text:
Does 1,3‐dichloroallene show geometrical isomerism?
Since the attached groups are present in different planes, $\ce{ClCH=C=CHCl}$ should not show geometrical isomerism, but my book say that it does. Why?

You are right, there is no cis-/trans- isomerism for allenes to speak of. Probably a typo, or the book author considers enantiomeric forms geometrical isomerism too. Feel free to provide an exact quote from your textbook alongside with bibliographical data.

Enantiomerism is a geometrical isomerism.It is so obvious that I do not check any definition. Just it is that often we intend geometrical isomerism as E/Z isomerism. Even more, isomerism of whatever forms implies a different geometry of stable molecules, though I would never call two chain isomers as geometric, as in this case it would bring more confusion than information.

@AbhishekKumar The compound shows axial chirality and hence has two enantiomers.

This is a chiral compound; it has two enantiomers.  This source defines "geometrical 8somerism" as isomerism involving different geometrical arrangements, and a pair of nonsuperimposable mirror images would seem to qualify.  But in the body they talk about about cis/trans or E/Z isomers, not chiral compounds.  The latter, which applies here, should be described by more precise terms like "enantiomers".

Since the attached groups are present in different planes,

This is exactly why is it has geometrical isomers. Going from one chlorine to the other or hyddrogen to hydrogen involves either a right-handed or left-handed rotation. I think the images below will allow you to see this relation.

Left-handed molecule $\qquad ~~~~~\qquad$ Right-handed molecule
Although this does not apply to this case a related concept is atropisomers which uses M or Δ for right-handed rotations and P or Λ for left-handed rotations which would apply to these types of molecules. This convention is also used in coordination chemistry.


text:
In Lassaigne's test, the organic compound is first fused with Sodium Metal instead of any other Group 1 Metal, Why is this so?
I know that Sodium has high polarising power than potassium so it can convert the respective functional groups more readily than Potassium and both Lithium and Sodium salts are soluble in water. This indicates that both Sodium and Lithium can be used but this is not the case as in each case the organic compound is first fused with Sodium to make respective ionic compounds and then identification tests are done. Why is this so? Why any other metal is not used?

You asked this historical question before (check Sci math history which is pretty good). For historical searches, Google scholar works the best. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=lassaigne+test&btnG=&oq=lassaign
The very first article is very useful on the history of sodium fusion test.
"A lost centenary: Lassaigne's test for nitrogen. The identification of nitrogen, sulfur, and halogens in organic compounds" J. Chem. Educ., 1945, 22 (5), p 212
The first few lines of the article clearly mention that Lassaigne used potassium rather than sodium. He must have the same question his mind which you are asking now. Surprisingly, it was another person named Jacobsen who suggested using sodium. The original test indeed used potassium.
Why sodium is used today in some teaching labs? Lower price and less reactivity than potassium. Essentially nobody uses this historical test in analytical laboratories anymore. Today, an very small quantity of organic compound is burned and the gaseous products are analyzed by gas chromatography.

Historical availability and procedural compliance. Why to use potassium or lithium if the procedure says sodium ? Plus, lithium is significantly more expensive and reacts with gaseous nitrogen. Potassium may be too reactive for manipulation and reaction with some samples.

source: Stack Exchange

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

sort: votes


text:
Changing the order of the reaction with change in concentration
The reaction
$$\ce{2 A + B + C -> D + 2 E}$$
is found to follow the rate law as
$$r = k[\ce{A}][\ce{B}]^2[\ce{C}]^0$$
If the concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ increases two times then the rate of reaction becomes eight times higher:
$$r' = k\cdot 2[\ce{A}]\left(2[\ce{B}]\right)^2\left(2[\ce{C}]\right)^0 = 8k[\ce{A}][\ce{B}]^2$$
Note: there is no change in the order of reaction. I want to change the order of this reaction by changing the concentration of reactants. Can I do this?
Because if we double the concentration of $\ce{A}$, $\ce{B}$ or $\ce{C}$, it does not change the order. But if we increase  the concentration of $\ce{A}$ by 4 times, then we can change order by
$$\left(4[\ce{A}]\right)^2 = \left(2[\ce{A}]\right)^{2+2} = \left(2[\ce{A}]\right)^4$$

You would change the reaction speed, but not the order.
The order of the reaction can be changed by reagent concentration, if it is result of multiple reaction steps with different orders. By variation of concentrations you can manipulate the relative reaction speed,
as the order is often ruled by the slowest reaction.
Let suppose the 2 step reaction
$$\begin{align}
\ce{2A + B &-> C} \\
\ce{A + A &->[slow] D} \\
\ce{B + D &->[fast] C} \\
\end{align}$$
For small $c_A$, the reaction speed
$$\frac{dc_C}{dt}=k_{low A}.{c_A}^2$$.
For high $c_A$, the first reaction becomes faster then the second.
$$\frac{dc_C}{dt}=k_{high A}.{c_A}^2.c_B$$.
You can also change the effective order of reaction, if some reactant is in about constant abundance and is not explicitly considered.
E.g. The speed of the reversed esterification
$$\ce{R1COOR2 + H2O -> R1COOH + R2OH}$$ is the reaction kinetics of the 2nd order. But in abundance of water, it is effectively a reaction of the 1st order:
$$\frac{dc_{\ce{R1COOR2}}}{dt}=-k_1.c_{\ce{R1COOR2}}$$.

You shouldn't be able to change the order of the reaction or the powers of the concentrations in the rate equation (which lead to the overall order). Those powers correspond to the rate determining step, so changing the rate equation would mean you have a different mechanism and potentially a different reaction.
Your substitution of different concentrations into the rate equation is imprecise which leads to error. [A] corresponds to the concentration of A in general. You should only get [2A], or rather 2[A], if you have a different chemical species, e.g F, which is produced from two moles of A. If you double the concentration of A, what you have is double a specific concentration of A, e.g 2[initial A].
To answer the edited question, your maths seems flawed as you take the power which applies to 2 only and apply it to the whole term, which would only work if they were same base (but for most cases of [A], 2 =/= [A]). So as well as the chemical intuition supporting not being able to change a reaction's order, there doesn't seem to be any mathematical way of doing so.
$$ \left(2^{2}[\ce{A}]\right)^{2} =/= \left(2[\ce{A}]\right)^{2+2} $$

Conceptually the easiest way to change the rate law in your example is to lower the concentration of C. In your rate law, the reaction is zero-order in C. However, if you lower the concentration of C to zero, you get a new rate law:
$$ \text{rate} = 0$$
instead of
$$ \text{rate} = k [A]^2 [B]$$
The reason the reaction rate did not depend on C in your rate law is that C does not come into play until after A and B have reacted to form some intermediate. If  the reaction of the intermediate with C is much faster than the formation of the intermediate, the concentration of C has negligible effect on the overall rate (if there is sufficient C available). If you leave out C, the last step obviously can't happen, so now the concentration of C has a significant effect.
To summarize this argument, having C be at zero order is only true for a certain range of concentrations of C, and if you go below it, a different rate law will be appropriate.

But if we increase the concentration of A by 4 times, then we can change order [...]

That argument is flawed. The defining equation for the order in A is
$$ \frac{\text{rate}(f \times [A_0], [B_0], [C_0])}{\text{rate}([A_0], [B_0], [C_0])} = f^\text{order}$$
In your example, you change [A] by a factor four, and the rate increases by a factor 16. That is second order. It is true if you had changed [A] by a factor of two and the rate would have increased by a factor 16, you would conclude that it is fourth order in A. However, you said you increased the concentration by a factor of four, so you have to stick with that.

If we take the concentration of C= 0 , then that will become the case of [0]^0 , the answer to this is controversial in maths I think @Karsten

@Beyond_Zero : Yes, but not controversial in chemistry. Without C, there will be no reaction. But I guess you are right, that this gives a hint that under certain circumstances, the rate of reaction might dependent on the concentration of C after all.

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


text:
Finding the repeating unit of polymerisation given two constituent molecules

The compounds $\ce{H2NCH2CH2NH2}$ and $\ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $\require{enclose}\ce{\enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCO\enclose{horizontalstrike}{)}}$
B. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2CO\enclose{horizontalstrike}{)}}$
C. $\ce{\enclose{horizontalstrike}{(}OCCH2CONHCH2NHCO\enclose{horizontalstrike}{)}}$
D. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NH\enclose{horizontalstrike}{)}}$

I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.

What you are mentioning is addition polymerization which malonic acid and ethylenediamine wouldn't dream of doing. Condensation occurs, rather.

Draw out the skeletal structures with different colors so you can see how the building blocks combine. If you have monomers A and B, and A can't link with A and B can't link with B, the only polymer you can get is A-B-A-B-A...

"As far as I know, there should be double bonds present at the ends of the repeating unit"; This is not correct.

You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:

The compounds $\ce{H2NCH2CH2NH2}$ and $\ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $\require{enclose}\ce{\enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-CO\enclose{horizontalstrike}{)}}$
B. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2CO\enclose{horizontalstrike}{)}}$
C. $\ce{\enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-CO\enclose{horizontalstrike}{)}}$
D. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NH\enclose{horizontalstrike}{)}}$

Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.

You got the answer you have looking for in Karsten Theis' answer. However, your statement of "as far as I know, there should be double bonds present at the ends of the repeating unit" is not entirely correct for all polymers in general (see A.K.'s comment). Yes, some have double bonds within the chain and at the end. But, most polymers forms without assistance from double bonds and, hence do not contain double bonds (unsaturations). Here is two best known polymers we used in our day to day life:

Both are fabrics and none has end unsaturation. The reaction shows you that none need unsaturation to condense.


text:
Taking kinetic readings for the Briggs–Rauscher oscillating reaction
I am current doing the IB diploma program where we are required to do an extended essay which is a form of a research paper on a specific subject. I chose chemistry.
I am analysing the kinetics of the Briggs–Rauscher reaction, adapting my experiment from the paper by Schmitz and Furrow [1].
The paper appears to use a spectrophotometer to measure iodine concentrations which I have tired, however the readings failed since the reaction needs to be continuously stirred. Another alternative they have used is to use an ion selective electrode to determine iodide ions in the solution however I don't have access to this.
I do have an AgCl reference electrode and was considering trying method from this YouTube video, however I lack a potentiostat and the ion-selective electrode. I was wondering if anyone could tell me any tricks on how to build this electrode if possible as I couldn't find anything online.
Alternatively, is there any other recommended way to determine the kinetics of this reaction. I would like to not have to use a stopwatch to clock the oscillations since that does not provide me with much detail.
References

Schmitz, G.; Furrow, S. Kinetics of the Iodate Reduction by Hydrogen Peroxide and Relation with the Briggs–Rauscher and Bray–Liebhafsky Oscillating Reactions. Phys. Chem. Chem. Phys. 2012, 14 (16), 5711–5717. https://doi.org/10.1039/C2CP23805E.

