text:
Reference states for molecular orbital energies in DFT calculations
When doing a population analysis using density functional theory (specifically as implemented in Gaussian), what are the orbital energies (i.e. alpha and beta eigenvalues) with respect to? If I have two different but closely related compounds, can I directly compare their HOMO (or LUMO) energies? It should be noted that I am not referring to comparing their HOMO-LUMO gaps, but rather the energy of a single molecular orbital.

As mentioned by others, the zero of the system is a free electron.
As for comparing energies, people do this, but they shouldn't. People love to compare HOMO-LUMO gaps and LUMO energies but they are almost entirely arbitrary. In DFT (and Hartree-Fock), the energy of the system is entirely a function of your occupied orbitals. You could delete the bottom 10 unoccupied orbitals from your basis set and you would get the exact same energy but your LUMO energies (and hence HOMO-LUMO gaps) will be completely different.  So making your basis smaller can increase your LUMO energies. What about the flip side? If you use a bigger basis, your LUMO will get lower and lower in energy. In the basis set limit, you will have a bunch of orbitals at the zero (you will basically be replicating the "orbital" that corresponds to a free, unbound electron). There's a reason that the electron affinities people try to calculate from LUMO energies are terrible. You are basically banking on the chance that your basis set is just the right size for your molecule in just the right geometry.
To summarize, you can basically tune your basis set to produce whatever LUMO energy you want. I wouldn't compare LUMO energies between the same molecule in different basis sets or even geometries, let alone different molecules.
All hope is not lost though, if you really want some sort of LUMO-like thing. You can try working in some sort of localized basis: NBO produces "anti-bonding" orbital energies that are no where near as sensitive to basis set as canonical orbitals. However, that might not suit your purpose.

this is what I also thought, but (unless I'm missing something) it seems to somewhat disagree with the answer provided by @user213305 . Could we try to reconcile the answers?

Yes, they are wrong. Some quotes from the wikipedia article cited in their answer: "The LUMO energy shows little correlation with the electron affinity with typical approximations.[7]" Zhang, Gang; Musgrave, Charles B. (2007). The Journal of Physical Chemistry A. 111 (8): 1554–1561. "Calculations of electron affinities using this statement of Koopmans' theorem have been criticized[11]" Jensen, Frank (1990). Introduction to Computational Chemistry. Wiley. pp. 64–65. It is an extremely common misconception, but LUMOs are meaningless for DFT/HF.

Unoccupied orbitals are important for correlated methods, MP2 (and therefore double hybrid DFT) because they are actively used in the determination of the energy. For normal DFT/HF, they are the part of the Hilbert space granted to the molecule by our limited basis set that was determined to be useless for improving the energy. All methods are basis set dependent to some degree, so it is not surprising that changing the basis set can change your answer. However, all decent methods converge in to some answer in the complete basis set limit; the LUMO of DFT/HF does not.

Unfortunately, for my systems, the NBO code often has a lot of trouble finding a sufficiently "optimal" Lewis-like structure such that NBO-computed properties are often unreasonable. Nonetheless, this answer is the one that makes the most sense to me.

In molecular electronic structure, electronic/orbital energies are measured with respect to an energy zero of an isolated electron in free space.
Hence a lower LUMO energy does correspond to an orbital that is more apt to accept an electron/a more negative electron affinity compared to another molecule.
Correspondingly a higher HOMO energy corresponds to an orbital that is more liable to give up an electron/a less positive ionisation energy compared to another molecule.
This is the basis of Koopmans' theorem in Hartree–Fock theory and it's equivalent in DFT.

From your link: "Ionization energies calculated from DFT orbital energies are usually poorer than those of Koopmans' theorem, with errors much larger than two electron volts possible depending on the exchange-correlation approximation employed."

Yes, that is true.  Koopman's theorem for Hartree-Fock is a rough approximation.  As the orbitals of DFT are only used to form a density (unless you are using some kind of hybrid functional) you would expect orbitals/HOMO/LUMO from non-hybrid DFT to be less accurate than those from hybrid DFT which are less accurate than those from HF.

EDIT: After some thought, I've summarized the following: we can use Koopmans' theorem (approximately) within DFT to compare single point MO energies, as a reference value is indeed used (that of a free, or fully ionized electron). Therefore, we can (somewhat justifiably) compare the LUMOs of two different systems, as they share the same reference energy.
However, single point energy calculations of other kinds cannot be directly compared in a valid or consistent manner between systems of differing number of atoms and/or electrons. Instead, reference values (from literature or another experiment) must be used to compare between systems.  This makes MO energies a special case.
That being said, DFT works with densities that happen to reflect MOs quite well, and so Koopmans' theorem is not exact for DFT and should be used carefully.

It should be noted that I'm not talking about the electronic energy of the system.


text:
Why is the rate of reaction calculated as d[A]/dt?
For example, for a simple chemical reaction like
$$\ce{A -> C}$$
the rate of the reaction could be calculated as the rate of appearance of $\ce{C}$ or the disappearance of $\ce{A}$, which makes a logical sense. But if the reaction is
$$\ce{A + B -> C}$$
as I read somewhere the rate could be calculated the same way as the first case either as the disappearance of $\ce{A}$ or $\ce{B}$, or the appearance of $\ce{C}$. Is it logical to calculate the rate as the disappearance of $\ce{A}$ or $\ce{B}$ as the rate?
Think about it, isn't the appearance of $\ce{C}$, the combined effect of the disappearance of both $\ce{A}$ and $\ce{B}$, which mean if the rate of the equation can be calculated as the change in concentration of $\ce{C}$ using the product, shouldn't the rate be calculated as the change in concentration of $\ce{A}$ plus the change in concentration of $\ce{B}$ through some time change $t$?

No it shouldn't.

You might want to look at this answer for a more basic approach https://chemistry.stackexchange.com/questions/81019/why-is-the-rate-of-a-reaction-proportional-to-the-concentrations-of-reactants-ra/81473#81473

Although the appearance of C is due to the combined disappearance of A and B, the relation is still 1:1 between A and C. Therefore, the rate that A disappears = the rate that B disappears = the rate that C appears.
Let's look at an analogy in stoichiometry. Say I have 1 mole of A and 1 mole of B. Will I make 1 or 2 moles of C? By your logic, I should make 2 moles of C. But I don't.

but as you said doesn't  the rate that C appears have a 1:1 relation ship with B, which means it have a combined relation of 2 with A and B. and does the stoichiometry analogy work for this kind of cases.

Consider if I have 10 red legos and 10 blue legos, and I make red-blue legos by sticking 1 red and 1 blue lego together. I will be able to make 10 red-blue legos, and let's say I do it in one minute.  The rate of disappearance of the blue legos is 10/minute, as is the rate of disappearance of the red legos. What is the rate of appearance of the red-blue legos? Is it 10/minute, or 20/minute?

Elementary reaction rates have stoichiometric ratios just like the reaction amounts themselves do. So, in your example, the rate of change in concentration of A, B, and C are all equal in magnitude - For every mole of A you lose, you also lose one mole of B and gain one mole of C. Other stoichiometric ratios would work the same way - you would use the coefficients to figure out how the reaction rates of each species were related.
In practice, you pay attention to whichever compound you care about or is most useful for solving your particular problem. For example, if you could easily measure A but not C, you might use A in your calculations. Or, if you were mostly interested in how much C you would produce, you might track C. They are all related to each other through the reaction equation, though, so you could substitute in a rate equation in terms of any species, as long as you get the coefficient right. This is very useful in cases where one thing is easy to measure but another is not.
Note that for non-elementary reactions the exponents will not be the stoichiometric coefficients, but the amount of any given species will still be related to the amounts of the other via the rate laws and mole balances, and you can still write the rate equations for whichever species or intermediates that you want to.

RE: "Reaction rates have stoichiometric ratios..." That is true for elementary reactions, but not reactions in general. The statement is thus a tautology since elementary reactions are those reactions where the powers of the coefficients in the reaction equation are the same as the coefficients in the chemical equation.

@MaxW that's a good point - I was assuming elementary reactions since they were giving them as examples, but should have made that more clear.

source: Stack Exchange

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

sort: votes


text:
What is p defined as for the expansion and compression of an ideal gas system in the equation ΔU = ΔH - pΔV?
$$\Delta U = \Delta H - p\,\Delta V$$
Consider a gaseous system in a container with a frictionless piston, and a pressure outside the container.

Is the pressure outside the container assumed to always be constant?
Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.
For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is $p$ the pressure of the surroundings as work is done against the pressure of the surroundings?

This doesn't seem to be the case in several examples in my lecture notes:
e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl.
Here, work is being done against the pressure of the surroundings, so I'd expect $p$ = pressure of surroundings. However, the notes say 'consider $p(\Delta V) = (\Delta n)RT$' and proceeds to input the number of moles of hydrogen gas. So here $p$ is taken as pressure of the hydrogen gas produced.

For compression of the gas, work is done by the surroundings on the system, so is $p$ the pressure of the gas in the container as work is done against this pressure?

while H = U + pV is true, which can be rearranged to give U = H - pV, it is not necessarily the case that $\Delta U = \Delta H -p \Delta V$. It is true to say that $dU = dH - d(pV)$, which you can say is $\Delta U = \Delta H - \Delta(pV)$. Take a look at relavant link to see what happens to $\Delta(pV)$ terms. I mention this because trying to explain things with an equation that is not general (which is the one you have given) can do more harm than good.

In the following I limit discussion to the particular expression $$ΔU = ΔH - pΔV$$
which follows (spoiler alert) from $$H\equiv U+pV$$ when the pressure $p$ of the system (the container) is  identical at the beginning and end of the process.

Is the pressure outside the container assumed to always be constant?

For an isobaric process such as the stated process involving zinc, by definition yes. Generally speaking the expansion work in the 1st Law is written  $$dw = -p_{ext}dV$$ where $p_{ext}$ is the pressure of the surroundings against which the system does work. If $p_{ext}$ is constant work is still against or by the pressure of the surroundings and the integral can be evaluated as $$w=-p_{ext}\Delta V $$ For a reversible process the pressure of the system and surroundings are equal and we evaluate $$w=-\int_{Vi}^{Vf} pdV$$ where $p$ is the pressure of the system.

Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.

In the case of a reversible process against constant external pressure, yes.
BTW Boyle's law ($p \propto V^{-1}$) does not consider changes in the number of particles during a reaction.

For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is p the pressure of the surroundings as work is done against the pressure of the surroundings?

