text:
Calculating internal energy change of a reaction
We are asked to find the relation between $\Delta H$ and $\Delta U$, and also, the sign of the internal energy change of reaction:
$$\ce{CO(g) +\dfrac{1}{2}O2(g)   ->  CO2(g)}$$
This was part (3) of a question, and this one explicitly mentioned: "reaction happens in an isolated container". The other parts had "closed container".
Approach: For ideal gases, we know that $U=nCvT$. So, we have:
$U_{rectants}:(1*5R/2*T) +(1/2*5R/2*T)=\dfrac{15}{4}RT$
$U_{products}:(1*3R*T)=3RT$.
Thus $\Delta U=U_{products}-U_{rectants}=\dfrac{3}{4}RT>0.$
However, the answer given is $\Delta U$ is $0$.
And the answer also claims that $\Delta H$<$\Delta U$ , which is expected from the relationship $\Delta H$=$\Delta U$+$RT\Delta n$ is not true.
Now I do not know exactly how do we account for "isolated system" in the formula for internal energy.
What exactly does isolated system refer to in this context? And how exactly is it affecting the internal energy?
A possible approach might have to do with the FLOT. I (guess) that isolated system means that there is no heat exchange with the surroundings, so the expression reduces to $dU+P_{ext}dV=0$. Does "isolated system" somehow also imply that $P_{ext}dV=0$?

Could you post the exact question?

Yes, please provide the exact statement of the problem.  It would help to know the initial state, for example.

"calculate sign of $\Delta U$, and the relationship between $\Delta U$ and $\Delta H$, for the reaction_________ in an isolated container"

Does it mention what the initial state is, such as molar quantities of reactants at 25 C and 1 bar in an isolated container?

@ChetMiller Ok, so I asked my friend (the source of this Q), for all the possible details: and it was a matching type question. We had to match reactions in Column 1 with column-2. The second column contained  the options: (A)$\Delta(H)<\Delta(U)$,   (B)$\Delta(H)>\Delta(U)$,  (C)$\Delta(H)=\Delta(U)  \neq0$,       (D)$\Delta(U)=0$.

As per the book , the only correct option is D. So it seems like the book claims that $\Delta(H)=\Delta(U)=0$ is the relation b/w H and U.

No other data is provided. Can you justify $\Delta H=\Delta U?$. My guess is that the definition is $H=U+P_{ext}V$. So for an isolated system, the second term becomes zero..

And another question. Where exactly did I go wrong in the approach i list in the question? I understand why Delta U=0 from your answer, but what exactly is wrong with my approach?

Have you had a chance to look over that expanded analysis I presented in the new Answer I provided?  I was wondering whether you were able to follow what I did, and if you had any questions.  I put a lot of time in on this, and would appreciate some kind of response.

@ChetMiller I did. However, I am afraid that we weren't provided with any kind of data.....and as per the book, the answer is,apparently ,$\Delta H=\Delta U$, So there might be some insight relating to "isolated system" that might be the clue.... But anyways, your answer gave my a new approach to solve problems, and I greatly appreciate the time.

Well, the book is wrong.

Ideally, if there was some issue with the posing of the question in the book, this question should have contained the question verbatim and with a proper citation. In its current state I don't find it very helpful for future visitors.

An isolated system is one that can exchange no mass, heat, or work with the surroundings.  So, from the first law, the change in internal energy of an isolated system in any process within the system is $\Delta U=0$.
Now for my analysis of this problem in greater detail to determine the enthalpy change.  I'm taking as the initial state of the system one mole of CO and 1/2 mole of O2 at 298 K and 1 bar in a rigid adiabatic container, and the final state as 1 mole of reaction product CO2 in this same container at whatever temperature and pressure result when the reaction occurs spontaneously in this same adiabatic rigid container.
My game plan is to examine the internal energy change for the process implemented in 3 successive steps, based on Hess' law to determine the overall internal energy change:

Reactants reacting at constant pressure and removing whatever heat is necessary to hold the temperature constant at 298 K.  So the final state in step 1 is one mole of CO2 product at 298 K and 1 bar, at 2/3 the original volume (since 3/2 moles goes to 1 mole).

Expanding the 1 mole of CO2 isothermally from 2/3 the original volume back up to the full original volume, while decreasing the pressure from 1 bar to 2/3 bar.  So the final state of step 2 is 1 mole of CO2 at 298 K and 2/3 bar pressure.

Adding heat to the 1 mole of CO2 at constant volume to increase its temperature to whatever value is necessary for the overall change in the internal energy for combination of the 3 steps to be equal to zero.

Analysis of internal energy change for step 1:
The process described for step 1 corresponds to the conditions for the standard heat of the reaction CO + 1/2 O2--> CO2 at 298 K and 1 bar, $\Delta H^0=-283000\ Joules$.  So the change in internal energy for this step is the standard change in internal energy $$\Delta U^0=\Delta H^0-RT^0\Delta n=\Delta H^0+\frac{RT^0}{2}=-283000-(8.314)(298)(-1/2)=-282000\ Joules$$
Analysis of internal energy change for step 2:
Step 2 involves an isothermal change for an ideal gas (without chemical reaction), so the internal energy change in step 2 for the system in this step is zero.
Analysis off internal energy change for step 3:
Step 3 involves the constant volume heating of one mole of the ideal gas O2 from $T^0=298 K$ to the final temperature T, so the change in internal energy from this step is $$(\Delta U)_3=C_v(T-T^0)$$ where $C_v$ is the molar heat capacity of CO2.
Overall combined internal energy change
It follows from all this that the overall internal energy change for this combined 3 step process is $$\Delta U=\Delta H^0+\frac{RT^0}{2}+C_v(T-T^0)$$But we know that this overall internal energy change must be equal to zero.  This enables us to calculate the temperature rise of the system in the adiabatic rigid container:  $$(T-T^0)=-\frac{\left(\Delta H^0+\frac{RT^0}{2}\right)}{C_v}$$The corresponding enthalpy change for the combined process is $$\Delta H=\Delta H^0+C_p(T-T^0)$$After some algebraic manipulation, this reduces to $$\Delta H=-(\gamma-1)\left(\Delta H^0+\frac{C_pT^0}{2}\right)$$Substituting literature values for the molar heat capacities and heat capacity ratio into this equation then yields the enthalpy change for the overall process:  $$\Delta H=+83000\ Joules$$ This indicates that answers B and D are both correct.
So, for this problem, we find that $$\Delta H>\Delta U=0$$
The main error you made in analyzing this (difficult) problem was failing to account for the heat of reaction (where the internal energy and enthalpy can change at constant temperature).


text:
Inconsistency in Elementary Bi-Molecular Reaction Rate Expressions
Assume the following bi-molecular reaction is elementary as written with rate constant $k_\mathrm{f}$:
$$\ce{A + A -> P}$$
This review, suggests to express the rate in terms of the production of product "P":
\begin{align}
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= k_\mathrm{f}[\ce{A}]^{2},&
\text{where }
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= \left(-\frac{1}{2}\right)\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t},
\end{align}
which gives:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}.\tag{1}$$
While this other review suggests that the rate be written in terms of the consumption of reactant "$\ce{A}$":
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k_\mathrm{f}[\ce{A}]^{2}.\tag{2}$$
Which is correct?

Read up on rate law, there is an issue there..

@Safdar I think I got it. Both expressions are correct, because 1) both rates quadruple when the reactant concentration is doubled and 2) the rate of consumption of reactants is equivalent to the rate of production of product. If we integrate both expressions and evaluate their concentration profiles w.r.t time, we will see that one has double the rate of reactant consumption compared to the other. However, vice-versa, the expression with the larger reactant consumption will have a larger rate of product formation. Therefore, as long as one is consistent, the approach is immaterial.

Related: https://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law

@CurtF. Oh wow, I should have been more diligent with my research before posting this question. Should I take my question down? Also, per the the link you shared, I guess my previous comment was correct? Thanks!

I don't think this is quite a duplicate, but the other question is related.  Just wanted to make sure you saw it!

And yes your comment sounds right to me.

If the reaction is
$$\ce{2A -> P},$$
then the rates are defined to be related as
$$
  -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
$$
and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with the rate constant. With any other usage an explanation should be added so as not to confuse things.
In the general case
$$\ce{aA + bB -> pP + qQ},$$
then
$$
  -\frac{1}{a}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = -\frac{1}{b}\frac{\mathrm{d}\ce{B}}{\mathrm{d}t}
  = \frac{1}{p}\frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
  = \frac{1}{q}\frac{\mathrm{d}\ce{Q}}{\mathrm{d}t}
  = \frac{\mathrm{d}x}{\mathrm{d}t},
$$
where at any time $t$, $ax$ moles of $\ce{A}$ and $bx$ of $\ce{B}$  are consumed and $px$ of $\ce{P}$ are produced, where $x$ is the extent of reaction.

Using the extent of reaction, then deducing an expression for reactant consumption/product formation makes the most sense. Thanks!

The first expression
$$\frac{\mathrm{d}[A]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}\tag{1}$$
is the correct one for the reaction given. As the OP mentions in the comments, both approaches describe the situation correctly. However, the values of $k_\mathrm{f}$ would be different by a factor of two.
You could rewrite the chemical equation by dividing by two (no longer an elementary reaction, so you have to specify that it is second order in $\ce{A}$):
$$\ce{A -> 1/2 B}$$
In this case, the correct expression for the rate of $\ce{A}$ reacting would indeed be the second expression:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k_\mathrm{f}[\ce{A}]^{2}\tag{2}$$
Because the two reactions are written differently, you again expect a different value for $k_\mathrm{f}$.

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text:
Does the temperature during ice melting change?
I did an online test for physics and one of the questions was: Salt was added to ice. Because of it the ice melted. What was happening with temperature DURING the process of melting? The right answer was that temperature was lowering.
Can please someone explain why? I always thought that during melting the temperature is constant and after all ice is melted then the temperature changes.

You have to account for freezing point depression of salt water.

I understand that salt lowers the freezing point of water but why is the temperature changing during melting. Why isn't working the rule that during melting energy is used for state transition and the temperature stays same?

That rule is only true for the melting of a homogeneous substance.

Oh and could be so nice please and explain me a bit more what happens during the melting in this case? Where is the energy used in this case when temperature changes?

Salt water freezes below 0°C, precipitating ice with significantly lower salt content. Hence, if ice melts in salt water, it cools down to a lower temp. While it melts, the salt concentration in the water would decrease of course, and the temperature would slowly rise towards 0°C again. The energy goes into melting the ice, obviously. The opposite of heat of fusion.

