text:
What does Bohr's model predict if the mass of an electron becomes 10 times its actual mass?

If the mass of an electron becomes 10 times its actual mass, which of the following statements is correct regarding Bohr's model:

Velocity of electron increases by 10 times.
Orbit radius decreases by 10 times.
Energy of electron increases by 10 times.
Wavelength of the electron remains the same.

(An objective question)
I first tried the rest mass formula, but it is not related to the options. So I tried the energy and radius formulae in Bohr's theory and they give option 4. But the wave nature of electrons is not discussed in Bohr's theory, is it?
The formulae:
Radius $r= n²h²/4π²mze²$
Total energy $E_n= -ze²/2r= -2ze²π²mze²/n²h²$
Velocity, from $MVR=nh/2π$ is $v=2πze²/nh$

But the wave nature of electrons is not discussed in Bohr's theory, is it?  Well, Bohr theory considers the length of an electron orbit is an integer multiple of the electron de Broglie wavelength.

@Poutnik I think the temporal order is reversed. Then the way the exercise should be seen or used to learn depends on the stage at which it is proposed.

For an actual example, look up muonic hydrogen.

Look for those quantities which may vary when mass of  electron may vary i.e those quantities which contain term $m_e$ ( where $m_e$ is mass of electron) in their expression.
Velocity in $n^{th}$ orbit= $\dfrac{2kZe^2}{nh}$ (independent of mass)
Radius of $n^{th}$ orbit = $\dfrac{n^2h^2}{4k\pi^2Ze^2m_e}$ ( varies inversely with mass)
Energy of electron $n^{th}$ orbit= $\dfrac{-kZe^2}{2r_n}$ (varies proportionally with mass)
Wavelength of electron = $\dfrac{2\pi r_n}{n}$ (varies inversely with mass)
Thus, the velocity remains unchanged, radius decreases by 10 times, energy increases by 10 times and wavelength decreases by 10 times.
Therefore, the final answer would be (2) and (3).

If you're answering this in an exam setting, you likely don't remember the formulae for velocity and radius of electron in $n$th orbit (you should, however, learn those for energy and wavelength). The answer given by kanchan is correct but relies on memorization of unnecessary formula, so let me propose a logical way of thinking about this.
In Bohr's model, electrons travel in circular orbits around the nucleus. As centripetal force is always perpendicular to the electron's velocity, its angular momentum doesn't change. That is to say,
$$mvr=\text{constant} \tag{1}$$
This means that if mass increases by $10$ times and velocity stays constant (I'll prove that it does in a while), the radius of the orbit should decrease by $10$ times.

The centripetal force acting on the electron is almost entirely electrostatic (you can calculate the gravitational force between the hydrogen nucleus and its electron to convince yourself of this). So,
$$\frac{mv^2}{r}=\frac{1}{4\pi\epsilon_o}\cdot \frac{Ze^2}{r^2}$$
Multiplying both sides by $r^2$
$$mv^2r=\frac{1}{4\pi\epsilon_o}\cdot Ze^2 = \text{constant} \tag{2}$$
If you divide $(2)$ by $(1)$ you get
$$v=\text{constant}$$
This means that the velocity of an electron won't change with mass.

The formulae for energy and wavelength of electron (which, as I said, you should remember) are
$$E=-\frac{1}{4\pi\epsilon_o}\cdot \frac{Ze^2}{2r}$$
$$\lambda=\frac{2\pi r}{n}$$
You can see that energy is inversely and wavelength directly proportional to radius. But you also know that radius varies inversely as mass, so you should reverse the proportionalities of radius with energy and wavelength to get the corresponding proportionalities for mass with those quantities.

To answer your question, when mass of the electron dectuples, radius becomes smaller by a factor of $10$, velocity stays the same, energy increases by a factor of $10$ and wavelength decreases by a factor of $10$.
The correct options therefore are  (2) and (3)

we should never call a formula unnecessary, formulas are meant to be remembered. @RayBradbury the questioner may not have enough time during his exam to think about such logics in exam , If one knows concept and has understood derivation, there's no harm in remembering formula if one can. The questioner himself has mentioned formulas and I think he has no harm in remembering them

@kanchan I can only remember a handful of formulae for an exam. I want those that I remember to be short and of wide applicability. Indeed, I never memorised the formulae for velocity and radius and seemed to do okay. In my experience, it is easier to remember a short proof than arbitrary trivia ("radius shrinks as mass increases, velocity doesn't change"). But you may be right, and OP is free to discard my answer for yours.

I have mentioned the formulae but I too have the inability to remember formulae if they're being very specific. Also, the above ones are quite simple to obtain using Coulomb's Law, not unlike @Ray's method here.


text:
Finding the weight of the final product in a reaction sequence
The question is:

What is the molecular weight of the final product?
$$\ce{Ph3C-Cl ->[NaCN] P1 ->[H3O+] P2 ->[Conc H2SO4][CH3OH]P3}$$

My products were:
$\mathrm P_1$ = $\ce{Ph3C-CN}$ (by SN); $\mathrm P_2$ = $\ce{Ph3C-COOH}$ (by hydrolysis); $\mathrm P_3$ = $\ce{Ph3C-COOCH3}$ (By esterification)
If this was correct, the answer should have been 302. But the correct answer is 274. What did I do wrong?

Your answer looks correct to me. The "official" answer looks to be in error

@Waylander, However, is the first step feasible? I was unable to find literature referencing the reaction between trityl chloride and sodium cyanide. Any preparation of triphenyl acetonitrile involved copper cyanide.

@SafdarFaisal  In the context of a question like this, I think the answer is yes. Like so many other questions, it has been set without reference to the literature by a teacher who thinks it should go like this. If it isn't this reaction, then what is P1? It becomes a much more difficult question.

@Waylander The answer might be incorrect but I doubt that is the case. It is supposed to be a ‘Spotlight’ question, which means it is tough and, by judging on previous spotlight questions given to us, unusual. All that I can think is there is something to do with Ph3C- group as it is not very commonly given in questions.

BTW I’ll get some hint (or solution) tomorrow. I’ll be sure to share it here.

@WhySee That will be interesting. I was playing around with the idea that P1 could be the isonitrile and P2 the N-formyl tritylamine, but I cannot make anything of MW284 from it by acid treatment - the iso-indolone is 285

Guys I’m extremely sorry. I just confirmed with my batch mates and the answer is 274 and not 284. I have corrected it in the question. I’m very very sorry for any trouble caused by the answer change.

274 corresponds to 3 phenyls + C + 31. 31 = 16 (O) + 15 (CH3). So it looks like if CN goes on, it comes off because Ph3C+ is so stable, but in conc H2SO4 - probably adjusted carefully - OCH3 can go on, while the H2O produced reduces the acidity of the H2SO4 somewhat. But I just played with the numbers.

So are we proposing that P2 is triphenylacetic acid and it undeergoes an acid catalysed decarboxylation?

You are correct in your first two steps. Cyanide adds to trityl chloride to give triphenylacetonitrile which is then hydrolysed in aqueous acid to triphenylacetic acid. In $\ce{H2SO4/MeOH}$ the acid is protonated and loses water to give the acylium cation ($\ce{Ph3C-CO+}$). This cation loses $\ce{CO}$ to give the very stable trityl cation ($\ce{Ph3C+}$) which is eventually captured by $\ce{MeOH}$ to give trityl methyl ether.

Wow. This is actually the correct explanation. I just confirmed today. Thanks.

source: Stack Exchange

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

sort: votes


text:
Is tetrachloridocuprate(II) formed when dilute hydrochloric acid is added to a solution containing tetraamminecopper(II)?
As observed in the lab, when concentrated $\ce{HCl}$ is added to a solution of deep blue $\ce{[Cu(NH3)4]^2+}$ a green color appears indicating the partial formation of the yellow colored $\ce{[CuCl4]^2-}$.
However, when diluted $\ce{HCl}$ is added, a light blue color appears because the tetrachlorocuprate complex doesn't form. Instead, $\ce{[Cu(NH3)4]^2+}$ reacts with $\ce{H+}$ to form free $\ce{Cu^2+}$ ions in water.
I wanted to prove this by calculations,  so I wrote the reactions happening.
$$
\ce{[Cu(NH3)4(H2O)2]^{2+} + 4H2O <=> [Cu(H2O)6]^2+ + 4NH3} \tag{1}\\
K_1 = 10^{-13}\\\\$$
$$4×(\ce{H3O+ + NH3 -> NH4+ + H2O}) \tag{2}\\
K_2 = 10^{9.25}\\\\$$
$$\ce{[Cu(H2O)6]^{2+} + 4Cl- -> [CuCl4(H2O)2]^2- + 4H2O}b\tag{3}\\
K_3 = 4×10^5$$
Overall:
$$\ce{[Cu(NH3)4(H2O)2]^{2+} + 4H3O+ + 4Cl- -> 4NH4+ +4H2O + [CuCl4(H2O)2]^2-}\\
K= (10^{-13}) \cdot (10^{9.25})^4 \cdot (4×10^5) = 4×10^{29}$$
But as I obtained, $\ce{[CuCl4]^2-}$ must form regardless whether the $\ce{HCl}$ is concentrated or diluted, because the constant of the overall reaction is very large. I know this is wrong but I don't know where have I mistaken.
I speculated that the stability constant of $\ce{[CuCl4]^2-}$ is indeed not $4×10^{5}$ but a number much, much smaller. In this case, the constant of the overall reaction will also be very small, and $\ce{HCl}$ has to be concentrated in order for the equilibrium to be shifted forward.
But actually I searched for long on the internet and that was the only stabilty constant that I found on a site that I don't  know whether is credible or not.

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it  can be challenge to decipher.  Consider  copy/pasting or rewriting of the essential parts and using of MathJax  as written above.

Have you considered lower complexes with 1-3 Cl- , which may be preferred at low [Cl-] ?

@Poutnik you mean that the stability constant of tetrachlorocuprate II depends on the concentration of Cl- ? Or why should these complexes be preferred at low concentrations?

Let us continue this discussion in chat.

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text:
molecular orbital theory and crystal field theory confusions
I am a high school student and I am very confused in Molecular orbital theory, it says that when two orbitals overlaps in the same phase, they form antibonding orbitals and when they overlap in opposite phase forms antibonding orbital, but my questions are:

how those orbital overlap in the same and opposite phases simultaneously, forming antibonding and bonding orbitals together?? or those are the electrons which are overlapping?

if orbitals are the regions where there is probability to find the electron then how two vacant orbitals can overlap each other? for e.g if there is a 3 electron system then two will be in bonding and 1 will be in antibonding then if I think of it as electrons overlapping then is this electron overlaps with the vacant orbital to form antibonding orbital? do vacant orbitals really have self existence(isn't it just the empty space)? if they exist independently then why do we take there energies to be 0 if there is no electron in them?