The pressure of the system (container) and surroundings are assumed to be equal (balanced) for a reversible process. However, if the pressure of the system is greater than that of the surroundings (the latter assumed constant) then the work is still $$w=-p_{ext}\Delta V$$

This doesn't seem to be the case in several examples in my lecture notes: e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl

$pV$ work involves a change in volume. Here the volume change is due to formation of gas which causes the pressure in the container to increase by an infinitesimal amount $\mathrm{d}p$. The gas pushes against the piston, increasing the volume and returning the system to mechanical equilibrium when the internal pressure equals the external pressure $p$.

For compression of the gas, work is done by the surroundings on the system, so is p the pressure of the gas in the container as work is done against this pressure?

Assuming no chemical processes, compression requires the surrounding pressure to be higher than that of the system. It suffices that the difference be infinitesimal in magnitude, resulting in a reversible process. During such a process, the pressure $p$  of the system may vary during the entire process, remaining always equal (to within an infinitesimal amount) to the applied pressure.

$\Delta U = \Delta H - p,\Delta V$ does not require that the pressure be constant.  Because U and H are state functions, this expression is independent of path, and thus only requires that the initial pressure be equal to the final pressure.

@theorist I was presumably assuming a reversible isothermal process but see that that was not justified. I'll update the answer.

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


text:
Lagrange Multipliers for the derivation of Maxwell-Boltzmann distribution
When deriving the MB-Distribution while taking the number of quantum states $g_j$ into account, the final equation I'd have to solve would be:
$$\ln(g_j)-\ln(n_j)+\alpha+\beta\cdot E_j=0$$
Where $n_j$ is the number of particles at energy level $E_j$. The $\alpha$ and $\beta$ are the applied Lagrange Multipliers restrictions.
I have understood that adding or subtracting the Lagrangian Multipliers should not matter for the end result.
Now, when I continue with this equation, I'd eventually get:
$$n_j=e^\alpha\cdot e^{\beta\cdot E_j} \cdot g_j$$
When I want to derive the probability distribution formula of Maxwell-Boltzmann, I'd have to consider energy to be continuous. Substituting $g_j$ with the energy density formula times $\mathrm dE$, the total number of particles $N$ would then be:
$$\int_0^\infty\frac{V\cdot 2^{2.5}\pi\cdot m^{1.5}}{h^3}\cdot \sqrt E\cdot e^{\alpha}\cdot e^{\beta\cdot E}\cdot \mathrm dE=N$$
However, the result of this integration up to infinity would simply give the outcome $0$. But when you subtract the Lagrange Multipliers in the initial equation, the constants would have negative signs and the integration would give the correct formula for $N$.
But this means that it does matter whether you add or subtract the Lagrange Constants. What am I doing wrong here?

There is a great deal of difference between "$\mathit 0$" and "integral diverges". On the other hand, addition and subtraction are pretty much the same thing.

@Johnny Gui I think your problem is that you skip the step of identifying the Lagrangian Multipliers with physical quantities. It is correct that when setting up your auxiliary function for the minimization using Lagrangian Multipliers it does not matter whether you give the Lagrangian Multiplier terms a positive or negative sign for the correctness of your end result. But your choice of sign will reflect in the (sign of the) physical quantity the Lagrangian Multiplier represents, e.g. $\beta$ will be $\frac{1}{kT}$ in one case and $\frac{-1}{kT}$ in the other.

@IvanNeretin Perhaps I did it the wrong way, but how should this problem with my integration then be approached if addition and subtraction are the same thing?

@Philipp But what if the constants can only be derived physically by integrating the formula? I have seen sources in which e^-a is derived by integrating the equation up to infinity and then rewriting the outcome. But this is only possible when the Langrangian Constants are subtracted. How would one know this is how it should be derived if he is taught that subtraction or addition should not matter for the Langrangian Constants?

@JohnnyGui Have a look at this Wikipedia article. In the derivation contained therein the Lagrangian Multipliers are identified by recognizing a correspondence of the resulting equation with the Euler-integrated fundamental equation of thermodynamics. And it is this point where you will see a difference depending on whether you chose to subtract or add the Lagrangian Multiplier terms.

@JohnnyGui Furthermore, as you stated e^-a is derived by integrating the equation up to infinity and then rewriting the outcome will only not diverge if $a$ is a positive quantity. Maybe this is a part of your confusion. The Lagrangian Multipliers in your equation are at no point defined as positive quantities. Thus $\int_{0}^{\infty} e^{\beta E} \mathrm{d} E$ can either diverge or give you 0, depending on the definition of $\beta$. Usually this definition is forclosed to be $\beta = \frac{1}{kT}$ with a positive sign.

@Philipp Thanks for your answer. Using the method of the Wikipedia article indeed gives me negative signs in front of the constants, but I can't grasp that the derivation of, say e^a depends on the method you're using while my method should be correct as well. What should the exact trigger be that would make someone using my method think "something is wrong here" even if adding or subtracting constants should not matter?

@JohnnyGui Have you read my comment after the one mentioning the Wikipedia article? You have to be mindful of the initial assumptions and definitions the equations and methods your starting with entail and those they do not entail, e.g. in your equations you don't have $\beta = \frac{1}{kT}$. If you start from the Lagrangian Multiplier equation you are starting from (both Multiplier terms with a plus sign) then this will give you for example $\beta = - \frac{1}{kT}$ and not $\beta = \frac{1}{kT}$ and your integral will be perfectly fine.

@Philipp Yes I have, this is indeed clear to me regarding $\beta$ since $\beta$ can be deduced from another formula. But for the derivation of $e^{-\alpha}$, sources just use the integration up to infinity without even the need to define $\beta$ first. And since subtraction or addition should not matter, I'd expect the same in my case without the need to derive $\beta$ first, but it does matter. How would one know whether you should define $\beta$ first or not based on addition or subtraction?

@JohnnyGui You seem to be under the strong impression that a priori $-\beta$ is something negative and $\beta$ is something positive and that therefore  $\int_{0}^{\infty} \sqrt{E} e^{\beta E} \mathrm{d} E$ must diverge. But as I'm trying to tell you the whole time this is not so. The "intrinsic sign" of the Lagrangian Multiplier is not known a priori. It will only be apparent once you fill it with physical meaning by identifying the physical quantities it corresponds to. Switching the sign in the initial Lagrangian Multiplier equation, i.e. subtracting or adding the Multiplier terms...

... will lead to a switch of the "intrinsic sign" of the Multiplier. But nothing more. If your sources derive $e^{-\alpha}$ by solving the integral they certainly just assume that their exponential term containing $\beta$ has a negative exponent, e.g. that $\int_{0}^{\infty} \sqrt{E} e^{\beta E} \mathrm{d} E = \int_{0}^{\infty} \sqrt{E} e^{-|\beta| E} \mathrm{d} E$. They might just have skipped this detail as is often the case. Usually identifying $\beta$ would have to come first.

@Philipp I am aware that $\beta$ may very well be negative intrinsically. The only struggle I had was how sources were able to continue on with the derivation without defining the constants first, while I couldn't. But the second part of your reply about these sources leaving out that detail (perhaps because they already know from previous derivations that $\beta$ must be negative) has cleared this up for me. Thanks a lot for your help.

Ignoring multiplicative constants, the required integral is (in general form)
$$ \int \sqrt{x} e^{ax} \mathrm{d} x = \frac{1}{a}\sqrt{x}e^{ax}+\frac{i\sqrt{\pi}}{2a^{3/2}} \mathrm{erf}(i\sqrt{ax})$$
where $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\mathrm{d}t$$
Setting $x=E$, $a=\beta$, and the integration limits as $0$ and $\infty$, gives
$$ \int_0^\infty \sqrt{E} e^{\beta E} \mathrm{d} E = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|_{0}^\infty + \frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|_{0}^\infty = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|^\infty +\frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|^\infty $$
since $\mathrm{erf}(0)=0$.
To make further progress, consider that the function $\mathrm{erf}(x)$ must have a real input $x$ (otherwise it will "blow up" to $\infty$), which implies that $\beta$ must be negative, or, explicitly, that  $\beta = -|\beta| $ (we could also argue that the second term on the r.h.s must be real and come to the same conclusion). With negative $\beta$ we have that
$$\mathrm{erf}(i\sqrt{\beta E})=\mathrm{erf}(i\sqrt{-|\beta|E})=\mathrm{erf}(-\sqrt{|\beta|E})=-\mathrm{erf}(\sqrt{|\beta|E})$$
With this condition we also see that the first term on the r.h.s. is zero and as a result that
$$ \int_0^\infty \sqrt{E} e^{-|\beta| E} \mathrm{d} E =  \frac{\sqrt{\pi}}{2|\beta|^{3/2}} $$

Thanks for your answer. But can you please elaborate further why a real input of x must lead to a negative beta? I'm missing that part.

@JohnnyGui I elaborated on how x becomes positive with negative beta.

Ok thanks, but one could ask, why is it "bad" for it to "blow up" to infinity? Why can't one then just accept that the integrand is divergent?

@JohnnyGui The point of the integration is to find the normalization constant such that the integral from 0 to $\infty$ over the MB distribution is 1. If the integral blows up you can't normalize the distribution.