This is called a "Colligative property" - which include depression of the freezing point and elevation of the boiling point (plus 2 others).  These properties increase in effect as more solute is added.  I believe adding salt to ice only melts it if there is already liquid water present, which there usually is (it might be a very thin  coat of water on the ice).  This thin layer dissolves some salt, which reduces the freezing point of the liquid water (salt solution), leading to the surrounding ice to melt, which dissolves more salt, melting more ice etc.  These properties are more related to the amount of solute, rather than the type of solute (at least in an ideal solution).
Reference

Colligative Properties of Simple Solutions, Frank C. Andrews,
Science  05 Nov 1976: Vol. 194, Issue 4265, pp. 567-571

Here is a naive interpretation, that may help you understand this phenomena.
In the presence of salt, water is attracted and "wants" to be adsorbed around $\ce{Na+}$ and $\ce{Cl-}$ ions. But this operation requires some energy for water molecules to separate the positive and negative ions. This is why the dissolution of $\ce{NaCl}$ in water is endothermic.
Now if you put salt in contact with ice, water in ice "wants" also to be attached to surrounding ions. But it cannot move. It first have to be a liquid. And there is no heat in the surroundings to make ice melt. Nature has found a way. Every piece of ice is divided in two parts. One part gives heat to the second part, that melts, makes water to dissolve the salt. But the remaining part has lost heat. So it gets cooler and cooler. The temperature of the remaining ice decreases from $0$°C to about -$20$°C.

After reading the wording carefully, I believe the question and the OP are asking, during melting, how does the temperature of system change (if any)? The question is not asking what happened to melting point of ice.
The temperature decreases is simply because melting of ice is endothermic (and assuming the system is adiabatic so the energy of system but not of surrounding is used for melting). It melts because the system is moving towards most thermodynamically favourable composition.
Considering pressure at 1 atm,

if you consider melting of pure ice at constant temperature (isothermal), the melting is endothermic too but the net effect is that energy used for melting is taken from surrounding not from system, so the temperature of system does not change, all ices melt into liquid water if that constant temperature > 0 °C;

else if you consider melting of pure ice adiabatically (no heat enter nor leaving the system), the temperature of system will drop as melting goes because the system uses its own energy for melting, until equilibrium temperature (0 °C) reached, and then the melting stop. Solid and liquid present at equilibrium, not all ices melt;

else if you consider the melting of ice in the presence of salt isothermally, melting happens if that constant temperature > -21.1 °C.1 Why -21.1 °C but not 0 °C ? That is a thermodynamic effect due to mixing, and apparently this effect is more significant than colligative freezing point depression. Of course the system temperature will not change, same reasoning as (1);

else if you consider the melting of ice in the presence of salt adiabatically, the temperature will drop to -21.1 °C as melting goes, same reasoning as (2).

I think the question is asking for (4), and you are confused because treating (4) as (3).

Reference
1 Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 189, (a) Eutectics

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text:
What is cn axis of symmetry
While reading about axis of symmentry and I'm not able to comprehend what does C infinite or rather n signify in a molecule

I know axis of symmetry is the axis along which rotation of molecule results in the same molecule so what does n or infinite signify. Animation could help i think so do let me know if theres an app for drawing 3d molecules

There are only two atoms in HCl, so you can rotate by any arbitrary number of degrees about the axis: infinitely many possibilities.

jmol or webmo(for easier UI), you can choose any, both show the symmetry elements, and perhaps you should ask this on Maths stack exchange

If there is an infinite number of proper rotations around an axis to match the original of an object with its image -- the example of $\ce{HCl}$ in the example by @EdV, than this axis is $C_\infty{}$ and thus (all altogether) $C_{\infty{}v}$ like $\ce{HCN}$.  Otherwise $n$ is an integer number for any rotation $360^\circ{}/n$ for $n > 1$ (a reference).  Training (including tests) is possible at Symmetry@Otterbein; actually both for molecular symmetry, as well for some examples in crystallography, too.

You may use Jmol to identify symmetry elements of molecules on your own, too.  Maybe you have a sketcher (like the NIH PubChem Sketcher) to draw a molecule, and export it in a format Jmol understands (e.g., MDL molfile .mol,  or .pdb frequently seen in biochem) which you may read by Jmol.  (The model kit included in Jmol may be a fallback to stitch the atoms together, too.)
Under File -> Console, you may access Jmol's text-based interface (CLI).  Let's assume we have a model of methane, $\ce{CH4}$, the analysis could proceed in the following sequence:

type command calculate pointgroup, to be confirmed by enter. Jmol will scrutinize the current model's symmetry, and display the result to the CLI as Td.

show poingroup will yield a listing of what Jmol identified.
# 5 atoms

Td    center  {0.0 0.0 0.0}

Td    nC3 4   2
Td    C3_1    {0.5773503, 0.5773503, 0.5773503}
Td    C3_2    {0.5773503, 0.5773503, -0.5773503}
Td    C3_3    {0.5773503, -0.5773503, 0.5773503}
Td    C3_4    {0.5773503, -0.5773503, -0.5773503}

Td    nC2 3   1
Td    C2_1    {0.0, 0.0, -1.0}
Td    C2_2    {0.0, -1.0, 0.0}
Td    C2_3    {-1.0, 0.0, 0.0}

Td    nS4 3   2
Td    S4_1    {0.0, 0.0, -1.0}
Td    S4_2    {0.0, -1.0, 0.0}
Td    S4_3    {-1.0, 0.0, 0.0}

Td    nCs 6   1
Td    Cs_1    {-0.7071068, -0.7071068, 0.0}
Td    Cs_2    {0.7071068, -0.7071068, 0.0}
Td    Cs_3    {0.7071068, 0.0, 0.7071068}
Td    Cs_4    {0.7071068, 0.0, -0.7071068}
Td    Cs_5    {0.0, -0.7071068, -0.7071068}
Td    Cs_6    {0.0, 0.7071068, -0.7071068}

Td    type    nType   nUnique
Td    E     1   1
Td    Ci    0   0
Td    Cs    6   6
Td    Cn    7  11
Td    Sn    3   6
Td        TOTAL    24

Command draw pointgroup triggers Jmol to display all elements identified at once, e.g.

To walk around the model, hold the left mouse key.  To change the scale of representation use (shift + left mouse button), to pan (Ctrl + Alt + left mouse button), etc.

to revert to the previous state, type undo.  You may display symmetry elements groupwise, e.g. by draw pointgroup Cn to see only the $C_n$ axes:

or draw pointgroup C3 2 to draw only the second $C_3$ axis

(The underlying syntax is compiled here.  To dive deeper into Jmol's syntax, the teaching set with examples related to biochemistry by Nathan Silva and David Marcey here offers a reference, too.)


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What would happen if equal parts of hydrochloric acid, mercury and methylene chloride were mixed?
I was going through the archive of Real Life Comics (first time reader, enjoying it so far), and in one of the comics, the main character of the comic reveals he has ordered hydrochloric acid, methylene chloride and mercury from an online chemist, and then says he ordered them to "pour them together and see what he gets".

I assume mixing these chemicals is probably not a good idea, but what would happen when mixing them? Like, I don't expect anyone to try this out (though if you can or have done this, your experience is definitely valued), but based on the properties of these chemicals, what would be the likely outcome?

Pretty much nothing, actually. They wouldn't mix together - there would be three layers.

They probably picked things that would not react on purpose.  I could imagine the emergency/fire calls if they had written pool chlorine, gasoline and chromic acid and some readers took it the wrong way.


text:
For IUPAC nomenclature, does the number of multiple bonds take precedence over the chain length?
I've received a few mixed answers to this question. Some sources say when determining the parent chain for unsaturated hydrocarbons, the longest chain should be considered before the chain with the most multiple bonds, for example:
https://www.acdlabs.com/iupac/nomenclature/79/r79_933.htm
While others say what appears to be the opposite, such as this excerpt from Wikipedia:

The steps for naming an organic compound are:
Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents or branches cited as prefixes.
It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
It should have the maximum length.
It should have the maximum number of multiple bonds.
It should have the maximum number of single bonds.

Would be great if I were to receive clarification regarding this.

Related: https://chemistry.stackexchange.com/q/67599/7951

In the 21st century, it is no longer important to create names that strictly follow IUPAC nomenclature. That made sense in the old days, when databases were printed on paper and came with thick volumes containing indexes. In the age of computer networks, unique human-readable names are largely irrelevant. Also chemists have always disregarded IUPAC in publications when an alternative order of precedence was more useful within a publication.

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes. Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Note that the new IUPAC recommendations (Blue Book in 2013) on order of chain length and number of multiple bonds have been changed (You can find more details here and here):

In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

For example consider following structures:

In all first three structures, the longest chain has chosen over the multiple bonds (the old names are in orange). In the typical case, the principle group is $\ce{-CHO}$ and hence hence to hexanal is the principal structure which has $\ce{-OH}$ substituent and longest chain. Thus, the name is 2-ethynyl-5-hydroxyhexanal.

How would it be named "tert-butyl" according to present IUPAC rules ?


text:
Is it possible for already charged metal ions to form a metallic bond?
I would like to know if it is possible for already charged metal ions to become part of a metallic bond.
I was thinking of a possible metallic bond formation with other already charged metal ions or a metallic bond formation between elemental uncharged metal atoms and already charged charged metal ions. By formation a mean the moment before the metallic bond has formed because my understanding is that in a metallic bond there are only charged metal ions and no elemental uncharged metal atoms.
If one of the two scenarios or even both could occur I would assume that the already charged metal ions would form or join a metallic bond by either giving up or donating additional electrons to the so called sea of electrons of a metallic bond or the already charged metal ions would form or join the metallic bond by not giving up additional electrons.
It would be very helpful if someone could provide a link for citation about whether or not charged metal ions can become part of metallic bond since I could not find anything explicitly ruling out that possibility.

"metallic bond there are only charged metal ions and no elemental uncharged metal atoms." - that is incorrect. There's no charge on atoms in metal, well perhaps there are some partial charges, especially in alloys, but you're view on metallic bonding is incorrect.

A bunch of metal ions aren't going to hang together without a bunch of electrons to neutralize the bulk to a great extent.

Your reasoning is correct. Such things do exist, and they came to be known as clusters. The link above does not quite do them justice; in fact, they are vastly more numerous and diverse.
Come to think of it, the clusters go beyond the scope of your question, in that many of them are not charged. To pick a few examples, see the osmium carbonyl (which, arguably, is too small to consider the bond really metallic, but its bigger brothers are known), or gold clusters which range all the way up to the immense five-layer icosahedron $\ce{Au561}$ with some ligands on top.
To pick something charged and at the same time big enough to be called metallic, have a look at this: $\ce{Ta6Cl18^4-}$.

