce{A -> C}$ (rate constants $k_1$ and $k_2$ respectively)
The overall rate equation of this process is said to be $-\frac{\mathrm d[A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$
I get the impression that this is derived from the addition of each individual rate equation, i.e. the addition of $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A]$ and $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_2[\mathrm A]$ would result in the overall rate equation shown above.
However, if this is the case, then why isn’t the rate equation actually $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$? I mean, you’re adding two $-\frac{\mathrm d[\mathrm A]}{\mathrm dt}$ terms together; and so considering basic algebra, surely this would result in $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt}$?
Or do differentials just not work in this way, and I’m simply confusing myself?

There is no need adding the number $2$ in front of the definition of the rate. So the rate equation is :
$$-d[A]/dt = k_1 [A] + k_2[A] = (k_1 + k_2)[A]$$


Stability of carbocation directly attached to a halogen atom I have two compounds..

How is the first one more stable?
As I have been taught that -I effect of Chlorine dominates its +R effect but to explain this case, My teacher is saying that because of the achievement of the complete octet, chlorine donates its electrons to the carbocation(showing +R effect). If that's the case then the nucleophilic addition of alkanoyl chloride should be the slowest but experimentally proven, they show the fastest nucleophilic addition reaction by increasing the electrophilicity of the carbon atom(showing -I effect). Moreover, on treatment of alkynes with HCl, we get gem dihalides(which also shows the +R effect of chlorine).

Please help me with this because I am really confused between the -I and the +R effect of halogens.

some people say that its because of octet completion of the carbocation
and the NS reactivity that you mentioned is due to the -I effect of chlorine which makes it the most reactive and not the least reactive


Is Kp and Kc only effected by temperature? Are Kp and Kc only effected by temperature?

So in exothermic and endothermic equilibrium reactions you are increasing or decreasing temperature, which will effect which side is favourable.

But then so does pressure if one side has fewer molecules?

According to LeChatelier's principle on increasing temperature in an exothermic reaction "which is at it's equlibrium initially" the reaction proceeds in backward direction and in endothermic reaction it proceeds in forward direction, this can be thought as in exothermic reactions the heat should be provided to the products to yield the reactants and in endothermic the heat should be supplied to reactants to yield the products!
And the value of equilibrium constants depend only upon temperature and the pressure only effect the equilibrium conditions and not the value of Kp because partial are so adjusted to provide the same value of Kp at the same temperature.


How does boiling water affect its hardness? Does boiling water reduce its hardness, or on the contrary, increase it?

It seems there are ambivalent phenomena at play here:

boiling will evaporate pure water, thus increasing the concentration of minerals including calcium, thereby increasing hardness
boiling will remove $\ce{CO2}$, making apparently $\ce{CaCO3}$ precipitate, thereby reducing hardness

This is based on rather uncertain sources, as I didn't find any better, and maybe I missed other aspects.

I would appreciate any clarification, but I understand that the answer might depend on the context and type of water considered.

You may need to think outside chemistry a little bit and pay a little bit attention to your everyday life. This is what happened when you boil water in a container:

Most hardness in water, calcium/magnesium, will precipitate onto the hot surface and is removed from water. Now if you can get more something by removing it, is it against the very fundamentals of science?

Boiling water will NOT reduce hardness of water, whereas heating hard water to the boiling point will reduce solubility of $CaCO_3$ (Edit: and other carbonates and bicarbonates), thus reducing hardness, if filtered before cooling back. Solubility of $CaCO_3$ reduces with increasing $T$ hence boiling will not help, as the temperature is constant.

The $CO_2$ removal would not have much effect unless you have a closed vessel, where you can substantially increase the $P[CO_2]$ by removing small amount of dissolved $CO_2$ in water.


Monochlorination of 2-chloropentane and possible enantiomer products I found a question to determine the number of possible chiral products of monochlorination of 2-chloropentane.So, first the products are (with possible number of enantiomers in the parentheses:

2,2-dichloropentane(no chiral product),

1,2-dichloropentane(4 or 2, I am not sure is there a plane of symmetry here)

2,3-dichloropentane(2 or 4)

2,4-dichloropentane(obviously none for a plane of symmetry)

2,5-dichloropentane(2,4?)

But my surprisingly the answer says it's only 5. That's even lower than my lower estimate. I am not sure how there are 5 enantiomers.

2-chloropentane is optically active compound hence in some cases we will get only diastereomers. For example 1,2-dichloropentane is single product because it is formed by attack at carbon no. 1 hence no change in chirality of carbon no. 2.  So product formed by attack at carbon 1 (one chiral product) carbon3 (two chiral product i.e. diastereomers) carbon 4 (one chiral product, i.e. one chiral and one meso) carbon5 (one chiral product)

The answer is 5.
 See-
1) 1,2 dichloropentane --- 1 chiral
2) 2,3 dichloropentane --- 2 chiral
3) 2,4 dichloropentane --- not chiral as there is plane of symmetry
4) 2,5 dichloropentane -- 2 chiral

Therefore 2+1+2 =5 Chiral Products


Why is the para product major in the nitrosation of phenol? When phenol reacts with nitrous acid ($\ce{NaNO2 + \mathrm{conc}\ H2SO4}$), 4-nitrosophenol and not 2-nitrosophenol is formed. I cannot understand why the para-isomer should be preferred over the ortho-isomer.

