text:
Are zinc & copper extracted from its ore using cyanide salt?
In my textbook, the extraction of Zinc is done from $\ce{ZnS}$ and copper from Cuprous oxide, so neither of extracting procedures involve the use of $\ce{CN^-}$ salts.
Thinking further, both Zn(II) and Cu(II) form complexes with $\ce{CN-}$.The complex of copper can be reduced by zinc (or any other more reactive metal) to obtain Cu(s)(very much like Mac Arthur Forest Process for Ag/Au). I do not think a more reactive metal than zinc can be used to reduce its complex as they'll react with water.
Accordingly, copper should be the answer but the answer is zinc. Can you please explain this?

Why would they need to be converted to cyanide complexes if the existing processes work? Note that cyanide are poisonous, so better to avoid them.

I apologise but I came across a question asking which of Cu,Zn,Ag or Au are not extracted using cyanide salts.It was a single answer correct type and hence I was confused between Cu and Zn knowing that Mac Arthur Forest Process is used for Ag,Au.Can you please help me with the comparison between Cu and Zn.

I have added an answer.

Zinc and copper are primarily consider contaminants in gold/silver extraction and during cyanidation, zinc and copper can complex with cyanide forming respective complexes thus wasting cyanide and technically, you can recover zinc and copper from this complex but the idea doesn't seem feasible and as the method are cost-heavy, ineffective and not to mention that cyanide is toxic in the first place. The below papers will help describe the point in details.

O. Alonso-González, F. Nava-Alonso, A. Uribe-Salas, Copper removal from cyanide solutions by acidification, Minerals Engineering, Volume 22, Issue 4,
2009, DOI: 10.1016/j.mineng.2008.09.004.
Feng Xie, David Dreisinger & Fiona Doyle (2013) A Review on Recovery of Copper and Cyanide From Waste Cyanide Solutions, Mineral Processing and Extractive Metallurgy Review, 34:6, 387-411, DOI: 10.1080/08827508.2012.695303
Bao Guo, Yongjun Peng, Yuliang Mai, The effect of zinc cyanide on the flotation of gold from pyritic ore, Minerals Engineering, Volume 85,
2016, 10.1016/j.mineng.2015.10.023.
E.Y. Yazici, F. Ahlatci, E. Yilmaz, O. Celep, H. Deveci, Precipitation of zinc from cyanide leach solutions using Trimercapto-s-triazine (TMT),
Hydrometallurgy, Volume 191, 2020, 105206, DOI: 10.1016/j.hydromet.2019.105206.

Even the idea of extracting zinc and copper from its ore through cyanide doesn't seem practical and there are already better ways known. Nonetheless, people still try. Here is a patent of extracting copper through cyanide.


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What does it mean when there is a term in parenthesis in an electronic configuration?
This may be a silly question, but I have not yet found an answer and I don't really know how to word my searches. I have been looking at the NIST Atomic Spectra Database (and also atomic physics papers) and I see configurations of excited states that have terms in parenthesis in them and I don't know what it means. For example, I might see a configuration for N II
2s2p2(4P)3p
and the term is listed as 3So.
What would the 4P term in the middle of the configuration mean? When would you write a configuration like this?

At least to me it isn't really clear what you are asking. It would probably be beneficial to view this in context. Can you provide a citation for a paper or a link to the database.

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Difference of reactivity between anilines and alkyl amines
Imine synthesis is somewhat common and straightforward, to achieve good yields a good practice is to remove the water formed from the condensation reaction by means of a desiccating agent or a Dean-Stark trap.
In the lab I've tried to synthesize the following imines from primary amines and aldehydes.

The first reaction proceeds with good yields, but in the second the product will not form. I have tried many different reaction condition but none seem to work.
The only explanation that I have come up with as for why this happens has to do with HSAB theory: 2 is resonance stabilized and thus is a soft electrophile, while aniline's free electron pair is delocalized in the aromatic ring making it a soft nucleophile and that's why both can react together to form 3.
On the second reaction 4 is a hard nucleophile because the electron pair of the amine cannot be delocalized and thus will only react with a hard electrophile.
Is this an acceptable reasoning?
I would be very grateful if anyone could provide references for this subject. Thanks in advance.

I don't think so. Maybe is a matter of higher basicity of benzylamine.

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Trouble analyzing amine using GC-FID
I am trying to analyze a sample containing hexamethylenediamine (HMD) dissolved in ethanol using GC-FID for an undergraduate chemical engineering project. The column used is Agilent DB-FFAP (high polarity). The program used was 95°C (hold for 6 min) followed by 7°C/min increase until 235°C (hold for 2 mins). Different concentrations of HMD were used, the highest being 0.1g/ml however none of the chromatograms showed peaks for HMD. There was always only one peak in each chromatogram. This was attributed to the solvent (ethanol).
Are there any potential solutions to this problem? Is the problem with the type of column used? Should higher sample concentrations be used? Thanks in advance.

For readers: Agilent J&W DB-FFAP is a nitroterephthalic-acid-modified polyethylene glycol (PEG) column of high polarity for the analysis of volatile fatty acids and phenols. It is a close equivalent to USP phase G35 and replaces OV-351.

It may matter the analyte is basic amine and not an acid or phenol. It may sit on column, being eluted too slowly.

First of all, run the column QC experiment under identical conditions. If you can nearly reproduce the chromatogram, then everything is working fine. Otherwise, you will have to consult someone experienced in GC in your academic department. Amines are a classic example of compounds which interact very strongly with any silica based column in both liquid, gas and subcritical fluid chromatography. One should not run free amines in GC.

Silanols are acidic and amines are basic. You will have to quantitatively derivative them to amides or some other inactive group. As some said "An hour spent in the library is worth a month in the laboratory", so start with a literature search and find a reagent that is used to convert amines to something for GC purposes.


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Is this a Pyrite crystal cube?

There is a metallic shine on it that does not clearly show in the photo
This cube weighs 458g
Dimensions: 55mm H x 45mm W x 39mm D
Is this a Pyrite cube?
Thank you

It could be block of pure iron. How can we confirm if this is pyrite just from the photo? Have you conducted any physical/chemical test?

Working out the density would probably be simplest.

It isn't even clear in the photo what colour it is. Pyrite is brassy/golden in colour. Is this cube that colour?

@NilayGhosh the density is too low for iron and galena, which I also checked, but comes close for pyrite. See below for suggested additional tests.

Iron is out if it does not pass the magnet test, i.e., if a magnet is not attracted to the specimen.

@EdV the magnet test would also rule out magnetite, which in any event is black (photo rules that out) and has too high density to match the measurements.

I’m voting to close this question because the sides of a cube would have equal lengths.

Pyrite is a reasonably good guess. Calculating the density of the  cube from the reported measurements gives 4.7 g/cm³, using the same number of significant digits as those reported in the length measurements (and assuming a geometrically perfect block). Assuming the length measurements are correct to within $\pm0.5$ mm the possible range of densities is 4.6-4.9 g/cm³. This overlaps he reported density of pyrite 4.8-5.0 g/cm³. You may also check the hardness (pyrite = 6-6.5 Mohs, from the source cited above) and streak (greenish-black, Ibid.); the latter is often a better color measure than the apparent surface color because it is less subjected to surface impurities. If you have access to a lab that does X-ray diffraction, then pyrite has a distinctive and well-recognized pattern and the test, unlike the others I mention, is completely nondestructive.

I agree with your guess. If I had the specimen, I would test with an ohmmeter (to eliminate metal alloy of some sort), do the streak test you suggested, rub a specimen face with white paper, and add a drop of vinegar to the rubbed off stuff on the paper. Any rotten egg smell would likely mean pyrite, since galena is ruled out by the density.


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Why is this a non-reducing end?

I was told that C is a non-reducing end. To my understanding, only anomeric carbon involved in glycosidic bond cannot be in the linear form, and in other words, is a non-reducing end. But C is an anomeric carbon for that monosaccharide in that particular structure and it is a free and unsubstituted anomeric carbon which should be the reducing end, so why is C a non-reducing end instead? Or am I being told the wrong answer?

I'll give you my humble opinion, perhaps waiting for other more correct answers: I think carbon C is certainly reducing, as it is a hemiacetal, therefore it can de-cyclize, regenerate the carbonyl group, which can then be reduced (for example from Fehling's reagent). In my opinion the question of the non-reducing refers to the entire molecule: in carbohydrates with many monomers the reducing portion is infinitely tiny compared to the non-reducing portion, so the molecule can be considered overall non-reducing.

In fact, if I'm not mistaken, Fehling's reagent does not work on polysaccharides, even though they have a single reducing end.

I meant oxidized from Fehling*


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When balancing a half reaction by adding water and oxonium ions or hydroxide ions, how do we know that the balanced half reaction will occur?
Let's take an example. Balance the following redox reaction under acidic conditions: $\ce{Zn + BrO3- -> Zn^2+ + Br-}$.
The half-reactions are:
$\ce{Zn -> Zn^2+ + 2 e-}$ (oxidation half-reaction and already balanced)
$\ce{BrO3- + 6 e- -> Br-}$ ($\ce{Br}$ is reduced but reaction is not balanced)
When balancing the second half-reaction by adding $\ce{H2O}$ and $\ce{H+}$, it becomes
$\ce{BrO3- + 6 H+ + 6 e- -> Br- + 3 H2O}$.
I understand that the underlying idea is that we can add $\ce{H2O}$ and $\ce{H+}$ since both are at least relatively abundant in an acidic solution. However, when adding $\ce{H2O}$ and $\ce{H+}$ we modify the reaction. How can you be sure that this modified reaction will occur?

Reference for chem equation/expression formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

We do not modify the reaction, we fix the reaction, as not fixed one does not occur. Without nuclear reactions, there is law of element atom conservation. It is an equation after all. It means atom counts for all elements must be equal for both sides. This count balance is required but not sufficient condition for a formal reaction being a real reaction.

Yes, the non-fixed reaction does not occur. And, yes, it is an equation so we can add atoms as long as we obey the "law of element atom conservation", as you said. However, what says that the "new" reaction will occur?

Knowledge of chemistry. There are no formal rules for that. $\ce{He + 3 F2 -> HeF6}$ is formally correct, but otherwise nonsense.

OK! Can you provide some resources or anything for me to understand why that is? I don't want to "just accept it". What is that "knowledge of chemistry" you are talking about? Thanks!

See also CH SE resources // Rules No 1 - there are no shortcuts in learning chemistry. You either have to understand it, to remember it or to search it out.

Yes, I checked that. It is incredibly broad. Could you point to what this knowledge is called (something similar to "acid-base neutralization reaction")? Thank you!

Let us continue this discussion in chat.

There are three basic rules in enumeration of redox half reactions.
Two are consequence of the fundamental conservation laws:

Conservation of counts of element atoms
Conservation of total charge

One rule is a mental helper:

For the reduction half-reaction where an oxoanion loses its oxygen, imagine formally there are temporarily released anions $\ce{O^2-}$, which are much stronger base than $\ce{OH-}$. They immediately react:
In acidic solutions: $\ce{O^2-(aq) + 2 H+(aq) -> H2O}$
In any solutions: $\ce{O^2-(aq) + H2O -> 2 OH-(aq)}$

Note that a formally correct equation for a chemical reaction is not sufficient condition for such a reaction to occur. This has to be addressed by applying particular knowledge of chemistry of involved reagents.
For more, see also CH SE learning resources

An alternative "mental helper". In acid, $\ce{O = H2O - 2H^+}$. In base, $\ce{O = 2OH^- - H2O}$ and $\ce{H = H2O - OH^-}$.