How about a video camera? You should be able to get an estimate of the iodine concentrations, and the video will give you the temporal information as well.

The comment by K. T. gives me a nice idea.
Set up your reaction in an environment with stable illumination and against a white background or far from any specific one. Then you can take ordered photographs that you print on transparent slides.
You can also set a cuvette in a white box and fill it with sample from the reaction vessel, to take better photographs.
After you cut them in stripes with the cuvette face size and by inserting them in the slits of the spectrophotometer you get the absorbance spectrum of a sample that mimics your reaction.
You should easily get stripes of the cuvette size, if not you can always arrange something to be measured against the appropriate background (baseline).
You can also make standards or at least checkpoints using a single component of the reaction, and in all cases measuring at a relevant wavelength.
The only hypothesis is that the A of the stripes scale linearly with that of the reaction solution, and this can be confirmed as in the above paragraph.
I never thought before of this especially because my work required some accuracy. But I could have tried it indeed when the kinetic study apparatus was already in use (just to identify a change in colour, for instance). For your essay, it might be more than enough.
If a student of mine would circumvent the given limitations in experimental set-ups using a trick as I have described, I'd be rather impressed.
edit I must add that once, for fun, I used the smartphone to photograph my finger at different levels of blood oxigenations, and after emphasising the differences looked for a trend, which I have found.  This made me consider possible colour picker applications or photography suites as a tool to quantify the color (absorbance) without even going to a spectrophotometer. Digging Internet with this in mind I have found that my idea isn't original anymore and found project similar to that in the Q, and indeed conducted using a smartphone. Examples can be easily found and downloaded searching for "smartphone as a colorimeter" and similar search keys.
https://pubs.acs.org/doi/abs/10.1021/ed300567p?src=recsys&journalCode=jceda8
is an introduction in J. Chem. Ed.


text:
What are the products in the reaction of sulphuric acid and Group 2 metals?
When group 2 metals, says magnesium, is reacted with nitric acid, oxide of nitrogen is produced beside hydrogen gas. The standard electrode potential values agree with this:
$$
\begin{align}
\ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\
\ce{H+(aq) + e- &<=> 1/2 H2(g)} &\quad E^\circ &= \pu{0 V} \\
\ce{NO3-(aq) + 4 H+ + 3e- &<=> NO(g) + 2H2O(l)} &\quad E^\circ &= \pu{+0.96 V}
\end{align}
$$
Does sulfate ion from sulfuric acid is reduced too when reacted with group 2 metals? The standard electrode potential values suggest it is possible, but I can’t find any source from internet stating $\ce{SO2}$ is produced from reaction of sulfuric acid with, says, magnesium.
$$
\begin{align}
\ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\
\ce{SO4^2-(aq) + 4 H+ + 2 e- &<=> SO2(g) + 2 H2O(l)} &\quad E^\circ &= \pu{+0.17 V}
\end{align}
$$
If sulfate ion can’t be reduced by magnesium to sulfur dioxide, then what’s the reason?

We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be predicted by tabulated electrode potentials. These tabulated values are reliable for dilute solutions (~ 1 M or less).
So the short answer is that in dilute H2SO4, and Mg, we would see a classical reaction of Mg forming magnesium sulfate and hydrogen. Under concentrated conditions and heat, Mg, may be able to able to reduce it to SO2, just like copper.

I am not clear yet. You said that these electrode potential value are only reliable for dilute solution, doesn’t mean that under dilute solution of sulphuric acid, Mg will react with it to form SO2?

Practically not. Mg will form H2 in dilute sulfuric acid. You are forgetting another half cell 2H+ 2 e -> H2 Eo = 0.00 V. If there are multiple possible half cells, the one which has the highest tendency will proceed first.


text:
For azeotropic mixtures, is the total vapour pressure at equilibrium equal to the partial pressure of one of the components?
I know, a ridiculous statement. But here's how I reached at it .
Say we have a solution of two components liquids A and B.
By Raoults law,
$$p_A= p_A^o \times x_A \tag1$$
Also by Dalton's law,
$$p_A= P_T\times x^V_A \tag2$$
Now if the solution is azeotropic,
$$x_A= x_A^V$$
So that implies the total pressure equals the partial pressure of A.
Which is absurd.
Please tell me where I am going wrong.

What makes you assume that the total pressure equals the partial pressure of A just because the mole fractions in vapour and liquid are equal? I cannot seem to arrive at how you made that statement.

Both equations (1) and (2) calculate the partial pressure of component A.  So they can be equated. Since the mole fractions in both phases are equal, the ther variables, total reassure and vapour pressure of A must be equal?

Yes, I see what you mean. But you are getting confused for actually nothing! The total pressure is equal to the vapour pressure of pure A, not the partial pressure of A over solution, so that is not a paradox at all. Nice observation though.

A mixture that exhibits an azeotrope does not follow Raolts law.  Raolts law is only valid for ideal solutions.


text:
Distinguishing between a nucleophilic substitution and an electrophilic substitution
Why don't we consider all nucleophilic substitution reactions as electrophilic reactions and vice versa?
For example, look at this reaction I found in my textbook. (See picture)

This is an electrophilic substitution if we consider the H+ to be replaced by NO2+.
But if we look at it as the OH- ion being replaced by a CCl3- ion, then it can also be a nucleophilic substitution reaction! I'm so confused!
Any help on this would be greatly appreciated.

It depends on if it is initiated by a attack from an electophile or a nucleophile.

It seems to me that your textbook failed here. That isn't a simple substitution.

@Akari how do we deduce which attacks first?

See the mechanism of the concerned reaction

@Akari does it have anything to do with the nitric acid being concentrated?

@Akari Well the mechanisms exactly what I'm looking for. Can't find it anywhere


text:
Solubility of lead phosphate
I've read that lead phosphate is very slightly soluble and so any addition of protons shifts the equilibrium back to the solvated ions. What I can't understand is why the lead phosphate is more soluble in hydroxide or basic solutions also.
Source- Vogel's qualitative analysis

Just because it's more soluble in acid does not mean it's insoluble in base.

Is there a reason for it to be soluble in a base? Apparently the base would promote precipitation by removing protons from the solution.

In neutral solution, the number of protons is low enough that if the phosphate/phosphoric acid equilibrium were the driving force for solubility, lead phosphate would be insoluble. You're dealing with two separate facts: (1) lead phosphate is sparingly soluble in water, and (2) lead phosphate is more soluble in acid.

That's what I'm saying. Being precipitated in a neutral solution ( which happens) and being reversed in an acidic solution both indicate that the said equilibrium is controlling the solvation- this obviously isn't the case with bases- what happens with bases- why don't they promote precipitation- do they LEAD to some other reactions(obviously do)- what other reactions??

So, I think I was confused by the statement of your question. "Why lead phosphate dissolved in hydroxide" It dissolves in water, too. What you meant, judging by the accepted answer, is why is it more soluble in basic soluble as well as acidic solution.

I've edited it now

The precipitation of lead phosphate would be due to the following reaction.
$$\ce{3Pb^{2+} + 2PO4^{3-} <=> Pb3(PO4)2}$$
In acidic solution the phosphate ions would be protonated, thus reducing the concentration of $\ce{PO4^{3-}}$ and increasing the solubility of $\ce{Pb^{2+}}$.
But the precipitation of the phosphate is not lead's only reaction in alkaline solution.  Lead can also form a hydroxide precipitate. In strongly alkaline solution lead can also form various soluble hydroxide complexes which would reduce the concentration of $\ce{Pb^{2+}}$ thus increasing the solubility of $\ce{Pb3(PO4)2}$.


text:
Why isn’t there LDPP, HDPP?
Unlike polyethylene that there are LDPE, HDPE, or LLDPE, when it comes to polypropylene the issue becomes about copolymer (homo/ copo/ block copolymer) or location of methyl group (isotactic/ atactic/ syndiotactic)? Why does polypropylene not have designations based on density?

Welcome to Chemistry.SE! Thanks for such a great question, I (re)learned a lot about polymers answering it.

First off, lets understand how densities of poly(ethylene) are differentiated. The prototypical polyethylene is often represented as a simple repeating ethylene monomer units as shown below:

This is what high-density polyethlyene (HDPE) is constructed of; simple linear monomer units that form closely packed crystal structures. This polymer is synthesized by  free-radical polymerization with catalyses that encourages radical at the end of the chain, keeping the polymer linear. low density polyethylene (LDPE) is free-radically polymerized without this control, which causes branching at random points in the polymer chain. This branching prevents the crystallization that made polyethylene more dense, thus it is more amorphous and lower density. To make a linear low density polyethylene (LLDPE), ethylene is co-polymerized with longer olefins such as 1-butene, 1-hexene or other olefins but with catalysts controlling polymerization similar to polymerization of HDPE. A diagram demonstrating molecular structure differences is shown below:

Source: Plastics Technologies
Bringing this back to about polypropylene, the co-polymers are a different monkey that are used for precisely tuning properties and that is not useful for this discussion. But as you mention the tactility is important. There are two reasons for this. 1.) there is no reason to produce low-density polypropylene as LDPE and LLDPE are cheaper materials that are more mature in their development with plenty of material data available and developed recycling streams, thus there is little incentive to investigate LDPP. 2.) polypropylene is already controlled by its tacticity. Isotactic and syndiotactic polymers crystallize fairly well while atactic polymers do not. Since polypropylenes rigidity and density can be controlled by using certain polymerization catalysts there is again little incentive to produce branched polypropylene


text:
Derivation of equilibrium constant
In general chemistry, we learn that the equilibrium constant, defined in terms of activities, for the reaction
$$\ce{aA + bB -> cD + dD}$$
is given by
$$K_\mathrm{eq} = \frac{(\ce{C})^c(\ce{D})^d}{(\ce{A})^a(\ce{B})^a} $$
But I can't find any explanation as to why stoichiometric coefficients should be exponentiated rather than appearing as coefficients in the equilibrium expression or something else.

This question has been asked a number of times before, please see: https://chemistry.stackexchange.com/questions/6924/why-does-the-reaction-quotient-use-the-products-multiplications-of-reactants-a
https://chemistry.stackexchange.com/questions/14836/is-there-a-reason-for-the-mathematical-form-of-the-equilibrium-constant

There are indeed alot of answers to this question. In my opinion... ignore the answers that use rate constants. Thermo does not care about time.

