text:
Is this true for conductivity of mixture of two solutions equal to the sum of conductivities of each solution?
Suppose $\ce{Na2SO4}$ has a conductivity $\pu{2.6 \times 10^{-4} S cm-1}$ and on mixing  $\ce{CaSO4}$ to saturation, the conductivity becomes $\pu{7.0 \times 10^{-4} S cm-1}$.
So, Is the conductivity of $\ce{CaSO4}$  the difference of $\pu{7.0 \times 10^{-4} S cm-1}$ and $\pu{2.60 \times 10^{-4} S cm-1}$?
I got an answer of a similar question here. But It is explaining things opposite to what I said to get the desired answer.
My book also solved by subtracting the two conductivities.
So I am confused, Please help me reach out an answer.

This question is a part of a large question.
This is written in my book.

EDIT:- I am adding the original question for further clarity

You can edit this question, and it is needed. Please take a site tour.

At this level of dilution (calcium sulfate is only sparingly soluble, so it has little impact unless the initial sodium sulfate is also very dilute), you can just take a difference. More concentrated solutions you would get with salts having good solubility are complicated by ion-ion interactions.

As for editing, I would describe the large question to provide context. Such context will allow us to develop a  good answer.

@Oscar Lanzi I've added the original question

Is the given answer (d)?

@Mridul Kumar See the solution, https://www.doubtnut.com/question-answer-chemistry/the-conductivity-of-0001m-na2so4-solution-is-26-xx-10-4scm-1-and-increases-to-70-xx-chi10-4scm-1-whe-19293800

You are missing one critical point. All these additions and subtractions of molar conductivities are valid at infinite dilution. This is called Kohlrausch's law. If you know the molar conductivity of NaCl at infinite dilution and then you have a saturated $\ce{CaSO4}$ and it is still such a dilute solution that you can consider it at "infinite dilution". Your excercise book is using that approximation during subtraction.

@Mathew Mahindaratne No the answer is a)

@M. Farooq Do you understand Hindi? If yes then you can see that in the video (that you have given) doesn't explains the reason for the substraction.

@MridulKumar, I explained you the reason behind subtraction (Kohlrausch's law). See my comment above.

OP's question is: Is the conductivity of $\ce{CaSO4}$ the difference of $\pu{7.0 \times 10^{−4} S cm−1}$ and $\pu{2.6 \times 10^{−4} S cm−1}$?

In this question, $\pu{2.6 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous $\ce{Na2SO4}$ solution and $\pu{7.0 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous solution when saturated with solid $\ce{CaSO4}$. The initial question before edition, it doesn't make sense because of the lack of information (e.g., no concentration terms were given). I completely agree with Oscar Lanzi and M. Farooq with their valuable comments.
Now, since OP has edited the question and given the full description of main question, we can give an approach to solve the question. It is evident that this is not a high level of analytical question (which would attract M. Farooq). Therefore we can take certain assumption such as the solutions are dilute enough to be considered as those of infinitely dilute solutions.
Strong electrolytes are hypothesized to dissociate completely in solution. The conductivity of a solution of a strong electrolyte at low concentration follows Kohlrausch's Law (see M. Farooq's comment: You are missing one critical point. All these additions and subtractions of molar conductivities are valid at infinite dilution. This is called Kohlrausch's law. If you know the molar conductivity of $\ce{NaCl}$ at infinite dilution and then you have a saturated $\ce{CaSO4}$ and it is still such a dilute solution that you can consider it at "infinite dilution". Your exercise book is using that approximation during subtraction):
$$\Lambda_m = \Lambda^\circ_m - K \sqrt{c}$$ where $\Lambda_m$ is molar conductivity, $\Lambda^\circ_m$ is limiting molar conductivity (at the limit of the infinite dilution), $K$ is an empirical constant, and $c$ is the electrolyte concentration. At sufficiently low concentrations solutions (when $\Lambda^\circ_m \gt\gt K \sqrt{c}$), the observed conductivity of a strong electrolyte becomes directly proportional to concentration of the solution. As quoted by Wikipedia:

The specific conductance of a solution containing one electrolyte depends on the concentration of the electrolyte. Therefore, it is convenient to divide the specific conductance by concentration. This quotient, termed molar conductivity, is denoted by $\Lambda_m$,
$$\Lambda_m = \frac{\kappa}{c} \tag1$$

If the limiting molar conductivities of cation and anion are $\lambda^\circ_+$ and $\lambda^\circ_-$, for the diluted enough solutions,  $\lambda_+ \approx \lambda^\circ_+$ and $\lambda_- \approx \lambda^\circ_-$, and
$$\kappa = c_+\lambda_+ + c_-\lambda_- \tag2$$
Now, let's look at the question in hand. Since $\pu{2.6 \times 10^{−4} S cm−1}$ is the conductivity of original aqueous $\pu{0.001 mol L-1}$ $\ce{Na2SO4}$ solution (or $\pu{0.001 \times 10^{-3} mol cm-3}$):
$$\ce{Na2SO4 -> 2Na+ + SO4^2-}$$
and from the equation $(2)$:
$$\pu{2.6 \times 10^{−4} S cm−1} = c_\ce{Na+}\lambda_\ce{Na+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+} \tag3$$
$$\pu{2.6 \times 10^{−4} S cm−1} = 2 \times \pu{0.001 \times 10^{-3} mol cm-3} \times \pu{50 S cm2 mol-1} + \pu{0.001 \times 10^{-3} mol cm-3} \times \lambda_\ce{SO4^2-} $$
$$\therefore \  \lambda_\ce{SO4^2-} = \pu{160 S cm2 mol-1}$$
Similarly, when the solution is saturated with $\ce{CaSO4}$:
$$\ce{CaSO4 <=> Ca^2+ + SO4^2-}$$
And, from the equation $(2)$ for the diluted mixed solution (suppose $\alpha$ amount of $\ce{CaSO4}$ has dissolved):
$$\kappa = c_\ce{Na+}\lambda_\ce{Na+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+} +  c_\ce{Ca^2+}\lambda_\ce{Ca^2+} + c_\ce{SO4^2+}\lambda_\ce{SO4^2+}\tag4$$
Applying the equation $(3)$ and $\lambda_\ce{SO4^2-} = \pu{160 S cm2 mol-1}$ in the equation $(4)$ (avoid the units for convenience):
$$7.0 \times 10^{−4} = 2.6 \times 10^{−4} +  \alpha \times 120 + \alpha \times 160 \tag4$$
$$\therefore \  \alpha = \frac{7.0 \times 10^{−4} - 2.6 \times 10^{−4}}{120+160} = \pu{1.57 \times 10^{−6} mol cm-3} = \pu{1.57 \times 10^{−3} mol L-1}$$
If you take $K_\mathrm{sp} = \alpha^2 = 2.46 \times 10^{−6}$, there is an answer, but it was incorrect (that is the common trick for multiple choice questions). That's because there is a common ion effect you may need to consider here. Thus correct answer for $K_\mathrm{sp}$ is:
$$K_\mathrm{sp} = \alpha(\alpha + 0.001) = 2.46 \times 10^{−6} + 1.57 \times 10^{−3} \times 0.001 = 4.03 \times 10^{−6} \approx 4.0 \times 10^{−6}$$
Thus, correct answer is choice $(a)$.


text:
In crystallography, how does the real space R factor relate to the atomic B factor?
In assessing the quality of different parts of an atomic model derived from diffraction data, you can look at the real space R factor or at the atomic B factor. Do these give the same information or do they complement each other? In which scenarios would you look at one or the other or both?

To calculate the real space R factor, you need an atomic model plus the electron density map (e.g. 2Fo-Fc map) derived from the diffraction data and the model. You calculate the correlation of the expected density (derived from the model alone) and the 2Fo-Fc map for given areas in the model. For example, you could calculate a residue-wise real space R factor. The lower its value, the better the correlation. For macromolecular structures deposited in the protein data bank, residues with particularly poor correlation are listed in the validation report (see user guide section 6.1).
The B factors (also called atomic displacement parameters or temperature factors) are part of the atomic model. They get optimized during refinement to obtain the best fit between measured and calculated diffraction data. If the model is accurate, the B factors reflect the disorder of each atom. If there are glaring errors in the model (either misplacement of atoms, or spurious atoms in the model that are not really in the structure), refinement will inflate the B factors of these atoms (this de-emphasises the contributions of these atoms to the calculated diffraction data).

[...] how does the real space R factor relate to the atomic B factor?

Both measures give local information about the quality of the structure. B factors are easily obtainable (they are deposited with the coordinates of a structure) whereas the real space R factor requires a calculation using suitable software.

Do these give the same information or do they complement each other?

When the real space R factor is high in a given region, it means the atomic model is bad at representing the structure in this region.
When the B factor is high, it can mean one of two things.

The region is correctly modelled and has a high degree of disorder (the real space R factor would be low, showing adequate interpretation).
There are spurios atoms in the model. The B factor will be inflated during refinement as a way of "downplaying" this problem (the real space R factor would be low as well).

When you find a couple of high B factors in a region of otherwise low B factors, you can be pretty sure there is a problem with the model that could be fixed. In that case, it is time to look at the electron density.

In which scenarios would you look at one or the other or both?

The real space R factor suffers from bias because it compares something calculated from the model with something calculated from the model and the diffraction data (for 2Fo-Fc maps). If you are looking at cryo-EM densities, though, or any other technique where the map does not rely on the model, there is no bias.
The B factor is provided by the crystallographers who solved the structure, so you don't exactly know what they did during refinement to arrive at the B factors.
If you are looking at a structure you are not familiar, a quick look at B factors can tell you which regions were easy to model and which were tricky. Graphing the B factors or color-coding a 3D visualization by B factors might give you an idea of problems in building the atomic model.
One the other hand, a quick look at the real space R factor can tell you which parts of the structure are represented well by the atomic model and which are not. It is a bit easier to interpret, but much harder to obtain unless you have the computational infrastructure already set up.
Once you find red flags using either method, it makes sense to look at the electron density and the model on a 3D display to see what might be going on.

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text:
Exact nature of HOMO of square planar isomer of tetraamminemagnesium(0)
I need to calculate information about the (hypothetical) square-planar form of the tetraamminemagnesium(0) molecule as a reference to my personal QC research project; everything goes fine until we go to the HOMO and LUMO.
If I use Hartree-Fock with basis set 6-311G(2d, p) on Gaussian, I get an antibonding s-type orbital moderately localised on the magnesium as the HOMO. Same goes when I use MP2 with the same basis set. However, if I calculate the same calculation while using the keyword Guess=Huckel, I get a nonbonding p-type orbital highly localised on the magnesium as the HOMO. Moreover, in both cases, the HOMO is mostly consisted of the core and/or Rydberg orbitals of the magnesium and nitrogens; these two (or three) inconsistencies happen irrespectively of the basis set used, from STO-3G to cc-pVDZ.
While calculating, I noticed that the HOMO-LUMO gap is quite small(about 0.1 Ha, approx. 2~3eV), which means that any small error can lead to a "switching" of the HOMO and the LUMO.
My question now follows; which is the exact symmetry and (anti)bondingness of the square planar, Ci-symmetric isomer of tetraamminemagnesium(0)? The magnesium-nitrogen bonds are all 1.9 angstroms, the nitrogen-hydrogen bonds all 1 angstrom, and the three propeller hydrogens of each ammine ligand have dihedral angles of 30°, 90° and 150° (resp.) with respect to the plane consisting of the magnesium and the four nitrogens.