You can choose the signs either way round, the final result will be the same because the same physics is involved. You have presumably started by finding the number of distinguishable stationary states such as $\Omega=\sum_n N!/\prod_in_i$ subject to $\sum_i n_i=N; \; \sum_,n_i\epsilon_i=E$. As it turns out the maximum terms is greater than all the rest put together, i.e. the equilibrium condition is overwhelmingly probable, and so $\Omega_m= N!/\prod_in_i$ and from this we aim to find the maximum value which leads to $\displaystyle \frac{\partial \Omega_m}{\partial n_i}+\alpha+\beta\epsilon_i=0$ where $\alpha,\,\beta$ are the undetermined multipliers but they are not indeterminate. The $\alpha$ is related to the partition function and $\beta$ in a simple way to the temperature so they have a physical meaning so that their values are fixed because of that, thus it does not matter if the equation were $\displaystyle \frac{\partial \Omega_m}{\partial n_i}-\alpha-\beta\epsilon_i=0$
Mathematically it is easier to use the log thus, start with $\displaystyle \frac{\partial \ln(\Omega_m)}{\partial n_i}-\alpha-\beta\epsilon_i=0$ and after several steps and as each $n_i$ is independent then $\ln(n_i)+\alpha+\beta=0$. This gives $n_i=e^\alpha e^{-\beta e_i}$  which is the Maxwell Boltzmann distribution and $n_i=g_ie^\alpha e^{-\beta e_i}$ if degeneracy is included.
To go further the multipliers have to be found. As $\sum_in_i=N$ then $N=e^{-\alpha}\sum_n g_ie^{-\beta\epsilon_i}$ and the partition function is $Z=\sum_n g_ie^{-\beta\epsilon_i}$.
It is more difficult to calculate $\beta$ and the entropy is needed to do so. But we know that $\beta$ must be a positive number (as defined above in $ \partial \ln(\Omega_m)/\partial n_i-\alpha-\beta\epsilon_i=0$) otherwise the partition function would diverge towards infinity (and if zero the partition function would be an infinite sum). The same is true for your integral.
$\beta$ must be related to temperature because if there were two isolated systems that were then brought into contact and so exchange energy and come to equilibrium this then is possible only with the same value of $\beta$. Finally an increase in $\beta$ is associated with a decrease in temperature. The ratio of two distributions is $\displaystyle \frac{n_k}{n_i}=\frac{g_k}{g_i}e^{-\beta(\epsilon_k-\epsilon_i)}$. If $\epsilon_k>\epsilon_i$ when $\beta$ decreases the ration $n_k/n_i$ must increase. Hence the total energy of the system must increase as  population moves from lower levels $i$ to those of $k$. Thermodynamically such a change is associated with a rise in temperature.
As the entropy is defined as $S=k\ln(\Omega_m)$  and given by classical thermodynamics as $\displaystyle \left(\frac{\partial S}{\partial E}\right)_{V,N}=\frac{1}{T}$ we find that $\displaystyle \left(\frac{\partial S}{\partial E}\right)_{V,N}=\frac{1}{T}=k\beta$, where $k$ is a (Boltzmann's) constant.
Had we used positive signs in the starting equation $\partial \ln(\Omega_m)/\partial n_i$ then we would have to change the sign in front of each $\beta$ and end up with the same definition of the partition function $Z$.

Thanks for your explanation. I noticed that when I write out $ln(\Omega_m)$ I would get $\partial[\partial n_i \cdot k(\alpha + \beta E)] \cdot \frac{1}{\partial E}$. Could you please explain how $\partial n_i$ is removed from the equation so that the formula for $\frac{1}{T}$ would not contain $\partial n_i$?

I'm not quite sure where you are starting, but to get the entropy use Sterling's approx for the factorial and get $S=k(N\ln(N)+\alpha N+\beta N)$ then obtain $S=N(k\ln(\sum_i e^{-\beta \epsilon_i} +\beta E))$. Now the rest is messy and tricky; $\beta$ depends on $E$ as $\sum_in_i=N ; \sum_i n_i\epsilon_i=E,; n_i=e^{-\alpha} e^{-\beta \epsilon_i}$ so to get $\partial S/\partial E$ we find $k\beta +k\frac{\partial}{\partial \beta}(N\ln(\sum_i e^{-\beta \epsilon_i})+\beta E)\frac{\partial \beta}{\partial E}$. Evaluate this and only the $k\beta$ remains. (Please check my signs).


text:
Bibenzimidazole nomenclature: Indicated H and locants
I am trying to understand how the locants work for the following molecule (locants added by me).

I added the locants following IUPAC Blue Book rules from 2013, P-14.4. Basically, first numbering heteroatoms in the ring, assigning the highest locant to the one with an indicated hydrogen, and then number clockwise around the ring (red locants).
I assumed the same would go for the second ring, so the name would be:
$1^1H,2^1H-1^2,2^7-\text{bibenzoimidazole}$
However, Chemdraw says the name is:
$1H,3'H-2,4'-\text{bibenzo[d]imidazole}$
I get the differences between the prime and composite locants (P-14.3.1), however it shows a numbering more like:

Is Chemdraw wrong or am I missing something here? If Chemdraw is right then I'd assume that for some reason the ring assembly has preference for a lower locant, but I can't find the rule that states this in the Blue Book.

Your understanding of the numbering of the individual cyclic systems is correct. If we consider the following arbitrarily simplified separate compounds, the numbering is in accordance with your expectations; low locants are given first to the indicated hydrogen and then to the substituent according to the usual rule P-14.4.

In ring assemblies of two identical cyclic systems that are joined with a single bond, however, low locants are given first to the junction according to Rule P-28.2.1.

P-28.2.1 Ring assemblies with a single bond junction
(…)
Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
  (…)

After that, indicated hydrogen is named and numbered according to P-28.2.3.
Therefore, the systematic name is 1H,3'H-2,4'-bibenzimidazole and not 1H,1'H-2,7'-bibenzimidazole since the locants 2,4' for the junction are lower than 2,7'.


text:
Is basic solution an electrode?
Question:

Calculate the potential of hydrogen electrode in contact with a solution of pH 10.

My attempt:
Cell reaction: $\ce{ H^+ + e^- -> \frac{1}{2}H_2 }$
I've attempted to solve this myself by using Nernst equation which yielded the answer -0.591 V. But I believe that Nernst equation is used when there are two electrodes, but my friends are saying that
"how will you make a cell from one electrode?".
So guys, please rectify the problem.

Easy. Choose smarter friends. Half cell potentials are measured relative to the Standard Hydrogen Electrode which is arbitrarily assigned an EMF value of 0.00

@MaxW, can I assume the solution to be an electrode, and thereby making a Galvanic cell from the two?

"how will you make a cell from one electrode?".

There is no need to, because there is nothing which prohibits you from writing a Nernst equation for a half cell.* Remember the rule no.1 in electrochemistry, whenever you mention half-cell potential, your next automatic question should be "with respect to what?". If the textbook is quiet and you are required to look up tables, it is understood that it is with reference to the hydrogen electrode under standard condition which is set at 0 V (exactly).
Now at pH 10, your standard conditions have changed and the activity or hydrogen ion concentration differs. It is no longer zero. So proceed with the Nernst equation for the half cell and work it out.

*(Yes, there is no single electrode galvanic cell, but for fun and general knowledge, people have done single electrode reduction using electrostatics (Bard and others). Faraday did it centuries ago. It was published in a very reputable magazine.)

Thanks. One last question: In general, does basic solution acts as an electrode?

No. Half cell is hypothetical and it cannot exist alone, but Nernst equation can be written for a half cell.

Search the definition of an electrode https://en.wikipedia.org/wiki/Electrode

You can't make a cell with one electrode, but there is such a thing as a cell with only one half-reaction.  Combellas et al. [1] used such a cell to generate a solvated-electron solution of magnesium in ammonia.
Suppose you have a magnesium anode, a chemically inert cathode, and an electrolyte that is a dissolved magnesium salt $\ce{MgY2}$.  (You have to choose the salt with some care because you have a small, multiply charged cation in a low polarity solvent, which leads to a limited range of soluble salts.  Also in practice the authors generated the magnesium salts from commercially available ammonium salts that were converted to the magnesium salts in situ.)  You apply the current and the anode duly produces a reaction that is probably most accurately rendered thusly:
$\ce{Mg(s) + Y^- -> MgY^+ + 2e^-}$
(i.e. the magnesium ions are probably in the form of ion pairs, which being only singly charged would make a more stable solvated-electron solution.)  The electrons go over to the cathode and pour into the solution with no reaction at all, becoming the solvated electrons!
We are unfamiliar with such a situation in water solvent because the solvated electrons in water get consumed so quickly that they in effect react at the cathode surface.  Thus you are forced to include the cathodic half-reaction.  But in the ammoniacal case, these electrolytically generated magnesium solutions are finding use in activating fluorinated polymer surfaces, which is driving the research.
Reference
1.
Combellas, C & Kanoufi, Frederic & Thiébault, A. (2001). "Solutions of solvated electrons in liquid ammonia: Part 1. Chemical properties of magnesium solutions." Journal of Electroanalytical Chemistry 499, 144–151.  Link

The magnesium oxidation process in liquid ammonia is very interesting. Current must be flowing in the system. Did they mention what was happening at the cathode?

The electrons just keep going, intermingling with the cations formed at the anode.  They do not last forever (and to get them to last anyway, you need a sufficient amount of supporting electrolyte, which suggests to me the formation of ion pairs).  But when they do react it's in solution, not at the cathode.


text:
Cerium electronic configuration change from 0 to +2 oxidation state
Cerium in $0$ oxidation state has electronic configuration
$$[\ce{Xe}]\mathrm{(4f)^1(5d)^1(6s)^2}$$
But when it gets oxidised to $+2$ state, it becomes
$$[\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$
This is a change as in other elements in the lanthanides series get their $\mathrm{6s}$ electrons abstracted and the other shells remain at liberty.
Could you please explain this phenomenon to me? And also correct me if anything unsaid above was incorrect.

Related, but having no answers with positive scores:  https://chemistry.stackexchange.com/questions/83170/why-is-the-ground-state-electronic-configuration-of-group-10-ni-group-elements?r=SearchResults.

I think it's because that when we go across the lanthanides, at the first d orbital is penetrated more, but then the f orbital penetrates more and d becomes the outer orbital after s, and electrons like to be closer to the nucleus duo to columbic forces, and that's what's happened there.

The lanthanides are most stable in their 3+ oxidation state. For example, lanthanum is perfectly happy changing from
$$\mathrm{La}^0 = [\ce{Xe}]\mathrm{(5d)^1(6s)^2}$$ to
$$\mathrm{La}^{3+} = [\ce{Xe}]\mathrm{(5d)^0(6s)^0}$$
Then, as you go along the lanthanide series, you're adding more 4f electrons. For example:
$$\mathrm{Ce}^{3+} = [\ce{Xe}]\mathrm{(4f)^1(5d)^0(6s)^0}$$
$$\mathrm{Pr}^{3+} = [\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$
$$\mathrm{Nd}^{3+} = [\ce{Xe}]\mathrm{(4f)^3(5d)^0(6s)^0}$$
etc.
This electronic configuration is extremely stable for all lanthanides, and there are very few exceptions to it.
Ce(II) is extremely unstable, and if there's one way to stabilise it, that would be adoption of the Pr(III) electronic structure. This way Ce has no 5d or 6s electrons, which makes lanthanides happy (apologies for the non-scientific way of framing this).

Calcium in the zero oxidation state is highly unstable, whereas titanium in the +2 state is more stable as titanium monoxide.  So why does the calcium not adopt the $d^2$ valence electron structure seen in the latter?

@OscarLanzi because Ca prefers losing the two s electrons rather than having a partially filled d shell? Whereas lanthanides are perfectly happy with partially filled 4f shells, as long as their 6s and 5d are empty.