@user55119 You must mean $\ce{O^2- = H2O - 2 H+}$ and $\ce{O^2- = 2 OH- - H2O}$  - But a good one.

When you need oxygen (O) or hydrogen (H) to balance an equation, these expressions are what I use. Balancing the overall charge in the half reaction comes later.


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What is the equivalence point in the context of a redox titration?

We know the standard reduction potential of the species: $\ce{Fe^{3+}}/\ce{Fe^{2+}}$, $\varepsilon_1$ and the standard reduction potential of the species: $\ce{MnO_4}^{-}/\ce{Mn^{2+}}$, $\varepsilon_2$. If we know the concentration of the ferrous cation in the initial solution $c_0$ in sulphuric medium $c_s$, at equivalence point in titration with potassium permanganate $c_t$ how to determine the ratio of concentrations of ferric and ferrous cations?

What I have thought on the subject is that Nernst's potential equation is useful, but I don't really know what equivalence point should mean in a redox context. I know it only from elementary acid-base equilibria.

As the ongoing reaction is
$$\ce{5 Fe^2+(aq) + MnO4-(aq) + 8 H+(aq) -> 5 Fe^3+(aq) + Mn^2+(aq) + 4 H2O(l)}$$
the point of equivalence is when the initial molar amount of $\ce{MnO4-(aq)}$ is equal to 1/5 of the initial molar amount of $\ce{Fe^2+(aq)}$.
Another condition comes from the equilibrium requirement that (not standard) redox potentials of all redox systems must be equal
$$E = E^{\circ}_{\ce{MnO4-}/\ce{Mn^2+}} + \frac{RT}{5F} \ln{\left(\frac{[\ce{MnO4-}][\ce{H+}]^8}{[\ce{Mn^2+}]}\right)}=E^{\circ}_{\ce{Fe^3+}/\ce{Fe^2+}} + \frac{RT}{F} \ln{\left(\frac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]}\right)}$$
So the reactants do not react completely, but the reaction stops when both redox potential are equal. It is similar like when a general reaction stops (its net rate) when the reaction quotient reaches the reaction equilibrium constant.


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Gibbs Free Energy - Maximum Work
I've been having a lot of trouble trying to truly understand Gibbs free energy from a practical perspective. I have no background in physical chemistry, but I think I have a firm grasp on many of the concepts related to Gibbs energy. That being said, I'm going to try and reason using less rigorous mathematics. I'm having a hard time bridging the concepts of Gibbs free energy and the maximum useful non-PV work that can be extracted from an ideal chemical system.
My confusion may be due to some fundamental misunderstanding, but currently my confusion follows this line of thinking:
Most chemistry texts explain that $\Delta_\ G = wmax$.
The most common equations for calculating $\Delta_\ G$ are:
eq1. $\Delta_\ G = \Delta H - T\Delta S$
eq.2 $\Delta_\ G = \Delta G^\text{o} + RT\ln Q$
Both equations essentially make the same relationship, with the difference that $\Delta\Delta S_\mathrm{mixing}$ is baked in with $\Delta\ S^\text{0}$ in eq1.
Question 1.
This may just be a notation thing, but I want to confrim that $\Delta_\ G$ is just an alternative notation for $\Delta_\mathrm{r} G$ and $\Delta\ G^\text{o}$ is an alternative notation for $\Delta_\mathrm{r} G^\text{o}$. Considering I've seen both notations used and I think its safe to assume they are.
Question 2.
If $\Delta_\mathrm{r} G$ or alternatively, $\Delta\ G$ is the slope of the Gibbs energy curve at some non-standard state defined by the reaction quotient $Q$ how can this value be equal to the maximum useful non-PV work. I mean it would make sense if the system were at some steady-state such that the reaction continued to generate products with no net change in product/reactant concentrations. But in reality the value of $\Delta_\mathrm{r} G$ will continue to change until the reaction reaches its free energy minimum.
Maybe I'm trying to get these relations to do more than they meant to do. I fully appreciate that the slope of the Gibbs energy curve will allow for determination of conditions when $\Delta\ G = 0$ which is useful for predicting spontaneity. I just figured (probably incorrectly) that you could for example take some known mass of reactant and from there calculate the theoretical maximum non-PV work harvested from  100% reactants --> equilibrium.
Many Thanks!


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Why does an exothermic/endothermic reaction have a negative/positive enthalpy
So regarding change in enthalpy and exothermic/endothermic reactions, I think I have some conceptual understanding but I'm missing some things.
My current understanding is that:

Given a reaction, if the energy added to break the reactant bonds is greater than the energy released when forming the product bonds, the reaction is endothermic / +Δ.
Opposite of above is true; energy added < energy released, then exothermic / -Δ
Enthalpy of formation is synonymous with bond energy

For example, if I have an exothermic reaction:

Energy added < energy released
Δ is negative because more energy was released than added
Δ is the negative of the difference between energy added and energy released

But where I'm falling short is with the equation Δ = ΔF(products) − ΔF(reactants)
In the example above:

Energy released is ΔF(products)
Energy added is ΔF(reactants)
Energy added < energy released

So how does Δ end up negative? I know it should be negative, but I'm not sure why it doesn't turn out that way in this equation.
Are some of the values supposed to be negative, or is enthalpy of formation not the exact same as bond energy?
I feel as if I oversimplified / dumbed down things too much when I tried to conceptualize the material and now I'm missing some fine print / technical knowledge.

Reference for chem equation/expression formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

To address the title itself:

Why does an exothermic/endothermic reaction have a negative/positive enthalpy?

it is because of the definition of terms enthalpy, exothermic and endothermic.

Exothermic or $\Delta H \lt 0$ means released heat at constant pressure.
Endothermic or $\Delta H \gt 0$ means absorbed heat at constant pressure.

Be aware that enthalpy change reflects the reaction net heat outcome at constant pressure, where part of the net energy change may be consumed or released by volume work. There may be a reaction that is slightly exothermic ( $\Delta U \lt 0$) at constant volume, but is slightly endothermic at constant pressure ($\Delta H = \Delta U + p \Delta V \gt 0$) if there is volume expansion and work done against external pressure. But that would be rather a rare case, as usually $|\Delta U| \gg |\Delta (pV)|$.
Enthalpy of formation is NOT synonymous with bond energy.
Enthalpy of substance formation is heat released at constant pressure if ( often just formally) prepared from chemical elements at their standard state. This enthalpy is negative if formed bonds are overall stronger than bonds within elements, so energy is released.
Imagine reaction $\ce{Cl2(g) + H2(g) -> 2 HCl(g)}$.
Imagine the formation of the bond $\ce{H-Cl}$ had released the energy equal the average of the respective energies of bonds $\ce{H-H}$ and $\ce{Cl-Cl}$ and then the enthalpy of $\ce{HCl}$ formation would have been zero. That would not have meant the bond formation energy was zero too.

$$\Delta H_\mathrm{r} = \Delta H_\mathrm{f,products} − \Delta H_\mathrm{f,reactants}$$

This equation means that the enthalpy of reaction
$$\ce{"reactants" -> "products"}$$
is the difference of heat released at constant pressure between
"formation of products from standard-state elements"
and
"formation of reactants from standard-state elements"
If formation of products releases more heat than formation of reactants, it is clear the reaction is exothermic.


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Degree of Unsaturation or Index of Hydrogen Deficiency
I know the formula for the DOU, but don't really understand why the formula works. Could someone elaborate that to me?

Simple things first. Is it clear why $2n+2$ works in alkanes?

@IvanNeretin Yes.

The rest is simple. How many hydrogens do you need to remove to turn a single bond into double?

@IvanNeretin 2 hydrogens.

Well, that's it. What else is left to understand?

@IvanNeretin I figured out I was not able to accept it without the derivation, the answer on this question linked a derivation after reading which I can now fully comprehend why the formula is as it is.

The formula is unnecessary. You may convert the formula of a given compound into CxHy and calculate the difference between number of hydrogen CxH2x+2 and divide it by 2 to get DOU.

The formula for calculating the Degree of Unsaturation (DU) is awkward. The derivation will come later. There is a better way to address this issue by what I shall call the Atom Replacement Method (ARM). The gist of the method is to replace all heteroatoms with C and/or H. This revised formula is compared with the most saturated hydrocarbon bearing the same number of carbon atoms. As an example, consider the molecular formula of the marine dye Tyrian purple: $\ce{C16H8Br2N2O2}$.

Replace halogens with the same number of hydrogens and ignore divalent atoms: $\ce{C16H8Br2N2O2}$ --> $\ce{C16H10N2}$
Replace each nitrogen with one carbon and one hydrogen: $\ce{C16H10N2}$ --> $\ce{C18H12}$ (1)
The most saturated $\ce{C18}$ hydrocarbon is $\ce{C18H38}$ (2)
Subtracting 1 from 2 gives 26.
DU=26/2=13. There are 4 rings and 9 double bonds in Tyrian purple.
The following derivation of the formula is located here. Here is a synopsis.
Let c=#C, h=#H, n=#N and x=#Br, where "x" is the number of halogens.
Ignoring divalent atoms and replacing xX with xH in CcHhNnXx
gives CcHh+xNn
Replacing nN with nC and nH yields Cc+nHh+x+n (1)
The most saturated hydrocarbon is Cc+nH2(c+n)+2 (2)
Subtracting 1 from 2 and simplifying gives 2c+2n+2-h-n-x
Further simplification affords 2c+n+2-h-x
Division by 2 gives
DU = c+1+(n-h-x)/2
Q.E.D.
For a related discussions on ChemSE, see this link and this one.

Aah! Got it after reading your answer and the derivation. Thank you so much!

This method is godly! No need to bend the brains.

@Laxmi Narayan Bhandari: This method may not work when N or S is at an oxidation state greater than -3 and -2, respectively. Compare nitrobenzene with aniline.

@user55119 so basically, for simple looking compounds right? with the usual covalent bonds, that is


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Which is more acidic, alcohol or phenol?
My teacher taught yesterday that though resonance happens in phenoxide ion, it is less stable because the negative charge gets distributed over the carbons in the ring. And hence, alcohols are more acidic than phenols.
His argument seemed convincing at the time but then I searched the Internet about it and everywhere it says that phenol is more acidic than alcohol.
So which is it?

Phenol is more acidic than ethanol. Sodium phenate may be obtained in water. Sodium ethanoate is destroyed in water.

An old alternative name for phenol is carbolic acid and that is a big clue.  Your teacher has it backwards and really should not be teaching chemistry if they get such a fundamental thing wrong.

Easy @Waylander. Nobody's perfect!

I'm with @Waylander on this. Fancy hand waving explanations are no good if they fly in the face of experimental facts.

The teacher misspoke. It's true that negative charge is relatively unstable in most cases when it's distributed over carbon atoms only, like let us say an allyl anion $(\ce{CH2=CH-\overset{-}{C}H2}).$ But if a more electronegative atom, most commonly oxygen, anchors the bulk of the negative charge then the anion is stabilized; and any delocalization onto carbon atoms only adds to this stability. Phenoxide will in effect concentrate as much of the negative charge on oxygen as it needs to be reasonably stable and delocalize the rest; one oxygen atom plus some carbon atoms beats the oxygen atom alone.