[source]

Upd. I thought I'd better stress this again. Yes, metal ions that still have some valence electrons can use those to form bonds, given the right circumstances. Whether those bonds qualify as metallic is a different question altogether, and depends on the number of atoms (or ions) in the cluster. Two is clearly too few. You can't possibly have a metallic bond between two atoms. A bond with electrons delocalized between two atoms is called covalent. That's the case of $\ce{Hg2^2+}$. Then you have clusters with 3 atoms, then 4, and so on, all the way up to 100 and beyond.
So it goes.

A somewhat exotic but interesting example is barium platinide $\ce{BaPt}$. Though there are several different compounds depending on the ratio between the elements, the idealised structure of $\ce{BaPt}$ consists of infinite chains of platinum atoms with one negative charge per atom. The distance between them is actually shorter than in pure platinum, so they are clearly tightly bound. The barium atoms are fully ionised, which leaves an extra electron to float around. The compound is also a 2D metal, though not through the platinum chain axis.

Thank you very much for the answer. After reading the answer I’m guessing that my question might have been imprecise in regard to what I was thinking about. By metallic bond I only meant the bonding metals do where their valence electrons are delocalized and part of the sea of electrons. Therefore the question should have rather been, can already charged metal ions join or form metallic bonds with delocalized valence electrons? Do you know if there is an example of such a scenario? My understanding is that your examples do not have delocalized electrons or a sea of electrons. Thank you again

My examples do include seas of electrons, albeit small ones, to the point that you may refuse to call them seas. How many stones comprise a heap, they ask? How big should a puddle be for you to call it a sea? In my opinion, $\ce{Au561}$ is pretty much as metallic as any metal (on the inside, that is).

Ok. Thank you, that is interesting. Could you lastly tell me if my conclusion of your answer is right that metal ions, that are charged before they are part of a metallic bond, are able to join or form metallic bonds with delocalized electrons or sea of electrons, with other already charged metal ions or other uncharged elemental metal atoms? I was until now under the impression that only elemental uncharged metal atoms can join or form metallic bonds with delocalized electrons and that these elemental uncharged metal atoms only become charged ions after the formation of the metallic bond.

Yes they are able to do so (sometimes), but not all metals and not all ions, and it depends on other circumstances too.

And if metal ions that are charged before the formation of the metallic bond with delocalized electrons, join or form metallic bonds with other before formation charged metal ions or other elemental uncharged metal atoms, by giving up or donating additional electrons to the sea of electrons?

I don't quite understand what your question is. Well, yes, those metal ions form bonds using their remaining electrons, much like $\ce{Hg}$ in $\ce{Hg2^2+}$. When these bonds grow to span many atoms, we use to call them metallic. How many is "many" is up to you. 2 certainly isn't; 561 certainly is.

Ok. By the question I meant whether metal ions that are charged before the metallic bond formation, are able to delocalize electrons or donate additional electrons to the sea of electrons just as elemental uncharged metal atoms do in the event of a metallic bond formation. I thought metal ions would not give up or donate all their remaining valence electrons to the sea of electrons since they already gave up some electrons at the time they became ions. That maybe falsely assumes that in your Mercury example there are no delocalized electrons

Yes the ions are able to do so, provided they still have some electrons left and the circumstances are right. $\ce{Hg2^2+}$ surely has two delocalized electrons, but they are delocalized over two atoms. That's what we call a covalent bond. In bigger clusters, the electrons come to be delocalized over 10, then over 100 atoms. That sounds metallic enough to me.


text:
How do I get the smell of Febreeze (uncomplexed cyclodextrin composition) out of my carpet?
Help! I sprayed my old carpet, which had gotten a little mildewy from some wet towels left on it, with Febreeze original scent. Now I can't get that horrible scent out! It might also be mixed with some newly formulated Nature's Miracle which I sprayed on the couch above the carpet at the same time, but I don't know the chemical formula for that.
I have  tried the following: white vinegar in water, vodka in water with some grapefruit essential oil, bleach solution in water, Concrobium Mold Control spray (I can still smell the sour mildewy smell...or is that the vinegar?), ZEP Mold Stain and Mildew Stain remover, Folex carpet spot remover, baking soda, corn starch, vacuumed up the last two as best I could...
So my question is, what next? Hydrogen peroxide? Isopropyl alcohol? Unscented dishwashing detergent? I don't want to layer on any more scents, god forbid, in an attempt to get rid of this fake flowery sour smell. (Someone described the Nature's Miracle scent as making their cat smell like "candy-coated skunk." I think the Febreze might have bonded to the Nature's Miracle...)
I would enjoy understanding the science of this too!
My research so far online says that Febreeze cyclodextrin is cup shaped and surrounds smelly organic molecules, but doesn't break them down. And the scent that's added to the cyclodextrin is re-released through heat, so steam isn't recommended. And I'm a renter and I doubt my landlords will replace the carpet.
Thanks!
Annette

Test chlorine bleach on a hidden spot of the carpet and wait a few days to be sure the material is unaffected. If so, then test a dilute solution of chlorine bleach on a cloth with the horrid scent, to see if it improves things, or makes it worse, before tryin it on the carpet itself.

Waiting for a cast iron pan seasoning one :)

Do not use chlorine bleach products here! While possibly effective the chloro-organics are not something you want to have permanently around. However, percarbonate based products (those other chlorine-free products) may work here and you may be able to live, at least longer-term, with the results. A dry path, ozone or singlet oxygen, but there are short term toxicity concerns.

Alchimista: One could actually try a small surface area cast-iron source (the anode) and lots of activated carbon (cathode) in an electrolyte of baking soda and H2O2. Logic: one-electron reduction of H2O2 forms hydroxyl radicals which convert to carbonate radicals (from .OH + HCO3- = H2O + .CO3-) that are good at breaking down organics. You could forget the Fe/Carbon galvanic cell approach and just use UV light on the H2O2/NaHCO3, or B-12/ZnO in H2O2/NaHCO3 in strong sunlight also.

One should test all my recommendations on small areas first to determine both effectiveness and correct exposure time. Note: Overdoing the latter could progressively remove coloration (as in bleach) the carpet also.


text:
How long will it takes two miscible liquids to fully mix (homogenize) through diffusion?
This question occurred to me while I was pouring milk into my coffee and was too lazy to stir.
Assume you have some volume $V_a$ of liquid A directly on top of another volume $V_b$ of liquid B. The two liquids are fully miscible. With no stirring, how long until the solution becomes fully homogenous solely by diffusion?

There's no 'formula' per se, without knowing some more details eg the mass of the liquids, the viscosity of the liquids etc. It's easier for gases, where you can apply Graham's Diffusion Law. For liquids, you should use a stopwatch, conduct the experiment and share the results with us :)

@AniruddhaDeb Haha, I expected it to be a pretty tricky problem. If it helps, you can assume whatever else you like: the masses of the liquids, the temperature, force of gravity (or zero-g), air pressure, diffusion coefficients, etc. Even an answer to a much simpler problem (spherical cow) would be interesting.

Diffusion is a slow process, see Ficks's Laws of diffusion and look up diffusion coefficients, typically $D=10^{-9}\mathrm{,m^2,/s}$ and distance diffused $\ell=\sqrt{6Dt}$ in time $t$. In most cased, as it your example, convection dominates over diffusion


text:
How are diazonium salts prepared without self coupling
While preparing diazonium salts, what prevents the freshly formed diazonium salt from reacting with the still unreacted amines left in the solution and perform diazo coupling?
If this happened then there won't be an appreciable yield of the diazo salt.
So what's the reason?

Reaction do not have 100% yield in reality..

Yes but if coupling happens then percentage of diazonium salt left would be practically negligible right?

Why do you say so? Have you attempted to extract a diazo compound from a solution and faced this problem? (I assume this is a query for something you think should happen, I haven't done this either so I cannot answer for methods of extraction) But, a goal of synthesis is to maximise yield and so I would assume that a simple way would be that you add enough of the reagent to minimise the amount of amines present in the final solution, maximising yield of the diazo compound.

Yeah this was something I think would happen. So we add the reagents in excess and control it! Okay I understand that now. Thanks!

An example of a diazonium salt can be seen here.Also this youtube video shows preparation of the chloride..

The NO+ species is more reactive than the diazonium salt so the amine reacts with that if it is present.

The excess of nitrous acid and the order of addition are indeed important. But I am surprised nobody mentioned pH. It needs to be kept around 5-6 to allow enough unprotonated amine to be present and react with nitronium, while preventing side reactions, e.g. addition of water to the newly formed diazonium cation, or indeed addition of free amine to it. That, as far as I can recall from my organic chemistry days, is the most important parameter to consider.

During diazonium ion preparation, first, sodium nitrite is mixed with hydrochloric acid to produce nitrous acid. The nitrous acid can be protonated under acidic conditions, which undergoes the loss of water and produce the nitrosonium ion ($\ce{O=N+}$; this reaction is usually performed at or around $\pu{0 ^\circ C}$):

An aromatic amine can attack the electrophilic nitrosonium ion to form a nitrosoamine $(\ce{Ar-NH-N=O})$.
The nitrosoamine can undergo a proton transfer (tautomerism) to give corresponding imminol $(\ce{Ar-N=N-OH})$.
Under acidic conditions, this $\ce{N=N-OH}$ group can be protonated and left as a water molecule to give the diazonium salt, which is resonance stabilized $(\ce{(Ar-N#N)+Cl-})$.

The key word here is all of these happen under acidic conditions. Therefore, the aromatic nucleus of aniline derivative is not active because of protonation. Also, during the synthesis, the starting aniline derivative is used as the limiting reagent. In addition, many of diazonium salts are unstable and hence, they are always prepared as needed under acidic conditions with good stirring, kept at or near $\pu{0 ^\circ C}$ to minimize the side reactions such as the reaction with water to produce a corresponding phenol until used in the coupling reaction (usually immediately).
The diazonium salt reacts as an electrophile with an electron-rich coupling component (a phenol or an aniline derivative), through an electrophilic aromatic substitution mechanism under basic conditions (specially when coupling reagent is a phenol derivative). To complete the synthesis of some azo dyes, it is also required heating.


text:
What would be the effect of adding liquid beyond the calibration mark on a volumetric flask?
If this happens in the course of preparing a volumetric solution, what should be done?

Best solution : start again !

Secondbest soloution: Pour into a flask of twice the volume, and get it right there.

Volumetric flasks are TC(to contain) not TD (to deliver) type glassware, one cannot pour the contents into another flask, until and unless the first one is washed quantitatively.

