text:
Haloform reaction in acidic condition
I want to know about the reaction mechanism of haloform reaction in acidic condition. I am following a persons doctoral thesis where he used acid medium haloform reaction but unfortunately he has not given any references.

Do you have a link to the thesis?

Please give necessary references, it took me for a surprise and I'm interested to know more.

In the book by Clayden et al. you can find a very nice description of the mechanism of the halogenation of enols and the haloform reaction on pp. 533-536. Especially on p. 536 the authors give the mechanistic details of the acid catalysed halogenation of enols and some reasons why multiple halogenation steps on the same carbon atom (as happens in the haloform reaction) occur easily under base catalysis but do usually not occur under acidic conditions. Since I lack the time to write it up myself (and since the authors already did a marvelous job at describing it) I simply post the relevant passage here:


text:
What is the explanation of the odd even effect in liquid crystals?
What is the most accepted theory for explaining the odd even effect in liquid crystals? Can anyone explain and refer the corresponding paper?

Do you have access to a literature search tool like e.g. SciFinder, ISI Web of Science, or even Google Scholar? Have you already done some searching using any of those methods?

I hope you mean liquid crystals with LC. I also added the reference request to the question, if I introduced any mistakes, please [edit] yourself.

I have been busy in going through more and more about main chain LC and did not search rigorously. But I did search about odd even effect but did'nt get any good results. Searched in Google. I will try to look whether my Institute has SciFinder access or not

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Evaporation of acetic acid
I performed aan acid catalysed haloform reaction in acetic acid to an acid. I need to evaporate acetic acid which I used as solvent and It is around 150 ml. What are the alternatives , I have a rotavap in lab but 150ml AA is too much for a rotavap I guess.

Are you evapotating to dryness? Don't worry about alternatives - use the rotavap with a not too small wide-neck (NS29) flask.

Alternative: destillation followed by neutralisation/extraction. I used that to remove propionic acid (whose boiling point is a lot worse for rotavaps than acetic acid’s) from my porphyrin condensation.

Rotavap is fine even until > 500 mL volume.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Synthesis Challenge
I have been trying to make a monomer reached a dead end in the middle. A B C are the products I have made so far respective order. After reaching C I wanted  to prepare 2 but could not find a way to do so. haloform had a high chance of brominating Hydroxyl group at other end I speculate. Acid catalysed Haloform on A did not result in 1 but ended up with B again [with suprisingly 2 % more yield than I got with simple base hydrolysis. Now at this point I have no option but to produce 1 and then go for 2. I need references or suggestion to make 2 from C cause that will greatly reduce my work load. I am running short of time for my summer project.

Protect terminal OH, reduce ketone, eliminate to styrene, ozonolysis with oxidative work-up?

@Jan You could probably write this up as an answer

@punarbasu-roy I'm not sure which compounds you need help synthesizing. Correct me if I'm wrong, but you need help with making compound 1 and 2?

The route I would try to prepare 1 would be to purchase 4-Bromo-4'-hydroxybiphenyl which is commercially available (Alfa Aesar sells it - see http://www.alfa.com/en/catalog/A10819)
First Step:
1) Protect phenol moeity with benzyl group via refluxing a solution of 4-Bromo-4'-hydroxybiphenyl (1.0 eq, 1.00 g), milled potassium carbonate (4.0 eq), benzyl bromide (1.1-1.25 eq), and acetone (25-30 mL). Reaction will most likely take 10-20 hours to be complete. Once reaction is complete, cool to ambient temperature and filter reaction mixture. Concentrate the filtrate to dryness.

If you dont have Benzyl bromide, you can use benzyl chloride (1.1-1.25 eq) and potassium iodide (0.1-0.5 eq).

**Alternatively, instead of filtering the reaction mixture, you could add water (50-75 mL) to the mixture to cause the product to crystallize out (this will only work if the product is a solid at room temperature). If you dont see any solids after adding water, you can try adding an ice-water bath under the mixture to encourage crystallization.
Once isolated the material, allow the protected product (benzyl ether) to be dried in a desiccator or vacuum oven to remove traces of water.
Second Step:
2) Perform a Grignard reaction with protected product (above).
   The Grignard reaction can be prepared either in THF or diethyl ether using Mg metal/iodine as catalyst, or using isopropylmagnesium chloride solution in THF).
Once the Grignard reaction has been prepared, pour the reaction onto dry ice. This will form the Protected Compound 1 as the magnesium salt. Add water to solubilize the salt, then wash the mixture with diethyl ether (1-2 times to remove impurities) and then add concentrated $\ce{HCl}$ (37%) until pH 1-2 is acheived which will precipitate out the protected Compound 1.
Third Step:
3) Dissolve protected Compound 1 in ethyl acetate, add small amount of Palladium on carbon, and hydrogenate the mixture under hydrogen pressure (balloon pressure will work just fine too) at room temperature overnight.
Once reaction is complete, degas the mixture with nitrogen or argon, filter, and concentrate to provide the desired Compound 1.

I am not sure how feasable is your reaction scheme. It is just too big. Lately I have found a better way, You can have a look. journal of material chemistry 1991, 1(4), 611-619

If you have the capability,  convert the ketone to an enol ether and ozonolyse / oxidative cleave the alkene portion. You might want to cleave your acetate first, make your ether and then silylate , but you'd be silylating to make the enol ether anyway.


text:
What causes the iridescent colour in laboratory grown bismuth?
Artificially grown bismuth crystals are arguably magnificent, as seen in the image below:

Image source: Minerals.net
The website states that:

They have hopper-like growths in pseudocubic crystals and are usually coated with chemicals to prevent tarnish, thus maintaining the silver-white color. Sometimes the coating gives a colorful effect on the bismuth.

What coating gives the colourful iridescence? and what chemical processes cause the colourful iridescence to occur?

The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of thin film interference—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air interfaces. The interference is dependant on the wavelength and incidence angle of light as well as the film thickness, which results in different wavelengths of light being attenuated by different amounts, making the surface appear coloured and the colour changing with viewing angle.

Though not nearly as stunning, regular aluminium bars can sometimes also show the same effect with the correct illumination and viewing angles. However, thin film interference can be greatly increased in many metals by electrochemically forced oxidation, forming a layer of oxide with precisely controlled thickness. This process is called anodizing or anodization, and can be used to colour metals without using dyes. A great video explaining this process, by Bill Hammack the Engineer Guy, can be found here.

Yes, there are a few metals that make very nice colours from interference effects with oxide coatings, especially titanium and niobium.


text:
how to destroy excess iodomethane in reaction mixture?
I have used iodomethane to make a phenolic $\ce{OH}$ group to $\ce{OMe}$ group in Aq+Methanol medium. I have added water to precipitate the product of mine and will evaporate the solvents but I cant use rotavap until I am sure that any residual iodomethane is not contaminating the condenser of rotavap. How shall I destroy excess iodomethane? will aq. solution of $\ce{KOH}$ work?

Yes, you can destroy small amounts of methyl iodide by hydrolysis with an aqueous solution of sodium hydroxide. You can just add some sodium hydroxide solution to your reaction mixture, if your molecule is stable to basic conditions.
The reaction is: $\ce{NaOH + MeI -> MeOH + NaI}$
From: "A Comprehensive Guide to the Hazardous Properties of Chemical Substances" by Pradyot Patnaik, 2007


text:
How to analyze a acid and salt mixture using titrimetry?
Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.
\begin{align}
\ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } &
\mathrm{p}K_{\mathrm{b}1} &= 0.5 \\
\ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\
\ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}3} &= 9.0
\end{align}
All of the species in above equilibria are water soluble.
Progress so far:

$\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align}
\ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } &
\mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\
\ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } &
\mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\
\ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } &
\mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c}
\end{align}

Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.
Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

You mention both the titration curve and indicators. Are you planning to monitor the reaction with a pH meter? If so, an indicator might not be necessary.

@Molx No. No pH meters. Just want to sketch the curve before carrying out the titration.

What is the approximate concentration of the metal oxide? and of the hydrochloric acid?

@LDC3 No other information is given.

When you use the align environment, you can use & as an anchor: use odd appearances to align at that point, use even appearances  to separate equations in the same line. The \tag{...} will produce whatever you give as argument in text mode just left of the right margin.

Prologue
The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:

Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (1978): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.

There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.
Answer
Let's think a bit about your problem first to simplify it from scratch:

As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.
The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and
the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.

If we take a look at the titration curve of phosphoric acid, we can see what is meant by "too high to be determined". You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).

Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition (Protonen-Bedingung):
$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$
It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).
To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration)
$$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&~0\\c(HA)+c(A^-) =&~C(HA)\end{align} \tag{2.1}$$
needs to be solved to give
$$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$
This can be done analogously for the other acid and also for the base.
What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:
$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$
Combining everything so far we get the following proton condition:
$$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$
where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.
The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$
In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as
$$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$
To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get:
$$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$
If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give
$$\begin{align}C(HA) \rightarrow&~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$
Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation
$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$
As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.
Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.
The following figure shows some examples of your mixture using equation 8.

Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.

(+1) This is great! Glad I finally found it and it confirms the equation I have been using (your equation 8) for a long time.

authors: Диана Бауыржанкызы


text:
How to calculate the equilibrium pressure when ammonia and hydrogen gasses react in a closed system?

There is a container that has a volume of $5~\mathrm{L}$ on one side and $3~\mathrm{L}$ on the other and is separated in the middle. There are $2~\mathrm{mol}$ of $\ce{N2}$ on one side at $0~^\circ\mathrm{C}$ and $4~\mathrm{mol}$ of $\ce{H2}$ on the other at $0~^\circ\mathrm{C}$. When the valve is turned and the gases react, what will the total pressure be?

The balanced reaction is $$\ce{N2 + 3H2 <=> 2NH3}$$
My thinking was that the reaction occurs at STP so the pressure would stay at 1  atm?
Or that you could use
\begin{align}
pV &= nRT\\
p(8~\mathrm{L}) &=6(.082)(273.15)\\
p &= 0.06
\end{align}
The $8~\mathrm{L}$ comes from adding the two volumes, and the 6 from adding the number moles together.

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

You are neglecting the fact, that these gasses react with each other. Also the temperature is given as $0~^\circ\mathrm{C}$, so STP it is not. I feel like there is an equilibrium constant missing in the question.

Thank you for adding your attempt! This makes it much easier to help you.

That is all the information that was given for the problem. And I thought that STP was 1 atm and 0 degrees C, thats why I said that.

Note: Not all the reactants will react to form the products, but rather exist in equilibrium. However since the question doesn't mention anything about the equilibrium constant, I will assume that the reaction goes to completion as otherwise it will be impossible to do this question.
If you are aware of the various gas laws, doing this question is actually really simple and doesn't involved many calculations.
Step 1: Finding Initial Pressure
Firstly, the question doesn't specify what the initial pressure of the container is. Therefore you can't just assume that it is $1\ \mathrm{atm}$. So the first step would be finding the initial pressure: $$p = \frac{nRT}{V} = \frac{6\ \mathrm{mol}\times 8.3125\ \mathrm{L\ kPa\ mol^{-1}\ K^{-1}}\times273\ \mathrm{K}}{8\ \mathrm{L}} = 1702.4\ \mathrm{kPa} = 16.8\ \mathrm{atm}$$
Step 2: Find Moles of Products
Looking at the stoichiometry of the chemical equation, $4$ moles of gas react to form $2$ moles of gas. Therefore the ratio is $2:1$. In this case, we originally had $6$ moles of gas. Therefore at the end of the reaction, there should be $3$ moles of gas.
Step 3: Find Final Pressure
Now we know that the total number of moles of gas has halved (from $6$ moles to $3$ moles). One of the gas laws state that the number of moles of gas is directly proportional to the pressure of the system. This is clearly obviously by looking at the ideal gas equation. Therefore, by halving the original amount of moles of gas should result in halving of the original pressure of the system.
Therefore the final pressure will be: $8.4\ \mathrm{atm}$


text:
Stable natural helium hydride?
Reading the transcript of the Royal Society of Chemistry podcast Helium Hydride, they state that helium hydride is possibly the most ancient compound to form in the Universe.  They make the assertion that despite helium hydride being most stable as an ion, but:

The neutral molecule can briefly exist, but only as an excimer, where the excess electron has enough energy to keep it at a higher level.

Under what natural conditions could a stable neutral helium hydride molecule exist?

As explained in Emission Spectra of Bound Helium Hydride Physical Review Letters vol. 77 pages 2941-2944:

The ground state [of neutral HeH] is repulsive, as shown by
  numerous experiments and calculations.

but excited states of HeH that are bound exist (as first experimentally observed by this paper, but previously calculated).
Given that only excited states are bound, it don't think it proper to say HeH stably exists under any conditions natural or otherwise.
The paper does state:

Because of the abundance of He and $\ce{H_2}$ in stellar atmospheres
  and interstellar space, HeH is expected to
  be formed there. Therefore our data may be useful to
  find the HeH lines there.

However, that it exists does not mean it stably exists.


text:
Phosphate buffer from phosphoric acid and K2HPO4?
Is it possible to make a solution of ph 7 phosphate buffer solution using phosphoric acid and $\ce{K2HPO4}$ ? How can I calculate the amount of $\ce{K2HPO4}$ needed for 1L of phosphoric acid ?
Edit:
I have 50 mL of 0.2M $\ce{H3PO4}$ solution. How can I convert this solution into 50 mL of pH 7 buffer solution by adding (only) $\ce{K2HPO4}$ ? How can I calculate the weight of $\ce{K2HPO4}$ considering all the equilibria present in the $\ce{H3PO4}$ solution and by the application of Henderson-Hasselbalch equation ?

What concentration do you want? 50 mM or 1.0 M?

@LDC3 Say 1.0 M.

I suggest you first consider the following reaction:
$$\ce{H3PO4 + 3K2HPO4 -> 2HPO4^{2-} + 2H2PO4- + 6K+}$$
It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$
requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest.
At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M.
Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x.
At pH = 7.0: [HPO4(2-)] < [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:
7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.
Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.
Please check the calculations.

The edit of my answer does not look good. It should read HPO4(2-)!

Thanks for the reply. Can you please explain how that reaction happens ? I mean what about $\ce{H3PO4 + K2HPO4 -> 2 H2PO4^- + 2K+} $ ? And I want the pH to be 7.0 not 7.21.

@Bive I think thats the correct equation now isn't it?

You now tell us that the final concentration should be 1,0 M.
This cannot be right. Initially, you had 50 ml 0,2 M H3PO4, i.e. 10 mmole.
If you add K2HPO4 to reach a final concentration of 1,0 M, the pH of
the final solution will have a pH much higher than 7,0. If we approximate the volume of the solution to be constant, you have to add 5 mole equivalents of K2HPO4 to achieve 1, 0 M.
Initial: 50 ml*0,2 M = 10 mmole => Final: 50 ml * 1,0 M = 50 mmole?
I think you should stick to your original presented problem, which is interesting, since the problem does not state the final concentration.


text:
Are there any free phosphorus-31 NMR spectral libraries?
I know that there are some sites like Sigma Aldrich and AIST for $\ce{^1H}$ NMR and $\ce{^13C}$ NMR, but was wondering if there were any free sites to compare $\ce{^31P}$ NMR spectra?

There are quite a few it seems, a couple of examples are below:
University of Wisconsin Phosphorus NMR data, starts with a shifts overview.
Stanford's Magnetic Resonance Laboratory site P NMR at the SMRL, provides comparisons and a protocol for testing.
A general NMR site is NMR Wiki - has many links to many sources.


text:
Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?
I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).
$$\begin{alignat}{3}
\ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\
\ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\
\ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)}
\end{alignat}$$
Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)
To find out possible reactions, following steps are taken.
Equations $\text{(a)}$ and $\text{(c)}$;
$$
\begin{align}
&{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\
\ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\
&E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\
&\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F
\end{align}
$$
Equations $\text{(b)}$ and $\text{(c)}$;
$$
\begin{align}
&{-}\text{(b)} \times 5 + \text{(c)}: \\
\ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\
&E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\
&\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F
\end{align}$$
My questions:

Are there any errors in above calculations? If so, please suggest corrections.
According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
If not how can these results be explained?

Standard potentials are written for reduction reactions and you wrote the first one for an oxidation reaction.

(a) is written as a reduction.

$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $
$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.
Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.

Are there any errors in above calculations?

Yes, there must be.

According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?

Both of these cannot be correct.

Is it useless to use E values to predict the feasibilities of redox reactions ?

No, it is just as good and exactly as informative as $\Delta G$.

Should we always resort to ΔG ?

No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.

Is it possible for predictions based on E values and ΔG values to contradict ?

No.

If not how can these results be explained ?

UPDATE: Arithmetic error.
The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as
$$\ce{10M^3+ + 2 H2O + 2 MnO4- -> 2 Mn^2+ + 10 MO^2+ +4 H+}$$
Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale.  In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)
So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.

Thanks for the answer. Can you please point where the arithmetic error is ? Could it be in the provided $E^{\circ}$ values of imaginary reactions ?

Doesn't $\Delta G$ has the unit kJ or J because, $\Delta G = -n (mol) \times E (JC^{-1}) \times F (C mol^{-1})$ ?