In crystal field theory, we say that d orbitals splits due to the repulsion faces from ligands, but even if they are empty we still split them how is this even possible? for e.g  it is written that if there is say d1 configuration in octahedral complexes  then the d orbitals located in path of ligand will face  higher repulsion from the ligands so the  energy levels are not degenerate anymore and  but, isn't this statement very vague? if there is only 1 electron in the d orbitals then only that will face repulsion how can the empty space is facing repulsion?

there are several answers related to molecular orbital theory but I am looking for the one I can understand at my level as there are many difficulties to digest this.

@ArunBharadwaj I think (probably) the first and second points have already been covered on the site, thus I suggest you to research about those from the site itself. But the question 3 is a new question in the site(I believe). And it is to be noted that generally posting too many questions inside a single post is discouraged.

@Rishi  I have found a related answer but that I didn't understand very well, can u suggest me the answer?

https://chemistry.stackexchange.com/questions/16145/how-can-two-orbitals-constructively-and-destructively-interfere-simultaneously This and the the posts on the right side(if you are using desktop mode) or at the bottom(if mobile mode) will clear up your confusion in 1. And 2.

https://physics.stackexchange.com/questions/187269/molecular-orbital-theory This may help as well.

It doesn’t make even a teensy bit then, first check out quantum mechanical model of atom first, then Valence Bond theory and then molecular orbital theory(Atleast it worked for me this way)


text:
Why and how does ring expansion occur in the dehydration of (cyclobut-3-ene-1,2-diyl)dimethanol?
Can someone explain how does this ring expansion occur with the mechanism? I do understand that the hydride shift occurred because of carbocation stability but why did the ring expand? Can you please attach a mechanism if possible?

I tried writing the mechanism in my own way. Can someone verify if it is correct?

A 4 membered ring is highly unstable and moving to a 5 member ring reduces its ring strain. I presume the final answer is hexene.

Does this happen everytime that bigger the ring, the more stable it is?

This is true in the case of expansion to 5 member and 6 member rings.

You mean from 4 membered to 5 membered and from 5 membered to 6 membered?

Yes. Is the final product benzene?

Yes You are right

Would you please mind telling me how to approach these kind of question?

https://chemistry.stackexchange.com/questions/137215/formation-of-toluene-from-cyclobutane-1-2-3-triyltrimethanol-using-conc-sulph/137217#137217

Let us continue this discussion in chat.

I am not familiar with the arrows which looks like you are hanging the hydrogen, but  first the ring expansion step seems a bit off

By the arrows, I mean that hydrogen is attacking the OH part i.e protonation.

@Buraian : It is not hydrogen,. It is $\ce{H3O+}$ (proton from aqueous acid)., commonly known as $\ce{H+}$. Once protonated, $\ce{OH}$ group would leave as $\ce{H2O}$ (good leaving group).

The mechanism for the given reaction would be as follows.

Why does this happen? This is because a four member ring has a high amount of ring strain. Expanding itself to a 5 member or (if possible) a six member ring is highly advantageous as it helps alleviate some of this ring strain.
Now, in this case there is also the added advantage of the six member ring being aromatic (benzene).
A similar example has been discussed here: Formation of toluene from (cyclobutane-1,2,3-triyl)trimethanol using conc. sulphuric acid

A ring expansion typically occurs to overcome the high strain present in smaller sized rings.
Now strain isn't one single thing but a combination of many different factors which are nicely summarized by this[1] paper.

Types of strain
... the total strain of a conformation is the sum
of:

Bond strain—stretching or compression of chemical bonds. This type of strain is rather severe and is not encountered very often in
organic compounds. To minimize bond strain, a molecule adopts
conformations that have other, less energy-demanding, types of strain.

Torsional strain (eclipsing strain, Pfitzer strain) is caused by eclipsing interactions. Torsional strain is considerably higher
compared to steric or angle strain, which are explained next.

Steric strain (Van der Waals strain, Prelog strain) is caused by atoms forced too close to each other. Transannular strain (Prelog
strain) is a form of steric strain characteristic of medium rings.

Angle strain (Bayer strain, classical strain) is a result of deviation from the ideal bond angle. Compared to other types of
strain, increase in energy of a conformation caused by angle strain is
relatively low. As a result, a molecule can accommodate relatively
large deviation from an ideal bond angle and still be stable.

Different rings undergo expansion to overcome different types of strains. The major driving forces in smaller rings is the angle strain and the eclipsing strain.

Small rings are rigid and highly strained. They are characterized both
by a large deviation from the ideal tetrahedral valency angle—a high
angle strain, and eclipsing interactions—a high torsional strain.

For example, cyclobutane expands to cyclopentane because of reduction in angle stain and torsional strain.
But in case of cyclopentane's expansion into cyclohexane, the expansion occurs mainly due to reduction in torsional strain as angle strain actually increases in case of cyclohexane.

Angles in a regular pentagon are 108°. That is very close to the ideal
tetrahedral angle of 109.5° and, therefore, if cyclopentane were
planar it would not exhibit any significant angle strain. However, in
a planar cyclopentane there would be a high degree of torsional
strain. All ten of the carbon–hydrogen bonds would be in eclipsing
conformations.

Regarding the mechanism, the following reaction pathway is followed to obtain the product:

Later OP proposed a mechanism of their own and asked in the comments to verify it:
The mechanism you propose looks a bit incorrect. The arrows are pointing in the opposite direction. I have never seen an arrow hanging the $\ce{H+}$ to an $\ce{-OH}$ used to signify protonation.
I believe you are confused in the process of identification of nucleophile and electrophile. The arrows are pointed from the nucleophile to the electrophile. The arrows signify the movement of electrons. You might benifit from reading Drawing Organic Mechanisms.

Reference:
(1)   Dragojlovic, V. Conformational Analysis of Cycloalkanes. ChemTexts 2015, 1 (3). https://doi.org/10.1007/s40828-015-0014-0.


text:
What is the product of the reaction between alcohol and acetyl chloride in presence of pyridine?
What happens when butan-2-ol reacts(under heat) with $\ce{CH3COCl}$ in the presence of pyridine? Does it form $\ce{CH3COOCH(CH3)CH2CH3}$? I got the ester mentioned but the solution says it gives an alkene. What is the product when an alcohol reacts with an acid chloride in the presence of pyridine?

How much pyridine and at what temperature?

These are standard acylation conditions for alcohols done at ice-bath or room temperature, so unless it is being heating strongly in the presence of a large excess of pyridine I doubt you're going to get an alkene. What is the source that says you will?

@Waylander The amount of pyridine is not mentioned but the the reaction proceeds under heating.

In what solvent at what temperature?

I found that the reaction you describe is a special form of the Schotten-Baumann reaction.
The special form of the reaction is called Einhorn-Acylierung ("Unicorn acylation" in English). It is named after the chemist Albert Einhorn (Einhrorn means "unicorn" in German and hence the reaction name).
The reaction mechanism is as follows:
First, the pyridine forms an acetyl pyridinium salt. The alcohol can then perform a nucloephilic attack. The pyridine gets removed and can remove one hydrogen to reach the final product.

Image source: From the German Wikipedia article


text:
What is the effect of the oxidation number of iron on the properties of ferroin complex?
In our labratory expermient, we have determined the isobestic point for iron(III) complexation with ortho-phenanthroline, and we have also studied iron(II) complexation with ortho-phenanthroline by spectrophotometry. What does the isobestic point tell us and how can we answer the asked question?

It is Isosbestic point

Answers on CH SE site are figuratively paid by the user's own effort. When you ask, you are supposed to search and thoroughly think about the topic and to provide explicit compact summary of partial answers or at least ideas you have got until then. Effort not shown may be  considered as effort not done.


text:
What is the product of reaction between haloalkane and ammonia at different stoichiometries?
What is the difference in products when:

a haloalkane reacts with excess of ammonia

excess of the haloalkane reacts with ammonia?

I am sure with the 1st reaction, the mechanism will be nucleophilic substitution and the organic product will be a primary amine but what about the second?

I suppose you know what are the alternatives of "primary" in "primary amine"?

As per your guess an excess of ammonia when reacted with a haloalkane produces a primary amine via nucleophilic substitution.
$$\ce{R-X + NH3 -> R-NH2 + HX}$$
But when the haloalkane is in excess it subsequently reacts with the primary amine to form secondary and tertiary amines and eventually gives a quaternary ammonium salt.
$$\ce{4R-X + NH3 -> R4N+ + 4HX}$$
These quaternary ammonium salts give α-β alkenes on heating. This type of elimination is known as Hoffman elimination[1].

Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is
halide, this is often replaced by the more basic hydroxide ion through
reaction with silver hydroxide (or silver oxide). The resulting
hydroxide salt must then be heated (100 - 200 ºC) to effect the
E2-like elimination of a 3º-amine.
Example #1 below shows a typical
Hofmann elimination. Obviously, for an elimination to occur one of the
alkyl substituents on nitrogen must have one or more beta-hydrogens,
as noted earlier in examining elimination reactions of alkyl halides.

Reference:
(1) Quaternary Ammonium Salts: Hofmann Elimination  https://chem.libretexts.org/@go/page/32549 (accessed Jun 4, 2021).


text:
Why is ethanol more acidic than acetone?
The acidity of a compound is determined by the stability of its conjugate base and the major structure of enolate has a negative charge on the Oxygen atom. Moreover, the negative charge on the O-atom of ethoxide is increased by the +I effect of $\ce{-CH2CH3}$.

Image source: chem.libretext.org
So, how is ethoxide more stable than enolate?

If enol was the major tautomer, then it could be stronger, but it isn't.

@Neha if you got your answer, please feel free to self-answer it. Self-answering is always appreciated and it will help future readers.

@NilayGhosh Well... I think only good self answers are really appreciated. Usually, if more experienced user then one that asks, answers, then result is better.

Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is
$$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$
$$
\begin{array}{lc}
\hline
\text{compound} & K_\mathrm{eq} \\
\hline
\text{acetaldehyde} & 6 \times 10^{-7} \\
\text{acetone} & 5 \times 10^{-9} \\
\hline
\end{array}
$$
This means that acetone exists predominantly (if not wholly) in its keto form. For every one molecule of the enol form observed, we can expect $\pu{2E8}$ molecules of the keto form.
Because of this, any deprotonation that takes place in acetone will have to be via breaking a $\ce{C-H}$ bond, and that is not a simple thing to do when you compare it to deprotonating an $\ce{O-H}$ bond in ethanol. Thus ethanol is more acidic than acetone.


text:
Why does a methyl shift occur during the pinacol-pinacolone rearrangement?
Why does a methyl shift occur during the pinacol-pinacolone rearrangement?
I was told that                         methyl shift is done to stabilise the carbocation but here we are shifting the carbocation from secondary to primary.

The driving force is the formation of the carbonyl.