You can wash the first flask into the second though

If you have plenty of solution and reagents, the solution should be prepared again, because the volume is not the nominal volume written on the flask. However, the proper trick to fill the volumetric flask is to use a dropper when you are about to reach the calibration line. Experienced hands can use a squeeze bottle until the end but the chances of this mistake are high.
However, if the solution was was just a single sample solution (provided by somebody else, or the compound is too expensive or rare), slightly overfilling the volumetric flask is not the end of the world. You can mark the liquid level after proceeding normally (dissolve and invert several times). With a help of a high quality volumetric syringe or gas tight syringe (e.g., 2.5 mL Hamilton used in gas chromatography injections) draw the excess liquid until the liquid level is back to the calibration line. Measure that volume in the syringe and add it to the nominal volume of the volumetric flask.
If the volumetric flask had a nominal volume = 100.0 mL and you had to withdraw 1.80 mL then the total volume would be 101.8 mL for molarity or other concentration calculation.

(+1) Excellent answer! One semester I was teaching quant or instrumental analysis and I was in the lab with the students. One student went missing. Turns out they overshot the mark, so they went to the attached lab next door (the analytical balances were there), attached a hose to a water aspirator, put the other end into the neck of the volumetric flask and turned on the aspirator! When found, they said they were trying to get rid of the extra water by pulling a vacuum on it. Nice teachable moment! ;)

Evaporating the extra water is perfectly legitimate approach - as long as you don't induce boiling in the tiny flask neck.

Evaporation is a bad idea if your solute is also volatile as is the case in gas chromatography or sometimes in HPLC.


text:
Pyridine: Protic or Aprotic
I know pyridine is a polar aprotic solvent (owing to a permanent dipole generated), but if acidified it forms pyridinium cation. So, does it then function as polar protic solvent?

Pyridinium salts are generally solids. Are you refering to them as solvents in the molten state?

If you acidify pyridine with $\ce{HCl}$, for example, you form pyridinium chloride, which is a solid (m.p. $\pu{144 ^\circ{}C}$, same reference). Thus, at ambient conditions, it is not considered a solvent.  With a $pK_a$ around 5, it is about as (mildly) acidic as acetic acid, though.

pKa of conjugate acid of pyridine is 5.2, while pKa of water is 7. If you could get so many pyridinium cations in liquid form, then theoretically yes, it would be a polar protic solvent.

Ad hoc, because of Waylander's comment: as an example, $\ce{py.HCl}$ may be molten, then used as a solvent, e.g., 2004OrgProcResDev670.

Usually the designation of a solvent as aprotic is based on the unmodified solvent molecule, which in the case of pyridine is aprotic.
Pyridine is not unique in terms of this question.  We would generally expect polar aprotic solvents such as DMSO to accept a proton at their negative charge centers (such as the oxygen in DMSO) given a suitable acid.  Such a reaction enables polar aprotic solvents to preferentially conduct solvated protons versus other ions via a mechanism similar to the Grotthus mechanism in water.


text:
Experience with PTFE tubing & glass connectors?
I'm wondering which PTFE tubing size to get to connect to normal 10mm OD glass connectors. For a silicon hose I would take an ID of 8-9mm as it is flexible but I think PTFE is quite rigid.
Anyone has some experience with PTFE tubing?

In short: Do not attempt a direct connection of the PFTE hose with the glass.
Instead (assuming the tubing shall transfer gases with little pressure only):

Because to the (compared to silicon) lesser flexibility of the tubings made from PFTE, you should consider to use glassware with threaded joints with detachable connectors. This a theme already seen with condensers, e.g.

(source)
but may be equally applied to impingers

(from here)
or overhead exhausts:

(from here)
These connectors may be replaced with ones made mostly out of PFTE, too, e.g.

(source)
Thus you may connect the tubing to the connector at the bench (gentle warming of the tubing with a heat gun may help), secure this end with the clip, and then mount the connected hose to the reactor.  It is faster and more secure (should you need a bit of force, damages are to the tubing and the connector, but not to the reactor) than the direct mount of the tubing to the glass.

The other side of the PFTE tubing will connect with the metal ferrules to your gas supply, e.g.

(screen photo at 1:04 min from this video)
Depending on the dimensions, the fittings by Swagelok are sold in sets of 10; a typical mechanical workshop of your university may have them already in the shelf. The less pricey Bowden connections used for 3D printers (e.g., here) are designed for compressed air.

As for any equipment working with gas: monitor the gas pressure in your reactor and use a safety release / pressure valve.


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What are acid/base sensitive groups?
My textbook says that Clemmensen Reduction cannot take place in the presence of an acid-sensitive group and the Wolf Kishner reaction won't happen in the presence of a base sensitive group, citing $\ce{-OH}$ (alcohol) and $\ce{-X}$ (halogens) as respective examples.
However, I read from somewhere else that:

Acid sensitive groups are groups which are electron donors or have spare electrons, like $\ce{NH3}$. Base sensitive groups are groups, which are in need of electrons or have empty orbitals, like $\ce{BH3}$.

This seems incorrect to me as Halogens are clearly not in need of electrons. I can see why alcohols would be acid-sensitive but why would Halogens be base-sensitive?

Why do you think alcohols are acid-sensitive? By which I mean, what is going to happen to an alcohol in acidic medium? Then apply the same reasoning to 'halogens', as you call them, in basic medium, and you will have your answer.

A proton can attack an OH group so that the resulting group can leave as H2O. This happens because the proton requires electrons and the alcohol has extra lone pairs of electrons. In the second case, both the Halogen and attacking base have a surplus of electrons, so why would there a reaction?

What textbook is this from?

Indeed. OH gets protonated first. Then you say it can 'leave'. What causes it to leave? The initial protonation? What is the complete mechanism by which it leaves? Think about that and you will see where your confusion comes from.

What if you are a tertiary alcohol under Clemmensen conditions? The term "sensitive" means some group also reacts under conditions designed for some other group. If you want to effect a Wolff-Kishner, be aware of E2 elimination elsewhere in the molecule. If you choose to do an acid hydrolysis of an ester, be aware of the lability of acetals under these conditions.

Halogens can undergo substitution reactions under W-K conditions. The hydrazine will quite happily displace the halogen, indeed OH- can displace or eliminate the halogen too.

If a grp on addition of H+/acid it changes then it is called acid sensitive like OH grp on addition to H+ gives H2O+ hense acid sensitive and Cl grp is an EWG and withdraws e- from nearby grps due to -I effect and hense base sensitive and it withdraws e- from base(like OH-).
Explanation2:OH has -I effect and also +R effect and +R dominates over -I hense it acts as EDG and is acid sensitive while in Cl -I dominates and is base sensitive

You shouldn't use abbreviations in you comments. It does not add to the readability of your answer.

Alcohols show -I effect and are electron pair accpetors. They will be base sensitive. But halogens ACCEPT a pair of electrons to achieve stability and are also base sensitive.


text:
Maximum number of electron
How many electrons can fit in the orbital for which n = 3 and l = 1?
It is a NEET question which was asked in 2016.
And according to me answer should be 6 electron because maximum number of electron that p orbital can accommodate is 6 but the answer given is 2 electron.
Please explain how it is possible

It would be 2 electrons for l = 0.

Read the question carefully!!

Pay attention to "orbital" vs "subshell" vs. "shell".

It was just a word play.  In question they have used the word orbital.  It means they are asking for maximum no.  of electron in single orbital. The answer would have been 6 electron if it is "orbitals"  in place of "orbital".


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Arrhenius equation - Proper Interpretation of Activation Energy term
Consider the normal form of Arrhenius Equation:$$k=Ae^{-\frac {E_a}{RT}}$$
The term, $e^{-\frac {E_a}{RT}}$, is interpreted as the fraction of molecules having energy greater than Activation Energy.
My question: Isn't the term threshold energy better than activation energy?

Please provide a suitable interpretation of the terms involved.

In principle, the activation energy could have been called threshold energy. But nobody uses this word. It is a question of habit.

There is an answer here https://chemistry.stackexchange.com/questions/139196/why-does-activation-energy-not-change-with-temperature-change/139222#139222

@Maurice Sir does that mean it is in actual sense Threshold Energy?


text:
What is one formula unit?
I am really sorry if this is a beginner question but please still help me.
In this website, which explains balancing chemical equations mentioned the word formula unit here:

$\ce{3Ca(NO3)2}$ (just the oxygens) ---> There are 18. The 3 on the nitrate times 2 outside the parenthesis equals 6 oxygen in one formula unit. The coefficient of three times the 6 gives the final answer of 18.

My question: What does formula unit mean?

The formula $3$  $\ce{Ca(NO3)2}$ means that you have taken three times the same atomic assembly, or the same formula unit. Each assembly, or each formula unit, is made of one Calcium atom surrounded by two groups of atoms. And each group of atoms is made of one Nitrogen atom surrounded by three oxygen atoms. The formula unit is simply $\ce{Ca(NO3)2}$

Formula unit in calculations is one complete molecular formula. For instance, here it is $\ce{Ca(NO3)2}$. For example, in organic chemistry, empirical formula of benzene is $\ce{CH}$. However its  molecular formula is $\ce{C6H6}$, which is used as the formula unit in calculations.

"Formula unit in calculations is one complete molecular formula. For instance, here it is Ca(NO3)2.", that really helped explain it simply. Thanks a lot @MathewMahindaratne


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Difference between Oxidation Number and Oxidation State
I know the theoretical definitions and that they do not defer much, infact these terms are used quite interchangeably.
However, my chemistry teacher once taught me that Oxidation State of an atom in a compound is the average of the different Oxidation Numbers in the molecule. For example, in Acetic acid, the Carbon atom has oxidation numbers +3 and -3, so the Oxidation State of Carbon (i.e. the average) would be 0.
I've stuck by this for long, but yesterday I came across a question the answer to which implies that it is the other way round, as in ON is actually the average of all the OS of an atom. Several sites on the internet quote the same, and other references quote what my teacher taught me, which is the opposite.
What is the right answer?

There are no "right answer". Some teachers state that each atom has its own oxidation number or oxidation state. Other teachers admit that oxidation numbers can be averaged

@Maurice Okay, so you basically mean to say that it is the ON which is averaged and not the OS?

There's no difference between oxidation number and oxidation state.