@CharlieCrown True that thermo does not care about time, but if one is looking at relative rate constants, the time factor cancels, so there's no reason to ignore those answers. And I think many find the rate equation arguments more intuitive and easier to understand than the stat mech multiplicative possible states derivation.

@Andrew I don't think anything makes more sense than equivalency of chemical potentials :)


text:
Stoichiometry based on addition of simultaneous equilibria
I read on another equilibrium question that if two equilibria occur at the same time, then it is not accurate with actual stoichiometry to add the two equations and multiply the equilibrium constants $K_1$ and $K_2$ in order to get an equation with stoichiometric consistency. In , multiple ICE tables are used to get around the problem. This makes the most sense to me.
However, when dealing with solubility of compounds with complex formation, many textbooks (Atkins) and websites (libretexts Oxtoby) show that you just add the two equations together and calculate solubility from there. From the libretexts site, Example 16.5.2, Equation 16.5.15 - 17 does the same operation as what I thought was forbidden/not accurate above. Here, the website says $\ce{{AgCl} + Cl^- <=> [AgCl_2^{-}]}$, which I do not think is correct.
edit: I understand that this equilibrium is correct in the way that $\frac{\ce{[AgCl_{2}^{-}]}}{\ce{[Cl^-]}}$ is constant, but using the new equilibrium to solve based on stoichiometry (when the website says that
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0 - x}\approx1.9\times10^{-5}=x$
can be used to solve for $x = \ce{[AgCl_{2}^{-}]}$) is off to me.
Which one is right? Can you actually add two simultaneous equilibria together to simplify the problem with easier stoichiometry? Or is this some sort of approximation that works because of the specific conditions that most solubility problems have?
I feel like I have a big misunderstanding here.
EDIT: I figured it out. For solubility-complex reactions, the above is an approximation based on the relative magnitudes of the equilibrium constants. Because complex formation is so favorable the majority of the time, we can assume that almost every mol of solid dissolving into solution will be consumed to form complex ions. I didn't see this anywhere I looked so I figure I might as well post it here.

You add the chemical equations and cancel species on both sides of the chemical equation. You must multiple the mathematical equations and cancel like terms in the numerator and denominator.

@MaxW See my edit, my question was confusing. I know that your procedure is possible, but I really want to know the stoichiometry behind it.

You are allowed to combine equilibrium constant expressions. However, in the chem SE problem you linked to, the OP posted an equation that did not take proper care of mass balance. That was a somewhat tricky problem because $\ce{O2}$ is generated in two equilibria, so you can't equate the concentrations of $\ce{O2}$ with the product of just one one of the equilibria.


text:
To predict if a substance is deliquescent
FeCl3,FeSO4 and CuSO4 all are deliquescent solids.

Is there any way to predict this without knowing this fact?
  I am looking for a way to predict if a substance is deliquescent for my upcoming exams.

There is no general rule, just partial empirical rules for particular anions, which can be memorized from textbooks.
There is theoretical reasoning why salts  are soluble or not, concerning the energy of the crystal lattice, energy of ion hydration, entropy change etc. But easier way is to go for empirical rules and "chemical sense".


text:
Confusion on electrode potentials
Standard electrode potential values can be used to judge the reducing ability of a metal. When I was introduced to this topic a simple picture was painted: you put a metal into a solution of its ions. An equilibrium is set up. You do so for another metal. If you now hook up the two electrodes and measure the voltage, you can tell which one is 'more negative' hence more electron donating.
I was taught like it is covered here :
  https://www.chemguide.co.uk/physical/redoxeqia/introduction.html
My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals?

You just changed the question completely. Please revert back to the original question and ask another.

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.

That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:
$$\ce{Ag+ + 1/2Mg -> Ag + 1/2 Mg^2+}$$
has the same cell potential as
$$\ce{2Ag+ + Mg -> 2Ag +  Mg^2+}.$$
(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).

How has the voltage helped us determine the difference in the tendencies of these two metals?

What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.
You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.

"That is not a problem because the tendency to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. "   If one metal gives up electrons more easily than another metal, does that not have a bearing on their comparative reduction potential? This seems to be in contradiction to the link I've put in the question.

@Sal_99 - "If metal A gives up one electron more easily than metal B" has bearing on it, "If metal A gives up two electrons more easily than metal B gives up one electron" has not bearing on it.

I added a bit to that section because it was not clear.

"but the reduction potential is on an "per electron" basis." This cleared it all up. They should make this point more obvious in the books too.

"but the reduction potential is on an "per electron" basis." -- I think this has the wrong connotation. You measure the cell potential from outside of the cells. The voltmeter doesn't know how many electrons went into the complete chemical reaction. Each half cell has to remain electrically neutral however.

@KarstenTheis -- In your chemical formulas you forgot the 2+ exponent on Mg on the right hand side.

@MaxW, my understanding from this was that the values given for redox potential are for one electron i.e. they suggest the ability of the species to gain or lose one electron. Is this interpretation correct?

That is exactly my objection. The reaction is not $$\ce{Ag+ + Mg <=> Ag + Mg+}$$ but rather $$\ce{Ag+ + Mg <=> Ag + 1/2Mg^{2+}}$$ There must be an energy difference between the first ionization of Mg and the second. But you can't get one ionization because on the Mg surface $$\ce{2Mg+(surface) -> Mg(surface)) + Mg^{2+}(aq)}$$

@MaxW I see your point how my argument could be confusing as well. Are you saying the potential is related to the average of ionization energies (plus corrections for being in solution) because $\ce{Mg+}$ is not a stable intermediate?

@KarstenTheis - Not really. I was trying to point out that even though silver exchanges one electron, magnesium does two. So if you look in the table of standard electrode potentials you don't divide the magnesium EMF by two to get the EMF on a "per electron" basis.

@MaxW - No, but you do divide the Gibbs energy by z when calculating the cell potential from it. I would encourage you to write your own answer to clear up the question.

This is a great discussion. I would only add that reduction potentials are stated in V, which is J/C, an intensive property of the half cell. This may help in clearing up the “per electron” bit. It is per electron, but only because an electron’s charge is a fixed amount in coulombs.

My question is that suppose we have to metals, A and B. A produces +1
  ions. B produces +2 ions. Now even if the two have the same tendency
  to lose or gain electrons, the B plate would be 'more negative'( As in
  this plate would have more negative charge). How has the voltage
  helped us determine the difference in the tendencies of these two
  metals?

The premise of this question is incorrect. As per your assumption, the electrode potential of the half cell as shown in tables in reduction form A(+) + e-  -> A and B(2+) + 2e  -> B is identical. So far this is fine. Let us see the concept of electroneutrality.
The gist is that overall there is no net charge in a galvanic or electrolytic cell. Therefore, if one B(2+) ion gains two electrons, two A atoms must lose electrons according to the equation 2A -> 2A+ + 2e in order to maintain electroneutrality.
This is true the other way round as well. So it is basically electron book-keeping to avoid net charge build up in a cell. Keep in mind that the interfaces (electrode surface and solution boundary) are electrostatically charged.
Coming to your voltage part of the question: The Ecell is defined as Ecathode-Eanode, if they are same numbers, Ecell = 0 V.

yes in net charge is 0. But why does this mean plate B on its own cannot be more negatively charged than plate A?

If B on it its own is more negatively charged than A, and they are connected, it means current must flow in the external circuit. From your very first condition that electrode potentials are equal, this is incompatible with the first statement under standard conditions.

yes I see that now.

please see this better-worded problem https://chemistry.stackexchange.com/questions/112316/reduction-potentials

Which plate would be negative depends on concentration of ions, as the plate with 1 electron transfer would have double rate of the potential change with the ion concentration at the electrode.
But if at particular concentrations both electrodes have the same potential, the cell voltage will be zero and neither of plates would be negative wrt the other.

If both are put into a 1 M solution, how would it work? And what does "rate of potential change mean"?

If the standard potential is the same ( we would then need activities, as activity of multicharge ions decrease faster), the the potentials would be the same.

Rate of potential  change with concentration change.


text:
Heat of formation of aqueous H⁺
Heat of formation of aqueous $\ce{H+}$ is taken to be zero, by convention. Heat of formation is energy released or absorbed when 1 mole of a compound is formed from its elements in their standard states.
$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$
Since heat of formation of $\ce{H+(aq)}$ is assigned to be zero, this would imply that this reaction has $ΔH = 0$. Wouldn't this imply that:
$$\text{Enthalpy of atomization} + \text{Ionization energy} + \text{Heat of hydration} = 0~?$$
However, this doesn't seem to be the case:
$$435 + 1312 - 1091 = \pu{656 kJ/mol}$$
Where am I going wrong here?

Where am I going wrong here?

Your chemical reaction is not balanced. You wrote
$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$
but it should be
$$\ce{1/2 H2(g) -> H(g) -> H+(g) + e- (?)-> H+(aq) + e- (?)}$$
I put a question mark for the physical state of the electron because it's not clear where that electron should go or what enthalpy of formation we would assign it. You could instead involve a second element in your reaction, e.g.:
$$\ce{H2 + Cl2 -> 2H+ + 2Cl-}$$
You could assign an enthalpy of formation for the mixture of hydrogen and chloride ions, but you would not know how to assign it to the individual components of the mixture.

In my book it is given that by convention heat of formation of aqueous H+ is taken to be zero.

Yes, that's what we do. It would equally be possible to assign it a value of 100 kJ/mol. All the enthalpies of formation of ions would change (for example, that of chloride ions would by 100 kJ/mol lower to compensate). However, all the calculations of enthalpies of reactions would have the same result because it all cancels out in a balanced equation. Setting the enthalpies of formation of elements to zero is also a bit arbitrary (not once we specify that to measure them, we are making the compounds from the elements).
Updating the definition
Instead of saying the enthalpy of formation of a compound is the enthalpy of making that compound from the elements, we could widen the definition to say:
"The enthalpy of formation of a species is the enthalpy of the reaction making the species where all other reactants and products are either elements or the hydrogen ion""
This definition would include ions. For chloride ions, the following reaction would work:
$$\ce{1/2H2 + 1/2Cl2 -H+ -> Cl-}$$
For sodium ions, the following reaction would work:
$$\ce{Na + H+ - 1/2H2 -> Na+}$$
Why the hydrogen ion?
Again, it is arbitrary and any other ion would have been fine as well. The hydrogen ion is special for two reasons, though. It is the smallest ion, and it is fairly common in reactions, so it saves some arithmetic effort when doing calculations.


text:
Kb values of N,N-dimethylaniline derivatives
I need the $K_\mathrm{b}$ values for ortho-, meta- and para- positions of the following compounds:

I've looked and couldn't find them anywhere.