(Not too good with MathJax, so sorry about the lack of elegance. If someone could suggest a good tutorial, I'd appreciate it.)
Your methods are all correct, but it seems you are not comparing apples to apples. The $\Delta G$ of a particular reaction is going to scale with the amount of reactants you have. For example, the $\Delta G$ of $\ce{10N2 + 10H2 -> 10NH3}$ is going to be 10 times that of $\ce{N2 + H2 -> NH3}$. The same reasoning can be applied to your equations. So, in your first calculation of $-(a)\times5+(c)\times2$ the $\Delta G$ of $a$ and $c$ are 5 and 2 times greater than what they should, respectively. This is not a fair comparison.
What you need to do is to normalize your equations such that the same numbers of electrons are being transferred. You can do this with any integer, but it's probably easiest to just normalize to 1. In your first calculation, normalizing to an electron transfer of 1 leads to:
$$\ce{8/5H+ + 1/5MnO4- + 1/2X^3- -> 1/2X- + 1/5Mn^2+ + 4/5H2O}$$
$$E^\circ= 1.5\ \mathrm{V} + \left(-0.9\ \mathrm{V}\right) = 0.6\ \mathrm{V}$$
$$\Delta G^{\circ} =−nE^\circ F=-1\times0.6\times F=-0.6F$$
Applying the same normalization to the second equation yields:
$$E^\circ= 1.5\ \mathrm{V} + \left(-0.5\ \mathrm{V}\right) = 1.0\ \mathrm{V}\\
ΔG^\circ=−nE^\circ F=-1\times1.0\times F=-1.0F$$
The standard potential and the $\Delta G$ of the second overall reaction of half reactions b and c are in fact more negative than the first overall reaction. In case you're wondering why $E$ does not scale with the size of the reaction it's because $E$ is proportional to $G/n$. As you scale the size of the reaction, both $G$ and $n$ will grow or shrink in the exact same proportion and thus $E$ will always stay the same.

Welcome to Chemistry.SE. Our own in house tutorials are here and here. The second link has further reading suggested.

Also,  this page and this one are good complements to what @Ben suggested.

Yea, those links help a lot. Can't wait to make use of them next time. Thanks editors for cleaning up my garble.


text:
Condition For NaHCO3 to react to form CO2 with organic acids
My professor told us the following:

$\ce{NaHCO3}$ will react to evolve
  $\ce{CO2}$ whenever the acidic nature of the reactant exceeds the
  acidic nature of benzoic acid.

I want to know why this is. And is the given statement valid? He didn't give us the reason.  And in that case should $\ce{NaHCO3}$ react with 1,2-dihydroxy
cyclobutanedione to evolve carbon dioxide?

$\ce{NaHCO3}$ is the basic sodium salt of carbonic acid, a diprotic acid which exists only in small amounts in aqueous solution, because the equilibrium
$$\ce{CO2 + H2O <=> H2CO3}$$
lies mainly on the educt side (this is also the reason why $\ce{CO2}$ evolution is observed). When this fact is taken into account, a $\ce{pK_{a}}$ value of 6.3 at 25 °C is obtained for the first protolysis step (source). Benzoic acid has a lower $\ce{pK_{a}}$ value of 4.19 and is therefore able to liberate carbonic acid from its salts, like any other acid that is significantly stronger than $\ce{H2CO3}$.
$$\ce{C6H5CO2H + NaHCO3 -> C6H5CO2Na + CO2 + H2O}$$
Just in case that you mean 3,4-Dihydroxycyclobut-3-ene-1,2-dione, which is also known as squaric acid, then a reaction with $\ce{CO2}$ evolution is likely. Squaric acid is a fairly strong diprotic acid, with $\ce{pK_{a}}$ values of 1.5 and 3.4, respectively. The acidity can be attributed to the stabilization of the dianion by resonance:

If carbon dioxide evolution with 1,2-dihydroxy cyclobutanedione is possible depends on the $\ce{pK_{a}}$ value of this compound. However, I would expect it to be less acidic than squaric acid.

One could call the squaric acid anion aromatic...

@Martin-マーチン , How is squaric acid supposed to be aromatic without satisfying the Huckel (4n+2) pi electron rule?

@Aaron Hückel's rule is strictly applicable to monocyclic hydrocarbons only. Extending it to systems with heteroatoms is done frequently and  yields remarkable results by coincidence. (There really is not much rigour to it.) However, since the squaric acid anion has a very high symmetry (D4h) it works with the same premise. Depending how you would like to count the electrons in the pi-system, you can come up with 2, 6, or 10; all of which fit Hückel's scheme.

I'm sorry but thophene, furan, pyrrole etc all follow Huckel's rule, although they have got heteroatoms(not by coincidence). The lone pair on the heteroatom is considered to be two of those considered pi electrons, since they are delocalised.

@AaronJohnSabu no, these compounds do not follow Hückel's rule (by definition). They are indeed considered to have 4n+2 pi electrons, but there is more to these rules than this number. Just keep in mind that there is more to aromaticity than Hückel's rules.


text:
Distillation Heating Mantle Setup
In my lab courses, we were taught to use textbooks off the shelf as elevating props to adjust the heights of heating mantles to the right height during distillations, but I was wondering if there was a safer and better way to do so? I recently used a different type of heating mantle that was like a nest and charred the cover of a textbook as well as a heat glove that I put on the second time to prevent further damage to the book.

The use of any flammable material like that is a hazard. You are querying the practice because you've made a correct assessment of the situation. Why haven't those in charge done the same? I hate heating mantles anyway as they tend not to have as appropriate safety cut-offs as stirrer-hotplates.

You need to use something called a lab jack ... just search google to see what they look like.  They are quite cheap too.
You should tell whoever told you to use books about lab jacks, too.

Wow, didn't even know these things existed. I'll notify our chemistry department materials services person.

One supplier is www.fisher.co.uk but there are many others.


text:
Is there an abundance threshold of a 'synthetic' element for it to be considered natural?
I always (possible naively) thought that elements are either natural or synthetic, either one or the other.  But, according to the Wikipedia page about Synthetic Elements, it states that:

All elements with atomic numbers 1 through 98 are naturally occurring at least in trace quantities, but the following elements are usually produced through synthesis. Except for Francium, they were all discovered through synthesis before being found in nature.

These are shown in pink in the image below:

Image attribution: "Periodic table, good SVG" by Incnis Mrsi - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons.
I understand that the amounts of these elements exist naturally in trace quantities, not necessarily on Earth, for example, according to the Jefferson Lab page about promethium:

Promethium does not occur naturally on earth, although it has been detected in the spectrum of a star in the constellation Andromeda.

This leads me to my question:
Is there a natural abundance threshold of a 'synthetic' element for it to be considered natural?

Is there a natural abundance threshold of a 'synthetic' element for it
  to be considered natural?

$\mathrm{> 0}$
In other words, any element that can be detected outside of the laboratory (e.g. in the natural environment) is a naturally occurring element.  This would include the 80 stable elements along with the 18 radioactive elements with very long half-lives, and the shorter-lived radioactive elements produced upon their decay.
Therefore there are 98 naturally occurring elements; all of the others are synthetic, artificial, man-made, whatever you choose to call them.
This Wikipedia article provides nice background, especially the following sections:

the introductory section, especially the last paragraph therein
Occurrence and origin on Earth
List of the 118 known chemical elements - Occurrence

Now, if we lived at the center of a supernova, our threshold might change (not for long, though)...

Jon, the threshold would change or the the number of "naturally" occurring elements would change? Please elaborate if the former.

Well, we would be getting lots of every element and isotope conceivable (under, or in, the sun). For one, brief moment of glory... But, seriously, a supernova will make lots of everything. So, it has all happened "naturally", we are just billions of years removed, so what is left to be "natural" isn't a lot.

The radioactive elements which are regarded as occurring "naturally" have at least one of the following properties:

At least one isotope has a half-life sufficiently long that if it is created in a supernova, significant quantities will persist for a long time afterward.  Many isotopes of uranium fit this pattern.
They are decay by-products of other naturally-occurring radioactive isotopes.

Trans-uranic elements are considered artificial because the their short half-lives mean the probability of any remaining from a million or so years ago is essentially nil, and there are no heavier natural elements that could replenish the supply.  Radon is naturally occurring because, despite its short half-life, the supply gets replenished through the decay of heavier elements.  Technetium has a much longer half-life than radon, but since no naturally-occurring radioactive isotopes of anything else produce it when they decay, any supplies that may have existed will have dwindled away to essentially nothing by now.