@Mithoron No it does not :(

@Mithoron Okay, thankyou for that. I will just assume that the question is wrong:)


text:
What is the difference between free energy and standard free energy
As I understood $\Delta G^0$ is just $\Delta G$ but at standard conditions i.e. $\pu{1 bar}$ and $\pu{298 K}$. But formulas like:
$\Delta G^0 = \Delta H^0 - T \Delta S^0$
$\Delta G^0 = - RT \ln K$
make me uncertain about that as these are dependent on $T$ and don't seem to be used at standard conditions only i.e. $T = \pu{298 K}$.
Could someone explain to me what the difference is between $\Delta G$ and $\Delta G^0$?

$\Delta G^\circ$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and pressure 1 bar, to corresponding stoichiometric quantities of  pure products, each at temperature T and pressure 1 bar.  So $\Delta G^\circ$, $\Delta H^\circ$, and $\Delta S^\circ$ are all functions of T, but not pressure (which is fixed in the initial and final states as 1 bar).  The values of these function changes tabulated in your text book are all for 298 K.  But values at other temperatures can be obtained from integration, for example, in the case of $\Delta G^\circ(T)$ using the van't Hoff equation.
$\Delta G$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and its own arbitrary pressure, to corresponding stoichiometric quantities of  pure products, each at temperature T and its own arbitrary pressure. If these arbitrary pressures happen to match the partial pressures of the same reactants and products in an equilibrium mixture, then $\Delta G$ will turn out equal to zero.  Otherwise, $$\Delta G=\Delta G^\circ+RT\ln(Q)=RT\ln{(Q/K)}$$and, for equilibrium, Q=K.

A wonderful explanation by All difference between follows:

1. $\Delta$ G relate to these two quantities ΔH° and ΔS° while $\Delta$ G° does not relate to the quantities:
$\Delta$ G definitely does relate to those two quantities. ΔH° and ΔS° represent the change in enthalpy and entropy between product and reactant but they do not mean a “100% complete reaction.” They determine the energetic difference (at a given temperature). This difference in energy determines the composition at equilibrium
2. $\Delta$ G is not a non-zero value while $\Delta$ G° is a non-zero value:
$\Delta$ G° is a non-zero value and it can’t use ΔH° or ΔS° to find $\Delta$ G because either ΔH° or ΔS° represents 100% complete reaction.
3. At equilibrium $\Delta$ G is equal to zero while $\Delta$ G° is negative:
If $\Delta$ G° is negative at equilibrium, then we will have lots of products at equilibrium, which means that Q must be greater and also greater than 1 to approximate K. When Q gets bigger, it means more product is accumulated. The term ‘$RT \ln Q$’ gets increasingly positive, and eventually adding that term to a negative $\Delta$ G°, will make $\Delta$ G = 0, equilibrium will be established and no further change occurs.
4. Equilibrium is established when $\Delta$ G is at zero while lots of reactant at equilibrium when $\Delta$ G° is positive:
If $\Delta$ G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K. As Q gets smaller (i.e., as we get more reactants), the term ‘$RT \ln Q$’ gets increasingly negative and eventually adding that term to a positive $\Delta$ G°, will make $\Delta$ G = 0, equilibrium will be established and no further change occurs.
5. When $\Delta$ G is zero at equilibrium, it will define which way the reaction proceeds while $\Delta$ G° does not:
Note that it is $\Delta$ G and not $\Delta$ G° that will be zero at equilibrium and the sign of it generated by the combination of $\Delta$ G° and $RT \ln Q$, will define which way the reaction proceeds.


text:
Contradiction of general rules of bond character
The rules the ionic character of metal halides of halogens is given in order $\ce{MF > MCl > MBr > MI}$.

While judging polarisability we say that a compound that has large cation and small anion will have more covalent nature. So here for same cation, anion with small size should have more covalent character but the trend is totally reverse. Is there any explanation to this?

Although atomic bond in a compound such as $\ce{M+X-}$ (e.g., $\ce{Na+Cl-}$) is considered to be 100% ionic, in reality, it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Kazimierz Fajans in 1923 (Ref.1). According to Fajans, if two oppositely charged ions (say $\ce{M+}$ and $\ce{X-}$) are brought together, the nature of the bond between them depends upon the effect of one ion on the other. In inorganic chemistry, these Fajans' rules are used to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion, summarized in the following table:
$$
\begin{array}{c|c}
\hline
\bf{\text{Ionic}} & \bf{\text{Covalent}} \\
\hline
\text{Low positive charge }(\text{e.g., } \ce{Na+}) & \text{High positive charge }(\text{e.g., } \ce{Al^3+})  \\
\text{Large cation }(\text{e.g., } \ce{K+}) & \text{Small cation }(\text{e.g., } \ce{Li+})  \\
\text{Small anion }(\text{e.g., } \ce{F-}) & \text{Large anion }(\text{e.g., } \ce{I-})\\
\hline
\end{array}
$$
When the two oppositely charged ions, $\ce{M+}$ and $\ce{X-}$, approach each other, the positive ion ($\ce{M+}$) attracts electrons on the outermost shell of the anion ($\ce{X-}$) and repels its ($\ce{X-}$) positively charged nucleus. This results in the distortion, deformation, or polarization of the anion. If the polarization is quite small, the forming bond between them would be more ionic, while that would be more covalent, if the degree of polarization is large.
The power of an ion (cation) to distort the other ion is known as its polarization power and the tendency of an ion (anion) to get polarized by the other ion is known as its polarisability. Based on the (cation’s) polarizing power or (anion’s) polarisability, the formation of covalent bond is favored by the following factors:

Small positive ion (Cation): The smaller the size of cation, the greater is its polarising power. Due to greater concentration of positive charge on a small area, the smaller cation has high polarising power. Thus, the greater will be the covalent nature of its bond with an anion. This explains why $\ce{LiCl}$ is more covalent than $\ce{KCl}$.
Large negative ion (Anion): The larger the size of anion, the greater is its polarizability, i.e. susceptibility to get polarized. It is due to the fact that the outer electrons of a large anion are loosely held and hence can be more easily pulled out by the cation. Thus the greater will be the covalent nature of the bond with a cation. This explains why iodides (e.g., $\ce{Kl}$), among halides (e.g., $\ce{KF}$ and $\ce{KCl}$), are most covalent in nature.
Large charge on either of the two ions: As the charge on an ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases. As a result, the cation’s ability for forming the covalent bond increases (see the similarity with second point). For example, the covalency increases in the chlorides of $\ce{Na+, Mg^2+,}$ and $\ce{Al^3+}$ in the order of $\ce{NaCl < MgCl2 < AlCl3}$.
Electronic configuration of the cation:  For the two ions of the same size and charge, one with a pseudo-noble gas configuration (i.e. 18 electrons in outer-most shell) than a cation with noble gas configuration (i.e. 8 electrons in outermost shell) will be more polarizing. Thus copper(I) chloride is more covalent than sodium chloride although $\ce{Cu+}$ ion ($\pu{1.04 \mathring A}$) and $\ce{Na+}$ ion ($\pu{1.05 \mathring A}$) have similar size and charge.

References:

Kazimierz Fajans, “Struktur und Deformation der Elektronenhüllen in ihrer Bedeutung für die chemischen und optischen Eigenschaften anorganischer Verbindungen,” Naturwissenschaften 1923, 11, 165–172 (https://doi.org/10.1007/BF01552365).
Noam Agmon, “Isoelectronic Theory for Cationic Radii,” J. Am. Chem. Soc. 2017, 139(42), 15068–15073 (https://doi.org/10.1021/jacs.7b07882).


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Calculating the concentration of product in a solution based on the peak area of HPLC
Please help me resolving the following:

A sample solution, $\pu{0.24mg/mL}$ Compound A, is exposed to heat to form Compound B. From HPLC data report I have peak area for Compound A and Compound B.  The  STD  concentration is 0.3 Compound A. The concentration of Compound A in sample solution is $\pu{0.20mg/mL}$ Compound A calculated using STD Compound A information. Which is the correct way to calculate concentration of Compound B?

My advisor in the project I'm participating insist I must calculate the concentration of Compound B since is a degradation of Compound A  in term of the amount lost of Compound A is the amount gained by Compound B (she call it a indirect method for calculating the concentration). I'm at lost! Originally I performed the calculation using a STD Compound A at the LOQ level of the impurity:
$$\text{[Compound B]} = \frac{\text{Area Compound B} \times \text{[STD Compound A]}}{\text{Area STD Compound A}}$$
The new approach is calculate the [compound A] using a STD at the level of the sample and then indirectly calculate the [Compound B].

There is no way! You need to construct calibration curve for compound B or have a single standard injection for compound B.

@M. Farooq: I agree with you about calibration curves. Also, may need a calibration curve for compound B, if two compounds are structurally different (based on detecting method, usually an UV detector).

I think my advisor bold approach is based in Coumpond B is the enantiomer of Compound A (Doxycycline and 4-EPI-Doxycycline).

Okay, then his assumption is valid. The UV spectrum of enantiomers is identical.

Your advisor is making a pretty dangerous assumption (although sometimes pharma people do it). The assumption is that the impurity is structurally similar to the main compound and its UV-Vis spectrum is almost similar i.e., their extinction coefficients are the same.
The idea in your formula is that the calibration curve for A will be same as B. Well that might be a first order approximation when the impurity is known to be structurally similar.
If B is a known compound, you should buy a separate standard, and make a calibration curve.


text:
GGA functionals requiring second derivatives?
In "Lecture Notes in Quantum Chemistry II" from the European Summer School in Quantum Chemistry, the authors note that when discussing the LYP functional and other GGA functionals, that

Thus one can immediately see that an evaluation of $v_{xc}$ demands an evaluation fo the second derivatives of basis functions.
(pp102).

Indeed, when (for a functional like B88) I start with the functional itself and compute the functional derivative (either by hand or with a CAS), I arrive at a (messy) expression that involves second derivatives with respect to space. However, libraries like libxc only need scalar functions of the gradient to be passed, in particular $\nabla \rho \cdot \nabla \rho$ , in order to compute $v_{xc}$. Now libxc contains the expressions for its functionals in Maple and then performs symbolic differentiation on them and then uses the maple-to-C translator; thus the actual C code in libxc is uninterpretable by a human.
Is there a known expression for $v_{xc}$ for B88 or another GGA-like functional that might give me a hint as to how to simplify the functional derivative to only depend on these parameters?

The NWChem code for PW91 and B88 looks handwritten and may be understandable. I do not have the papers at hand, so I cannot judge.

It turns out that my original reference is now somewhat out of date, and it is quite possible to compute the relevant properties without resorting to second spatial derivatives. [Johnson1993] has a nice overview of the relevant math, including clean derivations of both YLP and B88 functional terms, and states:

Note that Eq. (7) does not require evaluation of the spin density second de- rivatives as is necessary in many other formulations, a major computational advantage [48]. The fact that these can be avoided has been previously noted by Kobayashi et al.