Is it possible that you misunderstood? If not, your teacher is absolutely wrong. The resonance in phenoxide ions stabilizes the negative charge compared to alkoxide ions.
Phenol has a pKa of 9.95 in water, while methanol has a pKa of 15.5.[1]
Whatever explanation we might use to explain it, the simple fact is that phenols are more acidic than alcohols.
[1] https://organicchemistrydata.org/hansreich/resources/pka/pka_data/evans_pKa_table.pdf


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How can I protect the ester group from reduction?
If I process this reaction the NaBH4 will also reduced the ester group to alcohol. I want to convert the Nitro (-NO2) functional group to Amine (-NH2) without disturbing the Ester group. So, my question is that, Which protecting agent can be used here to convert Me-3-Nitrobenzoate into Me-3-Aminobenzoate. I will appreciate, if someone could help me please.

I suggest you use SnCl2.2 H2O in EtOAc or EtOH as described by Bellamy & Ou as the ester group is untouched by the reaction conditions. I have used this method many times in high yield.
Various other methods, such as hydrogenation over 5% Pd-C or Pt in EtOAc or EtOH; or refluxing in ethanol with indium with ammonium chloride or sodium dithionite will also work.

@SUBHAN The workup into ice/water/EtOAc will initially give an emulsion. The trick to gettting it to separate is to keep adding conc NaOH soln until you get two phases


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Why does atomic radius decrease across periods?
My textbook says that this happens because the number of protons, and thus the total positive charge, increases - a greater attractive force acts on each electron. The book says that this is despite the fact that the number of electrons increases, as each new electron is added to the outermost shell and so the increase in the repulsive force between electrons (electron shielding) is small. But I do not see how this justifies the increase in the force per electron despite the increase in electrons. Help!

What have you done between getting confused by the textbook and asking for clarification?  // Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure.

The positive charge is concentrated at the center of the atom whereas the negative charges are spread in a much larger volume.

I'll try my best to answer from a conceptual standpoint followed by some well known formulas to help clarify the rationale.
I think your issue stems from the faulty idea that the attractive force of the nucleus is equally divided between the electrons. With this line of thinking I can see why you might think that as the number of protons increase 1:1 with the number of electrons that the overall 'pull' of the nucleus on its outer electron shell remains constant. This is not how the attractive forces between charged particles work. The attractive force between the positively charged nucleus and any given negatively charged electron is only dependent on the magnitude of the two charges and the distance between them. This relationship is described by Coulomb's law:
$F = k\frac{q_1q_2}{r^2}$
Where:
$F = electric force$
$k$ = Coulomb constant
$q_1, q_2 = charges$
$r$ = distance of separation
It's better to think of any given electron as a test charge at a certain distance from the nucleus. Let me illustrate this with an oversimplified example that doesn't take into account shielding from core electrons. Below is a diagram of oxygen showing the charge of its nucleus and its outermost shell of electrons (valence electrons).

We know the nucleus has a positive charge of 8+ and that each electron in its valence shell has a charge of 1-. Also, since valence electrons occupy the same shell, we can assume they are all equidistant from the nucleus. So the attractive force between the nucleus and any given valence electron (test charge) can be described using the above formula:
$q_1, q_2 = 8, -1$
$F_1 = k\frac{(8)(-1)}{r^2}$
Since $k$ is a constant and $r$ doesn't change within an electron shell, I'll leave these values as constants since we'll be comparing two values.
What happens when we move right one group to Fluorine?
Well the charge of the nucleus increases from +8 to +9 and another electron is added to the valence shell. However, since the electron occupies the same shell, its distance from the nucleus is unchanged.

$q_1, q_2 = 9, -1$
$F_2 = k\frac{(9)(-1)}{r^2}$
Comparing the values for Oxygen and Florine we can see:
$F_2 > F_1$,  $k\frac{(9)(-1)}{r^2} > k\frac{(8)(-1)}{r^2}$
But what about electron shielding?
We'll both Oxygen and Florine have the same number of core electrons (2). This means that amount of shielding is the same. It follows then, that Florine has a stronger attractive force between its nucleus and valence electrons, but the same amount of core electron shielding. Flourine will exert a larger net 'pull' on its electron cloud, resulting in a smaller atomic radii.

It is oxygen and fluorine, not Oxygen and Florine.

I disagree.You cannot treat electrons as classical particles when talking about orbitals....

I'm using these oversimplified equations to express a concept the question poster was struggling with. Of course Coulomb's law can't be used to describe actual electron clouds. It can be used to try and understand general trends and to dispel some of the confusion the original poster had. I'm of the opinion that if this Chemistry 101 concept can't be explained without wave functions, its a failure on the part of the teacher...not the student.

Still it is wrong.Chemistry is applied QM.

I would then dedicate this energy to answering the question.

Below is a link to a text book explanation of how Coulomb's law is used to illustrate periodic trends.
https://chem.libretexts.org/Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Readings/Week_1%3A_Analysis_of_Periodic_Trends/1.1%3A_Concepts_and_principles_that_explain_periodic_trends/1.1.1%3A_Coulomb's_Law

Despite believing my answer isnt correct , it is the leading method in calculating the wavefunction of orbitals so I will go mainstream for a while...If you compare the effective nuclear charge of the valence electrons from Wikipedia you will see that the effective charge is increasing as we go down in the periodic table.
This means that if we solve the time independent Schrodinger  equation the peak of the wavefunction $\Psi(x)$ of valence electrons will be closer to $x=0$ (the nucleus) as we go down in the periodic table which means that the average distance between the nucleus and the valence electrons will be smaller.


text:
If I know the freezing point of two solutions, and combine them into one solution, what is the resulting freezing point?
I am not knowledgeable in chemistry in any way, so I'm sorry if I seem stupid or something.
Suppose I have two solutions which I know the freezing point of, and I combine them into one solution. How do I find the resulting freezing point?
For example, let's say I have 4 parts water and 1 part maple syrup. Freezing point of water is $0 \ ^oC$, and freezing point of maple syrup is about $ -13\ ^oC$ according to a page I found from a google search.
Intuition tells me it would just be a weighted average of all the things you're combining, assuming that there is no chemical reaction when you mix them. So for my water plus maple example, it would be
$$
\left( \frac{4}{(4 + 1)}  * 0 \right)+ \left( \frac{1}{(4 + 1)} * -13 \right) = \frac{-13}{5} = -2.6 \ ^oC
$$.
Is this intuition correct or is it more complex than that?

It is more complex.  For volume ratios, it would have to be corrected by densities. Even then, freezing point as function of composition is not linear. So, it can be used only as rough estimation that would differ from reality.

Okay suppose I measure the parts by mass instead of volume. If the function of freezing point isn't linear, what is it?

Nonlinear, empirical function. It is approximately linear for very diluted solutions, what is not case of syrups.

The dream of many physical chemists would be to describe the properties of a mixture by the properties of the individual components. Unfortunately, thermodynamics of solutions is much more complex and you can't use simple rules like that. However, if you have some properties and data, we could see that the solid/liquid equilibrium tells us and get a rough estimation.

You need to clarify whether your question is about maple syrup specifically or mixtures generally.

In many cases, the freezing point of a mixture is lower than the freezing point of each component.
For example, table salt, $\ce{NaCl}$, melts at ~800°C. Water, $\ce{H2O}$, melts at 0°C. A mixture of salt and water can have a melting point as low as -20°C!. This is called freezing point depression.
Another example is the melting point of mixed metals, i.e., an alloy. Tin, Sn and lead, Pb, are used to make classic solder. The melting point of Sn is ~230°C, and of Pb, ~330°C, but a 60% Sn/40% Pb alloy melts about 180°C.
It is difficult to make a general statement about the melting point of mixtures. In some case, there might not be a freezing point depression, but in most, the freezing point does not vary linearly with concentration.
Don't despair though -- there are properties of sugar/water solutions that do vary comparatively linearly with concentration, such as the density and index of refraction. Those making alcohol from sugary solutions may use refractometers to check the amount of sugar. Cheers! Salud! L'chaim! Skol!

Dr Moishe Pippik states about melting points of mixtures. It has nothing to do with Logan Schlick's question.

@Maurice, it certainly does have to do with the question: interpolation of freezing point data: "Suppose I have two solutions which I know the freezing point of, and I combine them into one solution. How do I find the resulting freezing point?" If equal amounts of solutions of 20% ethylene glycol/80% H2O and 80% ethylene glycol/20 % water are mixed, the f.p. is not halfway between. https://www.chegg.com/homework-help/questions-and-answers/738-figure-732-shows-phase-diagram-h2o-ethylene-glycol-explain-mixture-approximately-50-50-q37847872


text:
Is CD4 an alkane?
Can $\ce{CD4}$ or $\ce{CT4}$ be considered an alkane?
I know that an alkane is $\ce{C_nH_p}$ where $p = 2n + 2$. An alkane is considered a "hydro"carbon, i.e., a compound containing purely carbon and "hydrogen" only. However, "hydrogen" has mainly three isotopes:
protium ($\ce{^1_1H}$ or $\ce{H}$), deuterium ($\ce{^2_1H}$ or $\ce{D}$) and tritium ($\ce{^3_1H}$ or $\ce{T}$).
Thus, in general, can $\ce{C_nH_xD_yT_z}$ be considered an alkane/ene/yne, where $x, y, z \in \{ 0, 1,\dots,2n + 2 / 2n /2n-2 \}$ and $x + y + z = 2n + 2 / 2n /2n-2$?

What is the meaning of $2/2n/2n$ ?

@Maurice Correlate it with alkane/ene/yne : 2n+2/2n/2n-2.

You are asking a question about CD4 before introducing D. This is bad.

Yes, they are still alkanes and they are named deuterated/tritiated alkanes. It's correct that alkanes are primarily considered hydrocarbons but if the hydrogen is replaced with deuterium/tritium, the names are also changed and this replacement is not a theoretical one. Practically this is possible (ACS Catal. 2018, 8, 3, 2296–2312).

Can we call them D/T-substituted alkane?

@CHEMUMAN  Is $\ce{H2^{18}O}$ water? Yes, it is. $\ce{CD4}$ is nothing else but $\ce{C^{2}H4}$. Do not let convention $\ce{^{2}H}$ being written as $\ce{D}$ confuse you. H in CH4 means hydrogen, not protium.

@Poutnik What do you mean is that $\ce{CH_4}$ is actually a mixture of various compounds of isotopes of carbon & isotopes of hydrogen(e.g.:  $\ce{^14_6C _1^3H_4}$ is also methane), isn't it?

@CHEMUMAN Yes, I do. There is nowhere said methane is a compound of carbon-12 and hydrogen-1.  Molecules differing just by isotopic composition are called Isotopologues // Hydrogen is exception just formally, having dedicated symbols for its isotopes, as they have relatively biggest differences due the biggest relative mass differences.  So H, D, T as hydrogen-1,  hydrogen-2 and hydrogen-3 formally look like 3 different elements, while being just one.

@Poutnik The link provided by you really helped. Thanks.


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Theoretical explanation for: Why is KMnO4 stronger oxidizing agent than K2Cr2O7?
I know that it can be explained using the Reduction Potential values. But, can it be explained from their structures or something like that ?

Redox potentials don't explain anything. By this point main effect of knowing std. potentials, or even knowing they exist, is mostly giving you false impression that there is some objective and universal scale to compare them.