In the "Chemistry of the Elements", By N. N. Greenwood and  A. Earnshaw, it is stated that the only lanthanides which exist under the oxidation state $+2$ are Samarium, Europium and Ytterbium. Only a couple of cerium(II) compounds are known. But sulfide for example is probably made of Ce(III) plus one electron present in the conduction band. The most stable compound is the iodide $\ce{CeI2}$. But this diiodide is very easily oxidized : It liberates hydrogen from water.


text:
Relation between electron gain enthalpy and electron affinity

The correct relation between electron gain enthalpy $(Δ_\mathrm{eg}H)$ and electron affinity $A_\mathrm{e}$ at any temperature '$T$' is
A) $Δ_\mathrm{eg}H = -A_\mathrm{e} - \frac{5}{2}RT$
B) $Δ_\mathrm{eg}H = \frac{A_\mathrm{e}}{RT}$
C) $Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT}$
D) $Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT} + \frac{1}{A_\mathrm{e}^2}$

I have read that electron affinity could be taken as electron gain enthalpy at absolute zero, i.e. 0 K. So, maybe we could use Kirchoff's law for temperature $T$, but why is $C_p$ taken and not $C_v$ or something else, is it from the definition?

You are correct, at absolute zero $Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e}$ for the gas-phase act of gaining an electron:
$$\ce{X(g) + e-(g) → X-(g)}$$
$$A_\mathrm{e} = E(\ce{X(g)}) - E(\ce{X-(g)})$$
The term $5/2RT$ arises from the so-called "electron convention" when electron is treated as ideal gas with corresponding heat capacity $C_p$ from Boltzmann statistics and the element of enthalpy
$$H_T(\ce{e-}) - H_0(\ce{e-}) = \frac{5}{2}RT,$$
resulting in exact relation
$$Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e} - \frac{5}{2}RT.$$


text:
Can we combine two reactions and then calculate the equilibrium concentrations?

At a certain temperature, $\ce{N2O5}$ dissociates as:
  \begin{align}
\ce{N2O5 (g) &<=> N2O3 (g) + O2 (g)} & K_1 &= 4.5
\end{align}
  At the same time, $\ce{N2O3}$ also dissociates as:
  \begin{align}
\ce{N2O3 (g) &<=> N2O (g) + O2 (g)}  & K_2 &= \mathrm{?}
\end{align}
When 4 moles of $\ce{N2O5}$ are heated in a flask of $\pu{1 L}$ volume, it has been observed that the concentration of $\ce{O2}$ is $\pu{4.5 M}$ at
  equilibrium. What are the equilibrium concentrations of the other products?

The combined reaction:
$$\ce{ N2O5(g) <=> N2O(g)  + 2O2(g)}$$
$$K_\text{combined} = K_1 \times K_2 = 4.5 K_2$$
There are 4 moles of $\ce{N2O5}$ initially. If $x$ moles of $\ce{N2O5}$ dissociate to give $x$ moles of $\ce{N2O}$ and $2x$  moles of $\ce{O2}$, then plugging in the values in the $K_c$ expression for this reaction we get
$$4.5 K_2 = \frac{4x^3}{4-x},$$
where
$$x = \frac{4.5}{2 \pu{mol}}.$$
Solving for $K_2$, I get $K_2=\frac{81}{14}$, which is incorrect.
Is it allowed to first combine the reactions and then use the combined reaction?

I think there is sufficient information about where the OP's attempt to solve this went awry. It might help to see the "combined reaction", though. And maybe which species are in the flask after equilibrium is established.

The misconception is the same as in this question. If there are two coupled equilibria, e.g. A reacts to B reacts to C, B might be non-zero, so you can't just combine the reactions to A reacts to C and neglect that there might be some B at equilibrium. I think this question merits an answer.

Does this mean that to apply stoichiometry to an equilibrium problem it is necessary for the reactants to directly convert to products (elementary reaction)?

You can, but you have to account for it. You can't say "x moles of N2O5 dissociate to give x moles of N2O", you have to say "x moles of N2O5 dissociate to give x moles of N2O3 or N2O", and use additional information to figure out how much is N2O3 and how much is N2O.

@KarstenTheis So should the reaction must be elementary for the OP's method to work?

@Anton The OP's method only works if the concentration of $\ce{N2O3 (g)}$ is neglible (when the second equilibrium constant is much higher than the first).

Different from Zhe's answer, I will use the approach of a combined ICE table using the model discussed in How can you use ICE tables to solve multiple coupled equilibria?. This results in a system with two unknowns that, in my opinion, gives a bit more insight into the problem as we are solving it. $x$ are changes due to dissociation of $\ce{N2O5}$ (first reaction), and $y$ are changes due to dissociation of $\ce{N2O3}$ (second reaction).
$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{N2O5}]  & [\ce{N2O3}]  & [\ce{N2O}]&[\ce{O2}]   \\
\hline
I &  4 & 0 & 0 & 0 \\
\hline
C &  -x  & +x-y & +y & +x+y  \\
\hline
E &  4-x & +x-y & +y & 4.5 \\
\hline
\end{array}
$$
I have two unknowns, $x$ and $y$. Unless one of them is much bigger than the other (in which case I can first neglect the smaller one and come back to it later), I have to solve a system of two equations for $x$ and $y$ simultaneously. The first equation is already buried in the ICE table (subscript $eq$ is for equilibrium state):
$$ x + y = \ce{[O2]}_{eq} = 4.5 \ \ \ \ \text{or solved for y: }\ \ \ \ y = 4.5 - x$$
The second equation is via the equilibrium constant $K_1$:
$$ K_1 = 4.5 = \frac{[\ce{N2O3}]_{eq} [\ce{O2}]_{eq}}{[\ce{N2O5}]_{eq}} = \frac{[\ce{N2O3}]_{eq} \times 4.5}{[\ce{N2O5}]_{eq}}$$
Canceling 4.5 and rearranging gives $ [\ce{N2O5}]_{eq} = [\ce{N2O3}]_{eq}$, and substituting from the ICE table gives a second equation for x and y:
$$4-x = x - y\ \ \ \ \ \ \ \  \text{(or simplified } 2x = 4 + y)$$
Substituting the first equation into the second, we get:
$$ 2x = 4 + y = 4 + 4.5 - x$$
$$ x = 8.5/3 = 17/6$$
Substituting the value for $x$ back into the first equation:
$$ y = 4.5 - x = 4.5 - 17/6 = 10/6 = 5/3$$
Now we can tabulate the equilibrium concentrations and (to compare with Zhe's answer) the equilibrium constant $K_2$:
$$ c(\ce{N2O5})_{eq} = 7/6 M = 1.167 M$$
$$ c(\ce{N2O3})_{eq} = 7/6 M = 1.167 M$$
$$ c(\ce{N2O})_{eq} = 5/3 M = 1.667 M$$
$$ K_2 =  \frac{[\ce{N2O}]_{eq} [\ce{O2}]_{eq}}{[\ce{N2O3}]_{eq}} = \frac{5/3 \times 4.5}{7/6} = \frac{45}{7}$$
We can check that the equilibrium concentrations add up to the initial concentration of 4 M, and that we obtain the correct value for $K_1$ when substituting our answers into the equilibrium expression.

Is it allowed to first combine the reactions and then use the combined reaction?

Yes, but we have to consider all species simultaneously because none of them are minor species. When calculating, for example, hydronium and hydroxide concentrations in a solution of a weak acid, we can set aside the autodissociation of water because hydroxide is a minor species, and adjusting the hydroxide concentration after calculating the weak acid equilibrium is fine because it does not change the hydronium concentration much, hydroxide being the minor species. Here, we can't do that and have to take the more comprehensive approach of considering the two equilibria simultaneously.

One suggestion is actually to have 2 rows in the ICE table for the changes. This way, you can write separately the changes for each equilibrium to draw a better analogy to the single equilibrium problem.

@Zhe That is a great suggestion, but we need a better table type-setting tool on StackExchange so it's easier to make subdivisions in tables.

Tables are kind of gross with every tool. :/

@KarstenTheis@KarstenTheis:I take each reaction as independent equilibrium,I calculate the amount of $\ce{N2O3}~\text{and the amount of} ~\ce{O2}$ produced in the first reaction where the concentration of each equal$\pu{2.55234M}$,then I found the amount of  the species in the second equilibrium where $[\ce{N2O}]=[\ce{O2}]=4.5-2.55234=\pu{1.94766M}$ and $[\ce{N2O3}]= 2.5534-1.94766=\pu{0.60468M}$ to calculate $K_2 = 6.3$. correct the solution.

Karsten's insight is correct. Your mistake is in assuming that you have either $\ce{N2O5}$ or $\ce{N2O}$ and $\ce{O2}$.
There are four concentrations of interest: $\ce{[N2O5]}$, $\ce{[N2O3]}$, $\ce{[N2O]}$, and $\ce{[O2]}$.
How can you relate these variables?
Well, you can apply conservation of matter in terms of oxygen and nitrogen.
For nitrogen:
$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$
For oxygen:
$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2 \ce{[O2]} = 5\times(4\ \mathrm{M})\tag{2}$$
Then, there are the equilibrium expressions:
$$K_{1}=4.5=\frac{\ce{[N2O3][O2]}}{\ce{[N2O5]}}\tag{3}$$
$$K_{2}=\frac{\ce{[N2O][O2]}}{\ce{[N2O3]}}\tag{4}$$
If you substitute $\ce{[O2]} = 4.5\ \mathrm{M}$ into (2), (3), and (4), you end up with:
$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$
$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2\times(4.5\ \mathrm{M}) = 5\times(4\ \mathrm{M})\tag{2'}$$
$$4.5=\frac{\ce{[N2O3]\times(4.5\ \mathrm{M})}}{\ce{[N2O5]}}\tag{3'}$$
$$K_{2}=\frac{\ce{[N2O]\times(4.5\ \mathrm{M})}}{\ce{[N2O3]}}\tag{4'}$$
That's 4 equations and 4 unknowns, and it's all algebra from there.
Assuming I did that correctly, I got $$K_{2} = \frac{45}{7}$$

kindly to write mass-balance equations for this problem:https://chemistry.stackexchange.com/questions/111041/how-can-you-use-ice-tables-to-solve-multiple-coupled-equilibria


text:
Qualitative analysis of iron(III) salt
Why will $\ce{Fe^3+}$ give brown ring test?
In my textbook it was given that ferric ion gives brown ring too. I am confused, if ferric ion is already present, then what will reduce nitrate ion to nitric oxide?