I understand the reaction mechanism proceeds by electrophilic substitution by the $\ce{NO+}$ cation, and that the $\ce{OH}$ group activates the ring, which leads to ortho and para substitution.

I don't understand why the para-isomer should be preferred over ortho?

Ortho and para positions are activated due to the +R effect of the $\ce{-OH}$ but the preference of the para product can be explained by the mechanism.

The mechanism proceeds through a dienone. The ortho dienone intermediate will experience strain as  $\ce{=O}$ annd$\ce{=NOH}$ lie in the same plane. Para nitrosated intermediate on the other hand is free from strain.

Reference: González-Mancebo S., Lacadena J., García-Alonso Y., Hernández-Benito J., Calle E., Casado J. Monatsh. Chem. 2002, 133 (2), 157-166. DOI: 10.1007/s706-002-8245-0.

Due to steric hinderance and also there is some electronegativity of alcohol group due to induction it takes electron from sigma bond though not dominating therefore

These are the 2 reasons for more p- substitution


How much can the pH change through dilution? Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it?

Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry?

Water undergoes autoionization, i.e., it reacts as follows:

$$
\ce{H2O + H2O <=> H3O+ + OH-}
$$

The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.

Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:

$$
[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}
$$

We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:

$$
\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}
$$

Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.

It can not be alkaline because the water is not really alkaline, more like neutral. But dilution in an alkaline solvent will easily make the solution alkaline.

EDIT: seems my answer was not considered an answer. Well since the question is composed out of more than 1 question...

Using water, you won't be able to get past aprox 7 PH. Using an alkaline non reactive solvent will allow you to go as far as the PH of the solvent.

Additional clarification: If no reaction happens, the PH limit of the solution is bound between the substance with the highest PH, and the one with the lowest.

No, an acidic solution can not be made alkaline by adding water.


Comparison Of Lewis Acidity Which is a stronger Lewis Acid $\ce{AlCl_3}$ or $\ce{SiCl_4}$?

I am confused as in $\ce{AlCl_3}$ , Al has an incomplete octet so it must be a stronger Lewis acid as compared to $\ce{SiCl_4}$, but the answer is just opposite.

AlCl3 + RCl → AlCl4(-) indicating that AlCl3 is a Lewis Acid
but
SiCl4 + RCl → SiCl5(-) ??????  So, this tells me that AlCl3 > SiCl4 in terms of electron accepting behavior.
Now, don't get me wrong: pentacoordinated Si is well known. One of the failures of the Lewis Acid concept is that, as far as I know, there is no unambiguous way to measure it. It is much more qualitative than quantitative, despite years of attempts to do so. You may know that Phosphorous is typically pointed to as the first example of the breakdown of the octet rule. (see Hypervalence) After Period 3, the concept has little use, imho. Since Lewis Acidity is tangled up with Lewis structures, perhaps it is not too surprising that the concept has some serious problems. Usually, we condsider both electronic effects as well as steric effects to 'explain' Lewis Acid/Base strength. One popular alternative is Hard.Soft.Acid.Base (HSAB) "theory". Anyway, comparing a tri-coordinated species with a tetracoordinated species and predicting which will form a bond with an electron donor is really difficult. Did you know that AlCl3 doesn't really 'exist' in the solid or liquid state? And even in the gaseous state, the dimer is common. The structure of the (solid) dimer (Al2Cl6) has two Al-Cl-Al bonds. Anyway, as far as I know AlCl3 is the stronger Lewis Acid in most apple-to-apple comparisons. To be honest, I don't think that if the Lewis Acid concept weren't a historically established meme, that it wouldn't be invented today. Personally, scales should actually be founded on measurement. The Lewis Acid strength scale is just sorta kinda correlated to what we can actually measure.


Deflection of Alpha Particles in Rutherford's model of atom Rutherford stated that deflection of alpha particles is due to repulsive positive charge of nucleus. Why can't alpha particles be deflected by attraction of so many electrons revolving around nucleus?

See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles.
Figure 1 shows a sketch of the apparatus.

Figure 1: depiction of the apparatus (courtesy of Wikipedia).
Now follows a simplified rationalisation.
Assume you have an atom-thick sheet of some material.
By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $n$ are there, and knowing the sheet's surface area $A$ allows you to calculate the atomic surface density $\sigma = n/A$.
Now if the atoms have radius $\rho$, an area $A_\rho = n \pi \rho^2$ of the sheet should be covered by atoms.
This means a ratio
$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$
is actually covered by atoms and $f = 1 - r$ is the ratio of "free space".
Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through.
Imagine $M$ particles were thrown and $m$ of those passed through.
If $M$ is big enough we can set $f_R = m/M$ as Rutherford's "free space" estimate.
Going backwards we may set the coverage ratio
$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$
were $\rho_R$ above is the measured radius of the collision cross section of an atom.
The experiment has shown that $r_R \ll r$ and thus $\rho_R \ll \rho$, which implies that most of the atomic mass is buried inside the atom.
In fact (Wikipedia),

The diameter of the nucleus is in the range of $1.75$ fm ($1.75 \times 10^{-15}$ m) for hydrogen (the diameter of a single proton) to about $15$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $23,000$ (uranium) to about $145,000$ (hydrogen).