I have never heard of a formal specification for such a threshold. Qualitatively speaking, though, I believe this threshold could be effectively considered "enough that we can make useful stuff with it." To my mind the key phrase in your blockquote is (emphasis added):

"... the following elements are usually produced through synthesis."

'Production' implies making stuff. 'Measurable in trace quantities by ICP-MS' is not helpful for making stuff.
Worth noting: since different amounts of different materials are needed to make useful things, it's possible that a hard, quantitative abundance number might not even be useful to define.


text:
Why do soft drinks fizz?
I know it has something to do with carbon dioxide, but what exactly happens and causes the characteristic sound we hear (especially after we shake the bottle)? How do we manage to dissolve a non-polar substance ($\ce{CO2}$) in a polar solvent ($\ce{H2O}$)?

Interesting question! Welcome aboard @AnthonyKelesidis! Just dive in with more questions and answers - take your time to learn the ropes, we are available in chat if you need help, or for a chat (a bit crazy in there sometimes!).

Decrease of pressure causes less-solubility of gas. the gas try to come out from liquid medium.

$\ce{CO2}$ is quite polar.

Because $\ce{CO_2}$ isn't exactly what's being dissolved. When it encounters water, it immediately encounters the following potential reaction:
$$\ce{CO2~(g) + H2O~(l) <=> H2CO3~(aq)}$$
Carbon dioxide is non polar, yes, but carbonic acid readily dissolves (and dissociates, but I didn't care to show that) in water.
The sound we hear is because when we open the bottle, the equilibrium above shifts to the left, creating more carbon dioxide that is being released. Why does the equilibrium shift? It has to do with the pressure in the bottle.
When the bottle is closed, it's under a certain set pressure. When it's closed at the same temperature and pressure for a long time, the reaction above reaches an equilibrium state, where the forward and backward versions of the reaction are happening at the same rate. However, when the bottle is opened, the pressure inside the bottle decreases, as the gas in the bottle suddenly has a lot more space to expand into. A decrease in pressure favors the side of an equilibrium that has more gaseous molecules, which in this case means the carbon dioxide. Ergo, when you open the bottle, the gas is formed, escapes from solution, and rapidly expands outward, making the infamous popping noise.

It may be a little too much hand-holding, but this answer does rely a lot on the meaning of $\leftrightharpoons$ which may not be clear to non-chemists. Explicitly stating immediately after the equation that it’s a two-way street would make the discussion about equilibrium more clear to those unfamiliar with the symbol.

You explain why decreasing the pressure makes it fizz, but now I'm wondering why shaking the drink makes it fizz more.

I think the main effect of shaking isn't to make the gas come out faster, but to make more liquid be carried along with the gas.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

There is barely any carbonic acid in carbonated water, the equilibrium constant is just too low
https://en.wikipedia.org/wiki/Carbonic_acid#pH_and_composition_of_carbonic_acid_solutions

@kmm: you didn't read to the part where it says "For a CO2 pressure typical for bottled carbonated drinks ( pCO2 ~ 2.5 atm), we get a relatively acidic medium (pH = 3.7) with a high concentration of dissolved CO2. These features contribute to the sour and sparkling taste of these drinks."

@Fizz a pH of 4 would taste a little acidic, but not very. Try diluting one part lemon juice in 100 parts water to verify. The real origin of the sour and sparkling taste is the action of carbonic anhydrase enzymes in our tongue and throat. We know that, because certain medications inhibit those enzymes, and one of the side effects of taking them is carbonated beverages tasting "flat".

Why do soft drinks fizz?

The key word for that is pressure
Manufactures close the bottles/cans by forcing carbon dioxide and water into your soda at high pressures around 40 psi.
When you crack open the can, you release the pressure and allow the gas bubbles to wiggle free from the liquid and rise to the surface. This requires energy because in order for the gas to break free from the liquid it has to overcome the force holding the liquid together.

Source: LIVESCIENCE


text:
Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?
This is quoted from Jim Clark's Chemguide

For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.

What is the reason stated above? Why does in only transition series, 4s has more energy than 3d & not in the preceeding elements?

There is evidence that the title statement is not true to begin with. See e.g. the links in the disclaimer in my (accepted) answer.

Disclaimer: I now believe this answer to be fully incorrect.
Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22.
However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles:

Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2.
Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498.
Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469.
Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481.
Some criticism of Atkins' books by Eric Scerri.

While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed:

The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d.

As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by
$$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$
where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation:

The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$, which makes the energy of the 4s orbital lower.

Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$. The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$, and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start.
However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ($\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$.)

I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells".

In the quest of re-reading old posts, I got somewhat baffled at one point here: you said up to $\ce{Ca},$ the $E$ for an electron in $\rm{4s}$ is lower than that of $\rm{3d}$ as the increase of $\rm{Z_{eff}}$ gets somewhat nullified by the increase in $n$ in the denominator. Okay. But I'm not getting why the same thing doesn't happen in elements after $\ce{Ca}$ eg. $\ce{Sc};$ $\rm{ 4s}$ is radially diffused even in $\ce{Ca}$ but does this affect the energy of the electron? Sorry, if I'm bothering you @Ortho, but would appreciate if you tell me why can't $\rm 4s\lt 3d$  in $\ce{Sc}$  [contd.]

for the same reason as in $\ce{Ca}$ viz. the increase in $n$ in the denominator is nullified by the increase in $\rm{Z_{eff}}.$ Also, there maybe a possible typo here: 3d > 4s but for Sc onwards, 4s < 3d. Thanks.

@MAFIA36790 Sorry I didn't get back to you earlier, I was travelling on that day and forgot all about it. To be honest, after a couple more years of chemistry, I am not entirely convinced how accurate a description this is. There have been criticisms of this particular graph in Shriver & Atkins, which I read before (various authors have written on it before), but I don't have the time to do thorough research into the matter right now. I will point you in the direction of these: http://pubs.acs.org/doi/abs/10.1021/ed055p2 and http://www.chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf

The idea in my post was that as atomic number increases, $Z_\mathrm{eff}$ of both the 3d and 4s orbitals increase. However, the 3d orbital is more greatly affected, i.e. $Z_\mathrm{eff}(\mathrm{3d})$ increases faster than $Z_\mathrm{eff}(\mathrm{4s})$. Consequently, there will be a crossover point where $$\frac{Z_\mathrm{eff}(\mathrm{3d})}{3} = \frac{Z_\mathrm{eff}(\mathrm{4s})}{4},$$ i.e. 3d and 4s have equal energies. Before this point, 4s < 3d, and after this point, 3d < 4s. Atkins' argument is that this point lies exactly between Ca and Sc.

Why do we look at the presence of radial nodes near the nucleus when measuring the amount of penetration of the electrons in a particular orbital. Shouldn't we look at the mean distance or most probable distance of the electron from the nucleus?

@TanYongBoon neither of those criteria will give you a measure of the penetration, which most critically depends on electron density near the nucleus. The bulk of the electron density is going to be on the outside, and doesn't contribute to penetration, so it can be anywhere it likes

May I refer you to this section of the answer: "The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better."

@TanYongBoon if you really wanted to be mathematical, I suppose it would be better to use some measure along the lines of $$\int_0^{r_0} |R(r)|^2r^2,\mathrm dr$$ where $r_0$ is some defined cutoff point which indicates the boundary between core and valence electrons. Obviously this is still quite wishy washy but I think it is more representative of the concept of penetration rather than a mean or a mode value of $r$ (which is predominantly influenced by the electron density outside the core).

That said, penetration is something that I've never seen put on a mathematical footing - it is pretty handwavy and I am not a huge fan of invoking it, which is partly why there is a disclaimer at the top of this answer, and this is my only answer on SE where I have used it. Perhaps not coincidentally, it was also one of my first answers.

This is a difficult question to answer. Following the Aufbau Principle and the n+l rule, the 4s orbital should fill before the 3d orbital. So why is 3d lower in energy? In short, the Aufbau Principle is not entirely correct. It is a guideline (like many things in chemistry.)
So, orbitals fill in order of stability. That is to say, electrons will go where they will be the most stable. It takes energy to hold electrons around the nucleus. The farther away they are, the more energy is needed to keep them. So the higher the principle quantum number, the higher the energy. I.e. 3s is higher in energy than 2s. At the same time, the principle quantum number is not the only number than needs to be considered. The quantum number l, for example, is also important. The higher value of l, the higher the energy. So 3d is higher in energy than 3p which is higher in energy than 3s. The 3d orbitals are more compactly placed around the nucleus than the 4s orbitals, so they fill first, even though this contradicts the Aufbau principle. This can be seen experimentally with the electron configurations for scandium:
Sc3+: [Ar]
Sc2+: [Ar]3d(1)
Sc+:  [Ar]3d(1)4s(1)
Sc:   [Ar]3d(1)4s(2)
Now, it is important to note that the 4s level does fill before 3d is entirely full. This is due to the 3d orbital's compactness. Electron repulsion "pushes" electrons into higher energy levels with less repulsion.
I would recommend reading this as it explains this in much more detail: http://www.rsc.org/eic/2013/11/aufbau-electron-configuration
I hope that helped!