The textbook must've been implying that the complex is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$, e.g. it's ferric iron that actually forms the complex. If nitrate is oxidized to nitrosyl by other means, then, yes, iron(III) will also give positive brown ring probe.


text:
No current in homemade galvanic cell
I built a galvanic cell using 0.01 M solutions of copper sulfate and zinc sulfate. I used copper and zinc electrodes and immersed them into the proper solutions (copper into copper sulfate, zinc into zinc sulfate).
I used approximately 30 mL of each solution and put them into separate beakers. I connected the solutions using a salt bridge that was filled with 0.01 M potassium nitrate solution. I used cotton to cover the ends of the salt bridge (a plastic tube).
I measured the voltage across the copper and zinc electrodes and I got around 1 V. However, there is no current being produced according to an ammeter. The galvanic cell cannot light up a small light bulb for instance.
Furthermore, I took voltage measurements over a 50 minute period and the the voltage looked relatively constant throughout the period, though very slightly lower at the end.

Your small copper-zinc cell may not light up a bulb as often shown in textbooks. Check your ammeter settings, it must show some current. How are you connecting it in series? voltmeter is connected in parallel. Now an interesting question. Imagine you join 12 (e.g., Duracells) in series to generate about 12 Volts. Can one start a car with 12 Duracells? Why do you need a large car lead-acid battery? The answer lies in current. Same potential difference does not mean same amount of current supply. You may think 1 V is close to an ordinary cell, your single copper-zinc cell cannot supply enough current to even light up a bulb. Cell potential is a thermomdynamic concept. Current supply is a kinetic issue-how fast the electrons transfer to or from electrodes. Now you can think of the reason why there is no commercial battery made from zinc and copper. See Daniell cell for an interesting history. Check the size of electrodes in Daniell cells.

Is there a way I could improve the current? Will increasing the size of the electrodes increase current?

Yes, increasing the size will increase current, or you can make two or three cell and connect in series.

How will connecting the cells in series increase the current? How is the kinetics of the electrons affected by adding cells?

The kinetics remains the same, however if you have 2 Duracells, each of 1.5 V, your connect them in series, the voltage is now 3.0 V. If the resistance of your load is constant (like a bulb), by Ohm's law, current also increases, V= IR.

Thanks for the information. Can you describe the kinetics of the electrons in the cell?

"Can you describe the kinetics of the electrons in the cell?" Not an easy task at all to determine electron transfer kinetics at electrodes. I assume that you are a high school student. There is a big handbook, called Handbook of Batteries by David Linden. See if you can get it from a library. Lots of info.

I added another cell to the battery. The voltage increased to 2V, but the current remained the same (nonexistent).  I decided to take one cell and combine the separate copper and zinc solutions into one beaker. Now, there is a measurable current (5-7.5 mA) with a 100-ohm resistor.

This is called serial connection and it was expected that you will find measurable current. It means that the high resistance salt bridge with cotton is gone.

you can't get the filament bulb glue even a LED doesn't
why :
there is 2 things:
1/ low voltage
the LED need minimum voltage of 1.2 v to 1.6 v to light up
2/ big internal resistor (low current)
the filament lamp need more current to light up; and in your case there isn't because of the internal resistor of the cell; ie your salt bridge contain
cotton to cover the ends of the salt bridge which is a big internal resistance (few ohms) in that way you can't get high current only a voltage between the electrodes

Good catch, the cotton balls must be huge resistance.


text:
Ammonium considered as metal
Why ammonium ion has very similar properties to the heavier alkali metals and is often considered a close relative?
Ammonium is expected to behave as a metal ($\ce{NH4+}$ ions in a sea of electrons) at very high pressures. Can anyone explain?

I'm confused about the charge here. Wouldn't the ammonium cation behave like a group I cation, and the ammonium radical ($\ce{NH4}$, neutral molecule) like elemental group 1 metals? Reference

Who needs high pressure?  Ammonium can be amalgamated into mercury like an alkali metal.  As with metals generally, there are not really any ammonium molecules in this phase; rather, the ammonium cations are intermingled with mercury cations and delocalized valence electrons.
Nor is all that high a pressure needed for unamalgamated ammonium, relative to metallizing other materials.  Bernal and Massey [1] report that a pressure of 250 kbar would be sufficient to metallize an ammonia-hydrogen mixture, which may be compared with the rocky material deep in Earth's mantle remaining nonmetallic even at megabar pressures. The authors suggest that the ammonia-rich giant planets Uranus and Neptune may contain this material in their interiors.
Ammonium is easy to metallize because it has, indeed, an alkali-metal electronic structure.  The tetrahedral structure common to ammonium and methane, $\ce{CH4}$, is most stable with eight valence electrons -- there are four bonding orbitals, corresponding to the four bonds in methane, and four antibonding orbitals.  In methane the neutral molecule matches this optimal electron count, but in ammonium the neutral molecule would have to include a ninth valence electron in its antibonding orbitals, like an alkali metal atom having one too many electrons for a stable noble gas configuration. Either this extra electron breaks down the proposed ammonium molecule (which is what we ordinarily see) or, under moderately high pressure or in a metallically bonded matrix like the amalgam, it goes into the metallic bonding.
Reference
1.
M.F.M. Bernal, H.S.W. Massey, "Metallic Ammonium",
Monthly Notices of the Royal Astronomical Society 114(2) (April 1954), Pages 172–179.


text:
How do salt additives and impurities affect a critical micelle concentration?
So I understand that a CMC typically decreases as more salt additives are added. What I am not clear on is how exactly.
The head groups are charged in this example (one anionic and one cationic) so I am not sure how the salts would impact this? There is a counter-ion present as well.
Thank you so much for your response, its really helpful. I'm just learning about CMC and micelle so please forgive my silly questions. When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller.
Would the presence of salt additives and impurities be of greater concern in measuring CMC by absorbance, fluorescence or conductivity???

In a solution with two dissociated ionic detergent molecules of equal charge (say two anionic detergent molecules $\ce{A^-}$), their interaction will be dominated by two effects:

Electrostatic headgroup repulsion
Hydrophobic interactions

The first contribution, electrostatic repulsion, will oppose the second and discourage association of the detergent molecules.
If you instead mix two oppositely charged detergents, $\ce{A^-}$ and $\ce{B^+}$, the electrostatic headgroup interaction will now encourage association. All else (e.g. tail length) being equal, this means mixed micelles will have a lower critical micelle concentration (CMC) than pure detergent at the same total concentration.
Adding salt (say $\ce{NaCl}$) to a charged detergent will also lower the CMC, for essentially the same reason as for mixed micelles (electrostatic repulsion is reduced).
Now comes the tricky part. If you add salt to a mixed detergent system, you set up possible competition between electrostatic attraction of salt and headgroups, with electrostatic attraction of oppositely charged headgroups. Individually, either attractive interaction has a tendency to lower the CMC. The question is what effect they will have together. My educated guess is that the effect of salt on mixed micelle systems will be subtle and complicated. The CMC will be further reduced, but interactions between oppositely charged detergent molecules will predominate so the effect of added salt will be minor.

When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller?

I don't think there is a universal answer to the first question.  I will have to wave my hands a little, and say it depends on the detergent, solvent, and temperature. However if you look at a table of aggregation numbers and CMCs for different detergents, you'll usually see that the ones with a lower CMC (for instance, nonionic detergents) also have larger aggregation numbers.
The second question is easier to answer than the first.  The size of a micelle can be measured by a number of different techniques. The aggregation number is often derived by making assumptions about packing in the micelle and starts from estimates of the sizes of the micelle and detergent molecule. So it's pretty safe to generalize that both the dimensions of the micelle (the effective diameter say) and aggregation number are reduced in smaller micelles.

Thank you so much for your response, its really helpful. I'm just learning about CMC and micelle so please forgive my silly questions. When you lower a CMC, are the micelles themselves now smaller or bigger? And when the literature describes micelles as smaller, does this mean the circumference is smaller, or the number of detergent molecules is smaller?

Won't mixing an anionic and a cationic surfactant result in precipitation of the two oleophilic parts as a fat or oil, while the smaller ions stay in solution?

@HarleyMcFarlen I added an answer to your question to the body of my answer.

@JamesGaidis While what you write may be true in some cases, for some detergents at lower concentrations it is possible to form stable mixed micelles near the isoelectric point.

Thanks for everyone's help! Ill keep picking away at it.


text:
Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride

Why does benzenediazonium chloride get converted to benzene upon
reacting with $\ce{H3PO2}$ and not $\ce{LiAlH4}$ ?
Also, what is the reaction mechanism of the reaction with $\ce{H3PO2}$
?

This Wikipedia page list all kind of nucleophilic substitution reaction on aromatic compound why can't $\ce{LiAlH4}$ / $\ce{H-}$ react like one of these to give benzene and expel nitrogen.
I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 / H-}$ can't reduce the double bond of Benzene, because when styrene reacts with $\ce{LiAlH4 / H-}$ it gives ethyl benzene.

Update
I was just researching (on google) for the mechanism where I came across this site suggesting the mechanism to be radical and as follow :

But still there is one problem the side products don't match this site states the side product to be $\ce{H3PO3}$ and $\ce{HCl}$ but as stated above the side products must be $\ce{H2PXO2}$.

Consider the possibility that the PCl(OH)2 undegoes hydrolysis - how stable is PCl3 in water?

@Waylander I know PCl3 reacts with water vigourously to give H3PO4 and HCl

So do you think it likely that other P-Cl species mat hydrolyse easily?

@Waylander yeah they may ........

![enter image description here](https://i.stack.imgur.com/jwac7.jpg this is radical substitution reaction ,to get idea for mechanism click image

I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 /
H^{−}}$ can't reduce the double bond of Benzene, because when styrene
  reacts with $\ce{LiAlH4 / H^{−}}$ it gives ethyl benzene.

That answer is as good as any for the first part of your question.  It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring.  Similarly there are aliphatic pi bonds in $\ce{Ar - N^+\equiv N}$, and so the LAH will reduce the diazonium group rather than displacing it.
For the hypophosphorous acid reduction, the mechanism you found has a problem.  The principle of microscopic reversibility argues against three components coming off at once in what is shown as the first step (it would be equivalent to three independent bodies colliding simultaneously).  The first step may be correct but it likely has more than one component to it.

I got the LiAlH4 part, thanks, can you please provide a viable mechanism for the reaction with H3PO2 ?