Johnson, B. G., Gill, P. M. W., & Pople, J. A. (1993). The performance of a family of density functional methods. The Journal of Chemical Physics, 98(7), 5612–5626. https://doi.org/10.1063/1.464906
and
Kobayashi, K., Kurita, N., Kumahora, H., & Tago, K. (1991). Bond-Energy calculations of Cu2, Ag2, and CuAg with the generalized gradient approximation. Physical Review A, 43(11), 5810–5813. https://doi.org/10.1103/PhysRevA.43.5810


text:
Hybridization state of oxygen in carbonyl sulfide (COS)
I've been scratching my head on this one for awhile here. In a carbonyl sulfide (COS) molecule, what is the hybridization state of oxygen?
Carbonyl sulfide is composed of a single carbon atom double bonded to oxygen and sulfur atoms. The sulfur and oxygen atoms each harbor two lone pairs.

Looking at oxygen, I count electron-dense groups:

There is one bonded atom
There are two lone pairs

This should amount to an sp2 hybridized orbitals for oxygen, before exceptions are accounted for.
Checking if any lone pairs are delocalized:

Oxygen both participates in a pi-bond and is adjacent to a carbon that participates in a pi-bond

This is where I'm stuck. Are any of oxygen's lone electron pairs delocalized due to these facts? Is the hybridized orbital not sp2 for oxygen?

In between, though generally considered as sp2.

@AdityaRoychowdhury Maybe, but they shouldn't https://chemistry.stackexchange.com/questions/80962/what-is-the-hybridization-of-terminal-fluorine-atoms-in-molecules-like-boron-tri


text:
Arrhenius Equation : Interpretation
Consider the relation:
$$E_{Activation}=E_{Threshold} - E_{Avg}$$
Here the $E_{Avg}$ refers to the Potential Energy of the reactants. Now in order of meet $E_{Threshold}$ , the molecules must have sufficient Kinetic Energy which is measured in terms of $E_{Activation}$.
Hence when we write Arrhenius equation:
$$k=Ae^{-\frac {E_{activation}}{RT}}$$
The term $exp(-E_{activation}/RT)$ means the number of molecules having Kinetic Energy greater than Activation Energy which coherently measures the minimum Kinetic Energy reactants need to have for reaction

Do my arguments make sense?

@Karl Sir I thought it was a famous relation from Potential Energy - Reaction Coordinate profile.

see the answer here https://chemistry.stackexchange.com/questions/139196/why-does-activation-energy-not-change-with-temperature-change/139222#139222

@porphyrin Sir I have seen it. I just want to be reassured that whatever I have understood is correct. Please do the same if you feel my understanding is correct.

The activation energy is $E_T$ which you can split into parts if you want to to include a threshold plus average energy in the transition state, but not as you have written it above.

@porphyrin Sir in the post that you referred to me https://chemistry.stackexchange.com/questions/139196/why-does-activation-energy-not-change-with-temperature-change/139222#139222 ,the same expression is in the question part as used as in my first expression. I have updated my expression accordingly.

That threshold energy makes no sense to me. The activation energy is the threshold. Please cite a source for this.


text:
What mechanism is this?
I was solving problems in Organic Chemistry related to Substitution and Elimination. This was a question asked in JEE 2019

This was the solution for above.

At last of solution, It has been mentioned that after rearrangement it undergoes"ArSE2 Pathway".
What is meant by ArSE2 pathway/mechanism?

https://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution

This looks like a standard Friedel-Crafts reaction, although it is intramolecular rather than intermolecular. The cation undergoes electrophilic substitution to give the product.
The choice of terminology used by the authors is strange: a search for "ArSE2" returns very few links, such as this one. There seems to be no commonly known reaction mechanism with this name, and the author was most likely referring to standard electrophilic substitution.

Electrophilic aromatic substitution is denoted as $\ce{S_{E}Ar}$.


text:
Reason for needing multiple radii for the same atom
Why do we need different types of atomic radii? Why can't there be just one radii for an atom?

Determining the atomic radii is rather difficult because there is an uncertainty in the position of the outermost electron – we do not know exactly where the electron is. This phenomenon can be explained by the Heisenberg Uncertainty Principle. To get a precise measurement of the radius, but still not an entirely correct measurement, we determine the radius based on the distance between the nuclei of two bonded atoms. The radii of atoms are therefore determined by the bonds they form. An atom will have different radii depending on the bond it forms; so there is no fixed radius of an atom.

It isn't about the "looking like a book problem" that is an issue here.. What have you attempted to find the answer before asking here?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed. If you explicitly write down your own effort to answer your questions, then probability to get satisfying answers raises significantly.

I remember Dudley Herschbach's experiments in Harvard University with crossed molecular beams in a vacuum. If the two beams are made of sodium or potassium atoms for the first one, and of chlorine molecules for the second, the reaction with formation of $\ce{NaCl}$ or $\ce{KCl}$ happens even when the distance between the beams is as high as $45$ Å or $4.5$ nm. This is more than $10$ times the usual dimension of sodium atom or chlorine molecule. Ref.: D. R. Herschbach, Disc. Faraday Soc. 33, 149 (1962)

It also depends on the state of the electron(s). When excited, they get carried away, a bit.


text:
How do you find the enthalpy of solution of a hydrated salt?
I am doing an experiment in school where I try to find how heating time affects the enthalpy of solution of a salt in water. I take a hydrated salt ($\ce{Na_2CO_3 \cdot 10H_2O }$), heat it for a certain amount of time (say 2 minutes) and then measure the enthalpy of solution of the salt.
Lets say after 2 minutes the fully hydrated salt turns partially hydrated (say $\ce{Na_2CO_3 \cdot 10H_{2}O ->  Na_2CO_3 \cdot 8H_2O})$. I know that for the solution enthalpy of solution of an anhydrous salt, I can find it out by adding the lattice enthalpy of the salt and the hydration enthalpy. But for hydrated and partially hydrated salts (such as  $\ce{Na_2CO_3 \cdot 8H_2O}$) how do you find the enthalpy of solution in water?

$\ce{Na2CO3·8H2O}$ does not exist. After heating $\ce{Na2CO3·10H2O}$, you will obtain the anhydrous salt $\ce{Na2CO3}$.

@Maurice, so when I heat 10 gm of $\ce{Na_2CO_3 \cdot 10H_2O}$, and lose 2 grams I get some proportion of anhydrous salt and some proportion of the fully hydrated salt?

@ Prasad.. Yes. If you only lose $20$% of the initial weight, it means that the déshydratation is not finished. The lost should be about  $63$% of the weight of the initial amount of $\ce{Na2CO3·10H2O}$

Why do you not record the thermogram while dissolving a known quantity of the water free form / the hydrated form for calorimetry and then apply Hess' law?  If your lab financially walks on a shoe string, a coffee cup calorimeter may help you to save some bucks (2017JChemEduc494, or open access 2020Heliyon03591).

@Buttonwood could you give an example? I'm not following. One equation would be $\ce{Na_2CO_3 \cdot 5H_2O -> Na_2CO_3 \cdot 10H_2O}$ (which does exist), but what would the other 2 equations be?

@Maurice that is exactly, what I am trying to find. If it is heated for 2 min, and 20% of the mass is lost, what would the enthalpy be. (Perhaps I should have worded it as the effect of water of crystallization on enthalpy of solution).

@prasadpradyumna The lower part of http://www.docbrown.info/page07/delta1Hc.htm showcases with section 7 and 8 the determination of enthalpies of hydration for $\ce{CuSO4 . 5 H2O}$ and $\ce{MgSO4 . 7 H2O}$ for with already determined enthalpies of solution of these, as well as their anhydrous analogues.

@Buttonwood thanks for the link. Will try out

@ Prasad. Why do you heat only 2 minutes ? It should last much longer !

@Maurice I will do it for longer. 2 min was just an example

@ Prasad.. The heats of solution of sodium carbonate and its decahydrate are taken from an old edition of the Handbook of Physics and Chemistry. Editions from $2002$ and later do not give these values. But in the old edition $1961$, the heats of solution of sodium carbonates (with or without water) are given, in kilocalories per mole. Transformed into kilojoule per mole, the heat of dissolution of the anhydrous salt is $23.6$ kJ/mol. For the decahydrate, it is - $67.6$ kJ/mol. The heats of formation are - $4085.2$ kJ/mol for the decahydrate, and -$1132$ kJ/mol for the anhydrous salt

@Maurice thanks for that. I'll use it in my calculations


text:
Does hybridization change atomic size?
I have come to know that nitrogen can undergo both $\mathrm{sp}^3$ and $\mathrm{sp}^2$ hybridization.
Does a different hybridization lead to a different atomic radius?

Consider that a molecule has its own structure. The length of the bonds to bound test atoms does change. You might keep a full space representation but the sphere(oid)s would compenetrate each other and do so differently.

Also related are this Q and the comments it has received https://chemistry.stackexchange.com/questions/139583/reason-for-needing-multiple-radii-for-the-same-atom

Let's be more general. Take any atomic size with a grain of salt.

How will you define a radius when sp$^3$ for example has four lobes ?

To put it bluntly: Hybridization is a stupid concept and we should finally let it die already, because otherwise we end with more generations of chemists who mistakenly think it's an actual physical thing that happens to an atom.


text:
Density of a sodium hydroxide solution
The specific gravity of a $50\%$ aqueous solution $\ce{NaOH}$ is $\pu{1.5298 g cm-3}.$
To my understanding, the $50\%$ means $50$ mass fraction $w,$ i.e. $\pu{500 g}$ $\ce{NaOH}$ in $\pu{500 g}$ $\ce{H2O}:$
$$w(\ce{NaOH}) = \frac{m(\ce{NaOH})}{m(\ce{NaOH}) + m(\ce{H2O})} = \frac{\pu{500 g}}{\pu{500 g} + \pu{500 g}} = 0.5.\tag{1}$$
However, when I try to calculate the density of that solution I end up with
$$
\begin{align}
\rho(\ce{NaOH}) &= \pu{2.13 g cm-3}; \\
\rho(\ce{H2O}) &= \pu{1.00 g cm-3}.
\end{align}
$$
$$V = \frac{m(\ce{NaOH})}{\rho(\ce{NaOH})} + \frac{m(\ce{H2O})}{\rho(\ce{H2O})} = \pu{234.74 cm^3} + \pu{500 cm^3} = \pu{734.74 cm^3},\tag{2}$$
$$\rho_\text{sln} = \frac{\pu{1000 g}}{\pu{734.74 cm^3}} = \pu{1.361 g cm-3}.\tag{3}$$
What am I missing here?