Maybe it is due to the chromium being at oxidation number +$6$ in $\ce{K2Cr2O7}$ and manganese at a higher oxidation number (+$7$ in $\ce{KMnO4}$)


text:
Can someone elaborate what it means when its said that electron shells do not exist in multi electron systems?
I'm an undergraduate student in chemistry. I'm trying to grapple with all the new stuff we're learning and making sense of it. Now I want to know if electron shells really "exist" in multi electron atoms and if they don't,  how / why do they still work to explain certain phenomena (atleast right now in our undergraduate curriculum)
On the chance of sounding a little stupid , I'm feeling a little lost with the very basics of atomic structures.

The answer may depend on what exactly you mean by the "electron shell" and "exists". As it is possible to have simultaneously true opposite statements about the same terms, if each statement means under the terms something different.

The nice hydrogen electron orbitals do not remain the same as more than one electron is attached to an atom - the three-body problem is not a nice solution, much less the 100-body problem. Nonetheless, with many caveats, the hydrogen-like orbitals are a reasonable first basis for looking at things. Just mind all the caveats, particularly in the d- and f-blocks...

That makes sense thanks

And consider a "shell" as a fuzzy probability cloud, that changes shape at higher energy levels. Look at this visualization of a carbon atom: https://www.labxchange.org/library/pathway/lx-pathway:21e1bc06-6002-4b02-addb-759cf78bc15b/items/lx-pb:21e1bc06-6002-4b02-addb-759cf78bc15b:html:cb8754f4


text:
Why does gallium expand on solidification
I've heard that Ga exist as Ga2 molecules in its crystal lattice. Does that have anything to do with expansion of Ga while solidifying? Why and how does it actually expand?

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text:
Why is Carbon marked as 1 acting like a pseudo chiral centre?

Since there is a plane of symmetry passing through the middle of the compound, the C (chiral centres) at 2 and 3 will rotate the light through angles opposite in magnitude, say, +A and -A respectively (since they have 4 same valencies). Hence 1 is always acting as a chiral centre in this compound, then why do we classify it as pseudo chiral and not purely chiral?

There is not always a plane of symmetry which you mentioned. Also, precisely when there is a plane, the carbon 1 can't be considered chiral, since the entire molecule isn't.

The methyl hydrindane you have drawn lacks stereochemistry and the question you ask cannot be answered as such. There are four possible diastereomers (red arrows) of the hydrindane: 1, 2, 3a(4a) and 3b(4b). Diastereomers 1 and 2 both have a horizontal plane of symmetry, a cis ring juncture and they are meso compounds. The ring fusion carbons are stereogenic and chirotopic (R/S) while the carbon bearing the methyl group is stereogenic and achirotopic (r/s).
Hydrindanes 3a(4a) and 3b(4b) have a trans ring juncture and they are enantiomers of each other. The carbon bearing the methyl group is both non-stereogenic and achirotopic. Hence, the lack of a stereochemical descriptor.
For related discussions on this topic on ChemSE, go here, here, here or here.

We do not have a plane of symmetry in a typical conformation. If we assume a typical chair conformation for the cyclohexane ring, then the hydrogen atoms attached to the brodgehead carbons are oppositely positioned above or below the plane of the paper, which then causes both bridgehead carbons to have R orientation or both S. Either the R, R or S, S configurations or the locations of the hydrogen atoms above or below the plane of the paper is then not invariant upon reflection through the proposed (but demonstrated false) mirror plane.


text:
How is the reason for “why neutralizing weack acids with strong bases gives less heat” that the heat is used in ionizing the acid?
In my lab manual this is written but I can’t see how this is true

The heat of neutralization of weak acids by strong bases is less than 13.7 Cal/mole because the weak acid becomes completely ionized before the neutralization reaction by absorption of heat.

If the weak acid can absorb heat before the neutralization reaction to completely ionize, then why do we say that it doesn’t fully ionize in the first  place? Before reading that text I thought that the reason was that less than 1 mole of H+ ions would be in the solution so less water will form therefore less heat will be released from the neutralization reaction, and that the difference in the heat released is the amount of heat that is needed to fully ionize the remaining weak acid. Where have I gone wrong?

Could you be more specific about the reaction you are referring to?  Remember solvent effects on both acid and base.  Sometimes dissociation into ions can be energy absorbing.  "Complete ionization" may be local (not all at once).

Note that effort to insert in the question title as much info as possible is contra-productive.

@RobertDiGiovanni the weak acid neutralization reaction is that of acetic acid and sodium hydroxide

It would appear that dissociation on acetic acid in water to acetate and H3O+ is "endothermic", or heat absorbing.

Interestingly, even dissolving common NaCl is slightly endothermic.  Solid  NaOH is quite the opposite.

Neutralization of strong acids and strong bases is the exothermic reaction
$$\ce{H+(aq) + OH-(aq) -> H2O(l) \label{R1} \tag{R1}}$$
Neutralization of weaks acids and strong bases is formally the reaction chain of endothermic reaction
$$\ce{HA(aq) -> H+(aq) + A-(aq) \label{R2} \tag{R2}}$$
followed by the reaction $\eqref{R1}$
The dominant reaction is more probably the faster direct reaction
$$\ce{HA(aq) + OH-(aq) -> H2O(l) + A-(aq) \tag{R3}}$$
but with the same energy outcome as the chain $\eqref{R2}$+$\eqref{R1}$, due the Hess law that is consequence of the fundamental law of energy conservation.
The net reaction enthalpy is $\Delta H_{\mathrm{R3}} =  \Delta H_{\mathrm{R2}}+\Delta H_{\mathrm{R1}} > \Delta H_{\mathrm{R2}}$
as some heat is absorbed by acid dissociation what diminishes the net heat release.


text:
Virial theorem confusion
Virial theorem says that $E_{t} = -\frac{E_{P}}{2}$ however this says that lets say for 2 gas molecules(Ne) the kinetic energy due to temperature a molecule of $Ne$ at 0K will have the same energy with a $Ne$ molecule at 300K which obviously isnt correct.So what is going on here?

The theorem implies the average values and does not say what you say it says. It is your false implication.

It's extremely poorly worded but what I think is being asked (but am not sure) is "Given there are no interparticle interactions in an ideal gas such as Neon doesn't the virial theorem predict the kinetic energy to be zero at all temperatures?" Which is actually a slightly subtle point. If that is the case and the poster clarify I'll provide an answer.

@IanBush yes thats what I am asking.

Kinetic and potential energy of electrons, on which the VT applies,  is non-zero at 0 K.  Non-zero is also kinetic and potential energy of Ne atoms in solid Ne matrix at 0 K (zero point quantum oscillations)

@Volpina Please make your question more clear, taking into account the comments. Also, check English, formatting etc. See the help pages for more information.

Also, why are you considering 2 atoms? And why do you write molecules of Ne instead of atoms (technically correct to say molecules but more accurate to write atom)? For formatting; use $ \ce{Ne} $ to show $\ce{Ne}$ (format a chemical formula). Use $ \pu{0 K} $ to show $ \pu{0 K} $ (format a quantity).

As noted in the comments I am taking the question to mean "Given there are no interparticle interactions in an ideal gas doesn't the Virial theorem predict the kinetic energy to be zero at all temperatures?" Hopefully the original poster will edit the question to improve the wording so I can delete this clarification.
The short answer is "because of the externally applied pressure". The longer answer below follows here
Firstly you have incorrectly stated the Virial theorem. As noted in the comments it applies to averages of the energies, but over and above that the form you have is for a very specific particle interaction, an inverse square law such as Coulomb's law or gravity. This is discussed in the link above and is clearly not appropriate for an idea gas! The general form of the Virial Theorem is
$$
\langle T\rangle = - {1 \over 2} \left\langle \sum_i {\bf F_i}\cdotp{\bf r_i} \right\rangle
$$
where $T$ is the total kinetic energy of the particles, ${\bf F_i}$ is the force on particle i, ${\bf r_i}$ is the position of particle i and the angled brackets indicate averages (note to poster: you really should define any symbols you use)
Now we know the average kinetic energy from kinetic theory, or equipartition of energy, or whatever is your favourite method:
$$
\langle T \rangle = { 3 \over 2 } NkT
$$
where $N$ is the number of particles, $T$ is the temperature and $k$ is Boltzmann's constant.
What about the RHS of the Virial Thorem? Well naively in the ideal gas model there are no interparticle interactions, so all the forces are zero, so the RHS is zero, so how can the Virial theorem be obeyed?
What is being missed is while all the interparticle forces are zero, these are not the only forces on the particles; the particles also interact with what in Chemistry is usually called the "wall" containing the gas, so establishing the pressure. And this interaction with the wall will causes forces on the particles, so the RHS above is not zero due to the confinement of the gas.
To procede it is better to think not of a real container in which the gas is being held, but an external pressure that confines the gas; at equilibrium the internal pressure of the gas will equal the external confining pressure. Thus as the external pressure causes a force per unit area, and as this pressure will equal the internal pressure caused by the particles we can write
$$
d{\bf F_i}=-{\bf \hat{n}}PdA
$$
where $P$ is the (external) pressure, $dA$ is an element of area, and ${\bf \hat{n}}$ is a unit vector normal to the containing surface. Hence, using the divergence theorem
$$
- {1 \over 2} \left\langle \sum_i {\bf F_i}\cdotp{\bf r_i} \right\rangle =
{P \over 2} \int {\bf \hat{n}} \cdotp r_i dA = {P \over 2}\int {\bf \nabla } \cdotp {\bf r_i} dV = {{3}\over{2}}PV
$$
Hence using equating the two sides in the Virial equation and cancelling the ${{3}\over{2}}$ constants one gets
$$
NkT=PV
$$
which is simply the ideal gas law, as expected.

The Virial theorem can be derived only when there is an interaction between each and every pair of 'particles' in a gas, i.e. between atoms and/or molecules. The particles could even be dust or stars. In the case of an ideal gas the only interaction is that due to gravity, in real molecules dispersion forces in the general sense will also exist.
With an inverse square force (such as gravity) the Virial Theorem has the form $2E_K+V=0$ where $E_K$ is the kinetic energy and $V$ the total potential energy. The potential terms are sometimes called the 'Virial of Claussius'.
If your particles are in space and there is a cloud then the Virial theorem will apply and can be used to determine if such vast clouds are stable or not.
In a gas in the lab the interaction has three parts, the forces on collision, those due to intermolecular forces and also due to collisions with  walls containing the gas. Collisions with the wall introduces the pressure and volume and so the total kinetic energy $\sum mv^2/2$ becomes  equal to $3pV/2$ plus any intermolecular interactions. As the average kinetic energy is also $3/2k_BT$, where $k_B$ is Boltzmann's constant and $T$ the temperature then $nk_BT=pV$ + intermoleculear energy and the connection to an ideal gas is clear.   The kinetic theory of gasses thus shows that the pressure comes from the kinetic energy of motion of the particles plus a small contribution from the intermolecular forces.


text:
Where can I find IUPAC's official definition of "green chemistry"?
I've been looking all around, including in IUPAC's color books, but could not find a definition of "green chemistry". Additionally, Wikipedia's article on green chemistry features a sidebar with a "IUPAC definition", although the reference (points to an Annual Review article) doesn't mention anywhere in it that the definition is from IUPAC.
Do note that I'm looking for a free online resource. I don't believe I'm the only one to think that such a definition should be open, accessible and clear.