Are you talking about the Hartree-Fock orbitals or the orbitals coming from NBO calculation? Small HOMO-LUMO gap is usually indicative of static correlation i.e. you would need a multiconfigurational calculation to actually model the system.

Hartree-Fock only.

I think you could try DFT and see what you get. Sometimes they perform better for systems with static correlation. (MP2 is very bad with those systems)

It says, "an ambiguous keyword was detected" when I tried to use the keyword DFT.

Just tried BLYP- looks like the HOMO is indeed of antibonding s-character.

Yeah you can't use DFT, you have to specify a functional. Why don't you try TPSS instead of BLYP? It's better for static correlation(https://aip.scitation.org/doi/am-pdf/10.1063/1.4991833). Also, if you don't know what DFT is, maybe you should brush up on the core areas before delving into such complicated calculations? I would not do multireference systems before getting the hang of single reference closed shell molecules.

Let us continue this discussion in chat.

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text:
Reaction of naphthalene with sodium dichromate/sulfuric acid
I recently came across a question where naphthalene undergoes oxidation with sodium dichromate in presence of sulfuric acid. The given options were benzoic acid, phthalic acid, decalin, and tetralin.

I assumed the reaction to be similar to that of the reaction between potassium permanganate under acidic conditions, yielding phthalic acid.
This reaction was described in
Oxidation of naphthalene with KMnO4.
Could anyone please clarify the reaction, whether my assumption is correct, and explain the mechanism?

It depends on the reaction conditions.
I suspect the answer the question setter wants to see is phthalic acid, but the true answer is none of these. I have found multiple references (such as this US Patent and this Org. Syn prep) that refer to the production of naphthoquinones by oxidation of naphthalenes with high valent Chromium reagents under acid conditions.


text:
Why is hydrogen peroxide acidic in aqueous solution?
This seemed like a trivial question to me... until I began to think about it.
The "usual" criteria for greater acidity are larger sizes of the atom attached to the acidic proton within the same group(e.g. hydrogen selenide over hydrogen sulfide over water) and inductive (and/or resonance, the latter of which usually dominates) stabilisation(e.g. phenol over ethanol); hydrogen peroxide has no reason to be more acidic than water by the first criterion and the second criterion would destabilise the anion instead, with the hyperconjugative electron donation from the oxygen-hydrogen bonding orbital far outweighing the negative-hyperconjugative electron withdrawal into the oxygen-hydrogen antibonding orbital.
So why is hydrogen peroxide acidic in aqueous solution, when it has no reason to be more acidic than water and every reason to be less acidic than water?

related https://chemistry.stackexchange.com/questions/154346/what-is-the-basicity-order-of-peroxide-superoxide-and-hydroxide-anions

Does this answer your question? Is there a general consensus on the causes of the alpha-effect?

It's too technical for me to understand properly...

Well, out of the top of my head, I don't see why we couldn't just say hydroxyl inductively withdraws better then hydrogen and call it a day. The more you'd want to get into details the more muddled it would become, I think.

Yeah that might be right... but doesn't resonance usually dominate over inductivity when it comes to cases like this?

The simple size comparison is not accurate. In hydra-acids you can often draw a correlation between the size and strength of $\ce{H-A}$ bond. However $\ce{H2O2}$ consists of only $\ce{O-H}$ bond with respect to hydrogen and hence has a comparable bond strength with respect to $\ce{H2O}$ regarding acid-base reactions.
$\ce{OOH^{-}}$ is not destabilized by resonance effects as $\ce{O^{-}}$ lacks vacant orbitals to accept $\ce{-OH}$'s lone pair. As a result the effect is mostly determined by inductive effect (($-I$) of $\ce{-OH}$ is larger than $\ce{-H}$.)
Negative hyperconjugative effect ($-H$) effect can't occur because it requires polarity of substituent here $\ce{H}$ to be more electronegative than $\ce{O}$ in $\ce{-OH}$.
One more interesting point (again in favor of making $\ce{H2O2}$ more acidic) is the gauche effect due to which lone pair of one oxygen donates electron density to $\sigma *$ of $\ce{O-H}$ bond thereby further weakening the $\ce{O-H}$ by reducing bond order.

He doesn't talk about "normal" hyperconjugation, but stereoelectronic effect, with lone pair and antibonding orbital. BTW why people say "no vacant orbitals" when there are antibonding orbitals, like, everywhere?

I upvoted your answer, but the gauche effect itself is due to hyperconjugation.

@JustSomeNerdyTeen I don't disagree though it is really just hyperconjugation but with ABMO. If you feel like this answered your question, consider ticking the check mark.


text:
A liquid which climbs out of its pot
My Russian born wife recently asked me to buy her a skin remedy widely used in Russia. In English this is the oil of a plant called Sea Buckthorn. It is a thin yellow liquid. My wife put some of this in a small spice bottle which had an open top. I kept finding yellow liquid up the sides of the pot, well above the surface of the liquid and eventually on the surface the pot was standing on.
I very carefully checked this glass pot for any sort of leak in the base but there isn't a leak path. It seems that the oil has climbed out of the pot. After thinking for a long time I seemed to remember that this is a known phenomenon. I can't remember what it is called or what liquids are involved. Please could someone enlighten me ?

This is known in superfluids, liquid helium at ultra low temperatures. I really doubt that you will observe this "creeping" behavior in ordinary liquids.

The well-known creeping of KCl out of its solution, even out of its beaker, is documented at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7000175/. Sea Buckthorn (Hippophae) https://en.wikipedia.org/wiki/Hippophae berry extract is complex: oily and aqueous. It could be that the aqueous material wets the glass, deposits a fat (like KCl comes out) and climbs thru it or over it!


text:
Why does LiAlH4 reduce the double bond of cinnamaldehyde, whereas it does not reduce the double bond in crotonaldehyde?
$\ce{LiAlH4}$ is a strong reducing agent capable of breaking the double bonds. It does break the double bonds in cinnamaldehyde.
Shouldn't $\ce{LiAlH4}$ be able to break the double bond in crotonaldehyde as well? Why does the double bond in crotonaldehyde remain unaffected?

Related: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not

α,β-Unsaturated carbonyl compounds, for the most stable intermediate with respect to $\ce{H-}$ ion prefer 1,4-addition. However, $\ce{LiAlH4}$ prefers 1,2-addition over 1,4-addition. This can be explained due to the following reasons.

It is more reactive so it prefers a kinetically favorable pathway (1,2-addition is often kinetically favored while 1,4-addition is thermodynamically favorable).

It is also plausibly explained by HSAB principle which also explains why $\ce{NaBH4}$ prefers 1,4-addition for α,β-unsaturated carbonyls.

Based on the above facts we should see that it is natural to assume that 1,2-addition is obeyed for both crotonaldehyde and cinnamaldehyde. However experiments contradict and instead show that for cinnamaldehyde 1,4-addition is more preferable even when $\ce{LiAlH4}$ is acting as the $\ce{H-}$ source. This can be easily explained by more stability of the carbanion by two-fold resonance, both by phenyl group and by the allylic (sort of) carbonyl group.
It would seem that stabilizing the β-carbanion can allow 1,4-addition to take place. The extent of stability required, however can only be predicted by experiments. It seems like phenyl group is sufficient to warrant 1,4-addition like it does for styrene (as it should since it distributes charge over larger set of nuclei).


text:
Confusion in R/S Nomenclature and Chirality
I have many confusions regarding optical isomerism:
What is the use of the R/S nomenclature? I understand that it is used to distinguish between the different spacial arrangements that are possible for a molecule, but i have few questions.

I believe R/S nomenclature can only be used for a chiral center. Now, say we have a two chiral center molecule, so there are 4 types of different molecules (because 4 different configurations are possible) are possible, according to me, (R,R), (R,S),(S,S),(S,R). Am i right? Are all these 4 molecules different?

I am getting confused when we introduce chirality here. If the molecule is achiral then will some of the above four become the same molecule? If the compound is chiral, will all the above 4 be different from each other?

Generalizing, if a molecule has n chiral center, will it always have $2^{n}$ different isomers, irrespective of whether it is chiral or not, or am I wrong in thinking so?
Please correct me

An environment around an atom may be chiral (stereogenic centre, or chiral centre).  A molecule including one, or multiple chiral centres may but does not need to be chiral.
You are right that the presence of two chiral centres ($n = 2$) may yield four permutations, namely (R,R), (R,S), (S,R), or (S,S).  However, depending on the overall structure of the molecule in question, $2^n$ is the maximum number of different molecules.
Take the example of tartaric acid with two chiral centers and start the permutation of the absolute configuration at C(2) and C(3).  The two configurations of (2R,3S) and (2S,3R)-tartaric acid are identical to each other and represent the meso form of tartaric acid.  Despite the environment around the atoms is chiral, the molecule of the meso-isomer is not chiral.

(derived from here)
If you have a model kit at hand, build the molecules in question.  For each conformation of the mesoisomer, the lower part of the Fischer formula is transferred into the upper part of the Fischer formula by a mirror plane.  The presence of either a mirror plane (in the drawing indicated by the horizontal dashed line), or a centre of inversion however is incompatible with the molecule's structure being chiral.  The principle identifying the mirror plane is illustrated below:


text:
what happens if iron clamp is attached to a copper pipe to a wall?
I'm guessing its not a good idea, since iron will then corrode easily (also depending on the location and environment), Although copper do not rust it will increase the rate of it from oxidizing. plus with the addition of replacing iron clamp quite often.
Or does copper aids in iron???

Your question is not clear. If an iron clamp attached to it creates so many problems, then don't attach it. Please describe your problem more explicitly. Also you can consider non-corrosive options like plastic or rubber (I'm not a plumber, so not sure if those are actually viable alternatives).

The question does not state whether the two metals (electrodes?) are immersed in an electrolyte. As stated, I see no problem.

You now have an iron clamp attached to a copper pipe on the wall. Please provide more details.

As you state, iron will  corrode more rapidly in contact with a more "noble" metal, e.g. copper. That is why plastic pipe hangers ("clamps") may be used. Also, iron pipe hangers are sold with rubber inserts, both to reduce vibration conducted to the wall and to reduce this galvanic corrosion.

Since iron is less noble than copper (in the electrochemical series), the former will act as sacrificial anode.