You don't calculate density.

Densities are rarely, if ever, directly proportional to mass percentage of a solute. You need to determine the values experimentally.

Plotting the densities to mass percentage creates an almost linear function (in this case). It is considered proportional in many cases. $ϱ_{sol}=0.0106W_{NaOH} + 1.0034$

Thanks for the input so far. I'm aware that density needs to be measured. I'm thinking of mixing ethanol and water, where the volume of the solution will change.  I was suspecting that I get the weight percentage part wrong. My calculations would yield that 1 L of 50% solution contains 680 g NaOH. According to the table I originally linked it would be 764 g / L. So basically my calculation of the solution is correct but I would need to look up the density in a table or use a correlation like @AndrewKovács suggested?

You have added volumes. This is wrong. Volumes in mixtures are never additive. Masses are additive, but not volumes. $234$ mL pure NaOH +$ 500$ mL water does not give $734$ mL solution.  It gives less.

@maurice that's what I suspected, hence the analogy to ethanol and water in my comment. If you or anyone else want to put that in an answer I'll happily accept it. Otherwise I would answer the question myself tonight.

@idkfa If you review some tabelized values of densities of NaOH ( or any other compound ) solutions, you can easily calculate the total volume of solute + solvent before dissolution, knowing densities of the pure compounds, and therefore you can get the volume deviations.

What's your first miss is the definition of $50\%(w/w)$ $\ce{NaOH}$ solution (although it doesn't matter here). Actual definition is $\pu{50 g}$ of $\ce{NaOH}$ in $\pu{100 g}$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $\ce{NaOH}$ are additive. Assuming $V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$ is a mistake.
Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Do you still think volumes are additive?


text:
How explosive is benzene compared to TNT?
Benzene is explosive within the explosive limits of minimum $\pu{1.2 \%}$ and maximum of $\pu{7.8 \%}$ in air. Source: Wikipedia More accurate, benzene itself is not explosive, but the fuel-air mixer or, fuel–air explosive (FAE), is. This fuel-air explosive is used, because of its power in Thermobaric weapons https://en.wikipedia.org/wiki/Thermobaric_weapon Benzene does burn and could be a fuel in a fuel-air explosive. So it would be, more specific, a benzene-air explosion.
I would like to know what is the explosive power of benzene measured in TNT-equivalent. Benzene can explode like in China in 2005 which was 'only' 100 ton benzene.
The reason why I want to know is that there was a huge explosion in Beirut on August 4th 2020, with the equivalent of 1.1 kiloton TNT. In my country, the Netherlands, benzene is transported in huge ships of 2000 ton. This could explode, spreading the benzene gas over the area, which is heaver than air. Benzene is highly carcinogen. I would like to know how a possible explosion could be compare with the explosion in Beirut. Benzene burns very easy, so may be only a fraction of benzene would explode, the rest would burn. I just want to know what we, as citizens, might expect when things goes seriously wrong.
An approach could be a list with the relative effectiveness of benzene compared to TNT. All kind of benzene derivatives are mentioned in that list but not benzene itself. I could not find  a list in which benzene itself is mentioned.
So my (much to simple) calculation is:

1 ton TNT: $\pu{4184 \times 10^{12} J}$ see: TNT equivalent
Energy [(Heat of combustion) / mol]: $\pu{3267.6 kJ/mol}$ see: Benzene
Energy / kg:   $\pu{41831 \times 10^{6}J/kg}$ ($\pu{\frac{3267.6 ~ kJ/mol}{78114 ~ g/mol} \times 1000 g/kg}$)
Mass: $\pu{2000 kg}$
Total energy: $\pu{83662 \times 10^{12} J}$ ($\pu{41831 \times 10^6 J/kg \times 2000 kg}$)
TNT-equivalent: $\pu{~ 20 kiloton}$ TNT ($\pu{\frac{83662 \times 10^{12}~  J}{4182 \times 10^{12}~  J/ton}}$ TNT)

My questions are:

Could all benzene explode at once? Because benzene might be mixid in
air within the explosive limits range, but this will be only for a
limited ammount of benzene. But because the explosion the rest of the
benzene will be blown into the air and possible then be again be
within the explosive limites ranges. So causing a chain-reaction. The
explosion in Jilin in China took 1 hour so that seems to be the case.
If not is it possible to calculate what is the maximum of benzene that
could explode?

Note: I have very limited chemical knowledge. I can passive read chemical formulae and make very simple calculations like above.

Benzene is not explosive. But if benzene is nitrated, it produces nitro derivates which are explosives. That was has happened in China. The explosion happened in a nitrate factory. Nitates can be explosive, like the recent explosion in Beirut harbor. But I repeat : benzene is toxic, but it is not explosive.

@Maurice Ok, but benzene does have explosive limits. What does that mean if benzene is not explosive?

@ Bernard. Benzene has explosive limits when its vapor is mixed with air or oxygen. But pure benzene is not explosive.

@Maurice. That is what might happen. First technical. If a material can explode without an other chemical it can explode, like TNT and ammonium nitrate. But also natural gas if mixed with air. Does these different kind of explosions have a different chemical name? To my example: If the ship has a serious accident and benzene leaks, then benzene can escape into the air. At first, very local this will have a high concentration which will get lower in time. Until it reaches the explosive limits. Than a little spark could be sufficient for a explosion. That is what I am interested in.

Check this pdf: https://www.nrc.gov/docs/ML0716/ML071650338.pdf

Yes, if you can get benzene vapour spread in a large volume of air at just the right concentration, a big explosion is possible. But this is very hard to achieve. Not least because you need the source of ignition to be present *only * at the point where the benzene reaches the right concentration or you get a fire not an explosion.

@matt_black Thanks. This does help. And how big the explosion could be, compared to TNT? Benzene has more risk then only explosions, I know. But I first really wants to figure out everything about the explosion. First off all the volume of the explosion. If this is solved may be I could start an other thread for the other items of risk(exact conditions for an explosion, quantitative calculation of changes, fire and carcinogen possible in combination of each other). But first just how big could 2000 ton benzene explosion be (with all required conditions) compared to TNT?

@NilayGhosh I started to read the document but hardly found something about benzene. What are you trying to say with that link?

See the table at page 17 of the pdf.

@NilayGhosh I have seen that table, and yes Benzene is mentioned in that table. But I still do not see the additional extra compared to other contributes which could solve my question.

Since the explosion (if any) is very much dependent on random circumstances like the concentration of the vapor, how closed space, I do not think people can give you a meaningful number. Also, such an explosion can happen with any kind of flammable materials, even solid powders like sugar. Do you really want to put a label on coal or sugar what is their equivalent TNT under circumstances that almost impossible to create?

Lets compare it to toxic with its median lethal dose . Even water can be toxic, called Water intoxication. A 28-year-old California woman died from drinking 3 liter water in a contest https://www.scientificamerican.com/article/strange-but-true-drinking-too-much-water-can-kill/  If she should have known this she still would live.
You ask if sugar should should be given a number for being explosive compare to TNT (or an other unit). If water can be given a number for being toxic, why sugar, or any other substance, can not be given a number for being explosive? A challenge for scientists.

All flammable organic liquids could, in principle, create large fuel-air explosions but the conditions are very hard to achieve accidentally
The first important point to note is that benzene should not be a particular worry. There are plenty of other, widely used, chemicals and mixtures that could, in principle, cause large explosions. Standard automotive petrol (gasoline) for example is present in most vehicles (non-diesel ones anyway) and is transported in 40 tonne tankers on most roads. Fires and explosions happen, but rarely, and they are never known to be of the thermobaric type you are worried about. Most motor vehicles have 40kg or so of fuel that is just as flammable as benzene but after accidents they rarely catch fire or explode and never do so with the power of a fuel air thermobaric explosion (the hollywood trope that all crashed cars explode is pure nonsense).
But big fuel air explosions are known to be possible and have been developed into some very effective military weapons. These are known as thermobaric bombs and the largest are just about the biggest non-nuclear explosions possible in existing military arsenals. But a quick read of how the weapons work (see this wikipedia entry) suggests that they require some fairly notable technology to get right. Largely this is because the fuel has to be dispersed at just the right concentration in the air to get a detonation or deflagration rather than a fiery but unimpressive "phut". It isn't easy to achieve this by accident with a large volume of fuel.
Bad maintenance and safety procedures in chemical and oil plants can lead to accidental fuel-air explosions. The Buncefield explosion in a UK oil storage depot was caused by one. It was a huge explosion probably involving about 300 tons of fuel (but that is only about 0.1% of the fuel stored in the depot a lot of which caught fire but didn't explode). Another famous UK example is the Flixborough disaster caused by a fuel-air explosion caused by a leaky pipe in a cyclohexane plant. In both incidents poor management and engineering contrived to make a dangerous event possible and standards were rewritten to prevent future mistakes. That such event are rare should be of significant comfort if you are worried about the dangers of transporting tankers full of benzene. It is worth noting that none of these (very bad0 accidents came close to involving the majority of the flammable compounds stored nearby. It is hard to get the right fuel-air mixture with very large volumes of fuel, thank goodness.
A more common form of fuel air explosion might also put the rare chemical plant incidents in perspective. This is explosions caused in flour mills and grain elevators. a mixture of finely divided flour or grain dust is explosive and plants that handle flour and grain have to be specially designed to avoid dangerous dust explosions. These are far more common that explosions in chemical plants. While this may not be much comfort, it does suggest that worrying about benzene is not that big a real world risk.

@mattblack Although not an exact answer to my question, "how explosive is benzene compared to TNT?" it gives sufficient information for me. To summarize your answer in my own words: Yes benzene has very strong explosive power, and you gave an indication with your link to https://en.wikipedia.org/wiki/Thermobaric_weapon But only in very specific conditions. Also you gave some examples with fuel air explosions, which where not benzene. In real life, just by accident, this hardly appear. Solved for me for this subject.

Could all benzene explode at once?

No, this is highly unlikely. While liquid benzene is flammable, it is not explosive. In the worst case, there will be a large fire, but chances of spontaneous combustion as compared to Trinitrotoluene are very, very low.

The explosion in Jilin in China took 1 hour so that seems to be the case.

According to the same Wikipedia article:

The accident site is a nitration unit for aniline equipment. T-102 tower was jammed up and was not handled properly, hence the blasts

Also, from China Daily:

The channel of one nitration tower of the benzene production branch got blocked at noon. A worker then bungled an attempt to unblock it.