How will we know that the driving force is carbonyl if we are just seeing this mechanism for the first time without knowing the final product.

but here we are shifting the carbocation from secondary to primary

Firstly the shift is not from secondary to primary but from tertiary to secondary.

I was told that methyl shift is done to stabilise the carbocation

Yes, a methyl shift is done to stabilise a carbocation. And it is done in this case for the same reason, to increase the relative stability.
Now you might think, "How can a secondary carbocation be more stable than a tertiary carbocation? Even after having a highly electronegative atom next to it. "
Throughout all this thinking you are forgetting the fact that a carbocation is an electron deficient species and an oxygen atom has some lone pairs.
When an carbocation is formed next to an oxygen, the oxygen atom donates its lone pair via the +R effect which is very much stronger than its -I effect.
This fact can also be verified from this article[1]:

Because heteroatoms such as oxygen and nitrogen are more
electronegative than carbon, you might expect that they would by
definition be electron withdrawing groups that destabilize
carbocations. In fact, the opposite is often true: if the oxygen or
nitrogen atom is in the correct position, the overall effect is
carbocation stabilization. This is due to the fact that although these
heteroatoms are electron withdrawing groups by induction, they are
electron donating groups by resonance, and it is this resonance effect
which is more powerful. (We previously encountered this same idea when
considering the relative acidity and basicity of phenols and aromatic
amines in section 7.4). Consider the two pairs of carbocation species
below:       In the more stable
carbocations, the heteroatom acts as an electron donating group by
resonance: in effect, the lone pair on the heteroatom is available to
delocalize the positive charge. In the less stable carbocations the
positively-charged carbon is more than one bond away from the
heteroatom, and thus no resonance effects are possible. In fact, in
these carbocation species the heteroatoms actually destabilize the
positive charge, because they are electron withdrawing by induction.

Reference:
[1]: Carbocation Structure and Stability https://chem.libretexts.org/@go/page/31459 (accessed Jun 4, 2021).


text:
Does Raoult's law hold for immiscible liquids?
I know that Raoult's law holds true only for a non volatile solute in a volatile solvent mixture wherein the vapour pressure of the solution gets lowered due to the addition of solute.

Now, first of all does it hold for a mixture of immiscible liquids ?

Does it hold for a mixture of immiscible liquids undergoing steam distillation? I came across that it does but can someone explain how ? I am just not able to understand.

My understanding :
(1) Raoult's law is used to calculate relative lowering of vapour pressure of solvent when a solute is added. In a mixture of immiscible liquids the vapour pressure of the solution increases and is equal to the vapour pressures of the two solvents independently.
So there is no lowering in the vapour pressure of the solution. Based on this can I say that Raoult's law does not hold ?
(2) A mixture of immiscible liquids is not ideal right ? So Raoult's law does not hold.
(3) If the liquids are volatile, it won't hold.
But despite all this a problem was solved this way:

When a liquid that is immiscible with water was steam distilled at
$\pu{95.2°C}$ at a total pressure of $\pu{99.652 kPa}$, the distillate
contained $\pu{1.27 g}$ of the liquid per gram of water. What will be
the molar mass of the liquid if the vapour pressure of water is
$\pu{85.140 kPa}$ at $\pu{95.2°C}$?
Solution:
Total pressure, $$P_{\text{total}}=\pu{99.652 kPa}\\
P_{\text{water}}=p_B=\pu{85.140 kPa}\\
P_{\text{liquid}}=p_A=\pu{(99.652−85.140) kPa}=14.512 kPa\\ \text{and}  \frac{m_a}{m_b}= \frac{1.27}{1}\\ \text{or }=  \frac{m_a}{m_b}=
\frac{p_AM_A}{p_BM_B}\\ \text{or,}
M_A=(\frac{m_A}{m_B})(\frac{p_BM_B}{p_A})\\ M_A=1.27\times
(\frac{\pu{85.140kPa} \times \pu{18 g mol−1}}{\pu{14.512 kPa}})\\
M_A≃\pu{134.1 g mol−1}$$

Just can someone explain point 2) intuitively and in basic terms ? This sum and the usage of Raoult's law in steam distillation is killing me here.
(Maybe if someone feels that I need a brush up of steam distillation topic please gladly do so)
still confused here..assistance will be appreciated pls

A mixture of two immiscible liquids is treated as two seperate phases and thus the total pressure will be sum of vapour pressure of both the liquids.

@NisargBhavsar...okay but that was not the required clarification

My comment meant that Raoult's law is not applicable for immisible liquids. Both act independently.

@NisargBhavsar Then why was the problem solved that way ?

And a source (reliable) told me that raoult's law is true for immiscible liquids undergoing steam distillation...what exactly is going on ?

That makes no sense immiscible means doesn't make homogenous mixture, raoult's law is about ideal solutions and Ideal solutions are always missicible

You bring us a solution that never mentions Raoult's law, never uses it, and even couldn't possibly have used it. Then you ask to explain why it uses Raoult's law. That's what is going on. It doesn't. That's the explanation.

@NisargBhavsar if we have 2 immiscible liquids the vapor pressure should be the vapor pressure of the less dense liquid right? Why should it be a sum? Can you explain a bit more elaborately?

@puma toppr is not a reliable source. If you want to say any other source, please mention your reliable source.

I believe this excerpt from sciencedirect may answer your question:
"Vapor Pressures and Boiling Points in a Mixture of Immiscible Liquids
When two immiscible liquids are present in a mixture, the total vapor pressure of the mixture is equal to the sum of each liquid's vapor pressure, independently of their concentration, as long as there is enough of each liquid so they can both reach equilibrium. The condition to reach this state is that the mixture must be in constant agitation. If the mixture is let to rest, both liquids will form layers and the vapor pressure of the mixture will equal the vapor pressure of the liquid forming the top layer. The bottom layer will not have any vapor pressure as it is in contact with the top layer, thus preventing any evaporation. Therefore, it is necessary for both liquids to be in constant agitation to let the phenomenon of added vapor pressures to take place.
Consequently, when two immiscible liquids are in a mixture, their boiling point can be greatly decreased. For example, the boiling point (vapor pressure of 101,325 Pa at standard pressure) of water is 100°C, and the boiling point of m-xylene is 138.7°C [3]. When these two immiscible liquids are present in a mixture, their vapor pressures add up to make the total vapor pressure of the mixture. The boiling point of the mixture is still at 101,325 Pa since it is operating at standard pressure, however the temperature required to reach the boiling point is now below 100°C. As a matter of fact, at 93°C the vapor pressures of water and m-xylene are 78,494 Pa and 24,431 Pa, respectively [3, 4]. This adds up to a total vapor pressure of 102,925 Pa, sufficient for the solution to boil. This is the principle on which steam distillation operates."
https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/vapor-pressure

Now, first of all does it hold for a mixture of immiscible liquids ?

From Wikipedia on Raoult's law,

it states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. In consequence, the relative lowering of vapor pressure of a dilute solution of nonvolatile solute is equal to the mole fraction of solute in the solution.

Bold mine, now see this remark I have paraphrased from the section of consequences of ideal solution in wiki:

Free energy of mixing is always negative. Ideal solutions are always completely miscible.


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How would this reaction occur through free radical substitution?
So, I was trying to solve this through free radical substitution:

(Side note, the (ac) means (aq))
The best product I managed to do/think was the following:

However, according to what I've seen, it's supposed to end like this:

The problem is that I don't really understand why or how.
My reasoning for why it ended with the second image, was that the Br left the molecule, and through its free radical, took the hydrogen from PhCHO. Then, as both carbons where left with a free bond, they would react together.

I think you mean phenyl ring in the first product instead of cyclohexyl ring. If so, please correct it.

Zn reacted with the allylic bromide to create an allyl zinc species.
This is known as the zinc-mediated Barbier- Grignard reaction  wikipedia. The allyl zinc species then reacts nucleophilically with the benzaldehyde example here. Reaction can take place through either end of the allylic system with regioselectivity influenced by the choice of reaction conditions.
If this was a radical-driven reaction you would expect to get radical attack on the second double bond to give a cycloalkyl species.
According to this reference science direct

The nature of the reactive intermediate is not always clear.  The formation of an alkylzinc halide is unlikely, since many of these reactions are performed in water. It seems also not to be a free radical (as Luche and co-workers had proposed), since 5-hexenyl halides do not undergo ring closure under these reaction conditions.

Yea, i was aware Luche and co-workers had proposed this reaction, and that the supposed end of this reaction was proposed by Wilson and co-workers. However, I couldn't find either proposal and just couldn't wrap my brain about this.
Thanks!

This reaction is an example Barbier reaction (Ref.1), which was discovered by Philippe Barbier who was the teacher of Victor Grignard, the discoverer of Grignard reaction. The main difference other than metal used between Barbier and Grignard reactions are the former is one-pot reaction while the latter is two pot reaction as ref.1 states:

A synthesis with the aid of an organometallic compound is usually carried out in two steps: 1) the organometallic reagent is prepared from an organo halide and a metal in a solvent, 2) a chosen substrate is added to this reagent. In a Barbier reaction these two steps are combined: a mixture of the organo halide and the substrate is added to a suspension of the metal-in an appropriate solvent.

The other differences include ability to use water as the solvent, making it more greener (Ref.2). First, the allyl halide reacts with metallic $\ce{Zn}$ to give the corresponding allyl zinc species, which then reacts with a ketone or aldehyde (the electrophile) to give corresponding alcohol (nucleophilic addition). Therefore, your first suggestion is not correct. The correct major product is the second one (Ref.3 & 4):

I said it is the major product, because the conjugate addition predominate under these conditions and with certain allylic or propagyl halides as given in the following example (Ref.4):

When $\ce{R = H}$, product $\bf{II}$ predominates while when $\ce{R = n-Hexyl}$, both products, $\bf{I}$ and $\bf{II}$, formed in significant yields (Ref.4). This suggest that the nucleophilic addition to carbonyl compound is the slower step and therefore, attack on more stable carbocation would predominates. In your case, it is secondary allylic carbocation over initial primary allyl carbocation. Also, considering resonance structures of the initial carbocation, the larger contributors are secondary and primary allylic carbocations because they are more stable than the other contributor, secondary cyclohexyl carbocation. The appearance of the product through that carbocation is based on the reaction rates of nucleophilic addition.
However, keep in mind that the regioselectivity of this reaction is highly dependent on the nature of the alkyl halide (usually, allyl or propargyl halides) and carbonyl compound, steric hindrances, solvent effects, and the metal used (note that Barbier reaction uses other metals such as $\ce{Zn, Sn, In, Sm}$, etc.).
References:

Cornelis Blomberg, In The Barbier Reaction and Related One-Step Processes; Springer-Verlag: Berlin, Germany, 1993 (ISBN: 978-3-540-57169-8).
Feng Zhou and Chao-Jun Li, "En route to metal-mediated and metal-catalysed reactions in water," Chem. Sci. 2019, 10, 34-46 (DOI: 10.1039/C8SC04271C).
Tak Hang Chan and Chao Jun Li, "Organometallic reactions in aqueous medium. Conversion of carbonyl compounds to 1,3-butadienes or vinyloxiranes," Organometallics 1990, 9(10), 2649–2650 (DOI: https://doi.org/10.1021/om00160a008).
Artur Jõgi and Uno Mäeorg, "Zn Mediated Regioselective Barbier Reaction of Propargylic Bromides in $\ce{THF/aq_. NH4Cl}$ Solution," Molecules 2001, 6(12), 964–968 (DOI: https://doi.org/10.3390/61200964).