I've redrawn the compounds of interest with chemfig to make them more readable. However, I'm not sure I recognized all groups correctly — feel free to proof-read the edit. $\mathrm{\LaTeX}$ source code is available at Pastebin.

chemicalize.com is  a cheminformatics website which is the kind of resource you are looking for.
You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.
Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are calculated / estimated, not measured $\mathrm{p}K_\mathrm{a}$ values.

Table 1. $\mathrm{p}K_\mathrm{a}$ values (of the conjugate acid)
$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & 3.71 & 5.19 & -0.98 & 1.44 & 2.66 \\
\textit{meta-} & 3.89 & 5.15 & 1.73 & 4.77 & 3.45 \\
\textit{para-} & 5.55 & 5.46 & 0.65 & 4.68 & 3.65 \\
\hline
\end{array}
$$
Since the question is asking for $K_\mathrm{b}$ values, I converted them too.
Table 2. $K_\mathrm{b}$ values
$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & \pu{5.13e-11} & \pu{1.55e-9} & \pu{1.05e-15} & \pu{2.75e-13} & \pu{4.57e-12} \\
\textit{meta-} & \pu{7.76e-11} & \pu{1.41e-9} & \pu{5.37e-13} & \pu{5.89e-10} & \pu{2.82e-11} \\
\textit{para-} & \pu{3.54e-9} & \pu{2.88e-9} & \pu{4.47e-14} & \pu{4.79e-10} & \pu{4.47e-11} \\
\hline
\end{array}
$$

Please be kind enough to explain your downvote.


text:
Computational program not using full cpu power
I started some computations on Gamess.
For instance, when I compute the frequencies of an optimized structure, I see in the terminal that the job only uses 40% of the CPU core most of time (sometimes it goes to 100%). When I decide to provide 2 cores for the job, it only uses something like 20-25% of each core most of the time (and sometimes 100% of each core too). Why would not the task take maximum power to achieve the computations ?

I probably won't be able to answer, but for those who can, it might be helpful if you could post the MWE (minimal working example) for the code you are running alongside with the hardware and OS specs.

I'd assume that you see a lot of disk activity and that fetching from disk isn't counted against the CPU. Also you have a multitasking system. A modern computer has dozens of processes running at any given time. So the other processes are not counted against the one you're monitoring.

Without knowing specifics / doing any profiling: sever bottlenecks often the disk reading-writing and memory use. Different steps of calculations can scale very differently, also can have different bottlenecks, so it is not rare that CPU use goes up at certain steps, while goes down in others.

Without knowing any specifics about Gamess: some programs allow for the explicit specification of algorithms concerning disk usage for storage vs. recalculation of integrals, "Conventional" and "Direct" are the terms used.

I'm voting to close this question as off-topic because it's not about chemistry.

What even makes you think it should? Any time there are many processes active and very rarely CPU is used in 100%.

Most applications in theoretical chemistry are memory and or memory bandwidth bound. Especially due to shared memory bandwidth it can be bad for your performance if you use too many processes.
In addition you benefit from additional processes only if your algorithm is more or less parallel. If there are too many serial parts/communication overhead you will have idle cores, if your process number is too high. (Amdahl's law)
Take home message:

Time the execution of your code with a non trivial example.
More is not always better.


text:
Factorizing Slater determinant into product of spin-up and spin-down slater determinant
I am reading the book on MONTE CARLO METHODS IN AB INITIO QUANTUM CHEMISTRY by Hammond and Lester. (Chapter 5-Variational Trial Functions, Section 5.3-Hartree Fock and Beyond, Sub-Section 5.3.3 Hartree Fock and Correlation Energies)In section 5.3.3, the book says the following,
"In Monte Carlo, we may factor the determinant into electron spin components, i.e., $\psi_{D}=\psi_{\alpha}\psi_{\beta}$. This factorization is justified because all possible permutations of this term that are present in the full antisymmetric $\psi$  are operationally equivalent: each term simply corresponds to a relabeling of the electrons. Expectation values, and even local values of any observable, are unchanged."
I couldn't understand how these two are equal. Is it really possible to factorize this into two slater determinants? How does one justify that this is true?

When exchange effects are not taken into consideration, the simplest possible description of a many-body wave function is simply the Hartree product, i.e. the product of individual-particle wave functions. This is reasonable for bosons, but electrons are fermions, which means that the exchange of any two must flip the sign of the wave function.
Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product. It is like up- and down-spin wave functions occupy different subspaces and interact via the electron–electron correlation term in the Hamiltonian.

"Since this is only relevant for electrons of the same spin" - I feel that this part deserves more explanation. So far I've always heard that the fermionic wavefunction must be antisymmetric under the exchange of any two electrons.

Gopal Iyer's answer is only partially correct and doesn't contain the right interpretation of the citation in the question in my opinion. First I would like to shed some more light on Gopal Iyer's answer and then I will try to answer the actual question. From Gopal Iyer's answer:

Slater determinants are used to antisymmetrize the wave function so that fermion antisymmetry (Pauli principle) is obeyed. Since this is only relevant for electrons of the same spin, it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product.

As Antimon's comment correcly points out,

the fermionic wavefunction must be antisymmetric under the exchange of any two electrons.

Likewise the bosonic wave function must be symmetric under the exchange of any two particles. Here exchange of two particles means that all of the particle's parameters (i.e. the spatial parameter $\vec{x}_i$ and the spin parameter $s_i$) are exchanged. For fermions we have for example
$$\psi(\vec{x}_1,s_1,\vec{x}_2,s_2)=-\psi(\vec{x}_2,s_2,\vec{x}_1,s_1).$$
In many cases $\psi$ can be written as a product of a spatial function $\phi$ and a spin function $\chi$ and we have
$$\psi(\vec{x}_1,s_1,\vec{x}_2,s_2)=\phi(\vec{x}_1,\vec{x}_2)\chi(s_1,s_2),$$
$$\phi(\vec{x}_1,\vec{x}_2)\chi(s_1,s_2)=-\phi(\vec{x}_2,\vec{x}_1)\chi(s_2,s_1).$$
This means that if $\chi$ is a symmetric function, $\phi$ must be an antisymmetric function and vice versa. If both fermions occupy the same spin state $s$, then $\chi(s_1,s_2)=\delta_{ss_1}\delta_{ss_2}$ is a symmetric function and as a consequence $\phi$ must be an antisymmetric function. However, if the fermions don't occupy the same spin state, $\chi$ and $\phi$ can have any symmetry as long as the product is an antisymmetric function under exchange of the particles.
This is probably what Gopal Iyer's meant with:

Since this is only relevant for electrons of the same spin, [...].

However, the conclusion

it is sufficient to describe the overall wave function by giving each spin its own determinant and then combining the determinants by taking the product

is wrong. To correctly describe fermions, one has to use the full Slater determinant including the spin functions
$$\psi(\vec{x}_1,s_1,...,\vec{x}_N,s_N)=\frac{1}{\sqrt{N!}}\det\begin{pmatrix}
\psi_1(\vec{x}_1,s_1) & \dots & \psi_1(\vec{x}_N,s_N)\\
\vdots & \ddots & \vdots\\
\psi_N(\vec{x}_1,s_1) & \dots & \psi_N(\vec{x}_N,s_N)\\
\end{pmatrix}.$$

So, is the citation from the book in the question wrong then? The answer is no. The short answer is that the book doesn't claim that $\psi_{D}=\psi_{\alpha}\psi_{\beta}$ is the correct description of the fermions, nor does it claim that $\psi_{D}$ and the full antisymmetric $\psi$ are equivalent. The statement is:

In Monte Carlo, we may factor the determinant into electron spin components, i.e., $\psi_{D}=\psi_{\alpha}\psi_{\beta}$.

But what exactly does that mean? Why and how does it work? In order to get an understanding for that, I will first clarify why the Slater determinent is necessary in VMC at all.
In general, if we are looking for solutions (eigenstates) of a Schrödinger equation, we don't have to look for antisymmetric solutions. If we find an eigenstate that is not (fully) antisymmetric, we can always apply the antisymmetrizer or antisymmetrizing operator $\mathcal{A}$ in the end, provided that the found eigenstate is not symmetric under any odd parity permutation. If we use a Hartree product to construct our eigenstate, this can be ensured by using pairwise distinct single-particle functions $\psi_i$ (pairwise orthonormal spin-orbitals). The Slater determinant can then be written as
$$\psi(\vec{x}_1,s_1,...,\vec{x}_N,s_N)=\sqrt{N!}\mathcal{A}\psi_1(\vec{x}_1,s_1)...\psi_N(\vec{x}_N,s_N),$$
where $\psi_1(\vec{x}_1,s_1)...\psi_N(\vec{x}_N,s_N)$ is the Hartree product. However, the problem is that the Hartree product isn't an eigenstate of the Schrödinger equation in general. This is only the case in very simple problems (i.e. if the Schrödinger equation only contains single-particle operators, it is separable and the Hartree product of single-particle solutions is an eigenstate).
I will now show specifically for VMC, why using the simple Hartree product instead of the Slater determinant leads to a problem. In VMC the goal is to minimize the energy expectation value $\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}$ to get an approximation of the ground state. What would happen to the energy expectation value if we use the Hartree product instead of the fully antisymmetric $\psi$? If the result were the same, we could generally use the much simpler Hartree product instead to minimize the energy. Well, let's see. The Slater–Condon rules tell us
$$\langle\psi|\sum_{i=1}^{N}\hat{f}(\vec{x}_i,s_i)|\psi\rangle=\sum_{i=1}^{N}\langle\psi_{i}|\hat{f}|\psi_{i}\rangle,$$
$$\langle\psi|\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\hat{g}(\vec{x}_i,s_i,\vec{x}_j,s_j)|\psi\rangle=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}(\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{i}\psi_{j}\rangle-\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle).\ \ \ (1)$$
On the other hand, for the Hartree product we have simply
$$\langle\psi_1...\psi_N|\sum_{i=1}^{N}\hat{f}(\vec{x}_i,s_i)|\psi_1...\psi_N\rangle=\sum_{i=1}^{N}\langle\psi_{i}|\hat{f}|\psi_{i}\rangle,$$
$$\langle\psi_1...\psi_N|\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\hat{g}(\vec{x}_i,s_i,\vec{x}_j,s_j)|\psi_1...\psi_N\rangle=\frac{1}{2}\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{i}\psi_{j}\rangle.\ \ \ (2)$$
The term $\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle$ in equation $(1)$ isn't present in equation $(2)$, which means that the energy expectation value of the Hartree product and the fully antisymmetric $\psi$ are not generally equal
$$\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}\neq\frac{\langle\psi_1...\psi_N|\hat{H}|\psi_1...\psi_N\rangle}{\langle\psi_1...\psi_N|\psi_1...\psi_N\rangle}.$$
If the goal is to minimize the energy expectation value $\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}$, we can't minimize the energy expectation value with respect to the Hartree product instead because in general this could lead to a different minimum.
Side note: Although the Slater determinant is much harder to compute than the simple Hartree product, there are even some advantages. Due to the introduced (anti)symmetry of the Slater determinant, the variance in VMC is much smaller. Also the fact that the Slater determinant vanishes when two particles with the same spin function approach each other, makes it much easier to fulfill cusp conditions.
Above we have seen that the only difference between equation $(1)$ and $(2)$ is the term $\langle\psi_{i}\psi_{j}|\hat{g}|\psi_{j}\psi_{i}\rangle$. However, this term vanishes if $\hat{g}$ is a spin-independent operator and $\psi_{i}$ and $\psi_{j}$ have different spin functions (i.e. $\alpha$, $\beta$). This gives rise to the assumption
$$\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}=\frac{\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle}{\langle\psi_\alpha\psi_\beta|\psi_\alpha\psi_\beta\rangle}.$$
I will now show this with an example Slater determinant
$$\psi(\vec{x}_1,s_1,...,\vec{x}_4,s_4)=\frac{1}{\sqrt{4!}}\det D,$$
$$D=\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_3)\alpha(s_3) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_3)\alpha(s_3) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_3)\beta(s_3) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_3)\beta(s_3) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix},$$
by using the hint in the citation in the question