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


text:
Near IR spectroscopy vs. IR spectroscopy
I am a backer of this Kickstarter: SCiO Near IR Spectroscopy
Here is a link to their technology: SCiO Technology
Question: How short of full IR spectroscopy do you think this device is? I have asked them whether we could program the SCiO to act as an IR spectrometer, and they have said this would require additional programming (which I am not qualified to do). But I am wondering if there would still be limitations. Conversely, Consumer Physics seems to imply that our taking more observations with the SCiO will make the device smarter and more useful.
This is obviously a question directed at experts in IR spectroscopy, which I am most definitely not.

This is more or less speculation, since they haven't released detailed technical specifications, but given the size of the instrument and the need to avoid moving parts to keep it durable, I would wager that it uses an IR LED for illumination, a diffraction grating for dispersion (although from the picture, it doesn't look like there's room for a conventional reflective diffraction grating setup so they might be doing something else), and probably some kind of silicon CCD or InGaAs diode array for detection. If this is the case, it's likely impossible to convert for use in conventional IR spectroscopy—not as a matter of programming, but simply that the hardware is not equipped to do it. Normal mid-IR spectroscopy (below about $4000\mathrm{\ cm^{-1}}$) involves much longer wavelengths beyond what LEDs can do, which generally means some kind of incandescent lamp, optics that don't involve glass (which blocks mid-IR), and a detector that can deal with such low energy photons (there are a few semiconductor-based sensors that looks promising, but most spectrometers use thermal detectors). There is also the matter of dealing with all the water in the air—many IR spectrometers have gas purge systems in their sample chambers.
There has been a lot of interest in standoff IR spectroscopy for things like explosives detection and advances in mid-IR optics and sensors may make this kind of small instrument possible, but we're not quite there yet. Certainly an instrument designed for NIR is not likely to have any usefulness for mid-IR, without being designed with it in mind.
A related technique that is much better suited to small handheld instruments is raman spectroscopy, since the light involved is actually usually in the NIR. I doubt this thing is sensitive enough to do it well, but it might be possible to add filters to the illumination and measurement ports to give it a shot. It's definitely possible to do raman with a handheld instrument though—I saw one of these at a conference recently. Quite expensive, but very nifty.

We had a sales rep come into the office this morning with a handheld XRF, not much bigger than a football. Ran some elemental analysis on a couple of our coatings -- impressive tech.


text:
Bonds and pairing of spins
I was reading Peter Atkins' Elements of Physical Chemistry where I came across this line:

Bonds do not form because electrons tend to pair; bonds are allowed to form by the electrons pairing their spins.

What does this line mean? What does Atkins want to say through this line & why did he italize the words I bolded above?

I dunno if the author meant exactly this, but it is a well-known fact that pairing of electrons actually consumes energy.
For example, free nitrogen atom in ground state has three unpaired electrons. To form triple bond in dinitrogen molecule, three pairs of electrons must be, well, paired. The process of pairing does consume energy.  However, the energy freed in the bond formation is so big, that it overwhelms pairing energy by huge margin. On the other hand, in oxygen molecule two highest-energy electrons has two orbitals to occupy, and they stay unpaired, so dioxygen is paramagnetic.
So, by no means it is the fact of being unpaired that drives electrons to form bonds. In fact, it prevents formation of weak bonds. However, when two atoms come closer to each other and have two half-occupied orbitals that can form a bond, the fact that electrons may be paired allows this two orbitals to combine into occupied bonding orbital, forming a bond.

How does the spin enter the picture?

@ToddMinehardt going with huge simplification, the pauli principle states that no two electrons can have exactly same state. Electrons on same orbital must have exactly same wavefunction, however as they have spin, they still can be in states, characterised by same wavefunction, but with different (opposite) spin orientation ('up' and 'down'). Thus, they may be 'paired'

The statement "Electrons on same orbitals must have exactly same wavefunction" is patently incorrect: That's not possible. The quantum number that distinguishes them for us is spin. But back to the OP's question: Why, in your opinion, does Atkins say what he does in reference to spin?

@ToddMinehardt Orbital != Spin-orbital ; Spin-wavefuction != wavefunction. Two electrons on same orbital have same '3d-wavefunction', or, if you are annoyingly nitpicking, same 3d component of the wavefunction but different spin component.  I've no idea what Atkins said and don't care about it at all, I wrote how the things are.

@permeakra This is coming across as somewhat rude.  Please be mindful of how your comments may be perceived.  Thanks!

Some actual numbers for the cost of electron pairing can be found in this answer of mine. The magnitude of the pairing energy is usually around 2 - 3 eV.


text:
How can hybrid orbitals be equivalent?
How can hybrid orbitals be equivalent? Take, for example, $\mathrm{sp^3}$ hybridization. The wave functions of each hybrid orbitals are given by:
\begin{align}
\psi_1 &= \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\
\psi_2 &= \psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\
\psi_3 &= \psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\
\psi_4 &= \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\\
\end{align}
No one of them has same linear combination. So, how can they be equivalent? I am not getting the sense.

Since the coefficients are different, the orbitals end up pointing in different directions. As for why they happen to point in nice directions, and why their energies are exactly the same, you kind of have to trust that the maths works out. Hopefully someone else can answer because I'm not too familiar with the application of MO theory to hybridisation.

see this related question and answers

Gosh, @ortho, you really didn't know anything at the end of first year, did you! They are equivalent, only in the same sense that the x, y, and z axes are equivalent in a cube...

You are right in that the sign of each orbital, especially the p ones, is different in every hybrid orbital.
Your equations are, however, lacking the normalised contribution factors that should precede the wavefunctions. A better equation for the first orbital would be:
$$\psi_1 = a_s \psi_s + a_{p_x} \psi_{p_x} + a_{p_y} \psi_{p_y} + a_{p_z} \psi_{p_z}$$
The point is, that the contribution factors are equal for every sp³ hybrid orbital — every hybrid gets ¼ contribution from every p and the s-orbital. That way, all orbitals are rendered equal because their atomic orbital contribution factors are equal, too.
The signs before each p-orbitals contribution do not matter for the orbital’s energy value, only for its orientation. So inspite of all orbitals having different combinations of signs, their energy can be equal (but they will ‘point’ in different directions).

(Note that this is a simplified view as are hybrid orbitals in general. Spectroscopically, you can measure two different energies of the bonding orbitals in methane — the explanation cannot be given by hybridisation theory but requires proper molecular orbital theory.)

The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them.  Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry.
Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.


text:
Why are real orbitals used to define the orbital of hydrogenic atoms instead of complex orbitals?
On the physics site, I asked the difference between real & complex orbital. Real orbitals are the superposition of complex orbitals having definite magnetic quantum number states.
\begin{align}
p_z &= p_0 \\
p_x &= \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\
p_y &= \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\
\end{align}
where $p_0 = R_{n1} Y_{10}$, $p_1 = R_{n1} Y_{11}$, and $p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $\ell = 1$.
My question is:
Why aren't the orbitals of hydrogenic atoms taken as complex orbitals instead of the real ones? They are valid states for an electron, aren't they? So, what is the problem in having complex atomic orbitals?

Quick and dirty answer: Because the complex orbitals don't have a direction, which likely helps calculation. But wait for one of the theoretical chemistry gurus to chime in here ;)

You can choose whichever you want or best suit for your purpose. For actual calculations often the complex ones used. I guess Chemistry textbooks prefer the one which is easier to draw and seems less complicated mathematically.

Indeed, one of the major pitfalls of using real orbitals is when students start asking what magnetic QNs they should assign to the p_y and p_x orbitals. I was never quite satisfied with the answer of "just pick one, it doesn't really matter." It wasn't until I took quantum that I really appreciated how ill-defined the question was. It still beats trying to explain complex orbitals at that level though.

AFAIK, when investigating a stationary state under Born-Oppenheimer, complex component may be separated into its own multiplier depending only on time and, since it has time-independent norm, it gives time-independent contribution for most stationary properties like energy. Since other multiplier depends only on space coordinates, for simplicity of math and faster computations,  almost entire computational chemistry uses real math only.

The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize.  For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few  advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular  and closed shell systems.
So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.

Totally agree, but I'd also add that because orbitals are not unique you can unitarily transform any real orbitals into complex orbitals. They only differ by a phase. Of course we must remember that orbitals are not physical but density is---which is always real---since the the inner product of the wave function eliminates any imaginary components.