These kinds of accidents in production plants are different from explosion of storage units: most likely there were other factors involved and it was not only benzene that was involved in the expolosion, but other chemicals as well, such as other nitrated benzene derivatives which are more powerful explosives than Benzene.
Finally, I would want to clear up your concepts of lower explosive limits and upper explosive limits[1]:

The lower explosive limit (LEL) is the minimum concentration of a specific combustible gas required to fire combustion when in contact with oxygen (air). If the concentration of the gas is below the LEL value, the mix between the gas itself and the air is too weak to spark. The upper explosive limit (UEL) is the maximum level of concentration of the gas that will burn when mixed with oxygen; when the gas concentration is above the UEL value for the gas/vapor, the mix is too “fat” to ignite or explode.

It is possible for Benzene vapours to explode when the air is saturated with them to a concentration that is between LEL and UEL; here, the oxygen in the air helps the already flammable benzene catch fire spontaneously, causing an explosion. However, there are several safety guidelines with reference to benzene transportation and shipping[2], [3]. Most notably, Benzene is transported in pressurized tanks and is subject to the requisite regulations for Hazard Class 3 liquids, which is why the accumulation of significant concentration of benzene vapours in the air (enough to cause a combustion) is unlikely.

Thanks for your explanation. So it is clear to me the explosion in In Jilin in China is not pure benzene. I know of those safety guidelines and indeed most often benzene will not explode. As most nuclear reactors will not explode because the safety guidelines. And still there where a few serious nuclear explosions. Partly because the safety guidelines where not followed. But what I am interested in is if still something goes wrong how strong the explosion of benzene would be compared to TNT.

@Bernard Benzene is not explosive. Benzene safety is at similar level as petrol safety, which is managed at much higher volume than benzene. Liquid methane would be even much worse, and also transported in higher volumes than benzene. These stuffs cannot be easily compared to TNT, as deflagration speed of their vapour/air mix is much lower than detonation speed of explosives, even if their specific energy is about 10x higher than of TNT. Also energy density of vapour/air mix is much lower than for as Solid TNT.

@Bernard There were NO nuclear explosions in the accidents with nuclear reactors involved. Explosions of hydrogen-oxygen gas mixtures, or any other CHEMICAL explosions, or intensive spills of radioactive materials, etc. - yes.

@AlekseyKuznetsov ' Charles Perrow theory predicts that failures will occur in multiple and unforeseen ways that are virtually impossible to predict.' see https://en.wikipedia.org/wiki/Charles_Perrow The change of nuclear power something seriously would go wrong is 1 time in 1.000.000 year (calculated for each nuclear plant). Still in 2011 in Fukushima in Japan there were 3 meltdown and in Chernobyl in 1986 there was a nuclear explosion. Much more than was predicted. So it is important to know how large the blast is for as Matt Black named a fuel-air explosion.

@Bernard You may wish to refresh your knowledge of the definition of nuclear explosion. Neither Fukushima neither Czernobyl were nuclear explosions. There were classical explosions triggered by heat released by nuclear energy, which lead to hydrogen/air ignitions and violent pressurizing by water vapour. what is far from being the same. Even if everything goes wrong with the reactor, slightly enriched uranium is not physically capable of nuclear explosion, as the fast neutron chain reaction is not possible there. It does not mean the total energy or max power has to be low without it.

@Poutnik Ok I do accept your explanation. Technical you are right. But a citizen living in 1986 in the village of Czernobyl nearby the nuclear plant and faced with an explosion from the nuclear plant how we have to call that explosion for that citizen? Without the nuclear plant the explosion would not occur. A correct name which both the citizen and the expert can use and understand is important. The same for benzene. Benzene by itself can not explode. But a benzene-air mixer between 1.2 % and maximum of 7.8 % can explode.

@Bernard Chemistry SE site operates within scientific context, therefore the primary goal is to describe what happens and how, not how it looks for people. If an explosion happens in an electric plant burning natural gas, it is not automatically the gas explosion either. Petrol vapours, LPG and liquid CH4 mixed with air can explode as well. Benzene or toluene are nothing special in explosion context, safe the fact explosives can be derived from them. OTOH, ammonium nitrate is different case.

@Poutnik I completely agree. But in the end it would be nice for a translation to common people, if possible. So they also can understand what is happening.


text:
Compare the boiling points of isobutane and 1-butyne

Question: Compare between the boiling points of isobutane and 1-butyne

The answer to this question is given as 1-butyne has higher boiling point than isobutane.

My approach: I have been taught a few concepts (which I learnt for granted), in my organic chemistry class:

The carbon with a longer chain has a higher boiling point than its isomer with more branching or a shorter hydrocarbon chain. (Kindly correct me if I am wrong here and provide an explanation)

From here I decided that butyne will have a greater boiling point (though I made an incorrect assumption), and luckily my answer came out to be correct, but I am not convinced about the reason for it. Please help with the correct explanation for this question.
I have a proposition that since the electronegativity of $\mathrm{sp}$ hybridisation carbon is very high (only a little bit less than nitrogen), will it show hydrogen bonding? This justifies the given answer.
If it does not show hydrogen bonding, why so? All conditions are fulfilled, a highly electron deficient hydrogen is present and electrons are present in the $\pi$ bond between the carbons. Please help.

The boiling points of non-polar hydrocarbons are determined by the extent of van der Waals forces in between them. More specifically, molecules with a larger surface area have larger van der Waals forces of attraction between the molecules.

From the structures, it is obvious that but-1-yne has a larger surface area than isobutane, and hence has the higher boiling point.
Your idea of H-bonding in but-1-yne is plausible, according to this source1:

Quantum chemical calculations indicate hydrogen bond energies in the range $\pu{1–2.2 kcal mol^–1}$. Interconnected contacts such as $\ce{C#C–H ··· C#C–H ··· C#C–H}$ exhibit a pronounced cooperativity effect, with energy increases of $\approx \pu{0.5 kcal mol^–1}$ compared with isolated contacts.

However, while this bonding may be sufficient to explain a higher boiling point compared to isobutane, it is still a very weak bond compared to the H-bonds in hydrogen fluoride or water.
References:

Thomas Steiner, Evgeni B. Starikov, Ana M. Amado, José J. C. Teixeira-Dias, “Weak Hydrogen Bonding. Part 2. The Hydrogen Bonding Nature of Short $\ce{C–H ⋯\pi}$ Contacts: Crystallographic, Spectroscopic and Quantum Mechanical Studies of Some Terminal Alkynes.” J. Chem. Soc., Perkin Trans. 2 1995, (7), 1321–1326 (doi:10.1039/P29950001321).


text:
Regarding the use of dilute sulphuric acid in this titration
I did a titration today for the first time.
A solution of Potassium Permanganate of unknown molarity was titrated against a solution of ammonium iron II sulphate of known molarity.
I have to say, seeing the purple solution turn clear was a lot of fun.
Dilute sulphuric acid solution was used twice during this experiment. First to prepare the ammonium iron II sulphate and sulphuric acid solution, and then 20 ml was added to each sample before titration.
Why did these have to be added ?
Also, could I have used pure sulphuric acid ? Why did it have to be diluted ?
I think the first usage provided the H+ for the reaction to occur but I am not sure. Would pure sulphuric acid not be more efficient ?

https://chemistry.stackexchange.com/questions/21696/where-does-the-feiii-ion-go

A little bit of sulphuric acid is used when preparing iron(II) solution. This acid prevents the possible hydrolysis of $\ce{Fe^{2+}}$ ion according to : $$\ce{Fe^{2+} + 2 H_2O <=> [Fe(OH)_2(s) + 2 H^+}$$ In case this reaction occurs, the added sulphuric acid will react and destroy the precipitate $\ce{Fe(OH)_2}$ according to the inverse reaction : $$\ce{Fe(OH)_2 + 2 H^+ -> Fe^{2+} + H_2O}$$
Now, during the titration, the ion $\ce{H^+}$ is needed to carry out the reaction of permanganate on the ion $\ce{Fe^{2+}}$ . Hopefully you know this equation.
And concentrated sulphuric acid is not wanted, because it reacts strongly with water just during its dissolution. It produces a lot of heat.

Why instead of adding the acid twice can I not just add it once ? Add to prepare and then add some more before titrating. Why not just add the whole lot when preparing ?

@ Kantura. The risk of precipitating $\ce{Fe(OH)_2}$ in the first phase is small. The solution must be slightly acidified. No need to add more. And the acidified solution may be kept many days in closed bottles. But a huge amount of $\ce{H^+}$ ions are needed during the titration with permanganate. The equation shows that 8 $\ce{H^+}$ ions are needed for each permanganate ion to react.  Indeed the equation is $$\ce{MnO_4^- + 5 Fe^{2+} + 8 H^+ -> Mn^{2+} + 5 Fe^{3+} + 4 H2O}$$ I repeat : $8$ $\ce{ H^+}$ are needed for one permanganate ion.

Ammonium iron(II) sulfate or Mohr's salt, is the inorganic compound with the formula $\ce{(NH4)2Fe(SO4)2(H2O)6}$. In analytical chemistry, Mohr's salt is the preferred source of ferrous ions because the solid salt has a long shelf life and the solid crystals resist to oxidation by air: The presence of $\ce{NH4+}$ ion prevents the oxidation reaction of $\ce{Fe^2+ -> Fe^3+ + e-}$, which slows down in lower $\mathrm{pH}$.
This stability of Mohr's salt extends somewhat to the solution phase reflecting the effect of aforementioned $\mathrm{pH}$ on the $\ce{Fe^2+/Fe^3+}$ redox couple. The ammonium ions make solutions of Mohr's salt slightly acidic, which slows this oxidation process (the oxidation occurs more readily at high $\mathrm{pH}$). Thus, during titrations, sulfuric acid is commonly added to solutions of Mohr's salt to reduce air oxidation to ferric iron, thus, the oxidation would take place only in the presence of permanganate solution.
The two half reactions are:
$$\ce{MnO4– + 8H+ + 5e– -> Mn^2+ + 4H2O} \tag1$$
$$\ce{Fe^2+ -> Fe^3+ + e-} \tag2$$
Equations $\left((1) + 5 \times (2) \right)$ would give the net equation:
$$\ce{MnO4– + 8H+ + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O} \tag3$$
As evidence in the equations, $\ce{KMnO4}$ solution need acid to function smoothly as oxident while Mohr's salt need acid to stay un-oxidized until it meets $\ce{KMnO4}$. The usual acid used in these situation is dilute $\ce{H2SO4}$, which did not interfere with the redox reaction. Keep in mind that dilute acid is usually $\pu{6 M}$, which is good enough for the purpose.