If is is dry - nothing. However even in a typical wall temperature changes could possible cause condensation which would preferentially corrode the steel ( "iron" is rare today). I had a house once with emt ( galvanized steel thin wall tubing for electric wiring). Because of an unusual heating situation, moisture condensed in the emt in the walls. The zinc galvanizing was completely corroded away in some locations, the steel was ugly but no significant metal loss.


text:
Find the ion with mass number 37 if the ion contains 11.1% more neutrons than the electrons

An ion with mass number 37 possesses one unit of negative charge. The ion contains 11.1% more neutrons than the electrons in its nucleus. An ion with a mass of 37 has % negative charge compared to % positive charge.

With the aid of Periodic table, I doubt if the given ion is Chloride $Cl^{-}$, but I am not sure. My attempt is as follows:
Let the ion be $X^{-}$, and let the number of electrons be $x$, then the number of protons will be $x-1$ as the ion carries one unit of negative charge.
Now I know that $A=(2 \times Z)$ if $Z$ is even and $A=(2 \times Z) + 1$ if $Z$ is odd, where $Z$ and $A$ are atomic number and mass number respectively. Therefore, $x-1=(2 \times Z) + 1$.

Let the number of electrons in the ion carrying a negative charge bex
Then Number of neutrons present
$$
\begin{array}{l}
=\mathrm{x}+11.1 \% \mathrm{of} \mathrm{x} \\
=\mathrm{x}+0.111 \mathrm{x} \\
=1.111 \mathrm{x}
\end{array}
$$
Number of electrons in the neutral atom $=(x-1)$
Number of protons in the neutral atom $=(x-1)$
Therefore,
$$
\begin{array}{l}
37=1.111 x+(x-1) \\
2.11 x=38 \\
x=18
\end{array}
$$
Therefore no of protons $=$ atomic no $=x-1=18-1=17$
Therefore symbol of the ion is ${ }_{17}^{37} \ce{Cl}^{-1}$

You are overly complicating the solution with even or odd atomic numbers.
It is by definition that $Z+N=A$ where $A=$ atomic mass number, $Z=$ atomic number, and $N=$ number of neutrons. Thus,
$$Z+N=37 \tag1$$
If the number of electrons in the ion is $E$, then
$$E=Z+1 \tag2$$
because the ion has a negative charge (otherwise $E=Z$ for a neutral element).
And,
$$N=(1+0.111)E \tag3$$
This is from the condition.
You have three equations and three variables. If you solve the three linear equations, you will know the value $Z$. This defines the element. Look, it up in the periodic table. $A$ is already given and you have a negative charge. Remember mass number does not define the element, only $Z$ does.
Your intuition about chlorine is right. All you have to do now is to prove it.


text:
Is Copernicium a transition metal?
Zinc, cadmium, mercury and copernicium belong to the group 12 of the periodic table. In my textbook , it is mentioned that zinc, cadmium and mercury are d-block elements, but not transition metals. But there is nothing mentioned about copernicium regarding this. This makes me wonder if there is a specific reason why Cn has been left out.
So, is Cn a transition metal? Also, if it is, could anyone please explain why is it so?

Copernicium may be newer than the textbook. So the textbook didn't know about it.

Classified as a transition metal, Copernicium is a expected to be a solid at room temperature.

Does this answer your question? How many transition elements are there?


text:
Mechanism of reaction of ketones with isoalkanes under conc. sulfuric acid and heat
Taken from the book of GOC by Dr. O.P Tandon, Himanshu Pandey, Dr. A.K. Virmani,

Can anyone elaborate on the mechanism?

Conversion of alkanes to alkenes using conc. H2SO4 and heating, where have you seen this? Do you mean alcohols?

@Ashish Ahuja my mistake. I edited it out

It seems like an interesting reaction, if it is correct; I've never seen alkanes reacting in such a manner. But surely acetone should undergo acid-catalysed aldol, it seems weird that the book has not mentioned that at all.

This isn't acid catalyzed aldol, for that you need two carbonyl compounds. Isobutane is not a carbonyl compound.

In which section/context/chapter/topic is this reaction given in the book?

@TRC uhh acetone can undergo self-aldol?

@TRC Under "saturated aliphatic hydrocarbons"

@AshishAhuja Oh yes, of course it can. I meant that the final product depicted here certainly isn't acetone's self-aldol product, rather it looks like a sort of condensation between the two compounds.

Due to lack of much complexity there is only really 1 plausible way to look at this, you have to generate a nucleophile to make any headway. This acidic media makes it impossible to do that. $\ce{H2SO4}$'s most extreme reaction is charring action (It never generates nucleophiles anyway). I don't see anyway to facilitate any further reaction. Self aldol looks way more plausible.

This https://pubs.acs.org/doi/pdf/10.1021/ja01156a075 and this https://onlinelibrary.wiley.com/doi/10.1002/9780471678656.ch12 may help.

This was going to be a comment but it got too long.

TL;DR: - This is not an answer, rather a justification for why the question is (probably) wrong.
This paper (linked by @Rishi) gives us experimental evidence that under action of concentrated $\ce{H2SO4}$ hydrogens are exchanged from paraffins in the following fashion:
$$
\begin{align}
\ce{(CH3)2CHCH3 + 2H2SO4 -> (CH3)C^{+}OSO3H^{-} + SO2 + 2H2O   & \tag{R1}}\\
\ce{(CH3)3C+ + (CH3)3CH -> (CH3)3CH + (CH3)3C+ & \tag{R2}}\\
\end{align}
$$
R1 is the slow step and chain initiating step. R2 is the chain propagating step (the hydride transfer here creates a chain of approximately 20 tert-butyl carbocation.
Conc. $\ce{H2SO4}$ is a well known dehydrating and oxidising agent and tert-butyl carbocation has considerable stability due to extensive hyperconjugation. So it is possible that the reaction mechanism is of the following manner:

From the paper by Otvos et.al.1:

Chains are terminated by some irreversible side reactions of carbonium
ions leading to nonvolatile products. This picture would account for
the lack of exchange between t-isobutane hydrogen atoms and sulfuric
acid, since any molecule dissolving in acid, reacting and re-entering
the vapor phase, would have received a new tertiary hydrogen from
another isobutane molecule.

The chain termination step is where we run into a problem. Carbonyl group cannot act as a nucleophile at the carbon center. There is one way to do this but it is extremely unlikely

This is not a good mechanism due to the excess of $\ce{H^+}$ present. In fact, any mechanism for the given product wouldn't be satisfactory because any nucleophile produced at carbonyl carbon would be would be quickly protonated (tert-butyl carbocation can't compete with $\ce{H+}$ due to steric reasons). Also it is highly unlikely that $\ce{SO2}$ would participate in the reaction rather than escape to atmosphere (I suppose you can use sealed tubes).
Further problems are the much better suited side reactions such as acid-catalyzed aldol and attack of $\ce{C=O:}$ lone pair on the carbocation.

Citation

The Behavior of Isobutane in Concentrated Sulfuric Acid1
J. W. Otvos, D. P. Stevenson, C. D. Wagner, and O. Beeck
Journal of the American Chemical Society 1951 73 (12), 5741-5746
DOI: 10.1021/ja01156a075

I fixed some punctuation and citation. Feel free to roll back if you don't like the edit. Tip regarding formatting: Instead of spamming ~, use \begin{align}. It looks pleasant.


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Tautomerism between cyclohex-2-en-1-one and cyclohex-3-en-1-one
I think cyclohex-2-en-1-one and cyclohex-3-en-1-one should undergo tautomerism as shown below:

I'm calling it tautomerism because of it's similarity with the tautomerism mechanism in basic medium. I can't draw reaction mechanism arrows using the software I'm using, but what I've shown is that the base abstracts the hydrogen at the alpha-position of the carbonyl group in the first step to form a carbanion, which has multiple resonance structures. The structure I have shown can abstract a proton from water to form cyclohex-2-en-1-one, which I believe should be the predominant form among the two (and among the enol forms as well).
Is this correct? I've never seen anything similar before so I'm a bit skeptical. A similar mechanism can be shown in acidic medium.
As an example, it would be expected that 4-chlorocyclohexanone on reaction with alc. $\ce{KOH}$ and heating would undergo E2 to give cyclohex-3-en-1-one. Assuming that alc. $\ce{KOH}$ is present in excess, if what I've shown above is correct then cyclohex-3-en-1-one will further undergo tautomerism to give cyclohex-2-en-1-one as the major product.

Well, yes, you're right, but then again what answer you expect? "Am I right" questions are very troublesome when OP is right...

@Mithoron ahh well.. if what I've shown is right, I don't think there is much scope to answer. I think I'll still leave the question because I couldn't find anything online or in the books I have similar to this. Nevertheless, thank you for the confirmation.

Add some follow up question, or maybe someone is just gonna elaborate on that. BTW what you say say about not finding stuff, only says about your searching. This is pretty elementary thing and I'm pretty sure was asked already on this site.

related https://chemistry.stackexchange.com/questions/139860/find-the-major-products-when-these-cyclohexenone-bromides-react-with-caustic-pot


text:
Why do purple Pedialyte freezer pops turn green when frozen?
I bought some Pedialyte freezer pops for my kids:

And I've noticed that after spending a long enough time in the freezer, the purple "grape" pops change color from purple to green:

After thawing (but before melting), the color changes back from green to purple. The dramatic color change happens to all the purple freezer pops, and only the purple freezer pops. They taste the same afterwards and are safe to eat.
I couldn't find any decent explanations for this effect searching Google. This Reddit thread is about the same effect happening to purple Powerade:
https://amp.reddit.com/r/mildlyinteresting/comments/4pi97r/grape_powerade_turns_green_when_it_freezes/
There are hundreds of comments, but they are mostly jokes. The only thing I learned was that it was not related to Anthocyanin because purple Powerade does not contain that.
These are the ingredients of the freezer pops:

Except for perhaps "artificial and natural flavors", the only difference between the different types of freezer pops are the coloring: "grape" has Red 40 and Blue 1. The others are different combinations of Blue 1, Red 40, and Yellow 6.
I initially had a naive hypothesis that maybe Red 40 turns yellow when freezing (since blue + yellow = green). But the Orange and Cherry pops also have this ingredient and don't change color.
Some answers on the Reddit thread suggested that the color change is due to the combination of electrolytes and blue coloring, which is plausible given that Pedialyte has a lot of electrolytes. But "Blue Raspberry" also has Blue 1 and doesn't change color after freezing.
So, my question is:
What specifically causes the purple freezer pops to change color, and why doesn't it happen to the other flavors of freezer pops?

I don't have access to a lab, but I would be happy to do any kitchen experiments within reason to help find the answer.

My guess would be the slight thaw-freeze cycle of a frostless freezer results in a segregation of the red dye deeper down, and the move to the surface of other components.