all possible permutations of this term that are present in the full antisymmetric $\psi$ are operationally equivalent: each term simply corresponds to a relabeling of the electrons.

The same steps in the following calculation can be used to formulate a rigorous proof. $\hat{H}$ can be an arbitrary spin-independent operator in which all particles are indistinguishable.
So, how do we get the said factorization and the terms? We can do this by using the generalized Laplace expansion formula, which allows us to expanded along multiple rows or columns at once. We choose to expand along the rows with spin function $\alpha$ and define $I^\alpha=\{1,2\}$ to be the indices of those rows. The universe is $U=\{1,...,N\}=\{1,...,4\}$, an apostrophe $'$ denotes the complement of a set, vertical bars $|\cdot|$ denote the cardinality of a set and $D_{IJ}$ denotes the submatrix of $D$ with rows $I$ and columns $J$. Thus $I^\beta={I^\alpha}'=\{3,4\}$. Then the expansion is given by
$$\det D=\sum_{|J|=|I^\alpha|,J\subseteq U} (-1)^{\sum I^\alpha+\sum J}\det D_{I^\alpha J}\det D_{I^\beta J'}.$$
Using this formula we obtain
$$\det D=\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_2)\alpha(s_2)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_2)\alpha(s_2)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_3)\beta(s_3) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_3)\beta(s_3) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
-\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_3)\alpha(s_3)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_3)\alpha(s_3)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_1)\alpha(s_1) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_1)\alpha(s_1) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_2)\beta(s_2) & \phi_3(\vec{x}_3)\beta(s_3)\\
\phi_4(\vec{x}_2)\beta(s_2) & \phi_4(\vec{x}_3)\beta(s_3)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_3)\alpha(s_3)\\
\phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_3)\alpha(s_3)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_4)\beta(s_4)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_4)\beta(s_4)\\
\end{pmatrix}\\
-\det\begin{pmatrix}
\phi_1(\vec{x}_2)\alpha(s_2) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_2)\alpha(s_2) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_3)\beta(s_3)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_3)\beta(s_3)\\
\end{pmatrix}\\
+\det\begin{pmatrix}
\phi_1(\vec{x}_3)\alpha(s_3) & \phi_1(\vec{x}_4)\alpha(s_4)\\
\phi_2(\vec{x}_3)\alpha(s_3) & \phi_2(\vec{x}_4)\alpha(s_4)\\
\end{pmatrix}\det\begin{pmatrix}
\phi_3(\vec{x}_1)\beta(s_1) & \phi_3(\vec{x}_2)\beta(s_2)\\
\phi_4(\vec{x}_1)\beta(s_1) & \phi_4(\vec{x}_2)\beta(s_2)\\
\end{pmatrix}.$$
We immediately see that the first term is $\sqrt{2}\psi_\alpha\sqrt{2}\psi_\beta$ and all terms are equivalent up to relabeling of particles. From determinants we can factor out the spin functions per column
$$d:=\frac{1}{2}\det D=\\
+\psi_\alpha(\vec{x}_1,\vec{x}_2)\alpha(s_1)\alpha(s_2)\psi_\beta(\vec{x}_3,\vec{x}_4)\beta(s_3)\beta(s_4)\\
-\psi_\alpha(\vec{x}_1,\vec{x}_3)\alpha(s_1)\alpha(s_3)\psi_\beta(\vec{x}_2,\vec{x}_4)\beta(s_2)\beta(s_4)\\
+\psi_\alpha(\vec{x}_1,\vec{x}_4)\alpha(s_1)\alpha(s_4)\psi_\beta(\vec{x}_2,\vec{x}_3)\beta(s_2)\beta(s_3)\\
+\psi_\alpha(\vec{x}_2,\vec{x}_3)\alpha(s_2)\alpha(s_3)\psi_\beta(\vec{x}_1,\vec{x}_4)\beta(s_1)\beta(s_4)\\
-\psi_\alpha(\vec{x}_2,\vec{x}_4)\alpha(s_2)\alpha(s_4)\psi_\beta(\vec{x}_1,\vec{x}_3)\beta(s_1)\beta(s_3)\\
+\psi_\alpha(\vec{x}_3,\vec{x}_4)\alpha(s_3)\alpha(s_4)\psi_\beta(\vec{x}_1,\vec{x}_2)\beta(s_1)\beta(s_2).$$
Our goal is to calculate the expectation value
$$\langle\psi|\hat{H}|\psi\rangle=\left(\frac{2}{\sqrt{4!}}\right)^2\langle d|\hat{H}|d\rangle=\frac{1}{6}\langle d|\hat{H}|d\rangle.$$
Since $\hat{H}$ is a spin-independent operator, it doesn't effect the spin states when we apply it to $|d\rangle$. When $\langle d|$ is applied from the left side, we see that each cross term contains at least one $\langle\alpha(s_i)|\beta(s_i)\rangle=0$ or $\langle\beta(s_i)|\alpha(s_i)\rangle=0$. Thus only the direct terms remain with $\langle\alpha|\alpha\rangle=1$ and $\langle\beta|\beta\rangle=1$ and we get
$$\langle\psi|\hat{H}|\psi\rangle=\frac{1}{6}\\
(\langle\psi_\alpha(\vec{x}_1,\vec{x}_2)\psi_\beta(\vec{x}_3,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_2)\psi_\beta(\vec{x}_3,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_1,\vec{x}_3)\psi_\beta(\vec{x}_2,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_3)\psi_\beta(\vec{x}_2,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_1,\vec{x}_4)\psi_\beta(\vec{x}_2,\vec{x}_3)|\hat{H}|\psi_\alpha(\vec{x}_1,\vec{x}_4)\psi_\beta(\vec{x}_2,\vec{x}_3)\rangle\\
+\langle\psi_\alpha(\vec{x}_2,\vec{x}_3)\psi_\beta(\vec{x}_1,\vec{x}_4)|\hat{H}|\psi_\alpha(\vec{x}_2,\vec{x}_3)\psi_\beta(\vec{x}_1,\vec{x}_4)\rangle\\
+\langle\psi_\alpha(\vec{x}_2,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_3)|\hat{H}|\psi_\alpha(\vec{x}_2,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_3)\rangle\\
+\langle\psi_\alpha(\vec{x}_3,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_2)|\hat{H}|\psi_\alpha(\vec{x}_3,\vec{x}_4)\psi_\beta(\vec{x}_1,\vec{x}_2)\rangle).$$
Since $\hat{H}$ doesn't distinguish particles, all six expectation values are equal to $\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle$ and we have
$$\langle\psi|\hat{H}|\psi\rangle=\langle\psi_\alpha\psi_\beta|\hat{H}|\psi_\alpha\psi_\beta\rangle.$$
So, what the book tells us is that we can either use the full antisymmetric $\psi$ or the factorized form $\psi_D=\psi_\alpha\psi_\beta$ in order to minimize the energy expectation value and we will get the same result. However, $\psi_D$ isn't the correct description of fermions. To get the correct fully antisymmetric description, one has to calculate $\psi=\frac{\mathcal{A}\psi_D}{\langle\mathcal{A}\psi_D|\mathcal{A}\psi_D\rangle}$.


text:
Autoionization state
I am reading this paper:
"Role of Autoionizing State in Resonant High-Order Harmonic Generation and Attosecond Pulse Production"
In the paper a transition from the $\ce{In+}$ ground state to an autoionization state is discussed:
$$4d^{10}5s^2 \, {}^1\!S_0 \longrightarrow 4d^95s^25p\,(^2\!D)\,{}^1\!P_1$$
I am trying to understand why this is an autoionization state.
If I follow the Aufbau principle, an electron in the ground state $4d^{10}5s^2$ that is excited should be in the state $3d^95s^25p$.
If this is just the lowest excited state, why is it considered an autoionization state?

Welcome to Chemistry.SE! Please note $\LaTeX$ is preferred over images when reasonable.


text:
How to report a triplet of septets in NMR tabulation?
Does anyone know the proper format to report an iso-butyl group, eg. $\ce{R-CH2-C\color{red}{H}(CH3)2}$, in $\ce{^1H}$ NMR tabulation? The proton in red should be a triplet of septets, so I've written:
δ 2.72 (d, J = 5.5 Hz, 2H), 1.80 (tsep, J = 5.5, 5.3 Hz, 1H), 1.02 (d, J = 5.3 Hz, 6 H).
Is the use of "tsep" acceptable here?