In Rutherford's own words (p. 68),

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
--- Ernest Rutherford

Thus, nuclei are simply very small, but very dense, as stated by Enigma and Ilmari Karonen.

Well gentlemen we can't deny that fact that there might be some forces of attraction between the alpha particles as well as the electrons,because the alpha particles passed through the atom by small distance margin from nucleus and stroked the zinc sulphide coating but it wasn't clearly stated by Ernest Rutherford that whether the deflection was also caused due to alpha particle attraction and election attraction. Also, attraction (interaction)is between comparable things like a heavy nucleus and heavy alpha particle but it's silly to think about alpha particle and an electron , the alpha particle is approximately 4000 times heavier than electron and it is travelling with a high speed (energy) so we can say that the interaction time might be very much less in order to show its effect and we can underestimate it . though most of the physics practicals are based on assumptions and we have to underestimate those assumptions that's way because the reason for performing the experiments was to propose a model of an atom (accidentally) , to find where is the mass of the atom is concentrated. that's why we got the  so we don't have anything to do with the alpha particle and electron attraction.


Over potential -what and why ? Value of standard electrode potential for the oxidation of Cl– ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl– oxidised at anode instead of water?

Standard electrode potentials describe the tendency of getting reduced.Positive value indicates that H2 can reduce that element or reduced form of the element is more stable than proton.
The answer to the question above is over potential
Please explain how it came into the story.My book explained it under "products of electrolysis",but the explanation is not lucid.
They explained it as :
Some electrochemical processes although feasible, are so slow that at  lower voltage these do not seem to take place and extra potential called over potential has to be applied which makes such process more difficult to occur.

When I took P-Chem, decades ago, one experiment we had was using a dropping mercury electrode, DME. I don't recall the details, but the set-up was fairly elaborate, especially compared to a Ag/AgCl electrode. The DME is fairly common (or was, before we decided that Hg was probably better avoided by the undergrad). The reason WHY we used it was it provided a CLEAN surface. Think about it. We wouldn't go to such great lengths if the problem of contamination wasn't significant.

All materials absorb contaminants, may chemically react with near-by chemicals (solvent, ions, gasses), and will form a charge polarization layer, especially in polar solvents like H2O. In fact some electrochemical methods rely on it. But for getting the "pure" ideal thermodynamic (reversible) numbers the large tendency of electrode surfaces to change is a real problem.

Overpotential is the term we use to describe the additional energy (voltage) we have to put in to these non-ideal systems (either at the anode or the cathode) to overcome these "side reactions". This energy is ultimately converted (mostly) to heat. Although maybe not an overpotential issue, surface area can be important, so smoothness does matter.

I can think of no area in chemistry (short of biochem) where you have to get into the weeds (details) so quickly as you do with electrochemistry. I wonder if its because chemistry is actually about electric interactions between atoms, molecules, ions, etc.? I should also mention that on other nice thing about the DME was that the surface was smooth - although I can't recall whether the surface was smooth at the atomic level, most solids are far from smooth, another problem in reproducibility.

This is a question of kinetics vs thermodynamics. From the perspective of thermodynamics equilibrium of water, HCl, O2 and Cl2 is shifted towards HCl+O2. However kinetics dictates Cl2 to form faster than O2.

Kinetic vs thermodynamic control is a commonplace in chemistry. Thermodynamics suggests that we all should be dead, but thanks to kinetics we are still alive. Why Cl- is kinetically favorable? Because 4 electrons should be taken from 2 H2O molecules to form O2. At the same time intermediates in this oxidation (atomic O, H2O2) can easily oxidize two chloride ions and yield chlorine gas.


Inorganic IUPAC for Compounds with bridged ligands? Can we use  bis(tetraammine) instead of octaammine is the 2nd example ?

As in the 1st example they used  bis(pentaammine)  and also in the 3 one the used bis(carbonylcyclopentadienyliron)

Here is an example of a similar ligand and it can be named in three ways.
All you have to do is to put hydroxido/hydroxo in place of superoxido.

So yes, bridged ligands can be named in a few ways, combining any amount of recurring atoms separately, but most common used would be (if the LHS and RHS are the same) μ-amido-μ-hydroxidobis(tetraammine)dicobalt.
Reference

Crystal structure of .mu.-amido-.mu.-superoxo-bis[tetraamminecobalt(III)] tetranitrate, $\ce{[(NH3)4Co(NH2)(O2)Co(NH3)4](NO3)4}$ by
Gary G. Christoph, Richard E. Marsh, William P. Schaefer, Inorg. Chem. 1969, 8, 2, 291-297, DOI: https://doi.org/10.1021/ic50072a022