Related: https://www.acs.org/greenchemistry/principles/12-principles-of-green-chemistry.html and this https://www.epa.gov/greenchemistry/basics-green-chemistry#definition

In Ref.1, the term green chemistry is defined as:

The invention, design, and application of chemical products and processes to reduce or to eliminate the use and generation of hazardous substances.

This definition of Green Chemistry was adopted by IUPAC in 2000 according to this article. It also defines closely related sustainable chemistry:

Sustainable chemistry is the design, manufacture, and use of environmentally benign chemical products and processes that prevent pollution, reduce or eliminate the use and generation of hazardous waste, and reduce risk to human health and the environment.

References:

Pietro Tundo, Paul Anastas, David StC. Black, Joseph Breen, Terrence J. Collins, Sofia Memoli, Junshi Miyamoto, Martyn Polyakoff, and William Tumas, "Synthetic pathways and processes in green chemistry. Introductory overview," Pure Appl. Chem. 2000, 72(7), 1207–1228 (DOI: https://doi.org/10.1351/pac200072071207).


text:
Preparation of alkane through decarboxylation by soda lime
Can we still prepare alkane through decarboxylation (using soda lime) if instead of using sodium    salt of carboxylic acid, we use the carboxylic acid directly.
Because I have always seen the equation being written like:
CH3COONa + NaOH --(CaO)(on heating)-> CH3COOH + Na2CO3
What if instead we just used: CH3COOH + NaOH --(CaO)(on heating)-> ???
What will we get then?

Yes you can  https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/The_Decarboxylation_of_Carboxylic_Acids_and_Their_Salts

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The decarboxylation equation is :  $$\ce{CH3COONa + NaOH -> CH4 + Na2CO3}$$ If you do want it, you can first synthesize $\ce{CH3COONa}$ from $\ce{CH3COOH + NaOH}$ just before heating it with an excess of $\ce{NaOH}$ so that the final and total reaction looks like
$$\ce{CH3COOH + 2 NaOH ->  CH4 + Na2CO3 + H2O}$$
The use of chalk $\ce{CaO}$ is not fondamental here. But if pure $\ce{NaOH}$ is used in the reaction before mixing it with acetate, $\ce{NaOH}$ will melt, flow and attack the glass of the tube used for the reaction, which will probably be destroyed by a reaction like $$\ce{2NaOH + SiO2 -> Na2SiO3 + H2O}$$ To prevent the contact between $\ce{NaOH}$ and the glass wall, a mixture $\ce{NaOH + CaO}$ is used. In this case, $\ce{NaOH}$ will not flow out of the chalk matrix. It can react with sodium acetate without touching the glass walls.


text:
Auxochromic action of an alkyl group attached to a double bond
In the UV-Vis spectra,
$\ce{CH2=CH2}$ has max absorbance at $\pu{171 nm}$.
$\ce{CH3-CH2-CH=CH-CH2-CH3}$ has max absorbance at $\pu{184 nm}$.
The alkyl groups decrease the π-π* gap by raising the π but not considerably the π*. The effect which I suspect is hyperconjugation, but when it happens between the σ-π orbitals, it should occur with the π* orbital; if anything, raising it even more? How does the gap decrease then?

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Thank you, that helps!


text:
Why do the successive ionisation energies of Ni and Pt diverge?
The data I am referring to is below on the graph. The explanation I have found is that the 5d orbital of Pt is more diffuse than the 3d orbital of Ni, so is less sensitive to the increases in Zeff as we remove successive electrons. This seems somehow incomplete though, so I was wondering if anyone has any further insight?


text:
Trend in formation enthalpy of metal fluorides
Please see below for the data I am referring to. I don't really understand why the trend in Cr ceases and reverses.
I think I understand what the logic for WFn is - while the cost of forming Mn+ increases with each successive ionisation, this is outweighed by the increasing lattice enthalpy.
However, I don't get why the same doesn't hold for Cr. I was wondering if it has to do with d orbitals on W being more diffuse and thus less sensitive to increasing Zeff as we remove electrons, which would mean that the successive ionisation energies would diverge, similar to the second image. But then this doesn't explain the irregularities in the Cr trend.
I was wondering if anyone might have some insight into this?

I know this second image is for Ni and Pt, but couldn't find any other.

P.S. I have also included the atomisation enthalpy trends, but am not sure how useful they will be since the difference between Cr and W seems insignificant.


text:
Deduce the shapes of stretching modes of a tetrahedral molecule
This question is in relation to the [FeCl4]- molecule, which is tetrahedral. I previously deduced, using symmetry analysis, that the vectors corresponding to the bonds of the molecule will be A1+T2, which, according to the table below and my scarce knowledge of spectroscopy, should mean that we will observe 1 IR band and 2 Raman bands. More specifically, we should see 3 IR stretching modes and 4 Raman stretching modes where 3 of the Raman ones will coincide with 3 IR ones and 1 will be Raman-unique (see character table at the bottom of the question).
This is a question from an assignment, which confirms my prediction with the data below and asks to sketch the stretching modes associated with each band. This is where I am stuck - I am yet to find a proper procedure to do this as most textbooks simply give the shapes of the modes with no explanation. I know that an IR stretch should involve a change in dipole moment, which rules out the 'breathing' mode, but this still leaves me guessing as to which ones will be correct as there are only 3 IR modes, but there are surely more ways than that to change the dipole moment?

you are on the right track. You can see that the sym stretch ($A_1$) is only seen in Raman, but its hard to see what $T_2$ will look like. You need to use projection operators. There is an example in this online book https://applying-maths-book.com/intro.html.   see  the section in  chapter-7, matrices part 6   normal-mode-vibrations   for a worked example.


text:
2Pb+O2+2H2SO4=2PbSO4+2H2O how many electrons are transferred in this process?
How many electrons are transferred in the process of this REDOX reaction? What would be all the oxidation numbers assigned to all the species in this equation?

Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure.
//  Note that questions, looking like copied  homework/test/exam/lab tasks without significant explicit resolving attempts, are explicitly off-topic on this site and will be closed.

Why won't you assign the said oxidation numbers and see what they are?

Isn't that similar to a lead/acid battery?  Read on.

The reaction is cited as follows:
$\ce{2Pb(s) + O2(g) + 2H2SO4 (aq) -> 2 PbSO4(s) + 2H2O}$
$\ce{2 Pb -> 2 Pb++ + 4e-}$
$\ce{4 H+ + 4e- + O2 -> 2H2O}$
SO4 does not react.  4 electrons are transferred.
Oxidation number of S in SO4 is +6, the 4 Os are -2 each. H is +1, Lead Pb goes from 0 to +2.  Oxygen in O2 goes from 0 to -2.
Further checking has yielded this possibility:
$\ce{Pb(s) + PbO2(s) + 2H2SO4 (aq) -> 2PbSO4 (s) + 2 H2O}$
... what powers a lead/acid battery cell, + 2.05 V.
The charges on PbO2 are +4 for lead and -2 for the oxygens.  The number of electrons transferred is the same, 4.

Interestingly, one could say only 2 electrons are transferred from Pb 0 to Pb+4 -> 2 Pb+2, because oxygen has already "taken" 4 electrons from Pb.


text:
Why I am unable to detect nystose & kestose using hplc and refractive index detector while could see peaks of glucose,sucrose,fructose, and raffinose
Mobile phase is 60% acetonitrile and temperature of the column is 25 degrees (supercosil amino from sigma). RID is also set up at 25.

For readers, see www.researchgate.net  - Structures-of-inulin-type-fructan-1-kestose-A-nystoseB-and-inulinC

You might also consider trying a resin that is more suitable for oligosaccharides. Shodex, for example, sells a number of different saccharides analysis columns. Here is a comparison of their fructooligosaccharide column to an amino column showing the peaks for nystose and kestose: https://www.shodex.com/en/dc/03/03/46.html

Thank you for the suggestion. The same column and the mobile phase concentration was used by one of our scientists in lab with the same column. They quantified the same sugars.So I thought of moving forward with the same. It's bit strange that they are not working for me.

I haven't tried 95% water. That's a good idea. Although I used 75% acn and the results were same.

@Andrew, Gradient cannot be used with RI detector.

Please include additional information into your post.

@AChem - I do have personal experience running RI detectors for carbohydrate analysis for about ten years in my lab, but I only have changed the solvent composition once or twice out of desperation, and looking back, it's likely that we used the inline UV detector in those cases rather than the RI. I'll remove that comment. But I still believe that OP's likely problem is that the oligosaccharides are being retained on the column for longer than the run under the given conditions.

Thankyou for your comments. Is there any chance that the brand new column could be faulty? People in my lab used to do sugar analysis like in 2016. So, I got a new column for myself. Since then I have been running  exactly the same protocol. The old protocol run time was 12 min and I have been running for 25 minutes and still no luck. So, just assuming maybe the manufacturers have made some change in their resin.

I will definitely try more water in my mobile phase composition and increase my run time specifically for those oligosaccharides.

Bonding alkyl amino groups on silica is just like an auto-immune disease. The basic amine groups help to dissolve very silica substrate in aqueous mobile phases. Amino-silica columns are typically not very stable (hydrolytically). This is why many amino-phases have a polymeric substrate. When an HPLC column misbehaves, it is time run the quality control (QC) test sheet that comes with the column. Make sure the column is within its specs. Since one cannot run a gradient with a refractive index detector, it is possible to use relatively deep UV on an ordinary HPLC UV detector and a gradient method, with wavelengths around 210 nm etc. Many older methods that analysed sugars used this trick. The UV cut off of ACN is 190 nm. It is likely your sugar has a high retention time under your HPLC methods. In the so-called hydrophilic interaction liquid chromatography (HILIC) mode, analytes follow a U-shaped curve with respect to retention factor and ACN percentage.


text:
How do I find entropy change for surroundings?
I know how to calculate entropy change of the system but am not able to calculate entropy change for surroundings. The question is as follows:
Q: At $100°C$, water vapour at $1$ bar is in equilibrium with the liquid.
Given: $\ce{H2O}\mathrm{(l,1\,bar, 373\,K) } \to \ce{H2O}\mathrm{ (g,1\,bar,373\, K) }$ ;
$\Delta_{r} H$ $=$ $40.639 kJmol^{-1}$, $C_{p}(g) = 30.305 JK^{-1}mol^{-1}$, $C_{p}(l) = 75.31 JK^{-1}mol^{-1}$. Compute $\Delta_{r}S$ for the reaction: $H_{2}O{(l,1 bar, 273K)}$ $\rightarrow$ $H_{2}O(g,1 bar,273K)$ and also $\Delta S_{surr}$ and $\Delta S_{total}$.
I was able to calculate the entropy change of system. It came out to be,  $\Delta S_{sys} = 123JK^{-1}mol^{-1}$. But I am stuck at change in entropy of surroundings.
I calculated the change in entropy of system by following this sequence:
A) $H_{2}O{(l,1 bar, 273K)}$ $\rightarrow$$H_{2}O{(l,1 bar, 373K)}$
B) $H_{2}O{(l,1 bar, 373K)}$ $\rightarrow$ $H_{2}O{(g,1 bar, 373K)}$
C) $H_{2}O{(g,1 bar, 373K)}$ $\rightarrow$ $H_{2}O{(g,1 bar, 273K)}$
Any guidance is appreciated. Thanks!