I personally use a space, so "t sep" as long as both "t" and "sep" are defined. I think it's slightly clearer than "tsep", but I don't think there's any official adjudication on what's a good or bad acronym.
Also, of some interest to me is how you manage to distinguish 5.5 and 5.3 Hz. If it looks quite like an nonet, I would probably prefer reporting it as an apparent nonet, although different people may have different opinions... I know some groups will have "in-house" rules / guidelines for reporting NMR spectra, so it might be a good idea to check with your supervisor on this front.

I used to call it heptet, but nowadays, those have changed. For example, now they call all pentets as quintets. :-) About 5.5 vs 5.3, I think OP has been using $\pu{400 MHz}$ or bigger machine so they can measure up to $\pu{0.01 Hz}$.

@MathewMahindaratne, it's not realistically possible to measure couplings to an accuracy of 0.01 Hz. If you take a 400 MHz machine, a spectral width of 20 ppm (= 8000 Hz), and 64k data points during acquisition (32k real + 32k complex), then the resolution is ~0.25 Hz. Even this is not enough to accurately distinguish a coupling of 5.5 and 5.3 Hz, you can see this for yourself at this website, or using e.g. MestReNova - it is indistinguishable from a perfect nonet.

You would also need to take into consideration linewidth contributions from e.g. field inhomogeneity and relaxation. It's very, very difficult to distinguish two couplings differing by 0.2 Hz.

Sure, but computer gives your peak picking to 4 decimal places, so I think OP might think it's okay to report that way without concerning the uncertainty. Nonetheless, I wouldn't report that way in a journal article.

@MathewMahindaratne, sorry, I think I might have misunderstood you, actually; I think it's fair to say that you can measure 5.5 and 5.3 for two different multiplets, i.e. the CH2 and the (CH3)2 signals, although the uncertainty is definitely still there. I was more concerned about the CH peak, where the shape of that multiplet (on its own) won't allow you to extract two different couplings of 0.2 Hz difference from that one multiplet.

Yes, you understood me now. And also, splitting for a single proton signal is hard to distinguish. I'd call it a multiplet in my publication.:-)


text:
Reduction Potentials
Standard electrode potential values can be used to judge the reducing ability of a metal. When I was introduced to this topic a simple picture was painted: you put a metal into a solution of its ions. An equilibrium is set up. You do so for another metal. If you now hook up the two electrodes and measure the voltage, you can tell which one is 'more negative' hence more electron donating.
My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions.
A ---> A(+1) + e
B ---> B(2+) + 2e
Suppose these two processes have the same tendency ( it is just as energetically feasible for both to form their respective ions) So in a half-cell of B and its ions, plate B should be more negatively charged compared to plate A in its own half cell. Charge should flow from B to A in the external circuit. The electrode potentials will be different when measured practically. I think this is all true and fine.
Now, we can use electrode potentials to tell us about the relative ability of a metal to lose or gain electrons. But in this case, the difference isn't due to a tendency difference: only the number of electrons is different.
How does this line up with the oft-taught idea that we can use reduction potentials to judge the ability to gain or lose?
I am aware I've asked a similar question, but apparently, I did not realize my wording was off and then it was too late to edit it all out

Well please conform to standard terminology. (1) Write the equations as reductions since all electrode data tables are formatted that way. (2) What do you mean that when you say "Suppose these two processes have the same tendency." -- Do you mean the two half cells have the same half cell potential.

Please revise your question. "Suppose these two processes have the same tendency." Tendency to do what? "So in a cell, plate B should be more negatively charged. Charge should flow from B to A in the external circuit. The electrode potentials will be different when measured practically. I think this is all true and fine. " This is still not correct.

@Max, no that is exactly what I don't mean. .

@M.Farooq, as in it is just as energetically feasible for both to form their respective ions.

@M.Farooq, A better way to say it maybe that the two form ions just as readily. Then why should electrode B itself not be charged comparitively more negative?

Are you thinking that because the half-cell reaction with A involves one electron and the half-cell reaction with B involves two electrons that the A half-cell must have a lower EMF than the B half-cell?

@MaxW, B should have a lower one (as in more negative) since it will have more electrons on its plate if the two metals form ions just as readily. This is my thinking

No, no, no no, no, ......

Also note that potential has nothing to do with charges on electrodes in a half reaction. Do not confuse the potential of a half-reaction as a direct measure of the spontaneity of the half-reaction.

Let's write the half-cell equations in standard form as half-cell reductions.

In a table of standard electrode potentials it would silly to write both the reduction and oxidation reactions since that would needlessly double the size of the table.

$$
\begin{align}
\ce{A+ + e- &-> A} &\quad V_\ce{A} \\
\ce{B^{2+} + 2e- &-> B} &\quad V_\ce{B}
\end{align}
$$
Now this assumes that each half cell has exactly a 1 molar solution of the cation. If the concentrations are not exactly 1 molar, then the voltage of the half-cell will vary according to the Nernst equation.
Now let's assume that we have used the Nernst equation and calculated the appropriate values for each half cell as $V^*_\ce{A}$ and $V^*_\ce{B}$. Now the values of  $V^*_\ce{A}$ and $V^*_\ce{B}$ can either be positive or negative. But let's just convert the chemical equation for B to an oxidation and get the overall reaction;
$$\ce{2A+ + B -> 2A + B^{2+}}$$
The EMF for that reaction will be  $V^*_\ce{A} - V^*_\ce{B}$.

If $V^*_\ce{A} - V^*_\ce{B} > 0 $ then the spontaneous reaction of the cell will be

$$\ce{2A+ + B -> 2A + B^{2+}}$$

If $V^*_\ce{A} - V^*_\ce{B} < 0$ then the spontaneous reaction of the cell will be

$$\ce{2A + B^{2+} -> 2A+ + B}$$
In either case the reaction will proceed until  $V'_\ce{A} = V'_\ce{B}$ at which point no current will flow between the half cells. ($V'_\ce{A}$ and $V'_\ce{B}$ being calculated by Nernst equation again with final concentrations of $\ce{A+}$ and $\ce{B^{2+}}$.)

I agree with everything you've said.That makes perfect sense. But my question is more about using these Voltage values to comment about the relative ease of forming ions for either. Let's say the second case in your example is what happens at 1M conc for both. My question is does this happening tell us anything about the relative ease with which A or B form their ions?

"The relative ease" is an ambiguous phrase. If the metal ion concentrations are exactly 1 molar then the EMF of each cell will start out at the reduction potential as listed in a table of standard reduction potentials. Then which reaction happens depends on which is greater $\ce{V_A}$ or $\ce{V_B}$.

I'll be more honest, but I'm risking confusing you totally. I wrote that the cells should have a concentration of exactly one molar. That isn't true. Rather the standard half cell potential is based on an activity of exactly 1. The activity is influenced by not only the concentration of the particular ion in the half-cell reaction, but also for example by the ionic strength of the solution. So a 1 molar solution of $\ce{BCl2}$ would have a different ionic strength than a solution of 1 molar $\ce{BSO4}$. Hence the activity of $\ce{B^{2+}}$ would be slightly different in the two solutions.

@Sal_99 I guess in part what I'm trying to say is that there is no function in chemistry to calculate relative ease. You have to be specific about what behavior you're trying to model mathematically.

@andselisk - Thanks for the edit. That did clean up the answer.

The main thing to point out is that your question doesn't actually make sense the way you want it to.
What does it mean if a reduction potential is zero? Or positive? Or negative? Is that favorable or not? It turns out we don't specify this at all!
As an example, consider the standard hydrogen electrode and it's reduction potential:
$$\ce{2H+ + 2e- -> H2},\quad \mathscr{E} = 0\ \mathrm{V}$$
Is this reaction on the borderline between spontaneous and nonspontaneous? I don't know the answer to this question, and importantly, this is the wrong question to ask. Why? Because we're doing electrochemistry by balancing an oxidation and reduction reactions. The individual potentials don't matter.
Take a reduction:
$$\ce{2A+ + 2e- -> A2},\quad \mathscr{E}=V_{A}\tag{1}$$
Its potential is relative to that of the standard hydrogen electrode. Can you tell from this value if it will be reduced or not? No, you cannot! What you can tell is its tendency to be reduced relative to the standard hydrogen electrode.
Let's examine why everything works.
Take this reaction:
$$\ce{2A+ + H2 -> 2H+ + A2},\quad \mathscr{E}=V_{A}\tag{2}$$
And if we have the reduction
$$\ce{B+ + e- -> B},\quad \mathscr{E}=V_{B}\tag{3}$$
Then we can also get:
$$\ce{2H+ + 2B -> H2 + 2B+},\quad \mathscr{E}=-V_{B}\tag{4}$$
Combining $\text{(3)}$ and $\text{(4)}$ and cancelling the hydrogens, we get:
$$\ce{2A+ + 2B -> 2B+ + A2},\quad \mathscr{E} = V_{A}-V_{B}\tag{5}$$
This is exactly the same answer as taking the two reductions half reactions, converting one to an oxidation, and summing the half reactions.
Now, consider reaction $\text{(2)}$ again. There is a free energy change associated with this reaction that is:
$$\Delta G = -nF\mathscr{E}$$
Note that this says absolutely nothing about the free energy change of equation $\text{(1)}$ because we have no idea what the free energy change for the reduction hydrogen
 is. That is:
$$\Delta G = \Delta G_{\ce{A+|A}} - \Delta G_{\ce{H+|H2}}$$
$$\Delta G_{\ce{H+|H2}} =\ ?$$
But the point here is that we don't care because we were able to obtain equation $\text{(5)}$ without needing this information.
As an exercise, you can confirm that the total free energy change is indeed the expected value because $\Delta G_{\ce{H+|H2}}$ cancels from the sum of $\text{(2)}$ and $\text{(4)}$.

+1 - Very well stated. When doing electrochemisty focus on the cell EMF.

@Zhe, okay. So everything in electrochemistry is a comparison basically. Nothing absolute. My book says "the more negative the value of the standard electrode potential, the greater the tendency to give up electrons". I understand that they were all compared to SHE and then put into an order. My core concern while phrasing this question was that should we care about the fact that some metals from +2 ions, some +1, etc?

@Zhe, this became a point since I was introduced to this concept using an equilibrium between the metal and its ions. I was then told that those that have a higher tendency to give up electrons will have equilibria lying to the left, and for the others equilibria will be  to the right. Then I was told that this results in a charges on the metal. Then it was that the metals with higher tendency to lose electrons will have comparatively more negative plates than their counterparts. Now. Two things with the same position of equilibrium but 1 losing 2 electrons, the other just losing 1.

How will they compare?