I think this is an optical phenomenon and no chemical reaction is taking place. When you are looking at a solution, you are basically perceiving its absorption spectrum. When you have a solid, you are looking at the reflectance spectrum and the story is more complex, if you have a heterogeneous solid like this stuff.
If you originally had the solution and you froze at home, this is slow freezing and the size of ice crystals could be larger (phase separation). If you had a chance to look at frozen Coke/Pepsi, it does not look like "liquid" Coke or Pepsi. It is much lighter, light brown rather. Ice crystals are to be blamed for this color change. If you froze Pedialyte very fast probably, the ice crystals would be smaller and the color change would not be that drastic.
The second effect could be the solubility of this particular dye at freezing temperature. If you had an optical microscope you would probably see the segregration of particles. Since you are saying that other "colors" do not show this effect, most likely, this pigment is not very soluble at low temperatures.
Also note that certain anthocyanins (common flower/fruit colors) change color when frozen because at a low temperatures, a different conformation of the dye molecule exists, but this is rare. This is an extreme example of a dye changing its color drastically upon freezing. See for example,
Unusual Color Change of Vinylpyranoanthocyanin−Phenolic Pigments, J. Agric. Food Chem. 2010, 58, 7, 4292–4297.Here


text:
Removing alkaline stain with acid
Professor Robert Winston once removed a blue alkaline stain using carbon dioxide from a fire extinguisher on James Corden's show. Here's the link for the episode
I would like to know if there is a way to recreate this experiment at home. I guess instead of carbonic acid I could use lemon juice or a vinegar. I'm really not sure about the base though. Would baking soda mixed with blue food coloring do the trick? I have very small chemistry background so any help is appreciated. I'd like to recreate this experiment for entertainment purposses

The problem is that there are not many alkaline stains you will find on household objects. Professor Robert Winston just did a demo for attracting attention of the public towards science (so-called outreach experiments). He simply used an acid-base indicator, perhaps thymolphthalein, which is blue in basic medium and colorless in slightly acidic or even in slightly alkaline solutions. It was a nice demo.

However for entertainment purposes, you can try red cabbage stains or beet root and see the effect of adding vinegar or lemon on a piece of white cloth.


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Reaction of phenol with phosphorus tribromide
I know that aliphatic alcohols react with reagents like $\ce{PBr3},$ $\ce{PBr5},$ $\ce{SOCl2}$ etc. to give alkyl halides as a product. But what will happen if we put aromatic alcohols like phenols with these types of reagents? Will haloarenes formation take place?

Please note that phenols are not aromatic alcohols.

Yes: https://chemistry.stackexchange.com/questions/148079/intermediates-in-aqueous-bromination-of-phenol

TL;DR: It is a possible but complicated reaction and you get a mixture of products but formation of haloarene is not significant in any reaction
Long answer:

phosphorus trichloride: form triphenyl phosphite

$$\ce{3 PhOH + PCl3 → P(OPh)3 + 3 HCl}$$
(The phenol is essentially acting as a nucleophile through oxygen, attacking the phosphorus.)

phosphorus tribromide give a similar reactions. See this
researchgate post for more information (Note: phosphorus pentabromide is already a reactive compound and it decomposes/dissociates to phosphorus tribromide and bromide. So, if you use it to react with phenol, the reactive species would ultimately be $\ce{PBr3}$)
phosphorus pentachloride: See this posts (here and here). In short, it is difficult and you get multiple products.
Thionyl chloride: Possible but very complicated. As noted from a paper:

Studies on the reaction with phenol and thionyl chloride are greatly complicated by the existence of readily-occurring nuclear [sic] substitution reaction. This probably account for the fact that the preparation of diphenyl sulfite by the direct reaction of phenol and thionyl chloride in the absence of hydrogen chloride acceptors had not been reported prior this time and that the preparation of phenyl chlorosulfinate  have been accomplished only recently.
A Study of the Reaction of Phenol with Thionyl Chloride, William E. Bissinger and Frederick E. Kung, Journal of the American Chemical Society 1948 70 (8), 2664-2666, DOI: 10.1021/ja01188a015

"readily-occurring nuclear substitution reaction." Nuclear? Huh? Autocorrect maybe? Recheck the original source.

@OscarLanzi First page, left column, third line (still first phrase) of the source cited indeed uses «nuclear».

[sic]: an expression used to indicate that a seemingly incorrect or paradoxical quotation is in fact rendered correctly.


text:
Setting up reactor in inert atmosphere
I am searching for reactor set-up that can maintain inert atmosphere, and found that many organic synthesis labs use Schlenk line. But organic synthesis is not main thing of my lab, so there is some difficulties (especially for vacuum pump) for setting up Schlenk line. Therefore I have searched for method that doesn't use vacuum pump, and I found this video.
https://www.youtube.com/watch?v=42_LPoWVJ1Q
There are some questions.

At 0:44, he pushes the red rubber-like thing to the oil bubbler and call it 'pipette bulb'. Pipette bulbs I know is two kinds. (see images below)

Is that second one? It seems a little bit different. And what is the role of pipette filler? Does it just let the gas to go to the oil?

At 1:01, he push Pasteur pipette to the adapter. What is that adapter? When I search for 'Pasteur pipette adapter', I found this.

It seems similar, but it doesn't have white screw-like thing.

At 2:06, he says that the end of the pipette should be under the surface of the liquid. I usually conduct synthesis in 10~100 mg scale, with about 10 mL of solvent. So it is hard for Pasteur pipette to reach the surface of liquid. Should I just find a smaller flask? (I usually use 250 mL three-necked flask)

At 2:06 he starts purging and at 2:35 he ends purging. How can I determine when to end the purging?

At 2:47, he removes the tube from adapter and connect it to the pipette bulb. Then is he supplying the inert gas to the oil bubbler continuously?

(Check again if your lab has not a Schlenk line or if one may be purchased, manufactured (your local glass blower), acquired (ebay-like second hand), or shared with your befriended group across the aisle.  Because it is a tool equally used in inorganic chemistry, catalysis, etc.  Then read a guide like this one to have some general idea, and get an instruction in person by an chemist more experienced with the design you will work with.)
The video omits one approach consisting of mounting the air dry glass ware in first place, to thoroughly dry the setup with a heat gun (a powerful hair dryer), and to allow the setup to cool to room temperature while being purged by inert gas.  The drying is more efficient if the setup is evacuated and if proceeding from the inlet of the inert gas to the venting outlet.  If the setup includes a condenser, then no water is allowed in the coil of the condenser during the drying.  If you have the option, argon may create a blanket on top of your reaction (thus, you may save some volume of inert gas), while the cheaper nitrogen by relative density goes off to the exhaust. Note, nitrogen equally may react with some metals (e.g., Li, Mg).
After this to your questions:

He briefly closes one branch of the T either with a stopper, or rubber bulb.  Don't use the Peleus sphere shown on the left hand side.  The oil in the bubbler is a visual for the pressure inside the setup, and a safety valve should the pressure in the setup increase too much.

The left-hand side of the adapter shown by you looks more like a silicon tubing, than glass.  Look for a thermometer adapter / inlet like the following:

(adapted from here)
Below the cap you find an o-ring.  With loosened cap, push the Pasteur pipette through, then tight the cap gently and mount the adapter.  But in many cases, a normal knee-shaped adapter may work fine to connect your setup with the inert gas, too.

One word: scale.  Except if you really need to work at low concentrations (say below $\approx \pu{1 mol/L}$), use a smaller flask. For safety (e.g., sudden evolution of gas, foam) and comfort (intentional quench of the reaction, start of workup), it is nice to use about 40% of the flask's volume by the volume of reaction.  Thus, consider a smaller flask e.g., $\pu{25 mL}$.  This equally reduces the space needed in the hood, eases drying the setup, monitoring temperature, etc.

If there is no flow indicator for the inert gas, connect a hose with the inert gas supply and a Pasteur pipette and set up once a pneumatic trough.  This gives you an estimate how much time is needed to replace x-amount of gas in the setup if no vacuum was applied earlier.   If you evacuated the setup (pump-thaw cycles, requires gentle greasing the joints), then the bubbler indicates when you reached normal pressure in the setup.

I can not recommend this change of the setup.  Mounting the inert inlet on one wing of the T, across the exhaust, no longer guarantees protection by  the inert gas.  Instead, it will flow the path of least resistance regardless what happens below the reflux condenser and round bottom flask.
Instead, 1) keep the inlet remote from the exhaust of the setup, 2) maintain a very low but steady flow rate to prevent air and humidity from entering the reaction flask.  If you don't aim to distill-off solvent (especially at reduced pressure), then you don't constantly blow nitrogen / argon into your reaction (it is not a fish tank).  You do not want to waste of inert gas, nor evaporate your solvent of reaction during the reaction.

Thank you for detailed answer. May I ask some more questions? 1) As I calculated, the concentration in my reaction volume is about 0.03 M. You mentioned that 'except I work at low concentration' I had better use a smaller flask. Then in this situation, I have to use a flask bigger than 25 mL? Is that okay to use 100 mL or 250 mL? 2) I'll consider with lab seniors to set up Schlenk line. But first, is it okay to use setup without vacuum like the video I uploaded, despite some low quality than Schlenk line?

Honestly, when I search for other Schlenk line videos, I don't understand heating glasswares (some heated with torch) is of importance. Is this procedure very essential and important? 4) Why does the constant flow of inert gas let the solvent to evaporate? (I don't understand the principle) Is it okay to remove the flow once the inside of the flask reach normal pressure? I thought that constant flow is necessary because anyway top of the reflux condensor is open to the air. Is this removal of constant flow same procedure when I use Schlenk line? (actually when I took over this synthesis,

senior supplied constant flow of inert gas to the septum, for 24 hours of reaction time) I think that inert atmosphere can be broken for 24h or more reactions.

I have flow indicator with cc/min unit. Then I don't need pneumatic trough? How much flow should I set up and how can I know if inside the flask reached normal pressure using bubbler? (sorry for too elementary question)

Just now, searching for oil bubbler, I found the text that if there is underpressure, oil can be sucked into the flask. During reaction, if pressure becomes too low (Does this correspond to infrequent bubbles?), then should I supply inert gas again?

@KrangLee Continuation in the chat room The Periodic Table.


text:
Etard's reaction
My textbook says the following on Etard's reaction:

Now, if the alkyl group attached to benzene is an unsymmetrical or branched alkyl group, such as this:

What will be the reaction product(s)? Which terminal carbon will be oxidised to a carbonyl carbon?

I don't know which book you used to get that information for ethyl benzene, but this is wrong. See https://chemistry.stackexchange.com/questions/140039/%C3%89tard-reaction-of-ethylbenzene (related) It is 1 of the products, but by no means major. Higher alkanes attached further undergo several different rearrangements giving chloride aldehyde and ketone isomers hence this reaction is only preferred for toluene

You'll get a mixture


text:
Why does tennessine have a larger (predicted) electron affinity than nihonium?
According to Wikipedia, tennessine has an electron affinity (EA) of $\pu{+166 kJ mol-1}$ and nihonium $\pu{+67 kJ mol-1}$.
Normally this would make sense since tennessine is in group 17 and nihonium in group 13. However, the key feature of the 7th period is that the group 14 element is a pseudo-noble-gas and the group 18 element a psuedo-carbon-group-element; flerovium has a negative EA, like a noble gas, while oganesson tetrafluoride has Td symmetry, like the carbon group tetrafluorides. Therefore, nihonium, being one electron short of becoming "flerovium", should behave more like a halogen (and therefore have a more positive EA) than does the "true halogen" tennessine.
My question now follows - is this description on Wikipedia true, and, if true, why is it so?