Reference for chem equation/expression formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Have you tried Delta S_surr = integral ( -dQ_rev,sys / T) ? That for 3 formal stages:  heating up, evaporation, cooling down.

@Poutnik,  I tried this thing $\Delta H $ $=$  $\int_{T_{1}}^{T_{2}} C_{p,m}\ dT$  for three stages and then I divided the three $\Delta H$ by temperature 273K, 373K and 373K for three stages respectively. Is this what you were trying to say? My answer came out different than what was given in the textbook though.

You have to keep 1/T in the integrals, as T changes in 2 of 3 stages. dS=dQ/T --> Delta S=Delta Q/T is valid only at constant T. (written on phone, no fancy formatting now)

Then what should I use for dQ? dQ= the values of enthalpy that were calculated before?

Are you aware that water vapor at 1 bar and 273 K is not an equilibrium state for water?

Isn't it 373K where the water vapour and liquid water are in equilibrium

Are you saying that water vapor at one bar and 273K is a stable speed?

I'm not saying anything. It is written like that in the book. And what speed?

I meant unstable state.

At 1 bar, 373K water vapour and liquid water are in equilibrium

Water is not a gas at 273K and 1 bar.

So my C) step is wrong?

As Chet Miller pointed in the comments, water is not in VLE at $P = 1 \; \pu{bar}$ and $T = 273 \; \pu{K}$. Thereby the entropy calculation you made needs some correction. Lets do that first and then we answer your question. Let me also name: $P_1 = \pu{unknown saturation pressure}$, $P_2 = \pu{1 bar}$, $T_1 = \pu{273 K}$, and $T_2 = \pu{373 K}$.
Calculation of Entropy

We estimate the saturation pressure at $T_1$ by the Clausius-Clapeyron equation
\begin{align}
  P_1 &= P_2 \exp\bigg[-\dfrac{\Delta_\pu{vap} H}{R}\bigg(\dfrac{1}{T_1} - \dfrac{1}{T_2}\bigg)\bigg] \\
  P_1 &= (\pu{1 bar}) \exp\bigg[-\dfrac{\pu{40639 J/mol}}{\pu{8.314 J/mol K}}\bigg(\dfrac{1}{\pu{273 K}} - \dfrac{1}{\pu{373 K}}\bigg)\bigg] \\
  P_1 &= \pu{8.230E-3 bar} = \pu{0.823 kPa}
\end{align}
Water tables inform that the exact value is $\pu{0.611 kPa}$ ($\pu{35
\%}$ relative error), but lets only work with the data given.

I will add two steps, A and E, to your thermodynamic path and take note of the entropies that you have calculated:
\begin{align}
  &A) \ce{H2O}(P_1, T_1, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_1, \pu{liquid}) \\
  &B) \ce{H2O}(P_2, T_1, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_2, \pu{liquid})
      \hspace{1 cm} \Delta S_B = \pu{23.50 J/mol K} \\
  &C) \ce{H2O}(P_2, T_2, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_2, \pu{gas})
      \hspace{1 cm} \Delta S_C = \pu{108.95 J/mol K} \\
  &D) \ce{H2O}(P_2, T_2, \pu{gas}) \rightarrow \ce{H2O}(P_2, T_1, \pu{gas})
      \hspace{1 cm} \Delta S_D = \pu{-9.46 J/mol K} \\
  &E) \ce{H2O}(P_2, T_1, \pu{gas}) \rightarrow \ce{H2O}(P_1, T_1, \pu{gas}) \\
\end{align}
The three entropies calculated sum to that value you posted. We need two additional ones. For a constant temperature process, the entropy change is
\begin{align}
  \Delta S = \int_{P_1}^{P_2}-\bigg(\dfrac{\partial V}{\partial T}\bigg)_P \; dP
\end{align}
With the data given, we cannot calculate the integrand for the liquid state (although we could try to use an EOS, if you want, tell me in the comments), so I will assume $\Delta S_A = 0$. For the gas state, application of the ideal gas law yields
\begin{align}
  \Delta S_E = -R\ln\bigg(\dfrac{P_1}{P_2}\bigg) \rightarrow \Delta S_E = 39.91 \; \pu{J/mol K}
\end{align}

Adding this to the sum yields
$$ \boxed{\Delta_\pu{vap} S = 162.91 \; \pu{J/mol K}} $$
Water tables state that $ \Delta_\pu{vap} S = 164.98 \; \pu{J/mol K} $. Quite a good guess IMO considering the simplifications.

Calculation of heat to the surroundings
Lets calculate all the heat exchanged with the surroundings. Steps B-D are constant pressure processes, and we know that $ Q = \Delta H $, so they are immediate
\begin{align}
  Q_B = \Delta H_B = c_\pu{p,l} (T_2 - T_1) &\rightarrow Q_B = 7531 \; \pu{J/mol} \\
  Q_C = \Delta H_C = \Delta_\pu{vap} H &\rightarrow Q_C = 40639 \; \pu{J/mol} \\
  Q_D = \Delta H_D = c_\pu{p,g} (T_1 - T_2) &\rightarrow Q_D = -3030.5 \; \pu{J/mol}
\end{align}
For steps A and E, for a constant temperature process, we know that $ Q = T \Delta S $
\begin{align}
  Q_A &= T_1 \Delta S_A \rightarrow Q_A = 0 \\
  Q_E &= T_2 \Delta S_E \rightarrow Q_E = 14886.08 \; \pu{J/mol}
\end{align}
And summing all the heats we have $Q = 60025.58 \; \pu{J/mol}$. This value makes sense, and is the heat that the surroundings must give to the system to make possible those 5 steps.
To be fair, this is as far as we can get; for the surroundings we need to make some assumptions. The simplest one is to consider it as a heat reservoir, i.e., a system that can exchange infinite amounts of energy without changing its temperature. However, there are three restrictions:

The temperature of the reservoir must be at least of $\pu{273 K}$ for step A, or we would violate the 2nd law.
Likewise, the temperature of the reservoir must be at least $\pu{373 K}$ for step C, or we would violate the 2nd law.
Point 2 applies equally to step E.

The best case scenario we can postulate, in terms of reversibility, is that of a surroundings at this last temperature, and thus
\begin{align}
 \Delta S_\pu{sys} + \Delta S_\pu{surr} \geq 0 \\
 \Delta S_\pu{sys} + \dfrac{Q_\pu{surr}}{T_\pu{surr}} \geq 0 \\
 162.91 \; \dfrac{\pu{J}}{\pu{mol K}} - \dfrac{60025.85 \; \pu{J/mol}}{373 \; \pu{K}} \geq 0 \\
 1.981 \dfrac{\pu{J}}{\pu{mol K}} \geq 0
\end{align}
So one possible answer is
$$ \boxed{\Delta S_\pu{surr} = -160.93 \; \dfrac{\pu{J}}{\pu{mol K}}} $$
Comments:

I don't like the symbol $\Delta_r H $ suggested by your textbook. That it not a chemical reaction, it is a phase transition. So I would write $\Delta_\pu{vap} H $ instead.
Reversibility can be attained by refining the heat exchanges for the heating (step B) and the cooling (step D). For example, heating some amount with a reservoir of $\pu{50 K}$ and then the other amount with one of $\pu{100 K}$.

For readers, VLE stands for Vapor Liquid Equilibrium

Thank you so much for explaining. I was stuck at it for hours.. Thanks again


text:
Is the hydrolysis of a salt produced by neutralisation a secondary reaction?
I understand that in a solution, the salt created by a neutralisation reaction can react again with water molecules, which can further affect the pH. Is it correct to say that the hydrolysis of salts is a secondary reaction after the neutralisation reaction?
Or, in the case where there is a strong base and weak acid (which are stoichiometrically balanced), it's just because the strong base fully dissociates and the weak acid only partially dissociates, so in the end there's just more $\ce{H+}$ than $\ce{OH-}$?

Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure.

I have read it before but I don't understand how else I can improve my question because I did research as much as I could before asking, and it's a general case so should I have provided an example or something? I would appreciate some specific tips to improve my questions to make it easier for people to answer in the future.

Research not mentioned in its summary cannot be distinguished from no research. Readers cannot know what you have reviewed enough but still do not understand, or   if you skipped that part and just asked. And some users have just habit to downvote questions without clear reasons  // Do not stay at questions, provide your reasoning for your current understanding or opinion.

Reactions of neutralization
\begin{align}
\ce{HA(aq) + OH-(aq) &-> A-(aq) + H2O} \tag{R1}\\
\ce{A-(aq) + H+(aq) &-> HA(aq)}\tag{R2}\\
\ce{H+(aq) + OH-(aq) &-> H2O} \tag{R3}
\end{align}
dissociation
\begin{align}
\ce{HA(aq) &-> A-(aq) + H+(aq)} \tag{R4}\\
\ce{H2O &-> H+(aq) + OH-(aq)} \tag{R5}
\end{align}
and hydrolysis
$$\ce{A-(aq) + H2O -> HA(aq) + OH-(aq)} \tag{R6}$$
always occur simultaneously.

Reactions can be rewritten as the half number of equilibrium reactions instead:
\begin{align}
\ce{A-(aq) + H2O &<=> HA(aq) + OH-(aq)} &K_\mathrm{b} \tag{R7} \\
\ce{ HA(aq)  &<=> A-(aq) + H+(aq)} &K_\mathrm{a} \tag{R8}\\
\ce{H2O &<=> H+(aq) + OH-(aq)} &K_\mathrm{w} \tag{R9}
\end{align}
bound by the equation of the equilibrium constant dependency
$$K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$$
Which reaction direction dominates depends on concentration of involved species and respective dissociation constants. As for all acid-base reactions, equilibrium is established very fast by approaching it from either left either right side of the reaction. At the equilibrium, the rates of respective forward and backward reactions are equal and the net reaction rates are zero.

The same occurs in respective modifications for cases of strong acids and weak bases, or weak acids and weak bases.

The various reactions happen concurrently [not simultaneously] depending on the relative rates, concentrations, activation energies and localized temperature differences. The main reaction in water is the self protonation:  2H2O = H3O+ + OH-.  An acid transfers a proton to water faster than another water molecule does. A "strong" acid transfers a proton to water faster than H3O+ does. A weak acid transfers a proton to water at an intermediate rate. A base works in reverse. A weak base reacts faster than H2O and slower than OH-. This means that the addition of a weak base to water will remove H3O+ faster than pure water raising the pH. If this is done suddenly, say by brute force addition or a severe temperature jump, the kinetics can be measured. At more reasonable rates, as in a proper titration rate, the differences seem simultaneous but are concurrent. [It is possible [really easy] to titrate too fast and miss an endpoint.]
Finally at equilibrium the various concentrations satisfy the pertinent equilibrium constants.