You've created more problems for yourself than you need to. On some level what you say is true because second ionization energies are higher than first ionization energies, but that's only the enthalpic component, and it's measured in the gas phase. We're in solution, so you need to account for solvation enthalpy and solvation entropy. That's extremely non-trivial. In order to not go crazy with those details, we abstract out what is important: spontaneity. And we use potentials and free energies to quantify it. @Sal_99

Quoting your comment

A better way to say it maybe that the two form ions just as readily.
  Then why should electrode B itself not be charged comparatively more
  negative?

Eo is not a direct measure of the tendency but rather Gibbs free energy. They are related as Delta Go= -nFEo; Write it the other way, you get
Eo = Delta Go/(-nF)
did you notice that, the number of electrons n, is already factored in? No matter how many electrons are gained or lost Eo is oblivious of the representation:
B(2+) + 2e = 2B
1/2 B (2+) + e = 1/2 B; all have the same electrode potential.
You are imposing a condition (1) now that there is a divalent metal B and a monovalent metal A. You have two beakers, one contains A dipping in A(+), and the other beaker containing B dipping in B(2+). We have not connected them as yet via an external circuit. Assume we have a electrostatic charge meter. You can test A and B separately. As per your logic, B should be more "electrostatically negative" than A because B is divalent. Let us assume this is true for a moment. The moment you complete the circuit, electricity should flow from B to A, because B had more electrostatic charge.
Now you impose another condition (2), which is quite ambiguous, i.e. both have the same tendency to "form ions just as readily". Translating this mathematically,
Delta Go values of the half cell with respect to SHE, A(+) + e -> A and B(2+) + 2e -> B are numerically the same. Hence Ecell = zero, which implies that there must be no charge build-up on B with respect to A.
Do you notice that both (1) and (2) are mutually incompatible?
Also think of the pair Zn(2+)/Zn and Cu(2+)/Cu, by condition (1), this cell should not work because both metals gain or lose equal number of electrons. In reality this pair works perfectly because the condition no 2 is not imposed on them.
It is good to have a thought process, however, when it defies the experiment, chances are that the line of reasoning is not correct. As I.M. Kolthoff used to say "Theory guides, experiment decides".

Unfortunately, this is not correct. You cannot translate a reduction potential to a simple $\Delta G$ because everything is referenced against the standard hydrogen electrode, and the $\Delta G$ you compute lost the value of the free energy change from oxidation in the standard hydrogen electrode. As a concrete example, what is the free energy change associated with the reduction of a hydrogen ion to hydrogen gas? This process does not have $\Delta G = 0$.

You raised a good point. We should always specify "with respect to" for Eo and delta Go. For the OP see https://chemistry.stackexchange.com/questions/4355/delta-g-for-standard-hydrogen-electrode-reaction

I was introduced to this concept using an equilibrium between the
  metal and its ions. I was then told that those that have a higher
  tendency to give up electrons will have equilibria lying to the left,
  and for the others equilibria will be to the right. Then I was told
  that this results in a charges on the metal. Then it was that the
  metals with higher tendency to lose electrons will have comparatively
  more negative plates than their counterparts. Now. Two things with the
  same position of equilibrium but 1 losing 2 electrons, the other just
  losing 1.

@Sal_99, You are trying to quantify the word tendency. If someone says that this vegetable soup tastes better than that chicken soup, how can we quantify this tendency to prefer vegetable soup? It is a qualitative descriptor at this end. Don't use electrode potential to correlate the amount of charge on accumulated an electrode at this moment and also don't impose the condition that both A(+) and B(2+) have the same tendency to lose electrons. Potential is work done per unit charge.
Now imagine another thought experiment, two metallic rods, A and B dipping separately in two beakers of A(+) and B(2+). They are not connected and the metal is in equilibrium with its own ion. Assume we have a charge meter (they exist in reality and they have two leads) and we can measure the amount of charge by connecting one lead to the test electrode the other end to the ground. In a pure electrostatic sense, if one A(+) ion is reduced or one A is oxidized at the A electrode, similarly, one B(2+) is reduced or one B is oxidized at B electrode, then the amount of charge developed on B will be higher. We cannot guess the sign of A or B with this thought nor we can say anything about this electrode potential, however, you are right in this pure electrostatic sense.

I went over what @Zhe and you said. I understand this now: The discussion with redox potentials is always in comparative terms. We use the SHE as a common halfcell to compare everything else to. The more negative E values simply suggest that that half cell has relative tendency to be oxidised when hooked up to the SHE. This is simply from energetics. Now about electrostatics: e.g. a half cell with negative E value hooked up to a SHE. Electrons flow from this half cell to the SHE. This must mean that the electrode is more negative (or less +ive) than the electrode in SHE.

One can rougly deal with this with position of equilibria and "tendency" to lose electrons ( @M.Farooq, you got exactly my mind's picture in your last para), but this is not a rigorous approach and should not be expected to stand rigorous critique since it is not really defined well. I use 2 texts and they both take this appraoch. Maybe it's simple and builts some intuition. But I now realise that I was attempting to define tendency on my own and then question theory. But why is it that the half cell that is relatively going to be oxidised always has an electrode more negative -or less +ive?

I can't seem to find a rigorous answer to this. Maybe dG is a better way of thinking what E values tell us ( not absolute dG, since we don't know the value for SHE) And now there is one thing I would like clarfication on: this equation linking E to G was new to me.Suppose we have the metals A and B as in my question. Suppose each half reaction has the same dG. So the dG when combined with SHE should come to be the same. How do I use this equation to show how the E s wil compare?


text:
Potassium ethoxide
Is it possible to make potassium ethoxide by mixing $\ce{KOH}$ and $\ce{EtOH}$ in the presence of a dehydrating agent like $\ce{MgSO4}$ or $\ce{CaO}$ or $\ce{K2CO3}$?
The easy way is adding metallic potassium to ethanol. But nobody likes to deal with potassium or hydrogen.
My understanding is that added $\ce{KOH}$ reacts with $\ce{EtOH}$ in a reversible way. But if there is a water absorber the product water is gone so the reaction moves to the right side.
$$\ce{C2H5OH + KOH <=> C2H5OK + H2O}$$

Better way is buy it: Sigma-Aldrich sells it in 95% pure form for USD 39.90/100 g.

What about ethanol being absorbed?


text:
Equilibrium constant of water
If $\ce{H2O}$ dissociates as follows: $$\ce{2H2O <=> H3O+ + OH-}$$ and not in this way: $$\ce{H2O <=> H+ + OH-}$$ why is the equilibrium constant: $$K_{eq}=\frac {\ce{[H+][OH-]}} {\ce{[H2O]}}$$ and not: $$K_{eq}=\frac {\ce{[H3O+][OH-]}} {\ce{[H2O]^2}}$$

Pure substances are excluded from equilibrium expressions. The last two equations are equivalent.

There is a similar issue with $\ce{CO2(aq)}$ and $\ce{H2CO3(aq)}$. Think of $\ce{H3O+(aq)}$ and $\ce{H+(aq)}$ as a shorthand for $\ce{H+_\text{total}(aq)}$, all forms of hydrogen ions associated with water.

I think it is quite clear what the question is asking.

In a strict sense, all quantities in the equilibrium constant are not molar concentrations but activities. Activity is a thermodynamic quantity which is proportional to concentration. Now by IUPAC definitions of standard states, the activity of pure liquids and solids is taken is unity (by convention and for reference purposes. Therefore in the following expressions, $\{\ce{H2O}\}$ or $\{\ce{H2O}\}^2$ are all unity, no matter what the power is (real number). Elementary textbooks tend to say a lot of wrong or oversimplified stuff.
$$K_\mathrm{eq} = \frac{\{\ce{H+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}}$$
and
$$K_\mathrm{eq} = \frac{\{\ce{H3O+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}^2}$$
Here the curly bracket indicate activities. The IUPAC notation is $a$ for activity. The species of interest is written as a subscript.
A more detailed answer as to why activity is taken as unity is already provided:
Why are activities of solids and liquids taken to be unity?

Well that is assuming that $\ce{[H2O]}$ and $\ce{[H2O]^2}$ denote the activities of water and not its concentration. Usually the bracket notation is used to indicate concentrations.

You are right, it should rather be a_H2O.

@M.Farooq If you like brackets, you may want to use the curly ones ({,}) for activities (see What do curly brackets {} mean?).

Interesting, I never saw curly brackets for activity. This must be rather recent. Is this common in European textbooks?

@M.Farooq Hmm, I never traced the statistics of usage of either notation for activity, I see both curly brackets and $a$ here and there occasionally — I'd say both are used fairly equally, suggesting I read mostly US/British, German and Soviet/Russian literature.


text:
Why is ethoxyethane the preferred IUPAC name for diethyl ether rather than 1,1'-oxydiethane?
Diethyl ether has a few systematic names. Its preferred IUPAC name is generated using substitutive nomenclature: ethoxyethane. But multiplicative nomenclature could be used, in the symmetric spirit of its common name, to generate the name 1,1'-oxydiethane. The Nomenclature of Organic Chemistry 2013 (blue book) makes it clear that multiplicative nomenclature is preferred to substitutive. So why is the preferred name substitutive?
The book says these about multiplicative nomenclature:

P-51.1.5: “Multiplicative nomenclature (P-15.3, P-51.3), as a subset of substitutive nomenclature, is preferred to simple substitutive nomenclature when the criteria for its use are met…”
P-51.3: “Substitutive nomenclature is used when the conditions for constructing multiplicative names are not met.”
P-51.3.1: “Multiplicative nomenclature is preferred to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical parent structures other than alkanes when
“(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical and
“(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
“(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.”

The deciding criterion is in the preamble to P-51.3.1, not in the numbered criteria: “other than alkanes.” The identical parent structure in diethyl ether is a straight-chain alkane, so multiplicative nomenclature does not apply.


text:
What exactly is the parasitic white layer that forms after iron parts are treated with ammonia?
I have little to do with chemistry and my only background is the inorganic chemistry I learned at school when I was 13-14 years old.
I need to have a basic understanding of the nitrating process (for an automation engineering application, more precisely an automated furnace used for nitrating iron or steels parts with the purpose of improving their mechanical and chemical properties).
A few things are not so clear for me:

What is the chemical formula of the iron lattice mixed with nitrogen atoms (see the picture)?
What is the unwanted white layer that I understand forms on the surface of the metal part treated with dissociated ammonia?
What is this dissociated ammonia?

Gas nitriding

(1) Natrium is the Latin name for sodium, hence the elemental symbol Na. (2) What "white layer"? -- Do you mean  the iron atoms that are shown as white circles?