It's not that cut and dried. Thanks to multiple relativistic effects on electron subshell energies, copernicium, flerovium, and oganesson all have configurations that may be potentially be called "closed-shell". No one of these elements is undisputably the "noble" one in Period 7. Indeed, the standard periodic predictions which are based on nonrelativistic quantum mechanics are breaking down at this point in the Periodic Table.

@OscarLanzi Is this only an issue for Period 7?  I.e., is radon unequivocally a nobel gas in the same way as those above it?

In fact it is not only an issue in Period 7. We see the mercury configuration developing a noble character in Period 6, where gold and even platinum can form anions with that configuration. Radon is also known in the +6 oxidation state corresponding to ... a mercury core. Period 7 is expected to be much more muddled, though, with its stronger relativistic effects


text:
Rotating orbitals and parameters for a many-body calculation
I am confused with some results I am getting. Given a molecule (In particular, I was toying with triangulenes, some benzenoids and things like that, but this is not very important), I perform first a tight-binding, Hartree-Fock or a DFT calculation to get some molecular orbitals $\phi_j$. Let me use greek indices $\mu$ to denote the sites on the lattice defining the molecule and latin indices to denote to molecular orbitals. I want to consider the Hubbard Model on the lattice
$$\hat{H}=-t \sum_{\langle \mu, \nu \rangle \sigma} \hat{c}^\dagger_{\mu \sigma}\hat{c}_{\nu \sigma} + \text{h.c.} + \frac{U}{2}\sum_{\mu \sigma} \hat{n}_{\mu \sigma}\hat{n}_{\mu \overline{\sigma}}.$$
I will solve exactly this model in an active space of a few  molecular orbitals around the Fermi level. Let's say I have the orbitals $\phi_1,\phi_2,\phi_3,\phi_4$ in my active space (for example, $\phi_2$ may be the HOMO, $\phi_3$ the LUMO, and the others are the HOMO-1 and LUMO+1). I can easily find a many-body hamiltonian for these four orbitals from the Hubbard hamiltonian using the truncation:
$$\hat{c}_{\mu \sigma}  \approx \sum_j \phi_j(\mu) \hat{C}_{j \sigma},$$
where the $\hat{C}_{j \sigma}$ are the anihilation operators for the molecular orbitals.
Now, the many-body model can be exactly diagonalized for small active spaces.
However, in many cases of interest, some of the molecular orbitals used in the active space are degenerated, so actually the representation of the subspace can be chosen in many ways. This shouldn't change anything but, actually, by rotating the degenerate orbitals I change the parameters of the many-body hamiltonian and not only the absolute energies, but also the excitation energies depend on the chosen rotations.
What is behind this confusion?  Can it be fixed somehow?
So far, the only thing that I came up with was trying to find the energy minimum in the space of all possible representations (i.e., rotations) of the degenerate orbitals, but this is in general a very heavy computation, and I am not sure it makes sense anyway.

If you don't get an answer here, you may want to ask this in https://mattermodeling.stackexchange.com/

Okey, thanks for the tip, Maiti. Should I wait? Should I close this question before opening it on other site?

Yes, definitely you should wait for a couple of days, there are some computational chemistry people here too. If there is no response, then you should ask a question on matter modelling. What to do with the one here is a tough question, as cross-posting is usually discouraged. Personally, I would say leave both questions open, but give a link of the other to each question to tell people that you have cross-posted. I am not a moderator however. And good luck with your question, it's quite interesting, but I don't know enough to answer.

@SRMaiti this question didn't end up getting any attention here, which was unfortunate for the OP since this was almost 6 months ago. In the future, perhaps I would be less hesitant to encourage migration earlier rather than later. In this case, the "couple days" wait didn't seem to do any good. Now a migration won't be possible due to the 60-day limit for migrations, so Qwertuy will have to post on MMSE manually.


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How Bread is made with yeast, sugar and luke warm milk?
Materials and Apparatus:

wheat flour
sugar
dry yeast
glass bowl
covering plate
milk

Procedure:

Lukewarm milk is taken in the glass bowl and sugar is added to it. Then, yeast is added to the same.
The mixture is left undisturbed for 10-12 minutes to activate the yeast
3 cups of wheat flour are added to the bowl containing the milk mixture.
The mixture is mixed thoroughly with 100ml of added water and the dough is kneaded well
The dough is placed in a bowl, covered with a plate and left undisturbed for 2 hours.

My query/confusion:

Why is milk needed?
"activated yeast"- what's the difference?
Can yeast work without sugar or milk.
Detail out the stages of the anaerobic oxidative process which takes place as a common first step in both aerobic and anaerobic respiration.
Finally, feel free to share anything I may be missing which should be here.

If you have any confusion regarding what I want to ask, please ask in the comments. Please upvote if you are curious about it too

This should be posted in seasoned advice.SE

What do you know about the topic already? When is your deadline?

milk is not needed, 'pure' bread is without milk
yeast is a fungus, therefore, it is alive. Its best to work with fresh yeast, which you find as small cubes in the refrigerated section. This one does not have to be activated. non-fresh yeast is dried, so in order for it to work properly, it has to be undried by adding water, which is called activation.
and 4. As said before, milk is not needed. Sugar however is the food for the yeast, without it, it does nothing. In aerobic breathing, the yeast metabolizes the sugar as we would: sugar + oxygen -> water + CO2. Without oxygen, the yeast resorts to ethanol fermentation: sugar -> alcohol + CO2 (this is, why it is used to make beer or wine). For making bread, we have a mixture of both respirations, which does not really matter, since we are only interested in the CO2, which makes the dough fluffy =) But without sugar, there is no CO2.


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Comparing strengths of σ-bonds formed by $\mathrm d_{z^2}$ and $\mathrm p_z$ orbitals
According to my professor $\mathrm d_{z^2}–\mathrm d_{z^2}$ σ-bond is stronger than $\mathrm p_z–\mathrm p_z$ σ-bond as the extent of overlapping is greater in $\mathrm d_{z^2}–\mathrm d_{z^2}$ overlap, because of directional nature and of $\mathrm d_{z^2}$ and its greater distance from the nucleus. There is a assumption that principal quantum number of both $\mathrm d_{z^2}$ orbital and $\mathrm p_{z}$ orbitals are the same.

But according to my intuition $\mathrm p_z$ orbital is also directional in nature, and $\mathrm d_{z^2}$ orbital has a ring in $x–y$ plane which can cause repulsion and the biggest reason that $\mathrm d$ orbitals are very diffuse so overlapping is not very good.

Is there any way we can compare their bond energy?
Which σ-bond will be stronger: $\mathrm d_{z^2}–\mathrm d_{z^2}$ or $\mathrm p_z–\mathrm p_z?$

It is possible that your instructor meant $\sigma$ bonding -- one lobe pointed straight at another lobe -- is stronger because it has more overlap than $\pi$ bonding, which would be them vertically aligned, but contacting "shoulder-to-shoulder" instead.
The energy term contains orbital overlap, and so it can be reasoned qualitatively that an interaction with more overlap (direct, $\sigma$) would be more favorable than the sideways version, e.g. $\pi$ bonds in alkenes.
Also, one has to remember that both of these are happening in the molecule at the same time, and these two states don't really "exist" in this "one or the other" form, but rather in some hybrid (e.g. 80:20 in the ground state, but 20:80 in the excited state).  Whether this basis needs $d$-like functions to make the energy lower, or perhaps more $p$ is situational.  One can post-hoc dig through orbital contribution terms, but there's some black magic here.
But otherwise, I would agree if there is ambiguity because technically d and p come from the same shell, and may have different field effects so that one would not know just how contracted in the bonding direction an orbital is to make a determination without resorting to quantitative calculation.
In my opinion, the fundamental problem here is using spherical harmonic (isolated atoms) models in a situation that is not spherical.  Its endemic to the chemistry curriculum, unfortunately.

Also, one has to remember that both of these are happening in the molecule at the same time, and these two states don't really "exist" in this "one or the other" form, but rather in some hybrid (e.g. 80:20 in the ground state, but 20:80 in the excited state). which two things, exist in hybrid i didn't get your point?


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How do I get calcium citrate to precipitate in experiment on extraction of citric acid from juice?
I was planning to extract citric acid from mango juice by first using Aspergillus niger fungus by fermentation to convert the mango juice to citric acid, then extracting by first adding sodium hydroxide until the fermented mango juice is somewhat alkaline.
Then adding the calcium chloride (at 70 °C), thinking the calcium chloride would first react with the hydroxide in the juice, producing calcium hydroxide, which would then react with citric acid to form calcium citrate by the reaction below:
$$\ce{2 C6H8O7 + 3 Ca(OH)2 -> Ca3(C6H5O7)2 ⋅ 4H2O + 2 H2O},\tag{R1}$$
then later adding sulphuric acid, to get citric acid
$$\ce{Ca3(C6H5O7)2 ⋅ 4H2O + 3 H2SO4 -> 2 C6H8O7 + 3 CaSO4 ⋅ 2H2O + 2 H2O},\tag{R2}$$
but I don't get any precipitate (calcium citrate) to continue experiment, just a dark slurry. Please, I need assistance.

Food grade citric acid is sold in food stores, where I live, as “sour salt”. It is sprinkled on food to give a tart or sour taste. Maybe buy a bottle (it is inexpensive) and test your protocol on that, just to get a bit of experience and debug the protocol without mango debris obscuring what is happening.

The remaining chloride ions from CaCl2 make the solution neutral, not basic. Start with calcium carbonate (limestone) and wait for a while for it to react, or use CaOH to start.

Usually ammonia is used to make the solution basic. Isolated calcium citrate salt is washed and then treated with sulphuric acid. Now after ion change between sulphate and citrate ions, citric acid dissolves and calcium sulphate forms the solid salt. From solution part, citric acid can be chrystallized by evaporation.


text:
Questions about molar ratio
I am new to chemistry so this might be about a very easy concept that I might be misunderstanding but I would appreciate your help!
I watched a video of combining hydrogen and oxygen and forming water. In the video I watched made sure they had a ratio of 2:1 (2 Hydrogen H2 : 1 oxygen O2) before they put on a fire to  cause the reaction. In my understanding the chemical equation tells us 2 hydrogen molecule and 1 oxygen molecule form 1 water molecule. Sure, you can put stoichiometric numbers to deal with moles of molecules, but even if you don't have this ratio of 2:1, as long as the hydrogen and oxygen molecule in present, shouldn't the reaction between hydrogen and oxygen still occur with having excess hydrogen and oxygen molecules leftover after the formation of water? Can you tell me if I'm wrong?
Also as a side question, is it possible to calculate the exact heat required to cause the chemical reaction?
Thank you.