To add my 5 cents to the discussion, note that we also usually use real basis functions ("atomic orbitals") in electronic structure calculations, although, molecular orbital expansion coefficients may be complex. So, each and every MO $\phi_{k}$ is usually expressed as follows,
$$
    \phi_{k}(\vec{r}) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(\vec{r}) \, ,
$$
where $\chi_{q}(\vec{r})$ are real functions and $c_{qk}$ may in general be complex numbers. The point is that there is no additional variational freedom in using both complex basis functions and complex expansion coefficients. Besides, the use of real basis functions reduces (due to permutational symmetry) the number of two-electron integrals to be calculated for a basis of $m$ real functions from $m^4$ to $M(M+1)^2$, where $M=m(m+1)/2$ is the number of distinct pairs of products of two basis functions.


text:
Hydration number: Carbonic acid in water
Carbonic acid ($\ce{H2CO3}$) is known to be soluble in water, and that this particular compound is unstable, so much so that according to the 2014 article from Berkeley Lab:
New Insights on Carbonic Acid in Water, it only exists for a fraction of a second (300 nanoseconds according the article).
Newly reported experiments and observations from the Berkeley Lab found that:

“Our results support an average hydration number of 3.17 with the acid’s two protons each donating a strong hydrogen bond to solvating waters, the carbonyl oxygen accepting a strong hydrogen bond from solvating water, and the hydroxyl oxygen molecules accepting weak hydrogen bonds from the water”

What is not clear is how is the hydration number calculated and how does this relate to the short existence of carbonic acid in water?

Please note, the first two links point to the linked questions indicated for this question.

You may wish to refer to the publication of Saykally et al., The hydration structure of aqueous carbonic acid from X-ray
absorption spectroscopy
Basically, to determine hydration number, X-ray spectroscopy was used. This relates to the short existence of carbonic acid as, since it is so short lived, it is necessary to analyze it in such a time frame. As such, this was "the first measurement of X-ray absorption spectra for aqueous carbonic acid, enabled by the use of rapid-flow liquid microjet technology" which allowed it to be captured.
Then, by analyzing the spectra, it is possible to calculate the approximate hydration number, as "the increasing H-bond strength is proportional to the hydration
number (measured for distances up to $2.5\mathrm{\overset{\circ}A}$): $\ce{Na2CO3}$ (5.55), $\ce{HCO3}$
(4.26), $\ce{H2CO3}$ (3.17)".

I will add that the radial distribution function ($g(r)$), which is used in this and similar studies to infer how many of (some species) are distributed around (some central point) up to (some distance), is tied directly to the hydration number. That quantity (as detailed in the reference given by Andy and in Kumar et al., referenced therein) is being interpreted to correspond to stronger hydrogen bonding if it's high, and weaker hydrogen bonding if it's low.

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one  on how to make your future posts better.

Brilliant answer! Thank you Andy (and thank you @ToddMinehardt) - this is much clearer now. Essentially, the hydration number is linked to the spectra.


text:
Why are there peaks in electronegativities in d-block elements?
Looking at the Pauling electronegativities in the Periodic Table (below, from ChemWiki):

Asides from the overall trend of increasing electronegativity across and up the Periodic Table (towards fluorine), there seems to be a minor peak in the first 3 periods of the $d$-block elements, this peak becomes more pronounced as you go down the $d$-block, with the most noticable peaks in Periods 5 and 6 - seemingly the reverse of the overall trend. This is shown in the graph below (Image source):

In the ChemWiki page, there is also a reference to the Allred-Rochow Electronegativity Scale, which also has peaks occurring most prominently in Periods 4 and 6:

Image source
The Wikipedia page Electronegativity details several methods of calculation, which brings into question how well-defined the electronegativity values are; however, there still seems to be the peaks occurring, particularly in Periods 5 and 6, and to a lesser extent in Period 4.
What is the reason for these electrongativity peaks in the $d$-block?

Perhaps a question of more immediate concern before tackling the title question is, just how well-defined are the electronegativities for the transition metals?

@NicolauSakerNeto that could be part of an answer

at your link they say nearly at the end (below slater's rules), that at the time allred and rochow constructed this scale the values for the covalent radius were wrong. they gave also a corrected table

Good points - I have added these details into the revised question

You should know that closeness to filling a shell and atomic size are the reasons for the overall electronegativity trend. This also applies to the subshells, how close is an element to filling, or half-filling, a subshell? How large is the atom?
The atomic size trend for the d subshell is opposite to that of the overall trend. In period 4 the 3d subshell is deep in the core orbitals and the electron density is high, whereas in the lower periods the d orbital is more diffuse and can more easily accommodate more electrons, leading to a better energy payoff for completing the subshell. For the elements that take s electrons to fill the d subshell, this energy payoff is even greater from then filling the s subshell as they are very diffuse.

The peaks in electronegativity for elements may be named as groups as next:
V Cr Mn has a peak and then Mn drops down. Iron is the key as shown next.
Zr Nb Mo has a peak and then a a plateau. Zirconium is the key as shown next.
My research is displayed :
Periodic Table of Elements : Shape of Each Nucleus affecting electronegativity

http://pyramidalcube.blogspot.com/2017/08/periodic-table-of-shapes-of-nuclei.html
The shape of the nucleus affects the properties of the element. Notice how
Manganese is the last in a series from the carbon core cube-2, then iron starts the cube-3 shape. That transition across the Table causes the abrupt change in electronegativity.
Alan Folmsbee

Do you have a reference for this beyond a blog entry that you wrote yourself (which subsequently links to another blog)?

Dear Jonska, Neon has a shape like in my Table. https://phys.org/news/2012-07-atomic-nucleus-fissile-liquid-molecule.html#nRlv and also Radon has a shape like in my table: http://ns.ph.liv.ac.uk/~lg/papers/StudiesOfNuclearPearShapesUsingAcceleratedRadioactiveBeams_Archive.pdf and also Barium has a shape like in the Table https://phys.org/news/2016-06-heavy-barium-nuclei-pear.html . I see a minus five insult from your welcoming committee. Progress is here, enjoy.

I did not downvote, but as I mentioned, it was probably due to having insufficient references . It may help others if you incorporated brief summaries of some of that research into your answer rather than only the chart.


text:
Could milk rust a steel teaspoon?
Recently, while cleaning a neighbour's fridge (turned off for a few weeks), I came across a cup (closed with a lid).  Inside the cup was, to my olfactory horror, congealed milk, with a steel (iron) teaspoon nicely wedged into it.
The milk had separated into a blue cheese-like solid, overlain with a semi-translucent liquid.
When I took the teaspoon out of the congealed mess, there was a layer of rust on the teaspoon where it was in the liquid. But, I observed almost no rust on the parts of the teaspoon in the blue-cheese-like solid and that exposed to the air.
The illustration below is a rough schematic of where the rust occurred (the 2 layers observed are labelled A and B, and the air in the top of the container is labelled C and the diagonal line represents the teaspoon:

A = congealed blue-cheese-like solid = virtually no rust on this part of the teaspoon
B = semi translucent liquid = heavily rusted part of the teaspoon
C = air = virtually no rust on this part of the teaspoon
How could the milk, or the liquid separation, cause rust on the teaspoon?
A hunch I have (and I could be way off the mark), is it possible that the corrosion could be due to, at least in some significant way, due to the fact that milk is slightly acidic $^{(1)}$?
$^{(1)}$ According to the International Livestock Research Institute, they state that

Fresh milk has a pH of 6.7 and is therefore slightly acidic

Related question and answer Silver and milk (or milk products)

Maybe, that steel was mostly iron and milk has H2O.

Yes, I am aware that steel is mostly iron, and water makes up most of water - but as I said, the rust was on the teaspoon within the liquid part of the separated mess - have clarified that there was virtually no rust on the other parts of the spoon.

Rusting requires oxygen (i.e. air) and water. The top of the spoon wasn't wet - no water. The base of the spoon was surrounded by cheese - no air. Hence only the part of the spoon at the water-air interface rusted. You can go and find this on steel pillars in the sea if you like - it rusts most quickly in the parts which are alternately wet and dry.

@AndyT it was not just at the interface (if that were the case I would not have asked, as I am well aware how rust forms there), as I said there was a layer of rust on the teaspoon where it was in the liquid. - the part of the teaspoon immersed in the separated liquid.

I don't have the chemistry expertise to formulate a full answer. I would guess that the spoon is actually stainless steel, and not just plain carbon steel. My understanding of how stainless steel works is that a passive film of chromium oxide is formed on the surface, which prevents further oxidization. Presumably something (perhaps lactic acid???) in the liquid part of the separated milk is able to react with the chromium oxide layer to the extent that this protective film is removed and then iron oxide is able to form.