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Why is salicylic acid less acidic than o- anisic acid?
The Pka of salicylic acid is 2.97 while that of o-anisic acid is 3.73.
Why is the latter less acidic if the order of +M effect states that -OH>-OR which should make the conjugate base less stable?

You might have messed up the data and thus in turn the entire post. $\mathrm{pK_a}$ of salicylic acid is 2.79 and that of o-anisic acid is 3.73.

@Rishi i corrected it

https://chemistry.stackexchange.com/questions/20190/why-does-hydrogen-bonding-in-salicylic-acid-make-it-more-acidic


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D and L configuration of monosaccharides
I got this question in the last week's test from Organic Chemistry.
I am a bit confused in assigning D and L configuration.
Help me understand the topic of D and L configuration and help with this problem
Question Type : MCQ

You need to rotate C5 into a Fischer conformation that has C6 in the vertical chain.

D and L configuration of sugars:
Draw the given sugar (aldose) in the Fischer projection with the most oxidized carbon at the top (i.e. aldehyde carbon). If it is a ketose, make the closest terminal carbon to the keto group at the top. Now:

if the $\ce{OH}$ on the bottom chiral center points to the right hand side, the sugar is referred to as D-sugar.
if the $\ce{OH}$ on the bottom chiral center points to the left hand side, it is referred to as L-sugar.

To learn some history of this nomenclature, read this article.
Now look at each of the given compounds:

In (A)-(C) of given compounds, the bottom chiral center is not set to reflect the required $\ce{H-C^*-OH}$ configuration. In order to get the required configuration, you need to switch each group in that chiral center twice as indicated in the image. The gray structures in the middle raw are after first switch (they are gray because, technically, they are epimers of the corresponding structures of first raw) and the bottom raw structures are after the second switch, which are the same as corresponding structures in the first row.
As evident, (A), (C), and (D) have the $\ce{OH}$ on the bottom chiral center points to the right hand side, thus, they are D-sugars. Only (B) has the $\ce{OH}$ on the bottom chiral center points to the left hand side, and thus, it is a L-sugar. Hence, the answer is (B), because it is not a D-sugar.


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UV-spectra of benzene
I measured the UV-absorption-spectrum of benzene vapor as well as solved in methanol and cyclohexane. The absorption of the vapor was much stronger than in solution. I also noticed that the peaks get slightly shifted to longer wavelengths - with cyclohexane more than with methane [sic] methanol - and I have no explanation for this. Any help is appreciated :)
 (vapor)
 (methanol)
 (cyclohexane)

Did you notice also that the vapour spectrum should have sharp lines and many more of them  that the solution spectrum ? if you did not then you need to increase the resolution of the spectrometer, i.e. decrease slit widths. The spectra have nothing to do with the air. In solution there are many more collisions with the solvent vs the vapour, these collisions change the energy levels slightly and so smear out the spectrum.

@porphyrin I added the spectra

I suspect that the spectral resolution is limiting the shape of the  lines in the vapour phase and possibly the others. Use as narrow slits as your instrument will allow and if the spectrum is noisy as a result average this out. If the instrument is at minimum slit width you should find  a better instrument :) Benzene is non polar so is little influenced by solvents (1 nm in 250 nm is not much of a shift in energy ) other than that they cause line broadening.

The shift to longer wavelength is due to solvent polarity. Shift to longer wavelength is known as a "red shift", or bathochromic shift. Methanol is polar and will interact with the polar excited state lowering the energy required to reach the excited state from the ground state, thus we see longer wavelength peaks. I am not sure why cyclohexane has shifted the spectrum, however, because it is non polar.


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Thickness of proton exchange membrane (Nafion)?
I am looking into using a proton exchange membrane for a bio-energy cell experiment. Previous scholars have been using Nafion as a PEM to place between the anode and cathode (air-cathode). the thickness of the material is about  (7-mil). I was wondering, would the ion transfer rate get better if I used a thinner version of the same material due to a decrease of distance! or would it cause electrons to go through!
I would really appreciate it if someone can point me.
Thanks

Thinner membranes may contain unwanted tiny holes.

Ideally, the ion transfer rate gets better with a decrease in distance. However, it is harder to implement thinner PEM depending on the application. The method of fabrication might cause smaller PEM to be less selective despite them shorting the distance between the electrodes.


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X-ray Diffraction Lattice Spacing
I have crystal structure data which I have indexed.  I am able to index the (010) (100) (011) (110) and (111) planes.  Is there a straightforward method to calculate the a,b,c lattice constants based on this data or is this analysis more complex.  The space group of the material is P2/c.
I understand that I will likely need to use Bragg's Law, but I am unsure exactly how this relates to the lattice constant.

You'll need the expression for $d_{hkl}$.


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Does incomplete combustion produce carbon dioxide at first?
I know that the incomplete combustion of a hydrocarbon produces $\ce{CO}$ or sometimes $\ce{C}$ if the amount of oxygen is very limited, but I was wondering if at first $\ce{CO2}$ is produced and only when there isn't enough oxygen, then $\ce{CO}$ and C are produced? Or is only $\ce{CO}$ and $\ce{C}$ produced from the beginning of the reaction?
My question if the answer is the latter case would be how would the reaction 'know' right from the start that there isn't enough oxygen?

https://chemistry.stackexchange.com/questions/150266/ratio-of-carbon-monoxide-and-carbon-dioxide-when-carbon-and-oxygen-are-made-to-r/150267#150267

Also: https://chemistry.stackexchange.com/questions/57423/incomplete-combustion?r=SearchResults

The link to look at is that given by Safari Faisal


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Spin Multiplicities of Ions
The multiplicity is fundamentally defined as $2S + 1$ where $S$ is the total spin.
From what I understand, the multiplicity corresponds with the number of unpaired/paired electrons. For example, in the case of $\ce{Cu^2+}$:

The single unpaired electron in $\ce{Cu^2+}$ means that $S=\frac{1}{2} \implies M=2$.
However, my professor has mentioned that the electrons that give the ion the final charge (positive or negative) can be considered as unpaired, thus in the case of $\ce{Cu^2+}$, there are 2 unpaired electrons giving it a triplet $M=3$.
I am having trouble understand how this arises, both physically and mathematically.
In addition, in the case of polyatomic ions, how do electrons pair? For example, in the case of $\ce{SO4^2-}$, looking at the MO digram, all the electrons are paired, including the 2 electrons that give it the final charge. However, I was told that the multiplicity should be 3 as it should be treated as having 2 unpaired electrons.

Source: Wikimedia Commons
I am more confused with regard to positive ions, such as $\ce{Al^3+}$, as I was told that the positive +3 charge can be regarded as having 3 unpaired electrons and thus $s=\frac{3}{2}$ and multiplicity of 4.*
Since the electrons have been removed, I am having difficulty understanding where unpaired electrons arise from.
From what I have read, coupling does not affect this, and I cannot seem to find any explanation or equations that might support this. May I know if anyone has an explanation for this? Thank you!

*more accurately, maximum spin where all electrons have parallel spin

Al3+ does not have any unpaired electrons, thus this cation exists only in the singlet spin state.


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Will addition of inert gas change the vapour pressure of a liquid?
From this question on Physics SE Thermodynamics of evaporation:

Now imagine the experiment is repeated but instead of vacuum conditions, the water is pressurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.

The accepted answer shows a kinetic perspective. I want to understand things from classical thermodynamics.
As even an inert gas is added to the container, there would be some increase of configurational entropy in the gaseous phase compared to the initial conditions. Won't this factor will decrease the chemical potential of the gaseous phase and thus increase the vapour pressure a bit ?.

If the nitrogen is truly inert, its presence will have no direct effect on the chemical potential of the water in the gas phase.  Essentially, the gaseous water is "unaware" of the nitrogen.  Consistent with this, in a mixture of ideal gases, the chemical potential of each gas is determined only by the temperature, and its partial pressure.  It is indepenent of the partial pressures of the other gases.
However, even an ideal gas can increase the vapor pressure.  It doesn't do so by reducing the chemical potential of the gaseous water.  Rather, by applying pressure to the liquid water, it increases the chemical potential of the liquid phase, thus increasing the equilibrium vapor pressure. This is a small effect.
Of course, gases aren't inert.  As you increase the pressure of the nitrogen, you have two opposing effects on the liquid water: (1) The effect described above, which increases the liquid's chemical potential.  (2) With increased pressure, the concentration of nitrogen dissolved in the water will increase, which will lower the water's chemical potential.  And of course one also has to account for the non-ideal interactions between the nitrogen and the water vapor.

Even if Nitrogen is inert it will surely affect the configurational entropy of the gaseous phase of water?

If we treat both the nitrogen and the water as ideal gases, then their configurational entropies are each independent of the presence of the other gas.

You're probably thinking:  If we have a container in which we have 1 L of ideal gas A on the left side at 1 atm, and 1 L of ideal gas B on the right side at 1 atm, and we remove the partition to allow them to mix, their entropies will increase.  Yes, but that's because each now has twice the volume to explore (correspondingly, their partial pressures are halved).   But suppose we instead squeeze A and B together into a 1 L space, such that their partial pressures don't change (only the total pressure changes).  Then their entropies will be unchanged.

I am trying and the derivative is not cancelling out for a closed container with a fixed amount of inert gas. It is something like 1-ln(x), x = Ratio of moles of water vapour with Inert gas.

No, I am using xln(x) + yln(y)...x and y mole fraction in the gaseous phase

For an ideal gas, adding an inert gas to the gas phase will not change the chemical potential of original species.  But, at total pressures beyond the ideal gas limit, the chemical potential of the original species will change.  The fugacity coefficient will not be equal to 1.0 for a species in a non-ideal gas mixture.

This is a small effect.  It is usually small effect, but may be substantial near critical point. E.g. $\ce{N2O}$ is liquid at 80 atm at room temperature, but it's mix 50/50 v/v with oxygen is at 200 atm gaseous.

@ChetMiller  That's just a restatement of what was already said in the last paragraph, which discussed what occurs when behavior is non-ideal:  "And of course one also has to account for the non-ideal interactions between the nitrogen and the water vapor."

@Poutnik   That statement ("this is a small effect") was explicitly specified to apply to the effect of an ideal gas upon the vapor pressure of liquid water.  Thus it will always be a small effect (relatively speaking, i.e., relative to the applied pressure) (until, of course, you get to pressures where liquid water no longer exists, and you have exotic phases instead).