You're not wrong.
The reaction would still occur if the amounts are not in the specific ratio. The ratio must have been maintained so that they could ensure that almost all the reactant is used. Or if it would be possible to collect the water this formed, the number of moles of water would equal that of hydrogen used (and double of oxygen used).
The particular ratio of reactants is significant when there is possibility of multiple reactions, a popular example being of burning of Carbon, where reducing the amount of available oxygen causes formation of CO as opposed to carbon dioxide (which forms when oxygen is available in high amount).
As for the side question, it is possible to know the amount of heat. You will study it in the thermochemistry topics as you further your knowledge of chemistry.

Thank you! Is it the same in acid-base and redox titrations? Why won't phenolphthalein react right away when titrant is being added?

OP is asking about the amount of heat to start the reaction. This is a chain reaction, so you need very little energy, but a high temperature, like a spark.


text:
Energy required to break apart water molecules in dissociation?
I'm learning about the process of dissociation. I'm aware in dissolution that essentially three processes occur (I know I may be simplifying it, but I am in Year 11):

Solute particles break apart (endothermic)
Solvent particles break apart (endothermic)
Solvent and solute particles bond/solvate (exothermic)

This much makes sense to me. However, when I get to dissociation my textbook only presented with two processes:

Lattice energy - energy required to break apart ionic lattice (endothermic)
Hydration energy - energy released by water molecules hydrating ions (exothermic)

What is confusing me is that the energy for the bonds (hydrogen bonds) between the water molecules to be broken apart doesn't seem to be accounted for?

The Lattice and  hydration energies applies to cases of dissolution of  ionic compounds like $\ce{NaCl}$.
For covalent liquid compounds with polar bonds like water, the lattice energy is replaced by the bond dissociation energy.
The value of the dissociation energy ( enthalpy ) -- involving hydration -- can be indirectly determined from the temperature dependence of the water auto-dissociation constant:

Water temperature
Kw / 10−14
pKw[12]

0 °C
0.112
14.95

25 °C
1.023
13.99

50 °C
5.495
13.26

75 °C
19.95
12.70

100 °C
56.23
12.25

using van't Hoff equation

$$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta_r H^\ominus}{RT^2}$$

My question is, is there no bond dissociation when salt is dissolved in water? Don't the water molecules also have to break their intermolecular bonds in order to hydrate the salt?

Braking intermolecular hydrogen bonds to form hydration bonds is implicitly involved in the hydration enthalpy. Note that it is only partial breaking apart, not the full one, that can be derived from the evaporation enthalpy.


text:
Acid catalyzed bromine substitution in an alcoholic ester. What is the major product and why?

This was in a paper I gave. I want to know the mechanism by which both of these products will be formed and why is the 1st one the major product.
My thinking is that the -OH group will be converted into a good leaving group by protonating it and then a 2-degree Carbocation will be formed which should give Product 2. But the answer given is product 1 (which will go through a 1-degree carbocation as an intermediate).
The 1st product:

The 2nd product:

Any input will be appreciated.

Consider the stability of the allylic cation and the size of the bromine anion.

I think if the reaction is under kinetic control (low temperature) then it proceeds with the most stable intermediate, so product $2$. But if the reaction is under thermodynamic control (high temperature), then the more stable product, the more substituted alkene $1$ will be formed. They have not mentioned the temperature, so not sure which one to choose.

In such papers if there isn't a symbol of 'Delta" i.e. heat then it isn't high temperature. So, according to your theory product 2 must form as a major pdt, which is what I had marked in the exam, but its wrong according to the paper setters. Any other logic? @TRC

Are you sure they mention heat every time? For instance, hydrolysis of the cyanide group to carboxylic acid makes heating essential - but I rarely see anything except $\ce{H3O^+}$ mentioned for the hydrolysis - in the same exam.

Besides, there is a famous question of addition of $\ce{HBr}$ to $1,3-$butadiene that most students come across while preparing for JEE. There the "high temperature" for thermodynamic control is $40^oC$. Do you think that temperature will be mentioned separately as heat?

I'm not saying it's your fault. I would probably have made the same mistake. Most of these data are experimental, and unless you have seen and memorized the results of the same reaction on that very same compound, it's nearly impossible to predict the product directly. It's all the fault of the paper setters.

How'd you know that this question was asked in JEE? @TRC bcoz I didn't mention that.

@GuyEternal most highschool questions here are from JEE students. But you can clarify if your exam mentions temperature for every reaction.

@GuyEternal The font of the question in your image is an exact match of the kind of font that NTA uses when it releases the response sheet, which it had done just the day before yesterday ;) So I guessed it was from there. Plus this is the kind of nonsensical question only NTA will ask in JEE Mains. Please correct me if I'm wrong - is it from some other exam?

Yeah Yeah it is from Mains @TRC

This reaction called Electrophilic addition to α,β-unsaturated carbonyl compounds.

Electrophilic addition to α,β-unsaturated carbonyl compounds is
analogous to electrophilic addition to isolated double bonds, except
that the electrophile adds to the carbonyl oxygen, the most basic atom
in the molecule. After that, the nucleophile adds to the β carbon, and
the resulting intermediate enol tautomerizes to the more stable
carbonyl compound.

The 1st & 2nd products are incorrect.

What is the major product and why?

The major product reaction mechanism should be as follows

and the reaction mechanism for the minor product is

Why?
As the carbonyl is protonated then the intermediate cation is a stronger electrophile and will react with the bromide ion.

I believe this might be incorrect - as far as I know the protonation occurs fastest at that position where the most stable carbocation is first formed. Your carbocation forming during Michael addition is primary and unstable. The carbocation formed by protonating and removing the alcohol group is secondary and allyl - more stable.

I would expect acid-base reaction leading to removal of water a much faster reaction. We do have concentrated HBr here, since acid-base is way faster than electrophlilic addition,  there may be a discrepancy with the source you have mentioned because it does not take into account the presence of alcohols.

@ThermEster Thankyou for your answer but I think that the effect of alcoholic grp isn't taken into account in your answer (as pointed out by @napstablook). I understood the answer after discussing in the comments.


text:
Friedel-Crafts acylation with withdrawing group?
Will a Friedel- Crafts acylation take place on ortho-bromotoluene? I do understand that Friedel Crafts reaction does not occur when there is an electron withdrawing group on the ring but is there any exception when it comes to ortho-bromotoluene? Please do help me understand this concept

Bromine is not very electron withdrawing, methyl is electron-donating, the two effect approximately cancel out so F-C will work on bromotoluene.

It will occur. Particularly favorable at the para location w.r.t to methyl group.


text:
Problem about TLC solvent and staining
I am conducting an organic synthesis involving 1-aza-18-crown-6-ether. I tried TLC, but they didn't work well. I have tried: (1) hexane+ethyl acetate in various ratio, (2) dichloromethane:methanol=3:1, (3) chloroform:methanol:ammonium hydroxide solution(28~30%)=5:3:1
In case of (2), as previously I asked here, aza-crown hardly drew up. ( Is it possible to run column chromatography in reversed phase? )
(additional question : Is it possible to show two spot in TLC by one material? TLC image I uploaded seems to have two spots, but it has only one material, 1-aza-18-crown-6-ether. Solvent is toluene, which evaporates fast.)
In case of (3), because aza-crown spot can't be observed by UV, I use ninhydrin as staining solution. I saw the paper which uses solvent system (3) (actually it used ammonia), but because of ammonium hydroxide, TLC plate turns all red after staining. Many people recommended adding ammonia in the previous question, but ninhydrin may stain it also. Should I change staining solution?

Throughly dry the tlc plate with a hot air blower before staining with ninhydrin. Also try a permanganate spray + heating for visualising it. Could also try adding triethylamine instead of ammonia.

@Waylander Thank you for comment. May I ask additional question? There are two methods of staining, dipping and spraying. What method would you recommend?

Whichever is easiest for you. I used to keep a pot of aq KMnO4 solution at the back of the fumehood for dipping plates. You can sometimes see the spots as soon as the solution is running off the plate.

@Waylander Actually I was worried about contaminating staining solution by multiple dipping. Does this problem have few effect?

Not really a problem, there's only a tiny amount of organic material in a TLC sample.

@Waylander Thanks a lot! I would try the methods you answered.

This may be useful http://www.cchem.berkeley.edu/rsgrp/TLCStainRecipes.pdf


text:
What is the coolest known flame?
Inspired by this youtube demonstration by ChemicalForce, I'm prompted to ask what is the coldest possible flame.
What the video seems to show is phosphorus vapour in an inert gas being blown into air. This seems to create a flame, but one so cool you can hold a finger in it and not be harmed.
So is this the coolest known flame or are there other known phenomena that are cooler?
NB I suspect that flame temperature depends not just on the stuff that is burning but also on the details of the flow and concentration, so please specify what they were if there are other candidates.

If you include plasmas among flames, then there are plasmas which can rather freeze water rather than boil it. There is a nice video "Fire That Actually Freezes Things Instead of Burns Them" https://www.youtube.com/watch?v=cUpv2AqbZ1E

If the rising air convecting from a warm object counts as a "flame" (it certainly is shaped like one and flows like one), then a flame can be formed at any temperature warmer than ambient.

I missed the duplicate question despite a search. Perhaps because the dupe is tagged [thermodynamics] and nothing else. Also, while there is a good answer, the accepted and most upvoted answer is blatantly wrong.

Also, and a possible reason this question is not a perfect dupe, the other question is theoretical in the way it is posed, this question is not about theory but about kown observed phenomena. None of the dupe's answers address the theory but would be better answers to this question.


text:
correcting Henry's Law coefficient for higher pressure
I have a bottle at room temperature that contains water, nitrogen, and a trace amount of vinyl chloride (VC). I'm trying to determine how much the Henry's law coefficient for VC (at 25 C and 1 bar, with units of pressure/concentration) should be corrected if the bottle is pressurized to 2 bar.
A thermodynamics textbook I've borrowed states that the Henry's law coefficient $H_2$ at pressure $P_2$ can be calculated with $H_1$ at $P_1$ as follows:
$$
H_2 = H_1 \exp \left [ \int\limits_{P_1}^{P_2} \frac{\bar{V}^\infty}{RT}dP \right ]
$$
where $\bar{V}^\infty$ is the partial molar volume at infinite dilution, $R$ is the gas constant, and $T$ is temperature. The textbook says that $\bar{V}^\infty$ can be approximated with the pure species molar volume $V^m$, which for VC is 45.2 mL/mol.
After making that substitution and simplifying I am left with the following approximation:
$$
H_2 \approx H_1 \exp \left [ \frac{V^m(P_2 - P_1)}{RT} \right ]
$$
which for my scenario works out to a negligible adjustment of $H_2\approx1.0018H_1$. Have I interpreted this material correctly?


text:
Hydrogen Bonding in case of nitro group
In many different cases of Nitro group I have seen that only one oxygen atom is particularly used for hydrogen bonding but not the other one.But I guess both the bonds should be polar and hence show hydrogen bonding.
Please explain.