I would have to guess.  All I can tell is that what is rendered as Step 1 must somehow involve multiple steps.  One possibility: the anion attaches to the hypophosphorous acid molecule making the phosphorous four-coordinate, then a single electron is transferred converting the diazonium ion to a radical, then N2 comes off that.  This avoids three-body interactions but is just a guess on my part, so I am unwilling to add it to the answer.


text:
Ab initio calculations
Why are quantum mechanical calculations so computationally expensive? What exactly is the process that makes it so time taking?
What I mean is, one can perform classical dynamics simulations on a large system like proteins with no difficulty unlike ab initio calculations. Why? What type of calculations are to be performed in ab initio that makes it expensive? Matrix diagonalization?

Any $n$ body problem is hard with strong interactions between all bodies.

Well, not every calculation is expensive and so time-consuming; it depends on the size of the system, level of theory, computer power, ... cheap calculations on small systems can be done in a matter of seconds, so it doesn't seem correct to call everything expensive.

@orthocresol What I mean is, you can perform classical dynamics simulations on a large system like proteins with no difficulty unlike ab initio calculations. Why? What type of calculations are to be performed in ab initio that makes it expensive? Matrix diagonalization?

@Chemist0304, yes, that makes sense! Could you please edit that into your question?

It basically comes down to scaling with the size of the system, which will correspond to the number of electrons (that is, occupied valence orbitals) and the size of the basis set (which must have a sufficient quality, that is, size to yield meaningful results).
Edit: I should note that not all quantum chemists consider DFT to be ab-initio. Given the prevalence of such calculations, I have included it in my answer.
Let's consider density-functional theory (DFT) and Hartree-Fock (HF) theory first. These are a ground-state methods that describe a system with a single "wave function" called a Kohn-Sham or Slater determinant, respectively. This is valid for many, but not all systems - especially when there are several d-block metals closely interacting, this can be a bad approximation. For DFT, the cheapest useful specific methods (called functionals) in the most common implementations (i.e. programs) will scale as $\mathcal{O}(N^3)$ (where $N$ is some measure of the system size) due to the calculation of the electron-electron interaction.$^1$ At some point, a matrix diagonalization is performed, which has the same scaling. For HF, the scaling is worse, because of the calculation of a term called exchange (that arises from the specific form of the wave function ansatz), which scales as $\mathcal{O}(N^4)$.
At this point, I should mention that efforts to lower the prefactor of the computational cost function of these workhorses of quantum chemistry are ongoing and some have reached maturity. For "small" systems, the cost is then often dominated by terms other than the highest scaling ones. However, you will eventually hit the scaling wall. There are also efforts to lower the scaling behavior by introducing more approximations, smarter discarding of near-zero terms etc. Again, you may find out the hard way, that in the system you wish to calculate, the approximations do not hold or are not effective.
For excited states, one can assume a $+1$ scaling behavior and a worse prefactor. When considering the ground state with more accuracy, one then turns to post-DFT methods such as double-hybrid density functionals or post-HF methods such as Configuration Interaction, Møller–Plesset perturbation theory, Coupled-Cluster theory (e.g. CISD, MP$n$, CCSD(T)). In the standard implementations, they scale at least as $\mathcal{O}(N^5)$, but can go up to 6 and 7. Again, smart selection of the involved terms can lead to significant speed-ups, leading to an empirical linear or near-linear scaling at the expense of increased prefactors and less certainty.
The comments to the question mention classical dynamics of proteins. The scaling wall even for a single-point energy is hit well before approaching 100 $\ce{CNO}$ atoms (in my personal experience, which is a bit dated. Make it 200, then). The necessary derivatives with respect to the nuclear coordinates also tend to add to the prefactor or the scaling in a significant fashion.

$^1$ This scaling applies if a particular additional approximation called density fitting is made, which is usually a very good one. Note that better DFT methods will have worse scaling.

Please see  "Elucidating the fundamental forces in protein crystal formation: the case of Crambin " Chem. Sci., 2016,7, 1496-1507 by Piane et al. which has a full structural optimisation using a B3LYP-D* functional of a small protein. It is a periodic calculation using a Gaussian basis set, with up to 1800 atoms per unit cell

It’s always surprising that a simple electron with no brain manages to do these calculations effortlessly...

Your sentence about the exchange term and the $\mathcal{O}(N^4)$ seems misleading to me. This scaling comes from the 2-electron repulsion integrals, where we have the Coulomb and Exchange term. The exchange integrals do not really add any additional complexity over the coulomb integrals. And in DFT we need the 2-electron integrals as well. So both methods have an $\mathcal{O}(N^4)$ when it comes to calculating these integrals.

@JonCuster The electron does not need to calculate these tings. It just does. The computational cost is related our mathematical description of the physics. It is not intrinsic to the actual physics.

@Feodoran - indeed, my tongue was firmly in my cheek. Just not a fan of smiley faces...

@JonCuster In that case, consider my comment not for you, but for anyone else who who misunderstood your comment, like me.

@TAR86 already described whats makes DFT and HF are computationally expensive. Here I would just like to add, that classical molecular dynamics simulations no explicit treatment of quantum mechanics is considered. This means the fundamental theory (Newton's equations of motion) are much simpler, and therefore easier to calculate. But of course this sacrifices accuracy. As a side note, those two approaches are actually simulating different things: DFT and HF are only about the electrons, while molecular dynamics is about where the atoms are (and move).
Also I want to try to answer the question from a more abstract perspective:
The problem about the equations we need to solve (in both, classical molecular dynamics and quantum mechanics) is that they aredifferential equations. In general, differential equations have no analytic solutions. There are some special cases which have (e.g. particle-in-a-box, harmonic-oscillator, Hydrogen atom), but  when it comes to multiple electrons and nuclei, we have a $N$-particle system and their pairwise interactions (be that the Coulomb force in chemistry/physics or gravity in astrophysics), it results in a type of differential equations which have no general analytical solution.
The only remaining option to solve this anyway is using numerical techniques. This essentially means splitting the whole problem in many tiny small steps. The more steps, the higher the accuracy and the higher the computational cost. A simple example would the integrate a 1D function, by splitting its area into many small rectangle or triangle (the above mentioned "steps"), for which we can use an analytic formula, and then add everything up. In HF and DFT the "steps" would be the basis set of orbitals, and in molecular dynamics small steps in time are considered, which yield small steps in the movement of the atoms.
Computers happen to be an excellent tool for numerical approaches, because for each step the same algorithm has to be repeated. So we can easily scale up the number of steps. However if the scaling for the method is higher then linear (which is usually the case), then the we are limited in the size of the problem. Faster or more computers can shift that limit a bit up, but fundamentally the limit remains.
This means we cannot just wait for better computers to tackle larger problems, we really need methods with a better scaling behavior.

We have linear scaling methods, at least in principle, see e.g. https://aip.scitation.org/doi/abs/10.1063/1.1330207

Yes, that is one example for what I mentioned in my last sentence.


text:
Determination of pKb of a mono acidic base

$20$ mL of a weak monoacidic base($\text{BOH}$) requires $12$ mL of $0.3$ M $\text{HCl}$ solution for the equivalence point. During titration, the pH of the base solution was $10$ upon the addition of $4$ mL of $0.3$ M $\text{HCl}$ solution. What is the $\text{pKb}$ of the base($\text{BOH}$)?

My attempt:
The concentration of the base is given by,
$c * 20 = 12 * 0.3$ -- (equating the moles of the acid and base)
$c = 0.18$ M
So, initial moles of the base in the container $= 20 * 0.18 = 3.6$ mmol
Moles of acid that is added $ = 4 * 0.3 = 1.2$ mmol
Since the base is monoacidic, they will react in a 1:1-mole ratio. The acid is the limiting reagent, so it will be fully consumed. Therefore the moles of base left is,
$3.6 - 1.2 = 2.4$ mmol
The concentration of the base is,
$\frac{2.4}{24} = 0.1$ M
Since the pH of the solution is $10$, therefore the concentration of $\text{OH}^- = 10^{-4}$
Applying the approximated formula for calculating the $\text{k}_b$ of a weak base,
$$\text{K}_b = \frac{x^2}{c}$$
Where,
x = concentration of $\text{OH}^-$ ions
c = concetration of the base
In our case,
x = $10^{-4}$
c = $0.1$
Plugging in the values I got $pK_b = 7$. But the answer is given $4.3$.
Any help would be appreciated.

I appreciate your effort you showed in your question, but you missed one important thing that when a weak base is reacted with a strong acid a buffer solution is formed and the $\mathrm{pH}$ is not only contributed by base but also by the acidic salt ($\ce{BCl}$ in this case).
So, simply applying the Henderson–Hasselbach equation (derivation can be found here)
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{B-}]}{\ce{[BOH]}}}$$
we get
$$4 = \mathrm{p}K_\mathrm{b} + \log{\frac{1.2}{2.4}},$$
which gives us
$$\mathrm{p}K_\mathrm{b} = 4.3010$$

I upvoted because you got the math right, but your assertion that this is a buffer solution is beside the point. The real point here is that you can't assume that $\ce{[OH^-] \approx [B^+]}$ at pH=10.00

@MaxW ok sir , thanks for upvoting , but will it not form a buffer ??


text:
Meaning of an ideal chemical system

What is meant by a an ideal chemical system?
Moreover, is it true that only in an ideal chemical system the equation
$$\mathrm{d}U = \mathrm{d}H - \mathrm{d}(pV),$$
i.e.
$$\Delta U = \Delta H - \Delta (pV)$$
can be written as
$$\Delta U = ΔH - pΔV?$$
If so, why?

I also presume that non-ideal chemical systems do not maintain a mechanical equilibrium. If I am wrong, please correct me.

U and H are functions of state, and their values are properties of the material, and not of the process that the material has experienced.  Their changes depend only on the initial and final thermodynamic equilibrium states of the material (i.e., the initial and final temperature, pressure, and chemical composition).

For myself, ideal means it is composed of ideal gas particles. This means that the ideal gas law applies. This means that there is no accounting for particles interacting which is a grave mistake for liquids and solids, and often for gases too.
$dU = dH - d(pV)$ is always true. In an real system however, the variables involved may take on different quantitative values for a given state point, thus the final change in eternal energy $\Delta U$ may be different.