Technically, even simple water can cause rust, so nothing surprising here.
However, spoilage of milk most probably produced a lot of organic acids (lactic acid and similar) which can speed up any corrosion process. Lactic fermentation is a natural fermentation process in milk, when bacteria start converting the sugar content of milk to lactic acid. It is not the only possible product, but a very common one. Link: https://en.wikipedia.org/wiki/Lactic_acid_fermentation
There are two important factors that help corrosion: First, acidic environment (low pH) itself generally help corrosion. Second, the organic acids produced in fermentation can often form metal complexes, therefore they help remove any protecting oxide layer from the surface of metal objects as well as shift the  redox equilibrium towards an oxidized complex form.

Thank you for this answer! I usually wait a few days before accepting an answer - in the meantime, could you please include some links, particularly for your second point in the second paragraph?

That is a kind of tricky: all I said is pretty common knowledge in chemistry. The fact that organic acids (carboxyl acids) can form complexes with metals is a trivial statement. Lactic acid silver or other metal complexes are not particularly famous or important. The complex of silver with lactic acid is sold by Sigma,  SID 24896481.


text:
Can molten sugar also rotate polarized light as sugar water does?
It is well know that the sugar water can rotate the plane of linearly polarized light, so light bends when travelling in sugar water. However, do the liquid state of sugar but not the sugar solution also have this property?

Molecules in every state - liquid, solid and gaseous - rotate the plane of linearly polarized light - it doesn't matter, because this rotation arises from inherent asymmetry of molecules, which is causing their optical activity
More to the point, not only water solutions of sugars, but also their crystals, molten sugars and even vapor, rotate the plane of linearly polarized light. If you're asking specifically about sucrose checking this may be problematic as it's decomposing during melting, as you can read in this question and here, but I think it still can be done, under vacuum.
You can also check these related questions about chiral gases and rotation of the plane of polarization by molecules

This is indeed a helpful answer! Thanks a lot, Mithoron. But a further question come to my head. For those chiral gases, do they have a non-chiral state? I.e., keeping molecules unchanged, if one heats the whole system, will the system in a chaos state that cannot bend light to a certain direction? (By checking the question you refer to, I didn't find the answer for the second question).

@hongchaniyi No in gas these molecules already point in every direction, the last question I linked is related to that imo. However in more obscure chiral molecules like some of those in this answer http://chemistry.stackexchange.com/a/31296/9961 highier temperature may cause losing of chirality (racemisation).


text:
In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack?
In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack ?
My approach:I considered hydroxy group as the stronger +M group.So my guess was para position w.r.t hydroxy group.But that's not the correct answer.
The answer is actually given as ortho w.r.t hydroxy and para w.r.t -CHO group!!

@yasir can you please explain why?

This is known as the ortho effect. According to it if an ortho/para and meta directing group are meta to each other, then the incoming electrophile prefers to come on ortho of meta directing group.
See the reference.

I wanted t0 know for m-hydroxy benzaldehyde

The same applies for m-hydroxy benzaldehyde too.In this case, you have $\ce{Cl}$ which is ortho/para directing meta to $\ce{NO2}$ group which is meta directing.In your case, $\ce{OH}$ which is ortho/para directing is also meta to $\ce{CHO}$ group which is meta directing.

@yasir Please edit in the complete reference. Give credit where credit is due.

So are you saying that the OP's stated 'correct' answer is wrong?

@Bon That's what i have learnt from my text.I would like someone to kindly elaborate on  ortho effect as my knowledge of it is limited.


text:
What causes the old book smell?
I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question):

One thing I notice is that there is a distinct smell that comes from the pages of older books. (the smell is not unpleasant) - it seems that the older the book, the more distinct the smell.
What is the chemistry behind the old pages in a book smell?

I love the smell of old books, but it's probably because of association with the pleasure of reading them.

This is a very interesting question. Provided that the materials used in making papers aren't the same around the globe, this is a very broad case of study. However, a study has been conducted in which the main goal was to identify the compounds that are the cause of the smell; VOCs:[1]

Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air.

So, quoting the article:

Using supervised and unsuper-vised methods of multivariate data analysis, we were able to quantitatively correlate volatile degradation products with properties important for the preservation of historic paper: rosin, lignin and carbonyl group content, degree of polymerization of cellulose, and paper acidity.
It is a result of the several hundred identified volatile and semi-volatile organic compounds (VOCs) off-gassing from paper and the object in general. The particular blend of compounds is a result of a network of degradation pathways and is dependent on the original composition of the object including paper substrate, applied media, and binding.

The 15 most abundant VOCs present in all chromatograms were selected for further analyses: acetic acid, benzaldehyde, 2,3-butanedione, butanol, decanal, 2,3-dihydrofuran, 2-ethylhexanol, furfural, hexadecane, hexanal, methoxyphenyloxime, nonanal, octanal, pentanal, and undecane.

Strlič, M., Thomas, J., Trafela, T., Cséfalvayová, L., Kralj Cigić, I., Kolar, J., Cassar, M. (2009). Material degradomics: on the smell of old books. Analytical Chemistry 2009, 81 (20), 8617-8622.
The main link to the article might be behind a paywall, so for the interested, the relevant ResearchGate article is accessible.

What's loading weight graph? PLS Calibration?


text:
What is the proper term for lines of constant volume and constant enthalpy on a thermodynamic diagram?
I know an isotherm is a line of constant temperature.
Isobar is a constant pressure line.
Isentrope is a line of constant entropy.
But what do you call a line of constant volume? An Isochore?
And what do you call a line of constant enthalpy? An "isenthalp"?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

But what do you call a line of constant volume? An Isochore?

Correct.

And what do you call a line of constant enthalpy?

The process is isenthalpic. More than one engineering/thermodynamic text I found by Googling indicates they are simply referred to as "isenthalpic lines." See this for example.

Logically, isenthalp would be the name for an isenthalpic line, but based on a quick Googling, the English world seems to be lagging the German world in this vocabulary innovation.

Agreed on all points @Curt F.


text:
What is actually the difference between valence bond theory and molecular orbital theory?
Recently I have read about both of the concepts in my book (Physical Chemistry by Atkins, Paula). It was literally a reading; though I could understand the language and superposition of orbitals, inversion symmetry, and so on, I never could really comprehend how these two theories differ (or agree) with each other. I have googled it but couldn't gain a lucid explanation between the differences in their approaches. So, now my question is:
How do these two theories differ from each other? What are the differences between their approaches?
Also, in VB theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why it is told that "MO theory provides a global, delocalized perspective on chemical bonding"? This portion of the question was split to a separate question: Why is the resonance concept not required in molecular orbital theory?
Edit: After viewing some of the answers, I think IMHO, I had to clear what I was really asking. By saying "difference between MOT & VBT", I am not intending to mean that just write the different predictions the two theories can separately succeed in making. What I really mean is the difference between their approaches & their way of explanation. Of course, the case is not like "Hey, look, one theory is outdated; just follow the other one..."; they are both important and hence do want to know the difference between their pathways of analyzing the cases.

It is a good question and a lot of books only scratch the surface of the theories anyway. Understanding both on a conceptual basis is necessary to distinguish the difference and similarities. There are plenty of resources of comparison available online, maybe UC Davis helps, I have not read it, but the following might also help: http://www.chem.wisc.edu/courses/562/fall08/PDFfiles/Hoffmann.pdf

UC Davis archive link. :-)

TL;DR
VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears little resemblance to the original atomic orbitals, or to the VB theory description.