I have just used nitro phenol for reference.

Yes, both N-O bonds are equivalent, so either can form hydrogen-bonds. However, both oxygens forming hydrogen bonds at the same time is probably unlikely due to geometry constraints.

The oxygen atoms are indistinguishable. That they look distinct is an artifact of the Lewis notation. There are examples when both are acting as hydrogen bond acceptor.


text:
Why does magnesium have a greater ionization energy than lithium?
I'm a high school student and I'm learning about ionization energy and atomic radius of elements. I want to compare the ionization energy of lithium and magnesium.
Here is the information provided in my textbook:

In periodic table, the atomic radius of elements gradually decreases from left to right of a period and increases from up to down of a group.
If the atomic radius of an element is greater than the atomic radius of another element, then the ionization energy of the first element is less than the ionization energy of the second element.

So according to my book, magnesium has a greater atomic radius than lithium which means magnesium has less ionization energy than lithium. But when I checked on the internet, the result was completely opposite. Is there an explanation for this exceptionality?
Please share an explanation that is suitable for a high school student.

Note that I have used a brief, simple explanation for a high school student asking such a question. As the more complete answer could be much more complex, involving quantum chemistry. There is no high math used in the answer.

The book has omited the other factors.
The short answer is: In spite of the bigger atomic radius, the valence electrons of magnesium are attracted by the greater net electrostatic force compared to lithium. As this effect is stronger than the radius effect, the ionisation energy is greater.

The ionisation energy depends on the atomic radius AND the effective nucleus charge.
It can be expressed as:
$$E_\mathrm{ion}=\frac 12 \cdot \frac {e^2 \cdot Z_\mathrm{eff}} {4 \cdot \pi \cdot \epsilon_0 \cdot r_\mathrm{eff}}$$
where

$E_\mathrm{ion}$ is the electron ionisation energy
$Z_\mathrm{eff}$ is is effective nucleus proton number
$r_\mathrm{eff}$ is the radius of a hypothetical circular electron orbit to have the same ionisation energy for given effective nuclear charge.
The factor $\frac12$ is due the electron kinetic energy contribution to the total electron energy.

The effective nucleus charge means what charge should the nucleus have to cause the same electrostatic effect on the particular electron, if there were no other electrons in the atom. It's value is affected by the nucleus charge being screened/shielded away by other electrons and by mutual electron repulsion.
Therefore comparison of ionisation energies based solely on the radius is applicable only if the difference of effective nucleus charges, acting on valence electrons, is negligible.
For more, see e.g. Wikipedia pages Atomic radius and Slater's rules (for the effective nucleus charge).


text:
How to proceed a reaction between piperidine and cyclopropyl methyl carbocation!

Since cyclopropyl methyl cation is highly stable, I proceeded through SN1 as elimination is not possible because of strain in the ring. Need help now in approaching the question.
Actually the given answer is 2, but I am getting no π bonds.

The cation created by SN1 is highly hindered for approach by the piperidine. Consider the possibility of nucleophilic attack at one of the methylenes of the [3] ring followed by ring opening.


text:
Does rearrangement occur in schmidt reaction?
I recently read about schmidt reaction in which a carboxylic acid is converted to an amine. While forming this, acyl carbocation $(\ce{RCO+})$ is formed, so does rearrangement takes place or reaction takes place without any rearrangement?

I am not sure where you read this( you have to show effort and give citations while asking questions) but schmidt reaction proceeds with hydrazoic acid forming geminal diol intermediate which undergoes wolff rearrangement and subsequent hydrolysis to give amine, not amide.

Take a look at this  https://www.organic-chemistry.org/namedreactions/schmidt-reaction.shtm

Oh sorry it’s amine only I didn’t see that I wrote amide

As napstablook pointed out in the comment section, I didn't understand where you got this erronous information about Schmidt reaction. For example, it definitely didn't give a amide with the reaction with carboxylic acid. The product is a corresponding amine with one less carbon.
Basically, the Schmidt reaction is the reaction of hydrazoic acid $(\ce{HN3})$ or an alkyl azide $(\ce{R-N3})$ with a carbonyl compound including a carboxylic acid, a ketone, or an aldehyde, often in the presence of a Brønsted or Lewis acid. Further, the reaction of $\ce{HN3}$ or $\ce{R-N3}$ with an alkene or an alcohol is also called the Schmidt reaction (Ref.1):

As shown by the products, all of these reactions have undergone a rearrangement to give perticular product except for the nitrile formation from the aldehyde.
The mechanism of the Schmidt reaction is well understood (Ref.2). In the case of a carboxylic acid reaction with hydrazoic acid, the initial formation of an acylium ion is followed by the addition of $\ce{HN3}$ to form an acyl azide intermediate. The possible migration of alkyl or acyl group produces the corresponding isocyanate, which undergoes hydrolysis to yield an amine (not an amide) and carbon dioxide:

References:

Karl Friedrich Schmidt, “Process of making derivatives of hypothetical imines including amines and their substitution products,” U.S. Patent 1,564,631, 1925.
Peter A. S. Smith, “The Schmidt Reaction: Experimental Conditions and Mechanism,” J. Am. Chem. Soc. 1948, 70(1), 320–323 (DOI: https://doi.org/10.1021/ja01181a098).


text:
Deactivate lectins in beans by baking
Having dried uncooked beans,
I read that the lectins can be deactivated by boiling and soaking for a while. What I could not find any information on is if dry baking beans would eventually deactivate the lectins too.
Does dry baking work to deactivate the lectins and if, how long at what temperature?

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

Previous studies on the effect of "dry heating" (baking in an oven) on the hemagglutinating activity of legume lectin (a proxy for or direct cause of toxicity) from  P. vulgaris suggest that this is not a good way to suppress toxicity. Citing Ref. 1:

Dry heating had little effect on hemagglutinating activity of certain
varieties of P. vulgaris, and some activity was still detectable after several
hours of heating (DeMuelenaere, 1964). Heintze (1950) found dry heating or
high frequency treatment ineffective for the cooking of beans. However, the
heating of soaked beans was fully effective. Too much heating reduces the
growth induced by bean diets, probably due to the impairment of the nutritive
value of the proteins (Kakade and Evans, 1965).

The precise answer probably depends on the identity of the bean.
References

Jaffe, W.G., Seidi, D.S., Toxicology of Plant Lectins, Ch. 11 in: Handbook of Natural Toxins, Vol. 7: Food Poisoning. Tu, A.T. (Ed.), Marcel Dekker (1992), ISBN  0824786521.


text:
What is an intuitive explanation for the exponential form in the Arrhenius equation?
What is an intuitive explanation for the exponential form $\exp(-\Delta E/(kT))$ in the Arrhenius equation?
While I know the reaction rate theory, I am puzzled by the origins of the form of the equation, esp. the exponential form. Please give references if any.

Are you familiar with this calculation in statistical mechanics? One intuition is that the reactant state forms at a rate proportional to its energy-implied probability.

As said by M. Farooq in the comments, The Origin and Status of the Arrhenius Equation[1] clearly describes the method by which Arrhenius came with the Arrhenius equation. An excerpt from the article:

Arrhenius considered eight sets of published data on the effect of temperature on reaction rates and showed (using  more currently conventional symbols) that in each case he  could choose a value of the constant $C$ such that $k$($T_1$, the  rate constant at a temperature $T_1$, was represented adequately  by the equation,
$$k(T_1)=k(T_0)\mathrm{exp}\left[C\left(T_1-T_0\right)/T_1T_0\right]\tag 1$$
where $T_1$ and $T_0$ are the temperature in Kelvin. This was  tantamount to showing that the rate constant could be represented as an explicit function of temperature, namely:
$$k(T)=A\mathrm{exp}(C/T)\tag 2 \label{eqn:2}$$
where $A = k(T_0)\mathrm{exp}(C/T_0)$ and both $A$ and $C$ are constants  for the particular reaction.
The reactions which provided these eight sets of data are  listed in the table, along with the respective equations put  forward by each author to inter-relate rate constant and  temperature, where $T$ refers to the Kelvin and $\theta$ to the Celsius  scale and $a$, $b$, and $c$ are constants.
$$
\textbf{The Data Considered by Arrhenius}\\
\begin{array}{cccc}
\hline
\text{Author} & \text{Reaction} & \text{Temperature range/°C} & \text{Author's Equation}\\
\hline
\text{Hood (1885)} & \text{Oxidation of }\ce{Fe}\text{ II by }\ce{ClIO3-} & 10-32 & k=a.b^\theta \\
\text{Warder (1881)} & \text{Alkaline hydrolysis of }\ce{EtOAc} & 4-38 & (a+k)(b-\theta)=c \\
\text{Urech (1884)} &  \text{Inversion of sucrose} & 1-40 & . . . \\
\text{Spohr (1888)} & \text{Inversion of sucrose} & 25-55 & . . . \\
\text{Hecht and Conrad (1889)} & \text{Reaction of ethoxide with MeI} & 0-30 & k = a{10}^{b\theta} \\
\text{van't Hoff (1884)} & \text{Decomposition of aqueous chloracetic acid} & 80-130 & \log_{10}k = \frac{a}{T} + b \\
\text{van't Hoff (1884)} & \text{Decomposition of chloracetate ion in alkaline solution} & 70-130 & \log_{10}k = a\theta - b \\
\text{van't Hoff (1884)} & \text{Dehydrobromination of dibromosuccinic acid} & 15-101 & \log_{10}k = a\theta - b \\
\end{array}
$$
... This leads to the logarithm of the rate constant being a linear function of temperature, which was the relationship used by van't Hoff in respect of two reactions. Thus, Arrhenius was the first to assert that eqn. $(\ref{eqn:2})$ was generally applicable to all reactions, and the equation is justly named after him, although he did not originate the relationship.

Reference:
(1)   Logan, S. R. The Origin and Status of the Arrhenius Equation. J. Chem. Educ. 1982, 59 (4), 279.


text:
respective posistion of d and p orbitals in a multielectronic species
we write the electronic configuration of atoms as [noble gas core]4s2 3d10.that is that due to energy considerations(aufbau principle) in multielectronic species.
what i want to know is whether the 4s subshell occupies a posistion before the 3d orbital in reality,or is this just written in textbooks to easily follow the aufbau principle.

Orbitals aren't situated at a certain distance, they are regions of probability distribution. Only distances to be considered are average, mean square average, modal probability distance etc. But comparing these distances actually does nothing good.

"whether 4s subshell occupies a position before 3d" - occupies a position where? In an atom, there are only probability distributions of electrons - there are no queues where orbitals can be assigned positions. In terms of shell number (a wrong concept when using orbitals), 4s is 4th shell, and 3d is 3rd - but as you said, due to energy differences, 4s is generally filled first.


text:
Chirality of conformations of butane
I have read that the conformations of butane are as follows:

I came across a question about number of chiral conformations of butane.
My Attempt :

The total or fully eclipsed is achiral because it has a plane of symmetry passing through the two methyl groups.
The gauche form is chiral as it neither has a plane of symmetry nor a centre of symmetry.
The anti form is also achiral as it has a centre of symmetry

But I am unsure about the eclipsed also known as the partially eclipsed form as it seems chiral to me but on searching on the net, I find that only the gauche form is chiral.
Any help would be greatly appreciated !