I did say it is usually small. :-)

@theorist, thanks for the wonderful comments. It is a pity they didn't teach us the Gibbs Theorem of mixing for ideal gas. It was Chet Miller comment in one of the Physics SE post which gave the hint.

At $\pu{46°C}$, the vapor pressure of water in $\pu{0.1 atm}$. It means that if some water is introduced into a vacuum, in an empty container at $\pu{46°C}$, the measured pressure in the container is $\pu{0.1 atm}$. Now if water is heated to $\pu{46°C}$ in an open flask at $\pu{1 atm}$., the composition of the vapor phase is $10$% water vapor, $90$% air. The vapor pressure is the same above the surface of water, whatever the presence of air. The total pressure has to be maintained at 1 atm. of course.

There will be a factor of  ΔSmix and its derivative will not be zero if we fix the inert gas amount in a closed container. That's why asked.

-1 This is wrong, putting pressure on water increases it's vapour pressure: Kelvin equation


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Does SIR(Steric hindrance in resonance) affect inductive effect?
In 2,6-dichlorobenzoic acid, the $\ce{-COOH}$ group is pushed out of the plane restricting resonance.

Question: Does it mean the -I effect of chlorine atoms also cannot operate and reduce the charge on the deprotonated acid?

No, Inductive effect does operate because it is not dependent on orientation

Simple answer is no.

The inductive effect is not dependent on the orientation of the compound. Hence it will still work


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Why is there a phenyl shift in this reaction?
Why is there a phenyl shift in this reaction. I believe that the +R effect by 2 phenyl groups would be better than the +R effect by oxygen because of its lone pair. Can somebody please confirm?

There's no methyl shift but a phenyl shift. What else would you do besides shifting the carbocation and forming a ketone?

If a reaction is taken from higher stability to lower stability,IMO the rxn shouldn't carry out.

@NisargBhavsar Can this be explained by the octet rule that a complete octet of all the atoms will be more stable than the resonance by +R effect?

@Ritviz Yes, the carbocation can complete it's octet if it gets a back bond from the oxygen next to it and this is what someone meant when they mentioned that ketone formation is the driving force of the reaction, in the comments of your previous question.

@NisargBhavsar Thanks for the clarification


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How can orbitals of Hydrogen and Chlorine atom combine to form molecular orbitals of Hydrogen Chloride?
Linear combination of atomic orbitals (LCAO) states that orbitals can only combine if the energies and shape of the orbitals are same. $3p$ orbital of Chlorine and $1s$ orbital of Hydrogen are the ones forming the bond by sharing electrons but how?

"Linear combination of atomic orbitals (LCAO) states that orbitals can only combine if the energies and shape of the orbitals are same." - Who said that? Is there a (reputable) source for this claim?

Heard it from my chemistry teacher and even some articles online. But after adequate research got to know that this is not correct. The actual conditions for orbitals of atoms to combine is having the same or nearly the same energy and having the same symmetry about the molecular axis. it indeed is possible for  s and p orbitals to combine but a Py orbital wont combine with a Pz orbital due to different symmetries. Got my doubt cleared thanks!

Yep, that's a better criterion, I agree.

There seems to be a confusion in this demand.
The atomic orbitals may combine if they have the same energy and if they belong to the same atom. This is the way hybrid orbitals are introduced. But this is not valid for atomic orbitals belonging to different atoms. The $1s$ orbital of hydrogen can combine with a $3p$ orbital from chlorine, or with a $4p$ orbital from bromine, to make a sigma bond, even though their energies are different.


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Why is standard state useful?
In order to calculate the equilibrium constant we must know the value of $\Delta_\mathrm{r}G^{\circ}$ (which is a function of $T$) so we can calculate $K$ as:
$$K = \exp\left(-\frac{\Delta_\mathrm{r}G^{\circ}}{RT}\right)$$
Now if the standard state is changed (for example from $\pu{1 atm}$ to $\pu{3 atm}$) would the equilibrium constant change? My intuition says no because we can experimentally measure $K$ which varies only with temperature. So the value of $\Delta_\mathrm{r}G^{\circ}$ must stay constant irrespective of the standard state.
What is the point then to specify the standard state?
Edit
I would like to add a specific example that troubles me. Suppose we have the reaction:
$$\ce{A (g) -> B (g)}$$
that takes place in a closed container (fixed volume). We fill the container  with $\ce{A}$ and after some time has passed equilibrium is established. We can predict the ratio of partial pressures of the two gases (omitting fugacity for simplicity) from the equilibrium constant $K$ which equals $\exp(-\Delta_\mathrm{r}G^{\circ}/RT)$.
What I don't understand is why if we change the standard state (picking a different pressure) must change the equilibrium constant. At least for gases it doesn't make sense because in equilibrium is their partial pressure that matters.
But for constant temperature the Gibbs free energy of the compounds must be:
$$G_i(P_i) = G(P^{\circ}) + nRT \ln \left( \frac{P_i}{P^{\circ}} \right) $$
which, as orthocresol said in the comments, shows that $\Delta_\mathrm{r}G^{\circ}$ depends on pressure. I want to find solution to that contradiction and understand the significance of the standard state.

The expression for K in terms of partial pressures inherently assumes that the standard state is 1 bar.  If the number of moles changes, and the standard state is changed to 3 bars, the value for K will change.

Your example probably won't be affected much. Now think of the reactions which have gas on one side and no gas on another.

@IvanNeretin In that case the standard state should be selected according to the pressure that the reaction takes place (which can be different that $1$ atm)?

The standard state is selected once and for all. When selecting it, you don't care about one particular reaction.

@IvanNeretin Is there a physical significance to choose $1$ atm over $3$ atm? Why we choose this specific pressure?

Ever heard about the planet Earth? Well, that's why.

So for a chemical reaction that is taking place at very high altitude (again closed container) should we pick a different standard state? How equilibrium constant is affected in that case. What about an open reaction vessel?

@Anton Please note that, since 1982, standard pressure is $1\ \mathrm{bar}$ and not $1\ \mathrm{atm}$. Also note that the use of the unit $\mathrm{atm}$ is deprecated.

We all are familiar with the standard state as 1 bar pressure and 273 K.The standard state can be any well defined state. You can define it in any way that is convenient for you. There is no universal definition. We may want to choose different references depending on the calculation we are doing, so we can choose different standard states.
In thermodynamics we are usually interested in changes in quantities like internal energy, entropy etc, and to measure a change we need to choose a reference that we can measure the change relative to. That reference is the standard state.
What happens in situations where Pressure and temp keeps on changing ?
In situations where a property like pressure or temperature changes we generally have to calculate things by integrating from some initial state to some final state which can be tough , a lot of calculations .
Therefore , you can define standard state as the most simplest state.

Also , I discussed regarding this Q with John Rennie sir.

Let's say you have a reaction that has reached equilibrium. The reaction quotient $Q$ will be equal to the equilibrium constant $K$, and the Gibbs energy of reaction will be zero. Now, you decide to change the standard state (without touching your experiment, which is still at equilibrium). For example, if you are a biochemist, you could say that the pH at standard state is 7 instead of 0, or define a magnesium concentration smaller than $\pu{1 mol L-1}$ as standard concentration because that is closer to the situation in the cell.
The reaction will still be at equilibrium, the Gibbs energy of reaction will still be zero, the equilibrium concentrations will still be the same. However, $Q$, $K$, and the standard Gibbs energy of reaction will be different.
To understand why this happens, you have to be exact when defining $Q$ and $K$. These quantities depend on the activity (or fugacity) of the species, which are dimensionless quantities and have the value of 1 at standard state. This means that the activity or fugacity (or "concentration relative to the standard state" if you want to use an approximation and neglect non-ideality) depends on the choice of the standard state. The activity of hydrogen ions in a neutral solution is about $\pu{1e-7}$ for the conventional standard state (standard pH = 0) but is 1 when we define the standard state pH to be neutral.
Your specific example
For $$\ce{A(g)->B(g)},$$
changes in the standard pressure don't affect $K$. However, if the reaction is not a 1:1 stoichiometry (or more generally, the amount of gas changes), i.e. for
$$\ce{A(g)->2B(g)}$$
or
$$\ce{A(g)->B(s)},$$
you would see a change in $K$ if you choose a different standard state. The most famous example is often labeled $K_\mathrm{P}$ vs $K_\mathrm{c}$.

My intuition says no because we can experimentally measure K which varies only with temperature.

The thermodynamic $K$ varies with the definition of the standard state (the equilibrium concentrations do not).

So the value of $Δ_r G^\circ$ must stay constant irrespective of the standard state.

No, $Δ_r G^\circ$ depends on the choice of standard state, whereas $Δ_r G$ for a given set of concentrations (and temperature) does not.

What is the point then to specify the standard state?

You define it such that the standard state is close to what you are interested in, or it becomes easy to calculate the activity of a species. For example, you choose the standard state of the solvent to be pure solvent because that is close to what you have in a dilute solution. You choose the standard state of a solute to be approximately $\pu{1 mol L-1}$ (extrapolated from infinite dilution) because it makes it easier to calculate $Δ_r G$ from $Δ_r G^\circ$ for dilute solutes. Another way to say it: you want the difference between $Δ_r G$ and $Δ_r G^\circ$ to be small so that in many cases, you can make arguments about the direction of reactions looking at $Δ_r G^\circ$ that carry over to states other than the standard state.

As Chet Miller and Karsten pointed out, the culprit is the simple definition of K which we use in textbooks.
As in your last equation $$Gi(Pi)=G(P∘)+niRTln(Pi/P∘)$$
(similar equation can be used for dilute solutions in liquid phase)
This equation (differential of this equation in the form of chemical potential) is the origin point of the relation $$ \Delta G = G∘ + RTln(Q) $$ 1
In this relation, we should not forget that all the concentration terms of Q are activities, ratios with respect to standard state (whatever that may be).
It is only when we take 1 bar or 1 atm or 1 molar as a standard state we neglect these denominators in every concentration term.
So what will happen if the standard state will be 2 bar rather than 1 bar, every concentration term will get divided by 2 instead of 1 and K will be different.
What about concentrations of real-world experiment which are related to K, of course they won't change, but now we can't use previous definition of K, now every term in K will be concentration/2atm. And again it will lead to the same concentration values at equilibrium.
So there is no contradiction.
We have defined 1atm or 1 bar as a standard state because our partial pressures are of the order of 1 atm or 1 bar. I can define X pascal or X molar as a standard state but then we will need to divide each term by X pascals or X molars.
Can we define 1 atm for one reactant and 2 atm for another reactant in the same reaction as standard states? Yes, Here we can define each standard state independently, require only that the concentrations are so dilute that your last equation holds true. Here $G∘$ is not standard Gibbs free energy of formation but Gibbs free energy of any specific state wrt which we are defining our activity. The value of which can be calculated from Gibbs free energy of formation $Gf$. The $G∘$ here shows how $G$ changes with concentration, $Gf$ shows how it is related with other substances (specifically, how it is related with free element's Gibbs free energy) through reaction.