Reviewing the difference,  simultaneously seems better than concurrently,, as the latter means, among other things, merely coincidence. See https://grammarhow.com/concurrent-vs-simultaneous/

The reactions have different rates.  The grammatical definition ignores molecular motions, the uncertainty principle and relativity. Thinking hard about it simultaneous events cannot be related or maybe not. [positron annihilation, electronic transitions, SN1reacttions etc are difficult]  I tried to give a reasonable simple explanation of how these reactions happen. Yes here are many simultaneous reactions going on.


text:
Melting point of various solids: should strength of ionic bond make ionic solids the highest melting
Arrange the following in the increasing order of melting point of different types of crystalline solids

covalent solid
metallic solid
molecular
ionic

According to me the arrangement should be:
ionic > covalent > metallic > molecular

Také tungsten versus ionic liquids with melting point below room temperature.

https://chemistry.stackexchange.com/questions/5394/is-melting-boiling-point-of-ionically-bonded-substance-higher-than-of-covalently https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds https://chemistry.stackexchange.com/questions/125702/which-bond-is-stronger https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds

https://chemistry.stackexchange.com/questions/91842/which-melting-point-would-indicate-a-metallic-structure https://chemistry.stackexchange.com/questions/127320/why-does-strength-of-bond-matter-if-melting-only-affects-the-the-intermolecular https://chemistry.stackexchange.com/questions/141/what-makes-some-metals-melt-at-higher-temperature https://chemistry.stackexchange.com/questions/163724/for-the-same-substance-which-are-stronger-the-bonds-in-the-solid-or-the-bonds

When solids are held together on a macroscopic scale by covalent, ionic or metallic bonds they are more or less equally bonded and thus equally likely to have high melting points. How high depends on structural features that do not directly correlate with whether the bonding is covalent, ionic or metallic.
We illustrate by comparing the melting points of some common materials. Magnesia melts at a higher temperature than silica not because the former is ionic and the latter covalent, but because magnesia forms a crystal structure with higher coordination numbers and thus more bonds to break than silica. Sodium chloride firms a crystal structure similar to magnesia but sodium chloride involves fewer electrons in the bonding and has larger chlorine atoms taking up most of the crystal space instead of smaller oxygen atoms. The combination of fewer electrons and greater distances for the electrostatic bonding interactions renders sodium chloride lower-melting than either magnesia or silica.
As these examples demonstrate, melting points of substances with macroscopically extensive "primary" bonds involve complex combinations of factors that go well beyond the type of bond formally assigned to them.

No universal ordering is possible as there are too many exceptions
It can certainly be argued that there is a broad ordering of melting point based on the type of bonding in a solid. But there are far too many exceptions for the ordering to be reliable.
And the question contains a subtle distinction that it doesn't explicitly call out: the difference between a molecular solid and a covalent solid. It also uses what I would call careless language: what is the difference between a crystalline solid and the other categories (all of which can form well defined crystals). (Note: If the original question said ionic not crystalline my interpretation is unfair.)
Some specific examples that illustrate the problems. There are ionic compounds that are liquids not crystalline solids, though many ionic compounds are, like table salt, high melting. There are several metals that are soft or melt easily (gold, sodium), some that are liquids (mercury, caesium) some that are strong and melt at high temperatures (tungsten, rhenium). Some molecular solids have high melting points (graphite, silica) and some that have much lower melting points (some glasses).
The only group that usually has low melting points are molecular solids where the forces that matter are typically weak intermolecular forces. Hence why things like menthol are volatile, low-melting crystals.
Given some of those observations, the type of bonding is not a good universal predictor of the properties of the compound.

Dunno why you bothered with answer such a question, but if you already did... NaCl has a lower melting point than gold; graphite and silica are covalent, but without discrete molecules; caesium is borderline, galinstan is a clear cut example of liquid metal; molecular solids can be pretty high melting...

I am not sure such a general classification can be made. Just for fun I would probably put both ionic and metallic solids together in the first place, followed by covalent compounds. But there are exceptions, like the mercury $\ce{Hg}$ (metal, melting at $\pu{-39 °C}),$ and the covalent silica $\ce{SiO2}$ melting at $\pu{1730 °C}.$


text:
Understanding how evaporation works
I am struggling to grasp the concept of partial pressure and vapor pressure.
Consider a sealed container filled with water and a bit of air that is sufficiently dry placed in a room of temperature 25 degrees celsius and pressure of 1atm. The container is initially in thermal and mechanical equilibrium with the surroundings.
The water in the container will spontaneously evaporate until water and steam chemical potentials match (diffusive equilibrium).
However, I don't understand why this process happens.
Before evaporation, the container with air initially starts at state 1 in the below phase diagram. In this state, water, air and surroundings are all in mechanical equilibrium with each other right? Since state 1 is located in liquid region, then all air particles should spontaneously condense to water particles?

This is the phase diagram for water. Not water plus air. Your assumption the air should condense is false. We can still use it to describe the behaviour when air is present if you recognise that, in the gas phase, molecules don't interact much. So the water vapour pressure is independent of the other gases present.

Not true an inert gas is an additional component in the phase rule.

The diagram is for the system without air, so at equilibrium, vapor pressure is equal to total pressure (if there is the gaseous phase).
At $\pu{25 ^{\circ}C}$, all water eventually evaporates, if pressure is kept below vapor pressure at $\pu{25 ^{\circ}C}$, or condenses if pressure is kept above the vapor pressure.
The boiling point refers to the total pressure regardless of the gaseous phase composition.
On the other hand, pressure of gaseous mixture cannot be generally used to determine position on the phase diagram. Water at room temperature would evaporate to ambient air, if water vapor partial pressure is lower than (saturated) vapor pressure, in spite of the total pressure being much higher than vapor pressure.
Total pressure can be used for equilibrium between condensed phases.

Reaction to feedback:
If there is solely water vapor in the gaseous phase of the container, and if external pressure is higher than vapor pressure at the given kept temperature, these processes happen:

External pressure would compress vapor until pressures equilibrate.
This compression adiabatically heats up the vapor.
As vapor is oversaturated, it starts condensing, releasing heat.
This may continue until it reaches temperature when vapor pressure equals to external pressure.
As heat dissipates to surrounding, vapor keeps being oversaturated and condensing, until there is no vapor left.

With air present, total system pressure keeps mechanical equilibrium with external pressure and liquid water keeps partial vapor pressure equal to (saturated) vapor pressure at the given kept temperature.
If there is initially air in the system with pressure equal to external ambient pressure $\pu{1 atm}$, water will evaporate until it's partial pressure reaches vapor pressure and system will expand to maintain the total pressure equal to ambient pressure (assuming isothermic + isobaric conditions).

At equilibrium, the air in the container is full of water molecules (ie. saturated), right? If water vapor and water coexist in the container, then the pressure of the whole container must be equal to vapor pressure, no? So, at equilibrium, the container is not in mechanical equilibrium with the surroundings since its pressure is less than 1atm?

You cannot assume there is and there is not air at the same time. It would be "double thinking" from the novel 1984. // BTW, TD equilibrium consists of thermal, radiative, chemical and mechanical equilibrium.

Answer updated.


text:
Spin-only magnetic moment of a coordination complex
While finding the spin-only magnetic moment of a coordination complex, we count the number of unpaired electrons on the central metal atom. What about unpaired electrons on the ligand? Say in the case of $O_2^{-}$ or NO? Won’t they contribute to the magnetic moment?

Good question, a paramagnetic ligand is an interesting thing. Maybe under some conditions coupling of the magnetic properties of multiple ligands might occur. Could do interesting things

Free ligand being radical doesn't mean it will be one in complex.


text:
Concept of Absolute Thermodynamic Activity
In the textbook Electrochemical Systems by Newman and Alyea, Chapter 14: The definition of some thermodynamic functions, chemical potential of component (ionic or neutral) is written as a function of absolute activity: $$\mu_i=RT\ln \lambda_i \tag{1}$$
where $\lambda_i$ is the absolute activity of the component $i$. This equation is referenced to the Guggenheim: Thermodynamics textbook. I checked this textbook, but it didn't really resolve my questions.
Author claims that absolute activity allows us to define chemical potential without reference to the standard state, as it can be seen by the equation 1. This means that absolute activity is independent on the choice of the standard state.
What I know from thermodynamics is that activity is defined as a quantity which allows us to express chemical potential for real systems in the same mathematical form as for ideal systems. Chemical potential for real systems is logarithmic function of activity as it is a logarithmic function of pressure, mole fraction, concentration etc. for ideal systems.
In the most general form, activity is defined as: $$a_i = \frac {f_i}{f_i^⦵} \tag {2}$$
where $f_i$ is the fugacity of the component in the system and $f_i^⦵$ is the standard state fugacity of the component.
Definition of activity is fundamentally tied to the chemical potential and its definition equation can also be written as: $$ \mu_i = \mu_i^⦵ + RT\ln \frac {f_i}{f_i^⦵} = \mu_i^⦵ + RT\ln a_i \tag {3} $$
My question is:
Concept of absolute activity doesn't make sense to me. Activity by definition expresses how much is a component thermodynamically active RELATIVE to the standard state. This is why equation 2 is written as a ratio and why in equation 3, standard state chemical potential $\mu_i^⦵$ shows up.
Given that, I don't understand equation 1 and where is it derived from as we can see a difference comparing equations 1 and 3.

As an aside, electrochemical books don't do justice in their chapters about thermodynamics.  Even Newman's book suffers from this. However, I remind you that the electrochemical potential (Newman doesn't like this term) is a definition. And as such, it is arbitrary but the goal is to be a practical for everyone to use it. Since you read the chapter, you can even see that Newman spends time dealing with several convenient/unconvenient definitions. You will not find a derivation of Eq. (1). I will try to answer the question when I have more time...

Yes, but he references this definition to the thermodynamics textbook by Guggenheim. I checked this textbook, but have to say I'm not a huge fan of its explanations regarding chemical potential and activity relationships. It's the only thermo textbook I came across which uses the concept ''absolute activity'' and says that it's independent of the standard state.

Using the definition of the absolute activity provided by IUPAC
$\mu_i=RT\ln \lambda_i \tag{1}$
$\mu_i^\circ=RT\ln \lambda_i^\circ \tag{2}$
Therefore,
$\mu_i = \mu_i^\circ + RT\ln  a_i \tag {3}$
becomes
$RT\ln \lambda_i = RT\ln \lambda_i^\circ + RT\ln  a_i \tag {4}$
or
$$ a_i = \frac{\lambda_i}{\lambda_i^\circ} \tag {5}$$
This is equivalent to your equation (2), except that $\lambda_i^\circ$, the value of the absolute activity in the reference standard state (at which $a_i=1$), need not be equivalent to the value of the fugacity in that standard state. It would appear necessary mathematically, however, that these not differ by more than a multiplicative constant, provided the same standard state is being referenced, that is $\lambda_i = \textrm{constant} \times f_i$.

$the$ $value$ $of$ $absolute$ $potential$ $in$ $the$ $standard$ $state$, you mean the value of absolute activity in the standard state? Constant you mentioned is equal to $\frac {\lambda_i ^⦵} {f_i ^⦵}$. Absolute activity and fugacity will thus be equal only if: $\lambda_i ^⦵ = f_i ^⦵ $.

I have one problem. Absolute activity should be independent on the standard state. That is the main motivation for using it. However, we previously saw that: $$\lambda_i = \frac {\lambda_i ^⦵} {f_i ^⦵} f_i$$
Both $\lambda_i ^⦵$ and $f_i ^⦵$ will depend on what standard state is. So, for absolute activity to be independent on standard state, we need to prove that $\frac {\lambda_i ^⦵} {f_i ^⦵}$ is a constant regardless of what standard state is chosen. I'm not sure I can prove this and I don't see why this would necessarily be the case.