In general $\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$
If pressure is constant during the change then $\Delta p \Delta V = V \Delta p = 0$ which then leads to the term
$$\Delta U = \Delta H - p \Delta V $$
If volume is constant during the change then $\Delta p \Delta V = p \Delta V = 0$ which would then lead to
$$\Delta U = \Delta H - V \Delta p $$
If neither volume or pressure is constant the simplest form is
$$\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$$
a.k.a.
$$\Delta U = \Delta H - (p_f V_f - p_i V_i) $$
see Andrew's answer for more details on the last step.
And I do correct you, non-ideal systems maintain equilibrium just as ideal systems do. If they didn't, we wouldn't be able to use thermodynamics on them since thermo only applies to equilibrium.


text:
Benzilic acid rearrangement

I'm in need for the exact mechanism of the above reaction. I attempted some tries(mentioned in the mechanism above) and I felt the yield quite unsatisfactory.
- the last step remains stuck in an equilibrium. Is the energy of the product sufficiently high to surpass the equilibrium created??
- the probability of the product formation (according to the mechanism I assumed) is too low as the pair of electron would hardly prefer an attack on the 5th carbon (as it doesn't possess any partial positive charge as the 6th carboxy carbon does)

Czakó/Kürti$^{[1]}$ suggest a [1,2] shift, which in your case would mean a shift of the 5-6-bond to the 1-carbon; in concert with a free electron pair of the anionic oxygen neutralizing the 6-carbon. The carbonyl oxygen becomes anionic and accepts a proton in the next step. So instead of the 1-6-bond breaking, the 5-6-bond breaks.
Responding to a comment regarding yield: Czakó/Kürti give several references and cite one example yielding 63% over the two steps (addition of base and acid).$^{[2]}$ Other examples are given, but without yield.
1) L Kürti, B Czakó: Strategic Applications of Named Reactions in Organic Synthesis, Elsevier (2005).
2) P A Grieco, J L Collins, J C Huffman, J. Org. Chem., 1998, 63 (25), pp 9576–9579, DOI: 10.1021/jo981419z (as given by Czakó/Kürti).

Thank you. But would the reaction (posted as the question) give a decent yield of the product?? Could it be used commercially well to yield the product? I suppose it to be a less productive method.


text:
Colour of colloids
Why colours of colloidal solutions are different when viewed along different directions?
For example Milk appears blue when viewed by reflected light and red when viewed by transmitted light
Is it  due to the irregular shape of the particles that they scatter different wavelengths along different directions?

Could you give an example? Like milk? Viewed with the naked eye? Perhaps it is just an apparent difference because of the distance the light must traverse.

something like a redshift?

I mean, what kind of experimental setup are you considering here? Or is this an observation you made at home? Why should a colloidal solution exhibit optical anisotropy?

This is mostly due to diffraction, which normally cancels out unless the particles are really  uniform. https://en.wikipedia.org/wiki/Mie_scattering

Also look for Tyndall effect https://www.quora.com/What-is-the-difference-between-Mie-scattering-and-Tyndall-scattering  and https://en.m.wikipedia.org/wiki/Tyndall_effect

related: https://chemistry.stackexchange.com/a/32133/23561

This is classic, the effect is called Mie scattering.
Very simply speaking, light is scattered on the whole surface of every (translucent) particle. For round particles, interference makes a diffraction pattern with conic symmetry, i.e. the intensity of diffracted light depends on the angle towards the incoming wave.
Obviously the pattern depends on the light wavelength, and so there is a dispersive effect. It's not very strong, because usually the particles are not all exactly of uniform size, and there is a lot of multiscattering.
Btw. it takes a keen observer to notice this without prior knowledge. Congratulations. ;-) It's easier to see if you dilute the milk (less multiple scattering). At least one good use for low-fat milk.
Warning: The explanation above is terribly simplified, to the point where it doesn't explain much. For example you'd expect the colours to change with every brand of non-/homogenised milk, because the size distribution varies. Not so much.
Also the effect is obscured by the fact that light scattering is generally much stronger for shorter wavelenghts. That also means that red light is more likely to pass through, and blue light to  get absorbed or escape on the front side. It's really hard to tell apart Mie scattering and the ordinary Tyndall effect.


text:
Mechanism for chloromethylation of Salicylaldehyde
Reaction mechanism for following reaction :
Salicylaldehyde + formaldehyde +con. HCl $\ce{->}$ 5-(Chloromethyl)salicylaldehyde

This seems like a homework question thus you must show some effort in the question since you are new please visit Homework Policy for more info .
So now on the answer,
this is a really simple and a nice case of Friedel-Crafts Alkylation . the mechanism go as follow :

The Oxygen atom on formaldehyde gets protonated, leaving a carbcation.
This carbcation gets added to the Salicylaldehyde by EAS (Electrophilic aromatic substitution).
The alchol so formed gets dehydrated to again get a carbcation which finally gets attacked by $\ce{Cl-}$ to form the final product .

The mechanism scheme is as follow :

There  is  a  formation  of  carbocation  but  not  carbanion

Ohh... sorry I wanted to write carbcation but wrote carbanion by mistake I will correct it


text:
Why is heat capacity defined as a partial derivative?

I understand that the heat capacity is the gradient of the slope of q against T, which implies that C = $\mathrm{ \frac{dq}{dT}}$, so how do we arrive at C = $\mathrm{\frac{∂q}{∂T}}$?
At constant volume, $\mathrm{C_V}$:

$$\mathrm{dU = dq + dw}$$
$\mathrm{dw = p(dV)}$, but $\mathrm{dv = 0}$, so $\mathrm{dU = dq}$
How do we make the jump to $\mathrm{C_V = \frac{∂U}{∂T}}$?

3. At constant pressure, $\mathrm{C_p}$:
$$\mathrm{H = U + pV}$$
so $$\mathrm{dH = dU + pdV = (dq + dw) - dw = dq}$$
hence $\mathrm{dH = dq}$
How do we make the jump to $\mathrm{C_p = \frac{∂H}{∂T}}$?

It is a matter of symbolism. Partial d's are used when a quantity depends on multiple variables and you consider all the variables as constant except one.

I think you are starting on the wrong foot.
$U$ and $H$ are state functions. They both depend on several variables. $U$ is a function of entropy $S$ and volume $V$ and composition. For a pure fluid there is no dependence on composition, so we can just say $U = f(S,V)$. Similarly $H$ is a function of entropy and pressure $p$ i.e., $H=f(S,p)$. You go from U to H by using a Legendre Transform (but ignore Legendre Transforms for now... I will give you a good reference to read for thermodynamics at the end, it will make your life better).
Important
Reality has a way of changing things. For instance experiments are done almost always at a constant $T$, and either constant $p$ or $V$. Therefore, we like thinking of things in terms of these variables. Lets face it... entropy is tricky.
Assumption
Okay, lets now say that $U=f(T,V)$ and $H=f(T,p)$ this is alot easier to handle. In thermodynamics we are always after changes in things. Therefore, the differential forms of $U$ and $H$ are desirable.
Also, this is not a haphazard switch, entropy and temperature are intricately related to each other $T \equiv \left( \frac{\partial U}{\partial S} \right)_{x's}$ where $x's$ are all other independent variables being held constant.
The differentials for both $U$ and $H$ are as follows (with respect to our new independent variables).
$$U(T,V)$$
$$dU = \left( \frac{\partial U}{\partial T} \right)_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$H(T,p)$$
$$dH = \left( \frac{\partial H}{\partial T} \right)_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$
Endgame
It is here that we now define what $C_V$ and $C_p$ are
$$C_V \equiv \left( \frac{\partial U}{\partial T} \right)_V$$
$$C_p \equiv \left( \frac{\partial H}{\partial T} \right)_p$$
These are fundamental definitions and can't be tampered with. In introductory thermodynamics teachers play games and try to frankly dumb things down, and this leads to, in my opinion, improper definitions and understanding. This link provides a definition (without much context) that ties $Q$ into the mix.
If we insert these definitions into our differential equations we have
$$dU = C_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$dH = C_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$
I leave it as an exercise, if you wish, to find the definitions for $\left( \frac{\partial U}{\partial V} \right)_T$ and $\left( \frac{\partial H}{\partial P} \right)_T$
Final Resource
Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd Edition
The man called Callen
The Textbook

In thermodynamics, the definition of heat capacity is somewhat modified and generalized from the version that we learned in freshman physics (i.e., in terms of heat q), which applied only to systems where no work was occurring. This modification is necessary because heat capacity is supposed to be a physical property of the material, independent of process path.  However, we know from thermodynamics that q actually is a function of process path.  So it is no longer appropriate to use to define heat capacity.  Now, in thermodynamics, the heat capacity Cv matches the old definition in terms of q only when volume is constant, and the heat capacity Cp matches the old definition in terms of q only when pressure is constant.  Otherwise, no.
The new definitions are much more general than the old definitions and, more importantly, because U and H are functions of state (and not path), so are Cv and Cp.


text:
The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?

The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?
  1. 2
  2. 1.58
  3. 1.6
  4. 1.31

I think the answer should be 1.6 km/h but the answer given in book is 1.31 km/h.
Most probable speed should be the mode of the given speeds, which is equal to 1.6 km/h. Am I correct?  If I am wrong, then what's the definition of most probable speed and how do I solve this problem?

I would give the textbook -1 on this question.  Most probable is ambiguous here, leading me to an answer I had to delete.

(-1) The question does not make sense. I don't see any compelling reason why one should assume that the question is asking for the mode of the population (1.40 or 1.31, which depends on how you calculate(!)) rather than the mode of the sample (1.60). Furthermore, it is illogical to claim that the answer is 1.31 simply because there is no answer corresponding to 1.40. We are supposed to answer based on chemical principles, not by reading people's minds and reverse-engineering the questions they set.

@orthocresol - I think we're having a violent agreement. I totally agree that this is a very lousy problem. It was more of a matter of trying to figure out the logic of the 1.31 answer rather than agreeing that 1.31 was "right."

I don't understand why this was closed as a homework question. The OP did state his analysis. Look at all the debate that it took us to come to a consensus on the book's "answer."

Mode. - The number which appears most often in a set of numbers. The OP was right the answer should be 1.6. The book answer is just wrong.

Following up on user andselisk's answer
...
The problem is weird for several different reasons.

First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules.
Second the answer depends on how you calculate to get a result.