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How to understand the role of the lone pair on the molecular structure when it participates in resonance?
I'm a little confused as to why we ignore the effect the lone pair has on the geometry of the molecule when it is participating in resonance. Wouldn't it be the case that the molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal? Besides, the resonance with trigonal planar shape has a less importance because it has formal charges and the other does not.

"molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal?" Pretty much that, but trigonal planar has more importance, as far as geometry is concerned.

https://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in

https://chemistry.stackexchange.com/questions/30943/is-nitrogen-of-aniline-sp%c2%b2-or-sp%c2%b3-hybridized

Molecular structure is determined by the balance of electronic effects. Resonance is a model necessary to address the shortcomings of Lewis structures; what is depicted as molecular structures are hypothetical electron distributions belonging to a single nuclear arrangement, the disposition of which lead to a good approximation of the electronic structure. See also: https://chemistry.stackexchange.com/q/51632/4945

Wouldn't it be the case that the molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal?

"Hybrid state" refers to the distribution of electrons, not nuclei. Different resonance structures can have different electron distributions, but they can't have nuclei in different places, which means that they can't have different geometries. Either you have both of them be pyramidal, or both of them be planar.

Besides, the resonance with trigonal planar shape has a less importance because it has formal charges and the other does not.

Let's briefly digress. At its core, the question is what geometry does the amide have at nitrogen: is it pyramidal or planar?

Most nitrogens prefer to be pyramidal because of lone pair–bond pair repulsion, i.e. classic VSEPR theory. If the amide nitrogen adopts a pyramidal geometry, it therefore gains some stability from having the lone pair further away from the bond pair.

On the other hand, if it is pyramidal then the second resonance structure can't contribute at all (the bond angles at a doubly bonded nitrogen must be planar). So the resonance can only truly kick in if it is planar.

So it boils down to whether the LP–BP repulsion is more important, or the stabilisation from resonance is more important. How do you tell which one is more important, especially when neither of these can be measured quantitatively (with your current knowledge)...?
There are two possible answers to this:

Come up with a quantitative way to measure the repulsion, as well as the stabilisation, so that you can actually say for sure which one is larger. This means lots of theoretical calculations and quantum mechanics.

Take an amide, go into a lab, and measure its bond angles (or some other useful molecular property). You'll find that it turns out to be planar at nitrogen. Logically, that means that the resonance must be more important. How much more important we don't know; but we know it is more important.

So, it doesn't really matter what you or I "think" about the importance of the resonance: the fact is that the amide is planar. The explanation using resonance is merely a way to rationalise the experimental observation that amides are planar.

Thanks for the thorough response. I still am not sure of "Different resonance structures can have different electron distributions, but they can't have nuclei in different places, which means that they can't have different geometries." But the lone pairs themselves contribute to geometry because of repulsive forces. So why wouldn't different electron distributions cause different geometries?

@AmejiB. Hmmm... That is true for molecules, but resonance structures aren't molecules which have come to equilibrium after figuring out all the forces involved. They are hypothetical, nonexistent things, which represent possible states of molecules, which in all likelihood don't represent equilibrium states. So, yes, the resonance structure with planar N and lone pair on N, is probably not very good. But this is not how the molecule ends up, the molecule is not the resonance structure.

Why can't you have nuclei in different places?

@Vikki-formerlySean, you mean in two different resonance structures?

@orthocresol: Yes.

@Vikki-formerlySean It's not really a rule, it's more about what resonance structures are. Resonance structures are not tools to explain the overall structure of a molecule incl. nuclear geometry. They are meant as tools to describe the electronic wavefunction of a molecule: the idea is something like this: resonance structure A is $\psi_A$, resonance structure B is $\psi_B$, and the real electronic wavefunction (aka the hybrid) is the superposition $c_A\psi_A + c_B\psi_B$ (where $|c_A|$ and $|c_B|$ are somewhere between 0 and 1).

Crucially, all of these $\psi$'s are wavefunctions of electrons only, not nuclei. The entire framework is contingent on having already chosen a geometry for the nuclei. I'm not really sure the comments are the correct place to get into this in detail, but you may want to look up the Born–Oppenheimer approximation, which puts all of this on a much firmer theoretical basis. (I was not sure whether to go into B–O in the answer: I thought it would be a bit too much.)


text:
Can ozone oxidise iodine to periodic acid?
I noticed that the reaction of ozone with $\ce{I_2}$ gives iodic acid and not periodic acid. Upon searching, I found this question:
Why does iodine get oxidized to iodic acid and not periodic acid by nitric acid?
The answer says, nitric acid isn't strong enough to oxidise it, but my question is, even ozone is also not strong enough? And, is there no other reagent that could oxidise iodine to periodic acid except the electrolytic oxidation?
The reaction is: $$\ce{I_2 + 5 O_3 + H_2O ->2 HIO_3 + 5 O_2}$$

See: https://chemistry.stackexchange.com/questions/140193/molecular-structure-of-iodine-nonoxide

Ozone is not mentioned in the synthetic techniques reported by Wikipedia, citing Greenwood and Earnshaw[1].
Summary:

Orthoperiodate ion, $\ce{IO6^{5-}}$, is obtained by alkaline oxidation of iodate either electrolytically on a lead dioxide anode or chemically with chlorine.  The product may then be acidi-fied to $\ce{H5IO6}$. Alkaline conditions favor formation of the oxygen-rich orthoperiodate ion; note the inclusion of hydroxide ions in $\ce{IO3^- + \color{blue}{4 OH^-}->H2IO6^{3-} + H2O +2e^-}$ (the periodate ion is listed partially protonated, as it appears at pH 14 in the Pourbaix diagram from [2]).

Metaperiodic acid, $\ce{HIO4}$, may be obtained by dehydration of the ortho acid.  This can be accomplished by heating to 100°C (but below 150°C) or by treatment with dilute nitric acid.

Reference
1.
Greenwood, N. N.; Earnshaw, A (1997). Chemistry of the elements (2nd ed.). Butterworth-Heinemann. p. 872. doi:10.1016/C2009-0-30414-6. ISBN 978-0-7506-3365-9.
2.
Voutchkova, Denitza & Ernstsen, Vibeke & Kristiansen, Søren & Hansen, Birgitte. (2017). "Iodine in major Danish aquifers." Environmental Earth Sciences 76.
Doi: 10.1007/s12665-017-6775-6.

There was a previous question in Chem.SE which discussed the reaction between iodine and ozone. The product was found to be tetraiodine nonoxide, $\ce{I4O9}$. This reaction was performed in gas phase and published in a paper1. The reaction was done in a flow system, with $\ce{N2/O2}$ mixture as carrier gas at a pressure of 100 kPa and temperature range of 293-370 K. It was noted that $\pu{3.9 \pm 0.2}$ molecules of ozone reacted with one molecule of iodine. The reaction was of first order with rate constant, $\pu{ln k = ( 14.7 \pm 0.6) -( 2050 \pm 230)T^{-1} dm^3.mol^-1.s^-1}$.
Reference

Reaction of iodine with ozone in the gas phase, A. C. Vikis and R. MacFarlane, The Journal of Physical Chemistry 1985 89 (5), 812-815, DOI: 10.1021/j100251a019

I have been taught that the product is $\ce{I4O9}$ when carried out in absence of water, and it forms $\ce{HIO3}$ in presence of water. Is that correct? I feel the OP was asking about the reaction in presence of water.

@TRC Yes, you are correct.


text:
How to draw the different butyl cations?
On the image , can someone help me how to draw the different butyl cations? I'm confused as to where to place the positive charge on the different butyl cations. Going from left to right, the compounds are: isobutyl cation, n-butyl cation, and sec-butyl cation. Thank you so much!

What do you mean where to draw the charge?? The diagrams clearly show where (on which carbon) the +ve charge is.

@ShoubhikRMaiti, yes, but this is already the answer. The question asked me to draw the different butyl cations. How do I know where to place the + charge?

Oh I see, the cations are named in the same way as the side chains are named in organic compounds (alkyl group names). I will try to add an answer soon.

As I mentioned in the comments, the carbocations are named in the same vein as the alkyl functional group/side chains are named. To some extent, you have to learn which structure the names correspond to. Look at Alkyl at wikipedia for the names of some alkyl groups.
For example:

This is the isopropyl group. When drawing the isopropyl cation, you will draw the positive charge at the carbon where the missing bond is (bond and then wavy line)
This is the isopropyl carbocation:


text:
Vapor Pressure of reaction mass undergoing an exothermic reaction
A vessel containing a reaction mass under going an exothermic reaction is under vacuum, with temperature and pressure constant. When the source of vacuum is closed, the reactor begins to increase pressure thus increasing temperature. This will continue until a pressure relief device is activated, equilibrium will not be achieved. If the source of vacuum is opened again and vapor pressure begins to decrease the temperature follows. Can someone explain the principle of why the reaction mass cooled when pressure is decreased using vacuum.

Answers on CH SE site are figuratively paid by the user's own effort. When you ask, you are supposed to search and thoroughly think about the topic and to provide explicit compact summary of partial answers or at least ideas you have got until then. Effort not shown may be  considered as effort not done.

The underlying principle why the reaction mass is cooled when the pressure is decreased is simply evaporation cooling. The relevant quantity is the enthalpy of vaporization $\Delta H_\text{vap}$ of the considered substance. In equilibrium, the observed quantity can also be the equilibrium temperature that corresponds to the equilibrium vapour pressure. Note that the vapour pressure increases non-linearly with temperature according to the Clausius–Clapeyron relation.
For example, assuming the relevant substance (a reaction product or solvent) in the reactor is water and the initial pressure maintained by a vacuum pump is constant at $p_0=0.030\ \mathrm{bar}$, the corresponding equilibrium temperature is $T_0=24\ \mathrm{^\circ C}$. The reaction mass is cooled by the specific enthalpy of vaporization $\Delta h_{\text{vap},0}=2443.9\ \mathrm{kJ\ kg^{-1}}$. If the power of the reactor increases, the resulting evaporation rate will increase, if the power of the reactor decreases, the resulting evaporation rate will decrease, thus keeping the temperature approximately constant as long as the vacuum system can keep the pressure approximately constant.
When the vacuum system is closed, the pressure will no longer be kept constant. The pressure will increase due to the evaporating water. Accordingly, the equilibrium temperature will increase, too. For example, when reaching a pressure of $p_1=3.000\ \mathrm{bar}$, the temperature has increased to $T_1=133.5\ \mathrm{^\circ C}$. The reaction mass is still cooled by the specific enthalpy of vaporization $\Delta h_{\text{vap},1}=2163.5\ \mathrm{kJ\ kg^{-1}}$; however, no heat is transferred from the reactor vessel since the steam cannot escape.
Assuming that after reaching a pressure of $p_1=3.000\ \mathrm{bar}$, a steam relief valve is opened so that steam can escape from the reactor to the environment, and assuming this valve is big enough, the steam pressure will quickly drop to the ambient pressure of $p_2=1.000\ \mathrm{bar}$. Additional water will evaporate until the temperature has reached the equilibrium temperature of $T_2=99.6\ \mathrm{^\circ C}$. After that, water will continue to evaporate with the specific enthalpy of vaporization $\Delta h_{\text{vap},2}=2257.4\ \mathrm{kJ\ kg^{-1}}$.
If the vacuum system is opened again (and the steam relief valve is closed), the pressure will decrease to the initial value of $p_3=p_0=0.030\ \mathrm{bar}$. Additional water will evaporate until the temperature has reached the equilibrium temperature of $T_3=T_0=24\ \mathrm{^\circ C}$ again.


text:
Why is potassium less dense than sodium?
Potassium has a density of $\pu{0.86g/cm3}$ and sodium has a density of $\pu{0.97g/cm3}$,even though potassium is below sodium and one might expect the alkali metals to exhibit monotonously increasing densities down the group. Why is this discrepancy the case? All of the answers I found were either unrelated or vague.