@DarioMirić You are right about the statement you point out, I edited that. I also agree with your definition of the constant . I don't think the absolute activity lets you avoid defining a standard state (call it a reference state instead if you prefer). It just makes the equation more compact.

Although the point I suppose is that a reference value of the absolute activity can be defined without defining an associated standard state (a specific ideal physical condition). (edit: no, that doesn't seem right; in fact it seems again that all you do is simplify the equation by multiplying the fugacity by a constant)

Yes, while I'm okay that the standard state needs to be defined (which means that absolute activity isn't really absolute), I'm not okay that in the textbook Electrochemical Systems, author claims that absolute activity can be defined without it. Some corrections are required in this textbook. It seems that @MetalStorm was correct about electrochemistry textbooks largely not being very accurate in using concepts from thermodynamics.


text:
Stability of benzyl free radical hyperconjugated with a methyl group

In this picture, which one is more stable?

My teacher says the second one as it has 4 alpha-Hydrogen whereas in the first picture the free-radical carbon has only 1 alpha-Hydrogen. That's why second compound is more stable than the first one.

But I am not convinced. My intuition says me that the benzene ring will provide more stability in the first compound compared to the stability of the second one.

What do you think of my proposed explanation?

I vote for the top radical being more stable. The hyperconjugation to the hydrogens and the C-C bond will have a very small effect when compared with the resonance delocalization of the radical into the benzene ring.

Sadly I think your chemistry teacher is wrong.

I agree. The benzylic radical is by far the more stable


text:
Is reaction between SiO2 and CaO is Lewis acid and base reaction?
In my textbook it is written that the slag forming reaction between silicate and calcium oxide is acid base reaction.
But this reaction is not Arrhenius definition of acid base reaction because this happens without water.
Also this reaction is not Brønsted–Lowry definition of acid base reaction because there is no proton donating.
I only know one more definition of acid and base, that is Lewis definition.
So is it okay to consider this reaction as Lewis acid and base reaction?
If that is the case, is it okay to assume that $\ce{O^2-}$ is the electron pair donor and $\ce{Si}$ is electron pair accepter?

This is a general question, which could be rewritten in a more generalized way, like this : Any oxygenated salt can be theoretically formed by a reaction between a metallic oxide and a non-metal oxide. Examples : $\ce{ CaO + SO3 -> CaSO4}$, or $\ce{Al2O3 + 3 N2O5 -> 2 Al(NO3)3}$. So the question becomes : Is it OK to assume that $\ce{O^{2-}}$ is the electron pair donor and the non-metal ($\ce{S}$ or $\ce{N}$) is the electron pair acceptor ?

You are right. Thank you for your meta-cognitive advice.

It is an acid-base reaction in the Lux-Flood definition: Hereby the acid is the oxide anion acceptor, the base is the oxide anion donator.

Yeah, it's Lux-Flood https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Lux%E2%80%93Flood_definition


text:
Is q=0 for irreversible adiabatic process?
Well, I am a little bit confused about this question. I learn that reversible adiabatic processes are isentropic. So $\Delta S=0$. Through $\Delta S=\frac{ q}T$, we can say that $q=0$. But if you take an irreversible adiabatic process, due to friction, $\Delta S\gt0$. Which makes:
$$\begin{align}
\Delta S=\frac qT&\gt0\\
q&\gt0
\end{align}$$
So is it safe to say $ q\gt0$ for the irreversible adiabatic process? If not, where I am wrong?

What do you mean by "Δq"? There is q, and δq, but Δq is (probably) not what you are referring to.

For most cases, using Delta q is wrong, as just q should be used, at q means system energy change, like W or Delta U or Delta H.  The only case where Delta q is justified, with proper legend, is in very specific scenarios the difference of two different q values. (I have it wrong myself in one of my comments - habits die hard.)

Are you saying that $\Delta S=\frac{\Delta q}{T}$ both for reversible and irreversible processes?

The definition of an isentropic process is $\mathrm{d}S=0$.
The definition of an idiabatic process is $\delta q=0$.
Be aware that $\mathrm{d}S \ge \frac{\delta q}{T}$. So even if $\delta Q=0$ as the adiabatic process requirement, there is still possible $\mathrm{d}S \gt 0$.
Imagine isolated system as a water reservoir. There is done mechanical non-PV work on this system by mixing water by the blender screw - like the classical Joule experiment about heat/work equivalence.
The process is adiabatic with irreversible friction - but no heat exchange, so $\delta q = 0$.
The process is not isentropic, as it is irreversible and entropy increases, with $\mathrm{d}S \gt \frac{\delta q}{T} = 0$
If the irreversible action like friction occurs out of the system, like at the piston with friction or mechanical machinery behind, the system may do non-PV work on the surrounding via friction, which converts work to heat.
$$\mathrm{d}S_\mathrm{sys} =  \frac{\delta q_\mathrm{sys}}{T} = 0$$
$$\mathrm{d}S_\mathrm{surr} \ge  \frac{\delta q_\mathrm{surr}}{T} \gt 0$$

Formally, process can be isentropic even if $\delta q <> 0$, if $\mathrm{d}S_\mathrm{sys}= \mathrm{d}S_\mathrm{sys,irrev} + \frac{\delta Q}{T} = 0$, so $\mathrm{d}S=\frac{\delta q}{T}=0$  is not correct definition of isentropic process.
E.g. Assume a system consisting of 2 reservoirs with temperatures $T_1$ and $T_2$ and there is  system surrounding of temperature $T_3$, when $T_1 \gt T_2 \gt T_3$. Entropy increase term of system will be caused by the spontaneous balancing of temperature in both system reservoirs, while the entropy decrease term will be caused by passing heat to the surrounding.
With properly chosen temperatures, there would be ongoing isentropic process for the system, with $Q \ne 0$.

It is just good to emphasize that generated heat is equal to the work done by the friction force and isn't connected to the PV work. This is to avoid confusions by using the same terms.

........Made fix

Note that by IUPAC Gold book, both q and Q are used for heat.

If process is irreversible than adiabatic process between states 1 and 2 will  generate entropy in the surroundings. Entropy is a state function which means that if system goes from state 1 to state 2, entropy change of the SYSTEM is the same regardless of process being irreversible or not.
For adiabatic process this change is equal to zero. However, if adiabatic state change is irreversible, entropy will be generated outside the system due to irreversibility causing entropy of the surroundings to increase.
Generally for irreversible process, 2nd law is:
$$ dS_{tot} = dS_{sys} + dS_{sur} > 0  $$
By definition of isentropic process:
$$dS_{sys} = 0 $$
This leaves us with: $$ dS_{sur} > 0 $$
Last inequality shows that irreversible state change generated entropy in the surroundings. This can be work done by friction when piston is moved, for example.

The way to do that is to exchange heat with the surroundings causing its entropy to increase.  For adiabatic process, $\Delta q_\mathrm{sys}=0$ by the definition. But, the system may do work on the surrounding via friction, which converts work to $\Delta q_\mathrm{surr} \gt 0$.

That is true if process is reversible. Irreversible state changes will generate heat which may be generated inside the system. If that is the case, system needs to discard it to be able to get to the state 2.

But then it is not an adiabatic process. //  OTOH, if it is an idiabatic process, but irreversible due not being in equilibrium during the transition, there is still $\Delta q=0$, but in absolute values, $W_\mathrm{irrrv,comp} \gt W_\mathrm{rev, comp}$  or $W_\mathrm{irrrv,exp} \lt W_\mathrm{rev, exp}$

That holds, if we don't allow any heat exchange with the surroundings (which is how adiabatic process can be defined). However, than we can't say that entropy change of the system is zero, it must be positive and surroundings have entropy change zero.
This discussion shows that we need to be careful what exactly we mean by adiabatic. Whether no heat exchange or constant entropy of the system. In thermodynamics and state changes, usually it is meant constant entropy, but that only works if system can discard generated entropy by heat due to the irreversibility.

Not "can be defined", but "is defined". Is says nothing about entropy change, that may or may not be zero and adiabatic process may or may not be isentropic.

Yes, adiabatic process will be isentropic only if it is reversible.

Let us continue this discussion in chat.

First of all, there is no such thing as $\Delta q$ unless you are referring to the difference in the heat evolved in two different processes. For an adiabatic process, q=0 by definition, and therefore the difference in the heat evolved in two adiabatic processes is and will always be, $\Delta q=0$, irrespective of whether these are reversible or irreversible.
Second, the second law of thermodynamics relates the heat exchanged in a process to the entropy change, with a reversible process as a limiting case for which the equality case in the following equation holds:
$$\Delta S \geq \frac{q}{T}$$
In other words,
$$\Delta S = \frac{q_{\textrm{reversible}}}{T}$$
That "reversible" condition is very important. The equality does not hold otherwise. The 2nd law of thermodynamics states that for any irreversible process at constant T between the same states, q is less than $T\Delta S$. This just guarantees that the entropy change of the system and surroundings offset properly to generate an overall positive entropy change for the universe. For instance, in a spontaneous exothermic process the change in the entropy of the system cannot decrease by more than $|\frac{q}{T}|$.


text:
How well the conclusion from the "Ignition!" book on rocket fuel science stood the test of time?
There was a wonderful book called "Ignition!" by John D. Clark, published around 1972, very interesting to read. That book covers the history of the rocket fuel development in the middle of the XX century, from an insider's standpoint. It ends with this claim, among others:

There appears to be little left to do in liquid propellant chemistry, and very few
important developments to be anticipated. In short, we propellant
chemists have worked ourselves out of a job. The heroic age is over.
But it was great fun while it lasted.

Author's conclusion behind this claim is that the following list of types of the fuel is the best which can be made and no further significant improvement seems possible:

Short-range tactical missiles - RFNA-UDMH, ClF5, hydrazine.
Long-range strategic missiles, lunar landers, service modules - N2O4, hydrazine.
First stage space boosters - liquid oxygen + RP-1
Upper stages of space bootsters - J-2 (hydrogen-oxygen combo)
Deep space - methane, ethane, diborane with OF2 or ONF3, NO2F as oxidizers.

What were the actual advances in the rocket fuel engineering since the 1972? Are the above claims stood the test of time?
To better illustrate what I'm talking about I'm including verbatim copies of the relevant pages from the book purely with the "fair use" intentions, but if it's inappropriate I'll take them down no questions asked.

I think he even managed to overestimate developments in here. BTW I don't see why post here and not on Space Exploration.SE

@Mithoron Well... :D This book left me with an impression of being a book about chemistry, and a pretty hardcore one at that. :D But why not, let me ask the same thing on Space Exploration.

Cross-posting is frowned upon. If the post gets closed than it can get migrated there, but you shouldn't repost it there.

@Mithoron lol, this question have actually already been asked on SE.SE!.. :D https://space.stackexchange.com/questions/19608/what-was-the-result-of-the-propellant-predictions-in-the-last-chapter-of-igniti Thank you so much for pointing me to that community.

Would be so cool to see B2H6 + OF2 flame, though!

For readers, RFNA-UDMH stands for red fuming nitric acid + unsymmetrical dimethylhydrazine

Agreed -- there have been minor improvements (e.g.,  ammonium dinitramide for more toxic hydrazine, https://cubesat-propulsion.com/comparing-cubesat-thruster-propellants-adn-hydrazine/), but to get much more specific impulse requires non-chemical thrusters, e.g. ion such as iodine, https://www.nature.com/articles/s41586-021-04015-y) or nuclear.