Long Version
The key difference between the VB and MO theories lies in their respective assumptions of the underlying mathematical/physical form of the electronic wavefunction.
VB theory holds tightly to the mathematical structure of atomic orbitals (AOs), assuming that the electronic structure of a molecular system is described adequately by retaining the atomic orbitals (possibly hybridized) from each constituent atom, and overlapping them to form the needed description of bonding.  VB calculations are computationally much more complex: the AO basis is generally non-orthogonal, and the linear algebra and other techniques required to cope with this non-orthogonality are much more involved.
More broadly, the need to invoke resonance (superposition of multiple valence-bond structures) and other concepts highlights that, in fact, often this adherence to atomic orbitals and their valence-bond overlaps is not adequate for a proper description, at least not without extensive use of methods analogous to the configuration interaction of MO theory.  VB theory does hold certain advantages, however—for example, I believe the AO basis does sometimes allow for a more direct linkage to fundamental chemical concepts such as hybridization and Lewis electon-pair theory. To note, Sason Shaik is one of the strongest modern proponents of the theory I know of—he describes some of these strengths and their applications in his recent retrospective article (doi:10.1016/j.comptc.2017.02.011).
MO theory, at its core, casts away most assumptions regarding the mathematical structure of the electronic wavefunction, as long as properties like indistinguishability and the Pauli exclusion principle remain satisfied.  Calculations within MO theory can use a wide variety of mathematical structures to try to represent the wavefunction, ranging from gridded numerical quadrature approaches (see here and here, and also orbital-free DFT) to Slater orbital computations to the Gaussian-orbital based calculations used most commonly today.
One potential point of confusion is that MO theory in its most common implementations makes direct use of atomic orbitals in composing the molecular wavefunction (see LCAO theory). The distinction to be made here is that instead of attributing significance to the atomic orbitals as an intrinsically good descriptor of electronic structure, they are used merely because they afford a convenient mathematical form for efficient computation, in trying to construct the MOs.  Particular advantages are obtained from the use of atomic orbitals constructed from Gaussian functions, since the product of two Gaussians is another Gaussian (even in multiple dimensions) and numerical integration of n-dimensional Gaussian functions is straightforward.
It should be noted that, for as much of an improved description it affords, MO theory still is an approximation. For a given system, the electronic wavefunction just IS the highly-abstract mathematical object that satisfies the appropriate Schrodinger equation.  MO theory is used extensively because it provides one of the most computationally tractable paths to getting in the neighborhood of the "real answer."
Practical Aspects of MO theory
The "simplest version" of MO theory for a time-independent system under the Born-Oppenheimer Approximation is the Hartree-Fock method, which provides a remarkably good description considering the significant simplification represented by its mean-field approximation of the electron-electron interactions.  Hartree-Fock is in general not good enough for computations at the level of chemical accuracy (~$1\mathrm{\ kcal/mol}$), however, and numerous methods have been developed to refine it: the most often used techniques are Moller-Plesset perturbation theory, configuration interaction, and coupled cluster.
In-Depth Reading on MO Theory
For a deep dive into the underlying mathematics of MO computations, it's hard to beat the classic Szabo & Ostlund. As a more readable, lighter-weight introduction to MO theory and its implementations, I would recommend Frank Jensen's Introduction to Computational Chemistry.

A lovely answer, but are you then saying that the only real difference is due to the choice of basis sets and that one (MO) is thus  more adaptable (and computationally convenient) than the other?

@porphyrin I wrote this answer before I'd learned much about the modern implementations of VB theory. I now know differently, and have plans at some point to revise this. That said, yes, I think both points of your question are correct. However, the VB analogues of correlated post-HF MO methods are able to provide quite good wavefunctions in many cases, so modern VB theory is actually much more useful than this answer makes it sound. IIUC, though, one has to make a deeper dive into VB as compared to MO before obtaining practically useful results.

@porphyrin To note one key difference, whereas MO theory works intrinsically with orthogonal MOs, IIUC the "working orbitals" of VB theory are not constrained to be orthogonal.

thank you, very clear. So I suppose the non-orthogonality must make diagonalising matrices very time consuming vs orthogonal ones for the same level of accuracy?

@porphyrin I have to assume so -- I'm not too familiar with the mathematical particulars. I think it also makes the equation systems dramatically more complicated when you can't assume $\left<i|j\right>=0$ for $i\neq j$.

Comparing modern valence bond and electronic structure theories one can argue that the generalized valence bond (GVB) wave function can be regarded as a special form of the multi-configurational self-consistent field (MCSCF) wave function.1
Thus, for instance, for the hydrogen molecule, the GVB wave functions has the following form (ignoring normalization hereinafter for simplicity),
$$
  \Psi_\mathrm{GVB} = \Big( f(1)g(2) + g(1)f(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, ,
$$
where $f = a + kb$ and $g = b + ka$ are non-orthogonal orbitals and $C$ is a normalization factor. This wave function is a linear combination
$$
  \Psi_\mathrm{GVB} = C_\mathrm{c} \Psi_\mathrm{c} + C_\mathrm{i} \Psi_\mathrm{i}
$$
of the covalent and ionic structures described by
$$
  \Psi_\mathrm{c} = \Big( a(1)b(2) + b(1)a(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, , \\
  \Psi_\mathrm{i} = \Big( a(1)a(2) + b(1)b(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, .
$$
The GVB orbitals $f$ and $g$ are determined variationally: $f$ and $g$ are expanded over a basis and expansion coefficients are varied to minimize the
energy of the wave function.
The simplest two configurational MCSCF wave function for the hydrogen molecule would also be a linear combination of two wave functions. But in MCSCF to avoid non-orthogonal functions (which complicates the matter) a bonding orbital $\phi_1$ and an anti-bonding orbital $\phi_2$ are introduced and the total wave function is written as a linear combination of two Slater determinants describing two different electron configurations, $\phi_1^2$ and $\phi_2^2$, i.e.,
$$
  \Psi_\mathrm{MSSCF} = C_1 \Phi_1 + C_2 \Phi_2 \, ,
$$
where
$$
  \Phi_1 = \left| \begin{matrix} \phi_1 & \bar{\phi}_1 \end{matrix} \right| = \phi_1(1) \phi_1(2) \Big( α(1) β(2) - β(1) α(2) \Big) \, , \\
  \Phi_2 = \left| \begin{matrix} \phi_2 & \bar{\phi}_2 \end{matrix} \right| = \phi_2(1) \phi_2(2) \Big( α(1) β(2) - β(1) α(2) \Big) \, .
$$
However different $\Psi_\mathrm{GVB}$ and $\Psi_\mathrm{MSSCF}$ appear to be, if the same basis set is used to expand the orthogonal MCSCF orbitals $\psi_1$ and $\psi_2$ as well as the non-orthogonal GVB orbitals $f$ and $g$,  it can be shown that these wave functions are identical. For details, see Exercise 4.9 in Ref. 1.

1) Szabo, A.; Ostlund, N.S. Modern Quantum Chemistry: Introduction to Advanced Electronic Stucture Theory; Dover: Mineola, NY, 1989 (revised in 1996). pp. 258-261.

Here, I am referring Valence Bond Theory by VBT and Molecular Orbital Theory by MOT.
1.In VBT, atomic orbitals of the combining atoms retain a large amount of their individual character. (For example, see Orbital Hybridization).
Whereas, in MOT atomic orbitals of the combining atoms lose their individual identity in the resulting Molecular Orbital.
2.Secondly, VBT fails to explain properties like para-magnetism of some compounds like O2, whereas MOT easily explains this.
3.Also, VBT fails to explain possibility and existence of ions like H2$^+$, whereas MOT is able to do this successful.
Why it is told that "MO theory provides a global, delocalized perspective on chemical bonding"?
Quoting from the Wikipedia page of MOT,

In MO theory, any electron in a molecule may be found anywhere in the molecule, since quantum conditions allow electrons to travel under the influence of an arbitrarily large number of nuclei, as long as they are in eigenstates permitted by certain quantum rules. Thus, when excited with the requisite amount of energy through high-frequency light or other means, electrons can transition to higher-energy molecular orbitals.

Why isn't such concept needed in MO theory?
See this question, which directly addresses this. However, I would recommend you to read How does molecular orbital theory deal with resonance?
Thank You!

I quoted the "MO theory provides a global, delocalized perspective on chemical bonding" from wiki; I couldn't understand the very lines of the wiki page that you have highlighted:(

In the words of a commentary in the Journal of Chemical Education:

At their theoretical limits, both VB and MO are equivalent; they simply approach that limit from different sides.

Source:https://doi.org/10.1021/acs.jchemed.1c00919
The authors argue that in introducing the VB method in General Chemistry, it is equated with Lewis structures, which does not fully represent modern VB theory.
In the article, they also describe the simplest forms of the two theories and how they differ:

As the names suggest, valence bond theory considers a molecule as a collection of bonds, whereas molecular orbital theory views the molecule as a whole. [...] VB is presented as a single configuration of electrons in hybrid orbitals, spin paired in bonds, while MO is presented as a single configuration of molecular orbitals arranged in order of increasing energy and filled with electrons in accordance with Hund’s rule.

One way of describing the different approaches is to look at the structure of methane. The symmetry of the atomic orbitals (the p-orbitals specifically) does not match the geometry of the hydrogen atom location. In VB, the carbon atom orbitals are mixed and matched (hybridized) to make localized bonds with hydrogen. In MO, the hydrogen s-orbitals are mixed and matched (linear combinations of AOs) to match the symmetry of the p-orbitals of carbon. At a higher level of theory, which recognizes that electrons are indistinguishable, you arrive at the same distribution of electrons in the molecule.