They can’t really be chiral, as they’re not stable.
The barrier to rotation is so low that the conformations are constantly changing

@NotEvans. they are interchangeable  at room temperature but maybe if we could create conditions in which they would not be interchangeable, then I guess we could comment on their chirality.


text:
Nomenclature of Colors of precipitates and solutions
I have just started learning salt analysis in inorganic chemistry. In it i came across the names of colors of various precipitates and solutions. Why do they have prefixes such as "scarlet" in "scarlet red"? Why does it need to be called "red litharge" and not just "red", "Prussian blue" and not just "blue", "crimson yellow" and not just "yellow", "apple green" and not just "green"?
I wanted to know if there are any rules for naming such colors, or they are just named randomly.

Crimson yellow is something I don't think exists. And it's Prussian blue, not Persian blue. Properly phrased questions are more favourably seen. 2. This is a trivial question of how colours are designated, and does not have anything to do with actual chemistry. 3. Colour is not something scientific that should have a systematic nomenclature. Since many compounds appear red, making it a bit more specific as to which shade of red it is, helps, specially while reporting the experiment. It all depends on what shade you see in the test tube. There's no systematic naming.

@TRC Persian blue is also a color but most likely the author meant Prussian.

They are just descriptors. Unless one is affected by color vision problems, the typical examples in text books have been observed so many times to be reliable. Then what is pale yellow can be different between you and me.... But I don't understand the questions really. It is not really different in normal day life. An accurate description of "colours" is something else, as said in other comments and answers.

There are many shades of "red", so for specificity, prefix are used. "Scarlet-red" sound very specific.

It is an awful practice to describe the colors of precipitates with object-colors, like apple green, strawberry red, olive green, etc. I have seen these weird descriptions of hydrogen lines in its spectrum in first-year lab reports. You will not encounter this language in professional work. Avoid such usage while writing lab reports.
You can undoubtedly use simpler expressions without using an object like an apple or a strawberry, e.g., deep blue, bright yellow type expressions. There is no hard and fast rule as stated in the comments.
If you were to discover a new precipitate whose color is hard to describe, then a picture is worth a thousand words for a formal report (assuming the camera won't distort the colors).
Those who work in the textile industry and deal with fancy dyes use more sophisticated instruments to "describe" color. Color measurement science is a whole field of science, and there you do not need the crutches of such words. See, for example, this essay on Wikipedia Colorimetry

+1 For the hint to just take a photo, which in some publications became part of the documentation (e.g., on OrganicSynthesis, example, instructions to the authors revised in 2019). And put a reference for scale and color next to your object of interest, too, the cap of your ball pen isn't the best yet better than nothing (because it is standardized) since cameras capture withe and colors differently in daylight/flashlight, etc.

This goes too far. If I see apple green I note in my lab journal apple green. All the rest is true. But has nothing to do with the actual question. If my crude is olive, that is the best description. Just we know that is the pictorial description of a single operator. To make a wise use of that is up to us reading that journal.


text:
Why are protons more common than hydride ion?
I'm a high school student. I noticed $\ce{H+}$ ion is commonly present in my books while I didn't find any presence of $\ce{H-}$ ions in my books. However, I found on internet that $\ce{H-}$ also exists but it is less common. Because Hydrogen has just one electron, it can either receive one electron to be $\ce{H-}$ or omit an electron to be $\ce{H+}$. So, both should have the same possibility to exist. Then, why is $\ce{H+}$ more common than $\ce {H-}$?
The answer to the question might be obvious to most of the users here with their knowledge. But please share a detailed explanation that is suitable for a high school student.

The answer is actually not so obvious considering that the energy required to eject an electron out of an H atom is 13.60 eV, but when an additional electron is attached to an H atom, 0.75 eV are released.

:/ Are you asking about actual separate entities existing in the world, or some letters in equations? As far as existence as chemical entities is concerned, both of them hardly exist at all.

@Loong: as it was pointed out in the comments below, no actual H⁺ (having presumed internal energy 13.60 eV/ion) exists in solutions or solids. Ions denoted as H⁺ are actually acids whose conjugate bases are (neutral) molecules. That is, presumed hydron ions are actually protonated forms. These acids have much lower enthalpy.

It's not only for H the case. Removing an electron from Cu(I) to form Cu(II) requires a lot of energy as well, still we find mostly Cu(II) in solution because the energy of hydration is large enough to 'stabilize' or compensate for it. For hydrogen it requires about +1312 kJ/mol to ionize it but dissolving that H+ in water then gives you about -1168 kJ/mol of enthalpy back.

This is because we live in a world dominated by oxygen and water. In other words, it is an oxidized world. Most metals occur naturally in the form of oxides, silicates, halides, or other derivatives. Hydrogen occurs as $\ce{H+}$.
In a hypothetical world dominated by metals, all that could have turned out otherwise. Oxygen would be a scarcity, and would come in the form of metal oxides. Nitrogen would be found in nitrides, hydrogen in hydrides (so, a lot of $\ce{H-}$), and so on. There would be no free water or free oxygen.
In our world, it is the other way around. Water is ubiquitous (that is, found pretty much everywhere); oxygen is even more so. $\ce{H-}$ can't exist in their presence. It will quickly react with either and cease to be $\ce{H-}$. It can only exist in an artificial environment.
So it goes.

Actually, $\ce{H^-}$ is relatively common in organic chemistry. Carbocation rearrangement is often done through a hydride ion transfer within the ion, and hydride-ion transfers are featured in organic redox reactions such as sodium borohydride reduction and the Cannizzaro reaction.

It's important to point out that these are merely formal hydride transfers; at no point is there truly any hydride ion present.

Not that hydrogen cations are found as such in their reactions either ... .

True, but at least, one could argue that it is solvated $\ce{H+}$. $\ce{H3O+}$ is probably "closer" to $\ce{H+}$, than $\ce{BH4-}$ is to $\ce{H-}$.

None is really close. In both cases we have to accept molecular fragments or nothing at all.

I don't agree with such an absolute, all-or-nothing statement. At the very least, borohydride anion acts as a source of hydride, so that is closer to the original intent of the question. But hydride migrations in rearrangements is more of a stretch. The electrons go from a C–H bond, into a 3c4e C/C/H transition state, then into another C–H bond. I don't have calculations to hand, but it doesn't seem sensible for the electron density to be polarised towards hydrogen at any stage of this (let alone localised on hydrogen, as is implied by "hydride").

Anyway, I don't think we disagree on any actual facts here; I just don't think that such a black-and-white stance is really mandated. So let's agree to disagree and leave it at that.

@orthicrrson carbocation rearrangement is 3c-2e, no?

Ah, yes—it is 3c2e.


text:
Why is effective nuclear charge used to explain the periodicity of size of atoms and ions?

Why is effective nuclear charge used for explaining the periodicity of size of atoms and ions, even though radius depends upon many other factors?

The size of $\ce{Na+}$ is smaller than that of $\ce{F-}$ and $\ce{Ne}$, as $Z_\mathrm{eff}$ of $\ce{Na+}$ is higher than that of other ions. This is written at this site. But I don't understand how $Z_\mathrm{eff}$ can be used as the only deciding factor for determining the size of the ions (or atoms for that matter).
It is clear that the size of an atom is determined by the net potential energy of valance electrons, with respect to the nucleus: More energy means less attraction hence bigger radius, and less energy means more attraction hence smaller size.
Now the formula of net potential energy for an atom is:
$$E_\mathrm{net} = \frac{KZ^2e^2}{r}-\text{potential energy due to electrons}$$ where $Z$ is atomic number $e$ is 1C charge and $r$ is radius of atom.
It is clear that the potential energy of two atoms (or ions) can't simply be compared on the basis of $Z_\mathrm{eff}=Z- e_\mathrm{insh}$ (the number of inner shell electrons is $e_\mathrm{insh}$). It also depends upon other factors as well.
But many textbooks explain the periodicity of the size of atoms on this basis, but why?

Na+, Ne and F- are isoelectric. While in general there is more factors affecting the radius, in this particular case are those factors constant.

P.S.: I have meant independent factors. As the bigger electron repulsion, weakening the contraction trend, is induced as the negative feedback by the primary reason for the contraction - higher Z_eff.

I think Poutnik's answer here would be helpful.


text:
Relative abundance of mixture in mass spectra
In MS, as far as I know, the $y$-axis, relative abundance means the number of ions detected. Then does ratio of heights of two peaks stand for molar ratio, not mass ratio
For example, given that I analyze a mixture of two components A and B using MS, and there appears only molecular ion peak (although it is unrealistic), then is the ratio of heights of two peaks equal to molar ratio of A and B?

No. There are all sorts of ionisation effects that influence the peak heights that have nothing to do with the molar ratios. Phosphine oxides for example ionise very well and a small amount can completely dominate a MS spectrum.

There are universal detectors (e.g. CAD) with which you can compare signals of different substances. But MS detectors are not universal — meaning that high intensity of the signal doesn't necessarily mean high concentration. And vice versa.
Whether the signal is high or not will depend on how well the compound is ionized. So in case of you A & B compounds:

A may be well-ionized and thus the signal will be high even though the concentration is low;
B is poorly ionized and you may not even see a peak even though the concentration may be high.

To identify how much of analyte you have you first need to run A with a known concentration so that you determine the relationship between area and concentration. Then if you use the same LCMS conditions you can run A with your to-be-determined concentration and by ratios of peak areas you can determine the unknown concentration.
Note, that this calculation depends on the linearity of the signal. Meaning: if the area is smaller by $x$ compared to the signal of known concentration, then the concentration will be also smaller by $x$. This isn't often the case and you may need to run multiple known concentrations to find out if the signal is actually linear. This allows you to build calibration curve which results in an $ax + b$ equation that you can further use to determine the unknown concentration.
To sum up: you can compare the signal of the same compound in different runs (given LCMS conditions and mixture matrix were the same). But not the signal of different compounds.

Oh.. I forgot that MS detector can detect only charged ion. Thanks a lot.

Minor correction: There are no universal HPLC detectors, even charged aerosol detector will give different peak heights and areas for simple anions. I do not know the exact reason but there are limitations.

@M. Farooq: In 1980s, Waters have made two different detectors with their HPLCs: UV detectors and refractive detectors. The refractive detectors are universal and detect every component in the mixture. However, their accuracy was very poor when compared to UV detectors.

@MathewMahindaratne, You are right that RI and CAD should ideally respond to every analyte "equally" but this is rarely the case. For example, if we have 0.1 mM of analyte A and B, the peak area will not be identical in RI or CAD. The RI has reproducibility problems and it cannot be used with gradient methods.