Googling leads quickly to things like https://www.quora.com/Density-increases-down-the-group-from-lithium-to-caesium-then-why-is-pottasium-lighter-than-sodium

I don't know why you were downvoted, although phrasing of the question could definitely be improved. I could not find an answer on the site (although perhaps should have looked a little longer, sometimes answers are hidden within posts that are actually asking something else). The answer (obviously) has to do with differences in electronic structure, and the presence of low energy 3d orbitals. It would be nice to see a more detailed answer to this, it is unlikely to be simple. To get to the heart of the question you need to look into the electronic structure of atoms.

@JonCuster after some 20 minutes searching I could not find a particularly good answer (other than trivial ones, or something like "3d orbitals are available" which is next to trivial). Answers on other sites are often terrible. A good answer might however require some discussion of electronic structure beyond an introductory chem 'level.

Related: How does density of alkali metals increase down the group?

@BuckThorn I doubt it is electronic structure, more likely it's just that K is just lighter than you expect - after all it is one of the elements that's put in the wrong place if you order things by atomic weight, being lighter than Argon. And I agree "3d orbitals are available" is dreadful, I see that more of an explanation for K to be denser than expected, due to the 3d contraction

Hmmmm, that said looking at the lattice constants at https://periodictable.com/Properties/A/LatticeConstants.html does suggest the unit cell for K is a bit bigger than one would expect. Hmmmmm .... looks like it could be a bit of both, light K and big K, but far from obvious. Of course the down voters will jump in now explaining what I have missed.

@IanBush a lot of properties suggest weaker bonding in K, from surface tension to density to viscosity. Atomic weight plays a role but I suspect electronic structure is important.

Although density may be a relatively easily measurable property of solid materials, it may not suggest the most fundamental relationship between mass and volume for the elements. Molar volume (the volume required to contain $6 \times 10^{23}$ atoms of each the elements) is pictured in the graph below (WebElements):

It is amazing that ~100 elements can be arranged in a table with so many regularities - and even the irregularities can be explained if you try hard enough. There are some that are more unexpected and greater than the $\ce{Na-K}$ difference.
It is good that both $\ce{Na}$ and $\ce{K}$ have the same crystal structure (body-centered cubic). To explain the $\ce{Na-K}$ difference, some other differences should also be taken into account. Both have one electron outside a completed shell system - but sodium's $\mathrm{3s}$ electron (2 nodes) is dealing with a different kind of shielding compared to potassium's $\mathrm{4s}$ electron (3 nodes). In fact, the outer $\mathrm{3s}$ electron of sodium is ionized at $\pu{5.139 eV} \ (\pu{496 kJ/mol})$, while potassium is ionized more easily, at $\pu{4.34 eV} \ (\pu{418.8 kJ/mol})$. Since potassium holds onto its last electron so weakly (relative to sodium), the "sea of electrons" that helps hold the crystalline metal together isn't attracting the potassium ions as well as the sea does for sodium - not binding the crystal together as tightly. Result: lower density. Complete explanation? Probably not, but it's a start.
Although the enthalpy of fusion for both $\ce{Na}$ and $\ce{K}$ is given as $\pu{20.5 kJ/mol}$, the melting points are $\pu{63.38 ^\circ C}$ of $\ce{K}$ and $\pu{97.7 ^\circ C}$ for $\ce{Na}$, suggesting that sodium is slightly more tightly bound in the crystal. Even the boiling points $\pu{759 ^\circ C}$ of $\ce{K}$ vs $\pu{883 ^\circ C}$ of $\ce{Na}$ suggest that sodium atoms are bound more tightly than potassium in the liquid.
Interestingly, if you eliminate the $\mathrm{3s}$ electron of sodium and the $\mathrm{4s}$ electron of potassium, you drop to neon, which has a solid density of 1.444 and argon, with a solid density of 1.616. These two densities are about what you would expect; i.e., increasing as you go down the column. Note that the corresponding molar volumes are $\pu{13.23 cm^3}$ for $\ce{Ne}$ and $\pu{22.56 cm^3}$ for $\ce{Ar}$.
The same trend is found in the molar volumes of $\ce{Na}$ $(\pu{23.78 cm^3})$ and $\ce{K}$ $(\pu{45.94 cm^3})$, and the "discrepancy" is obscured.

Although I liked James Gaidis's answer, I do not agree with some of arguments because they are all parts of one or more continuous trends. For instance, look at the melting points and boiling points trends of alkali metals and other trends as illustrated in the following table:
$$
\begin{array}{l|ccc}
\hline
\bf{\text{Physical property}} & \ce{Li} && \ce{Na} && \ce{K} && \ce{Rb} && \ce{Cs}  \\
\hline
\text{Boiling point, }\pu{^\circ C} & 180.54 & \gt & 97.72 & \gt & 63.38 & \gt & 39.31 & \gt & 28.44  \\
\text{Melting point, }\pu{^\circ C} & 1342 & \gt & 883 & \gt & 759 & \gt & 688 & \gt & 671  \\
\text{Heat of fusion, }\pu{kJ mol-1} & 3.00 & \gt & 2.60 & \gt & 2.32 & \gt & 2.19 & \gt & 2.09  \\
\text{Heat of vaporization, }\pu{kJ mol-1} & 136 & \gt & 97.42 & \gt & 79.1 & \gt & 69 & \gt & 66.1  \\
\text{First ionization energy, }\pu{kJ mol-1} & 520.2 & \gt & 495.8 & \gt & 418.8 & \gt & 403.0 & \gt & 375.7  \\
\text{Density, }\pu{g cm-3} & 0.534 & \lt & 0.968 & \gt & 0.89 & \lt & 1.532 & \lt & 1.93  \\
\hline
\end{array}\\
\text{Source of the data: https://en.wikipedia.org/wiki/Alkali_metal}
$$
As shown in the above table, only density has broken the trend. Therefore the reasoning for this density inconsistance using other physical measures in the table is somewhat valid but not solid (see Buck Thorn's comments elsewhere). I'd like to give a different angle on this uneven trend.
Density is generally defind as the mass of unit volume $(\rho = \frac{m}{v})$. When you consider the density of an element, it is better to consider the molar mass and molar volume in place of mass and volume of other solids, respectively. Since the question is about the density difference of sodium $(\ce{Na})$ and potassium $(\ce{K})$, let's concentrate on alkali metal group. As usual and easity understand, the mass of a metal increases going down on the group. It is also clear that the principle quantam number $(n)$ increases as going down the group and hence, volume should have increased as well (adding more shells). The question is, the rate of increment of mass is comparable with the rate of increment of volume in each element? Let's make a table for this trend:
$$
\begin{array}{l|ccc}
\hline
\bf{\text{Property}} & \ce{_{3}Li} & \ce{_{11}Na} & \ce{_{19}K} & \ce{_{37}Rb} & \ce{_{55}Cs}  \\
\hline
\text{Molar mass, }\pu{g mol-1} & 6.94 & 22.99 & 39.10 & 85.47 & 132.91  \\
\text{Percent mass increment} & - & \approx 231.3\% & \approx 70.1\% & \approx 118.6\% & \approx 55.5\% \\
\text{Proton/nutron} & 3/4 & 11/12 & 19/20 & 37/48 & 55/78  \\
\text{Calculated atomic radii, }\pu{pm} & 167 & 190 & 243 & 265 & 298  \\
\text{measured covalent radii, }\pu{pm} & 145 & 180 & 220 & 235 & 260  \\
a\text{ (Side of the unit cell), }\pu{pm} & 351 & 429.6 & 532.8 & 598.5 & 614.1  \\
\text{Atomic radii from $a$, }\pu{pm} & 152 & 186 & 231 & 242 & 266  \\
\text{Percent volume increment (exp)} & - & \approx 22.4\% & \approx 24.2\% & \approx 4.8\% & \approx 9.9\% \\
\hline
\end{array}\\
\text{Source of the data: https://en.wikipedia.org/wiki/Atomic_radius}
$$
In the table, the atomic radii computed from theoretical models (calculated atomic radii) are from published work of Enrico Clementi and others (Ref.1) while the measured covalent radii for the metals are from published work of J. C. Slater (Ref.2). The experimental mattalic radii are from published crystal data.
As illustrated in the above table, when going from $\ce{Li}$ to $\ce{Na}$, the mass increased by $231\%$ but volume increased by only $22\%$. Thus, without a doubt, $\ce{Na}$ should have higher density than that of $\ce{Li}$. Similarly, we can argue that same principle would apply when going from $\ce{K}$ to $\ce{Rb}$ and going from $\ce{Ru}$ to $\ce{Cs}$ to explain the trend. Howevr, to explain the broken trend when going from $\ce{Na}$ to $\ce{K}$, we shoul look at the rates.
When going from $\ce{Li}$ to $\ce{Na}$ and going from $\ce{Na}$ to $\ce{K}$, the percent volume change is basically the same (22% vs 24%). Yet, with that increment togethher with $+131\%$ mass increment (more than 2.3 time increment) has given only $0.964 - 0.534 = 0.43$ density change. Therefore, one can expect much less density increment (if there is any) when going from $\ce{Na}$ to $\ce{K}$ because the mass increment is minimal (less than 100%) when compared to the percent volume increment.
On the other hand, when going down through a group, each period of the periodic table would add one shell. However, when going down from $\ce{Li}$ to $\ce{Na}$ and from $\ce{Na}$ to $\ce{K}$, the relavent nucleous gets only 7 protons. After that the addition of protons increases significantly (from $\ce{K}$ to $\ce{Rb}$, gains 18 protons and $\ce{Rb}$ to $\ce{Cs}$, gain another 18 protons). Near equal percent volume increment in either case suggests that the effects of opposite charge attractions and electron repulsions ate minimal in both cases. Bottom line is there is a clear discripancy of percent mass increment and percent volume increment at transition from $\ce{Na}$ to $\ce{K}$ stage.
I have calculated the percent volume increment using experimental crystal data as follows:
All of alkali metals crystallize with same crystal packing called body-centered cubic (bcc), which consists of one atom in the center of the cube and eight other atoms at the eight corners of the cube surrounding the cetral atom, each of them touching the central atom. Thus, the dioganal of the cube is equal to $4r$ where $r$ is the radius of the atom. If the length of each side of the cube is $a$, then diagonal is equal to $\sqrt{3}a$. Thus,
$$r = \frac{\sqrt{3}a}{4} \tag1$$
Keep in mind that $a$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For instance, it has been reported that the closest $\ce{Na-Na}$ separation is $\pu{372 pm}$ implying a sodium metallic radius of $\pu{186 pm}$ while the closest $\ce{K-K}$ separation is $\pu{461 pm}$ implying a potassium metallic radius of $\pu{231 pm}$ (as both calculated by using the equation $(1)$). The molar volume of the metal is not just the calculated volume of an atom considering it is a sphere times Avegadro number. The void volume between the atoms must be put into concern, packing is a big factor in volume calculations, thus density of the metal. The following calculations would show you the effect of the packing on the density:
If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each center atom shares with only one unit cell. Thus, total atoms per unit cell is $8 \times \frac{1}{8} + 1 = 2$. The volume of the unit cell is $a^3$, which is considered to be the volume of $\pu{2 atoms}$. Now, you can calculate the molar volume of each metal ($V_\ce{M}$):
$$V_\ce{M} = a_\ce{M}^3 \times \frac{1}{\pu{2 atoms}} \times N_A \tag2$$
The units of the volume would be depend on the units you have used for $a$. For instance, for $\ce{Na}$:
$$V_\ce{Na} = (\pu{429.6 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{2 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} \\ = (429.6)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{23.87 cm3 mol-1}$$
Similarly, for $\ce{K}$ and other alkali metals:
$$V_\ce{K} = (532.8)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{45.54 cm3 mol-1}$$
$$V_\ce{Li} = (351)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{13.02 cm3 mol-1}$$
$$V_\ce{Rb} = (558.5)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{52.45 cm3 mol-1}$$
$$V_\ce{Cs} = (614.1)^3 \times \pu{3.011 \times 10^{-7} cm3 mol-1} = \pu{69.73 cm3 mol-1}$$
Once we know the molar volume of each metal, we can calculate the density ($\rho$) by using $\rho=\frac{\text {molar mass}}{\text {molar volume}}$:
$$\rho_\ce{Li} = \frac{\pu{6.94 g mol-1}}{\pu{13.02 cm3 mol-1}} = \pu{0.533 g cm-3}$$
$$\rho_\ce{Na} = \frac{\pu{22.99 g mol-1}}{\pu{23.87 cm3 mol-1}} = \pu{0.963 g cm-3}$$
$$\rho_\ce{K} = \frac{\pu{39.10 g mol-1}}{\pu{45.54 cm3 mol-1}} = \pu{0.859 g cm-3}$$
$$\rho_\ce{Rb} = \frac{\pu{85.47 g mol-1}}{\pu{52.45 cm3 mol-1}} = \pu{1.630 g cm-3}$$
$$\rho_\ce{Cs} = \frac{\pu{132.91 g mol-1}}{\pu{69.73 cm3 mol-1}} = \pu{1.906 g cm-3}$$
These calculated values are consistence with the experimental values. Therefore, it is safe to say that the increasing trend of alkali metals are broken at $\ce{Na}$ and $\ce{K}$. After that, it again continues as follows:
$$\rho_\ce{Li} \lt \rho_\ce{Na} \gt \rho_\ce{K} \lt \rho_\ce{Rb} \lt \rho_\ce{Cs}$$
References:

E. Clementi, D. L. Raimondi, and W. P. Reinhardt, "Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons," J. Chem. Phys. 1967, 47(4), 1300–1307 (DOI: https://doi.org/10.1063/1.1712084).
J. C. Slater, "Atomic Radii in Crystals," J. Chem. Phys. 1964, 41(4), 3199–3205 (DOI: https://doi.org/10.1063/1.1725697).

BTW, thanks for the edit on my answer! I totally flubbed the bcc.


text:
Why are f-f transitions spin forbidden?
I know that f-f transitions in lanthanides are Laporte and spin forbidden but I don't understand why they are spin forbidden actually... I tried to write it down on a paper but I didn't manage to understand. I used to have problem with d-d spin forbidden transitions so this is maybe why I can't understand it right now...
Thank you in advance!

See: https://chemistry.stackexchange.com/questions/105543/how-can-f-f-transitions-happen

I don't think f-f transitions are always spin forbidden. The f-orbitals are buried close to the nucleus so they doesn't feel the effect of the ligand, so it's term symbols are not split by the ligand field. There can be excited term symbols with the same spin multiplicity, and that would mean the transition is spin allowed.


text:
Do formaldehyde and ethylene glycol give positive Fehling's test?
Fehling's test follows a single electron transfer mechanism, with electron transfer form the substrate to $\ce{Cu^2+}$, which forms $\ce{Cu^+},$ finally giving reddish brown precipitate of $\ce{Cu2O}$. The mechanism is as follows:

It can be easily concluded from this mechanism that formaldehyde without α-hydrogen does not give this test as it is the enolate which forms the complex intermediate with $\ce{Cu^2+}$. Formation of enolate is only possible if α-hydrogen is present.
If we look at formaldehyde, there are no α-hydrogens present, and in ethylene glycol there is no aldehyde group, so neither should give a positive Fehling's test.
However, the answer key from my organic chemistry exercise inexplicably states that both formaldehyde and ethylene glycol give positive Fehling's test. Is this correct? If so, what would be an explanation for it?

See: https://chemistry.stackexchange.com/questions/127463/fehlings-solution-turning-yellow-and-then-black?r=SearchResults

Fehling’s test is given by aldehydes, hemi-acetals, hemi-ketals , and alpha hydroxy ketones (due to enolization, one form is like an aldehyde). Note that aromatic aldehydes do not give Fehling test. Also an alpha hydrogen isn't needed, what is needed is the carbonyl group to be present ($\ce{-CHO}$). The mechanism for aldehydes with no alpha hydrogen is given below:

As for ethylene glycol, It is sort of an exception. You can refer to this video. https://m.youtube.com/watch?v=Wt1VzZxWv3Y (it’s in Hindi though, but you’ll get the idea, reacting $OH^-$ and then enolisation)
Please see: Does an acetal give a positive Tollens test?. Also you can refer to Test for functional groups in organic compound, NCERT Lab Manual for Class XII.

"As for ethylene glycol, I believe it should not give the positive Fehling test, though I am not sure why it is given so in your textbook." Don't believe. If you know, you know. If necessary, provide a reference where you learned it and explain why OP's textbook is wrong. 2. Don't recommended NCERT textbook as references.

@NilayGhosh I actually found out that ethylene glycol gives positive Fehling test, not sure of the mechanism, it is a sort of exception


text:
Examples of substitution and elimination reactions
What would be the major product of 4-chloro-1 pentanol reacting with water? Would cyclization occur? Would this be a type of substitution or elimination reaction?

Are you using an acid as well?

In just water? No reaction. Introduce base then you'll get cyclisation to the tetrahydrofuran see https://pubs.acs.org/doi/abs/10.1021/jo00795a026#


text:
What happens to the 2s orbital in carbon-hydrogen bonds?
I was taught in high school that carbon atoms by themselves are capable of establishing covalent bonds with with as far as four hydrogens atoms.

Nowadays, trying to understand the topic in-depth, I am confused about orbitals regarding this case. As far as I know, in the carbon atom's outer shell, the 2s orbital is filled, and the 2p orbital has two lonely electrons, but at the same time, each hydrogen forms a bond with one of this outer shell electrons. Does this mean the 2s orbital splits, or is there any other explanation?


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Why is cyclopropyl carbanion more stable than propyl carbanion
A problem was given in which the order of rate of decarboxylation was asked for the compounds cyclopropane carboxylic acid and and butanoic acid. It's known that the rate here depends on the stability of carbanion intermediate formed during the reaction. The answer was given that the rate of cyclopropane carboxylic acid is higher than that of butanoic acid which means that cyclopropyl carbanion is more stable than propyl carbanion but I'm unable to figure out why? Please help me here. (Here I saw the +I effect on the cyclopropyl carbanion due to 2 methyl groups and, +I effect of ethyl group on propyl carbanion but then too couldn't conclude the exact reason for such an answer.)

Cyclopropyl ring has greater electronegativity than a normal alkyl chain carbanion. Because of this, the negative charge gets pulled, hence resulting in stability.

do you have a reference for this?

Did you search this site first? https://chemistry.stackexchange.com/questions/67704/why-is-cyclopropane-more-acidic-than-propane


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Conductance and concentration of a strong electrolyte
I learnt that  with the increase of concentration, conductance of tr electrolyte decreases. While looking for the reason I found in my textbook that if concentration increase, amount of ions increases too. And for that ion-ion interaction gets a bit increased and lowers the speed of individual ions. Opposite thing occurs in case of dilute electrolyte solution.
But why is speed of ions even a matter here for concentrated solution? If the solution is concentrated, then availability of ions near electrodes increases a lot which means electrodes get a lot of charges to transfer from one to another a lot more than in dilute condition. That is current increases.
Can anyone please explain why does conductance really decreasd for increased concentration?

There were several  closely related questions for this topic. You may have not searched for them hard enough.

In concentrated solutions, some ions are making groups of positive and negative ions which are not able to conduct the current, as if they would make a neutral assembly, as if there was a covalence between them.


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Which method has the highest yield of calcium sulfate?
I'm not sure which of following two methods will give better yields for $\ce{CaSO4}$:
A. Add dilute $\ce{HNO3}$ to $\ce{CaCO3}$, followed by dilute $\ce{H2SO4}$; or
B. Titrate aqeuous calcium hydroxide with dilute sulfuric acid.
I'm leaning towards Method B since it is easier to point out when the reaction is complete, and the yield would thus be close to 100%.
For Method A, however, any excess $\ce{CaCO3}$ would react with $\ce{H2SO4}$ and any excess $\ce{Ca(NO3)2}$ might undergo a displacement reaction where $\ce{NO3-}$ ions are displaced by $\ce{SO4^2-}$.

A if yield is based on precipitation (compare calcium sulphate and hydroxide solubilities), B if on evaporation.

Consider that in A. you add an extra species that is not needed in the end, and in B., the only product besides gypsum is water.

Are you going to evaporate the leftover solution?