Nitrogen, sorry! The "white layer" is not shown in the picture.

Steels are nitrided in ammonia gas at 900 to 1050 F ; It forms a very hard , very thin ( < 0.01 ") hard layer of iron nitride . An addition of aluminum alloy to the steel ( 0.5 to 1.0 % ) enhances the nitriding. Nitriding is applied typically to cutting tools like drill bits and wearing surfaces. At higher temperatures "carbo-nitriding " is done where C and N are diffused into the steel surface. The "white layer" is something seen in a metallographic sample at high magnification -100 x and higher. It is undesirable and apparently unidentified as the ASM  handbooks refer to it as only "white layer" ; they give procedures to remove it if necessary . The white layer should be thin , even compared to a 0.005 " thick layer of nitride. Plain ammonia or dissociated ammonia may be used with slightly different procedures.

ah... yes. I overlooked the "cation" Gas nitriding in the OP's post. So the picture is for creating a layer of iron nitride, not using Fe as a catalysts in reaction between hydrogen and nitrogen to make ammonia.

No guesses regarding what the chemical composition of the white layer might be?


text:
The major product of the following reaction

Here,1,2-dibromobutane is reacted with
i) alc. $\ce{KOH}$
ii) $\ce{NaNH2}$ in liq. $\ce{NH3}$

According to me, since alc. $\ce{KOH}$ causes dehydrohalogenation, using Saytzeff's rule the product obtained after reaction (i) should be one of the either:

And after reaction (ii), since $\ce{NH2-}$ is a strong base, an E2 reaction proceeds and again using Saytzeff's rule the final product should be:

However, according to the answer key the final product obtained is given as:

Can someone help by explaining where and how I went wrong in my attempt to solve the question?

An allylic bromide will react with KOH. What happens to terminal alkynes in the presence of NaNH2? Check pKa's!

Hey, all of your compounds except the first one are wrong as they have 5 carbons.

@AnubhabDas Yeah sorry, I've corrected them now.

When you form the allene, the methylene hydrogen has enhanced acidity because deprotonating it leads to a carbanion in which the negative charge is delocalized:
$\ce{CH3-CH=C=CH2} - \ce{H^+ -> CH3-CH=C=\overset{-}{C}H <-> CH3-\overset{-}{C}H-C#CH}$
The proton then reattaches to the opposite end of the allene chain to form a more stable product.  In effect this is a (strong) base catalyzed tautomerization.

Thanks @zhe.  How to get my resonance structures lined up properly?

Honestly, I'm not the best person to ask. I liked you what did you with $\equiv$, and that might be the best approach here. @OscarLanzi

Ok, but the equivalence sign did not really work.  The software put a line break into the contributing structure, which is bad, and I gave up.

@OscarLanzi For a triple bond mhchem uses # symbol, and yes, it looks somewhat misaligned and I guess one should just accept it. I also tried to improve the location of the negative charge on carbons with \overset{...}{...} so that it doesn't distance hydrogens away.

Thanks @OscarLanzi, but since the methylene H of the allene is acidic, shouldn't it react with NaNH2 instead of undergoing tautomerisation? And another thing, are alkynes, in general, more stable compared to allenes?

@asterix Wikipedia reports that in the parent compounds the equilibrium constant for converting methylacetylene to propadiene is less than $1$ and drops with lower temperature.  This shows that the alkyne is dominant and its formation from the allene isomer/tautomer is exothermic.

alkynes are of lower E than allenes. It might be questionable if allenes are formed at all in the reaction. Also allenes and alkynes are linear not angled molecules. Texts and web articles on this topic seem very inconclusive. If one is truly interested a trip to the library is in order.


text:
Distinguish 1-bromobutane and 2-bromobutane with IR
How can I distinguish 1-bromobutane and 2-bromobutane by their IR spectrums?
In the 1-bromobutane spectrum I think the $\ce{-CH3}$ peak is shorter, but I don't know if that is actually reasonable or just a fluke.
In the spectrum that I think is 2-bromobutane there is an additional peak at around $1290$. What might this be?
The only alkyl halide peaks we discussed in class was the $\ce{C-Br}$ stretch $600-750$. They both seem to have this (although they do look different). How can I distinguish?

Meh, with NMR it would be crystal clear.

Sure would! Unfortunately this is an IR problem set.

From NIST database 1-bromobutane and 2-bromobutane in the gas phase.

I did see the different spectrum on SDBS, but I'm more interested in given just the two spectrum and the two compounds, how can you tell which goes with which?


text:
Is temperature of crystallisation same as melting point temperature for polymers?
Melting point tends to be a temperature range. Is it true that the temperature of crystallisation is the highest temperature at which the polymer melts and thus there is some distinction between temperature of crystallisation, Tc and temperature at melting point, Tm?

Is temperature of crystallization same as melting point temperature for polymers?

In a word: NO.
If you undrestand what the glass transition, crystallization and melting temperatures are it will make sense why they are not the same. This paper does an excellent job at explaining the three concepts but I will summarize.
The glass transition temperature is the temperature where polymer chains become sufficiently mobile that the material transforms from being brittle to being soft and plastic (neither exo- nor endothermic, but does the heat capacity does increase). The crystallization temperature is the temperature where the molecules are mobile enough to rearrange into ordered arrangements (crystallize not decrystallize; this is an exothermic process; see plot below). The melting temperature  is where the polymer chains can freely move and ordered arrangements are disrupted (this annihilates any prior crystallization; an endothermic process).

Source: Plastic Technology Laboratories


text:
Reference Books on Electrolysis and Chemical Titration process
I'm in a General Chemistry course in college. We have laboratory, and then we are supposed to make a report on what we did, etc.
I'm not entirely sure what the last laboratory was about, but it was called Salt and Water Electrolysis
To sum up the procedures -two general procedures-, here's what we did:
1st Procedure:

We prepared 50 mL of a $\ce{NaCl(aq)}$ solution at 0.1 M (molarity) in a 100 mL beaker
We made a DC (direct current) circuit at 5V and 700 mA, with graphite electrodes
We turned the circuit on for roughly 20 minutes
We employed a magnetic stirrer to stir the solution
We added 3 drops of Phenolphthalein in the solution, and took notes of the phenomena. (I guess it kinda turned pink)

And that was that for the 1st procedure.
2nd procedure:

We measured the solution's pH with a pH meter
We measured roughly 15 mL of $\ce{HCl}$ (Hydrochloric Acid) at 0.5 M with a test tube
We poured the solution of $\ce{HCl}$ in a 25 mL burette.
We prepared the titration equipment
We were supposed to add the $\ce{HCl}$ (drop by drop) to the $\ce{NaCl}$ solution and at the same time be measuring the pH until it got to 7

And that was that for the second procedure.
Of course we kind of messed up, as we overshot the end point. The excess $\ce{HCl}$ brought it's pH to 3 point something.
The $\ce{NaCl}$ solution lost its pink color, and then turned yellowish.
So.. I guess I am asking for references to the equipment involved -I kind of have references for the pH and the properties involved, still, if you know of some references, let me know!-, I've looked for electrolysis processes in books but don't find any. Or they're too technical. Also, haven't found titration procedures in books.
Can you help me out, just so I can finish my report?

Welcome to Chemistry.SE! I wish to politely draw your attention to this post. Please make a note of it while creating future posts!

It is always good to read about the experiment before going to the lab. As a result, then you don't need to use expletives. There are very good Youtube videos https://www.youtube.com/watch?v=1nAe_T5mO2o

No expletive needed alright. My apologies. And thank you :D

I just want to add: the reports require bibliographical links, and I don't think they'll allow youtube videos as a reference..

I mean, I do kind of know what happens. The types of reaction taking place, but I have to back it up with bibliographical references.

You're much better served by asking your TA, professor, lab instructor, etc. for details regarding what the references should entail than ask a bunch of strangers about something that doesn't necessarily have any analogy to the standards of professional publications.

I guess you're right bro..

Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552
In the first part you electrolyze a solution of salt. Check the half cell reactions from the text: What happens at the cathode and what happens at the anode?
Water is reduced at the cathode and -OH ions are produced. This makes your salt solution basic. In the second part, your lab requires you to estimate how much base was produced.*
Essentially you do a titration of NaOH with standardized HCl. The end point is estimated by a pH meter. Look up any general chemistry textbook again for titration of a base with HCl using a pH meter. The technique is called pH titration. You did a very rough version of a pH titration.

Depending on the conc. of NaCl, either chlorine or oxygen is produced at the anode.

Thanks for the comment.
Just 20 minutes ago, I found a nice resource.
"Chemistry - The Central Science" 14th edition in SI units by Brown had all I needed, I think.
There's almost all of what I needed on pH and acid-base stuff on chapter 16.
Now, on electrolysis, I guess that's a little more advanced so it's until chapter 20, section 9 (20.9) that there's a whole section on electrolysis.
And yes!, now I know I titrated the solution with the Phenolph, pretty cool..


text:
Anhydrous magnesium sulfate to replace epsom salt?
Can anhydrous magnesium sulfate be used instead of epsom salt for feeding plants? Do hydrates and anhydrous salts become the same when dissolved in water?

$\ce{MgSO4}$ would get re-hydrated in even moisture. Anyway, dissolved in water, all that's left is sulfate and $\ce{Mg^{2+}}$ ions.

Yes, they become the same, with some rare exceptions. There is usually difference in the dissolving kinetics, but the final result is the same.
Sometimes there are different thermal effects, Hydrates have sometimes tendency to cool the solution, while anhydrates to warm it. It is caused by released hydration energy of the anhydrate.The typical example is calcium chloride.
The obvious thing is to recalculate the needed amount for the contained water.


text:
I have prepared ferrous sulfate solution in water. After some time, brown precipitates observed. How to prevent the same?
When ferrous sulfate solution is prepared in water, after some time, brown precipitates appear, how to prevent the precipitates.

Probably ferrous carbonate. Acidify the solution to avoid that problem.

Or ferric oxide/hydroxide from traces of $\ce{Fe^{III}}$ in the $\ce{FeSO4}$, or due oxidation. Another option, depending on the purpose, is to use the Mohr salt $\ce{(NH4)2Fe(SO4)2}$, that is less vulnerable to oxidation.

@Poutnik - Yes think you nailed it. $$\ce{FeCO3 + H2O <=> Fe(OH)2 + CO2}$$ $$\ce{4Fe(OH)2 + O2 <=> 2Fe2O3\cdot3H20 + H2O}$$