For the Maxwell-Boltzmann distribution there are relationships between the most probable speed $v_p$, the average speed $\langle v \rangle$, and the root-mean-square speed $v_{rms}$. Following the Wikipedia article:
\begin{align}
\langle v \rangle &= \frac{2}{\sqrt{\pi}}v_p\\
v_\mathrm{rms} &= \sqrt{\frac{3}{2}}v_p
\end{align}
Rearranging these two equations we get two different ways to calculate the most probable speed.
\begin{align}
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms}
\end{align}
Thus using the mean (average) speed we get:
\begin{align}
\langle v \rangle &= \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58\\
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40
\end{align}
However using the rms speed we get:
\begin{align}
v_\mathrm{rms} &= \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms} = 1.31
\end{align}
The difference is due to small sample statistics and not really a statistically significant result. However the difference is absolute proof that the sample doesn't follow the Maxwell-Boltzmann distribution.

I don't necessarily agree with the implications of the final statement. Like anything else in statistical mechanics or thermodynamics, the results of the MB distribution are not guaranteed to hold true for small systems. In the (admittedly meaningless) limit of having one particle, it is literally impossible to have a different rms, mean, and most probable speed, no matter how accurately you measure the speed of that one particle. So...

... the problem is not that the sample is obeying some different non-MB distribution, nor is it about measurement accuracy or confidence intervals. The problem is that the sample is too small, so the predictions of the MB distribution are not manifest.

@orthocresol - The point is that MB is a distribution which assumes a "large" number of particles. It is absolutely meaningless to discuss MB with only 5 particles. // If the 5 particles were from large number of particles which did obey MB, then the sample size just isn't large enough to get a decent measurement.

Yup, I agree with that assessment. Consider it just me being rather nitpicky about the wording, then.

@orthocresol The question is probably from a book for exam preparation, and as such wants the answers to be ~calculated~ according to a pre-mentioned formula. Yes, the problem doesn't make sense if we do about the definitions "strictly", but i don't think thats the answer OP wants. A lot of times on Chem SE such exam questions are shot down because they are based on outdated concepts, but simply stating "Well we don't believe that any longer" doesn't help anyone- not the other students having the same problem and wanting an explanation fitting the demands of the pattern, nor the asker, anyone

@Micelle, I cannot agree: we are not here to give the OP what they want. We are here to write good, sound chemistry in line with current scientific understanding.

The book is absolutely correct.
Average speed is not the same as rms speed which is defined as
$$c_\text{rms} = (3kT/m)^{1/2} = \left(\frac{\sum_{i = 1}^N v_i^2}{N}\right)^{1/2}, \tag{1}$$
where $v_i$ are the velocities of each particle with the total of $N$, and the most probable speed is even smaller:
$$c_\text{mp} = (2kT/m)^{1/2} \tag{2}$$
For the derivations of the formulas above see e.g. [1, p. 69].
Combining both equations
$$
\begin{align}
c_\text{mp} &= \left(\frac{2\sum_{i = 1}^N v_i^2}{3N}\right)^{1/2} \\
  &= \left(\frac{2 × \left[(\pu{2 km s-1})^2 + (\pu{1.5 km s-1})^2 + 2 × (\pu{1.6 km s-1})^2 + (\pu{1.2 km s-1})^2\right]}{3 × 5}\right)^{1/2} \\
  &= \pu{1.31 km s-1}
\end{align}
$$
References

Castellan, G. W. Physical Chemistry, 3rd ed.; Addison-Wesley, 1983. ISBN 978-0-201-10386-1.

Agree with the answer totally But when I started with the average velocty which comes out to be $1.58 Km/s$ now I must be able to multiply it by $\sqrt{\pi}/2$ (as $v_{av}=\frac{2}{\sqrt{\pi}} \sqrt{\frac{2RT}{M}}$) and should get the answer as 1.30 but the answer comes out to be 1.400 , I am not getting it please help

@AdvilSell I'm afraid I don't understand what you are using average velocity for (and how you determined it) and where $\sqrt{π}/2$ term came from.

Desmos can you please check this out I have shown calculations and formula ... please check it once

@andselisk - $\sqrt{\pi}/2$ factor came from relationship between average speed and most probable speed. See https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution#Typical_speeds

The gist is that the problem ignores the confidence interval of the measurement. So if you compute average speed and then use that to get most probable speed, then you get 1.40. However if you calculate rms and use that to calculate most probable speed then you get 1.31. There are so few measurements that the two values are not statistically significantly different.

@MaxW Aha, I see it. I thought of the average speed as of rms, and Advil calculated arithmetic mean value. You are absolutely right and you explained this quite well so that even I finally understood. Probably you both guys should put up your comments as answers as well to highlight the whole picture.

@AdvilSell Yep, your approach is also correct, feel free to post an alternative answer. Sorry it took me some time to realize how it's been deduced.


text:
How do we get the wave function of Ligand Group Orbitals with the same symmetry as that of the corresponding metal orbitals
In the book Principles of Structure and Reactivity by Huheey, Chapter 14, the author has taken linear combination of ligand orbitals to form ligand group orbitals and then further combination of LGOs(Ligand Group Orbital) with corresponding orbital of metal which has the same symmetry as that of the LGO.
I do not understand how they calculated the wave function for the Ligand Group Orbitals.

For this you need Group Theory to understand what exactly is going on. The symmetries of your molecule, e.g. mirror plane or rotational axis etc., give certain restrictions on how the molecular orbitals of the molecule can look, i.e. they will have to reflect the symmetries of the molecule in certain ways (the details of which you can be determined using methods from Group Theory).

There is something called the projection operator / formula in group theory which helps do this in a rigorous manner. Sometimes it is possible to figure them out "by inspection", but it can be tricky (or even impossible) with highly symmetric groups or degenerate irreps.

I have knowledge of group theory and SALCs, but nevertheless I fail to understand how to apply that here. To what function are we applying the projection operator? Like normally, we would apply it to a particular atom, bond, or some unit vector.


text:
Which one is the stronger Lewis base: diethyl ether or tetrahydrofuran?
I am trying to understand Lewis bases (bear with me, I am a physicist). I was asking myself how diethyl ether and tetrahydrofuran compare with respect to their properties as solvents in electrophilic additions. Or if there is any difference at all.
So, for example, is it better to use a (boron trifluoride + diethyl ether) complex or a (boron trifluoride + tetrahydrofuran) complex? As to my understanding this is mainly a question about how strong the solvent acts as a Lewis base. If the solvent (ether or THF) is a stronger base, it might inhibit the catalysis of the actual electrophilic addtion reaction we are targeting, right? So under this assumption it is probably better to use the weaker base.
My guess is that diethyl ether is the stronger Lewis base. It has two hydrogen atoms more, so it is also contributing two more electrons, and moreover due to the slightly lower electronegativity of hydrogen, these electrons tend to be dragged towards the oxygen. On the other hand, THF misses the two protons and electrons, and moreover it must be electroneutral in the symmetry plane where the opposite C-C bond is, so it contributes a little less electron stuff to the nucleophilic oxygen.
Is this argumentation correct?
And what about this: the THF molecule is a little more compact than the diethyl ether and moreover has a tensed 5-ring. So shouldn't this push the electrons a little more to the periphery of the molecule, especially to the oxygen, and thereby increase its strength as a Lewis base again? Is this effect relevant at all?
PS: After writing this question I found a table at Wikipedia (https://en.wikipedia.org/wiki/Lewis_acids_and_bases) which states that the enthalpy of formation of the BF3 adducts is

Et2O: 78.8 kJ/mol
THF: 90.4 kJ/mol

This is about the opposite of what I would have expected and seems to indicate that the effect of the two additional protons/electrons of Et2O is not as big as it might seem at first.
But is it really support for the "compactness of THF in the 5-ring" theory? Or are there other effects responsible for THF being the stronger Lewis base?

You call it compactness.  Think of the lone pairs.  In which case they are more exposed and thus readily form a complex? In THF.

THF is soluble in water in all proportions; diethyl ether is not.

Which is "better"? In some specific reaction, with all other circumstances identified. Ether is a little too volatile for my preferences, but THF is not bad. I like the smell of ether, but the odor of THF is OK. Little problem with oxygen (fire and peroxides) with both of them, so you must be careful with both.
The difference in enthalpy of formation (11.6 kJ/mol) is small compared with activation energies of typical organic reactions (~100 kJ/mol), so "better" might be related to human preferences more than than chemical performance.
A chemist might already have a feel for the molecular configurations of these molecules, but a physicist could benefit from making the models or looking at them:

The ability of THF to poke the BF3 molecule better than ether can is apparent. Whether this is the controlling factor in which is "better" remains to be seen. I'll go out on a limb and say that either one will work, although the results (convenience, yield, expense, etc.) could easily be +/- 20-50%, depending on the situation. In a lab, one might work better; in industry, the other might be preferable.
Either solvent could probably be optimized for best performance under slightly different conditions, and these optimum conditions will not be the condition you choose for the first experiment. Probably a dozen experiments will suggest which solvent could be the better choice. The literature should suggest the better (or more typical) choice for your needs.
Ref. https://en.wikipedia.org/wiki/Activation_energy


text:
What are the differences in shapes between molecular geometry and electronic geometry?
In my chemistry course we are told that molecular geometry considers only atoms as part of the shape, while electronic geometry considers atoms and electrons as part of the shape. However, I can't find any difference between the two shapes for methane online. All I can find is this:

It confuses me that there can even be another shape for this because isn't the molecular geometry already take into consideration the fact that each molecular wants to be as far away from one another as possible? What is the difference in shape between electronic and molecular geometry?

For methane, the molecular and electronic geometries are the same. For something like NH3, which has a lone pair of electrons, they are different. @matryoshka

The term electron geometry is something which you will find in general chemistry books only (good for VSEPR only). Don't take it too literally. The is that if electron pairs are being considered, you call it electron geometry.

Electronic geometry and shapes might be used loosely as many terms. The fact that you post the Q let me think that the meaning is somehow clear to you,  and you just make sure to not misunderstand something. At a more general lexical level, electronic geometry would better refer to the electron density size and shape around atoms/molecules, and this never show up in straight edges and vertices. Something like in the pic here https://www.researchgate.net/figure/a-The-total-electron-density-surface-3D-mapped-with-electrostatic-potential-of-this_fig6_301558792  anyway ans below are correct

Molecular geometry which is the arrangement of other atoms around that central atom.
the difference between Electron geometry and Molecular geometry is that electron geometry includes  the **lone pairs or electron around the center ** they still influence the overall geometry but when we name the shape of a molecule.
as in case of like central atom having no lone lone pair of electrons..in that case the Molecular and electron geometry of that atom would b same ; else due to presence of lone pair there would b generation of a bent
thats why Methane has same Molecular and Electron geometry
 hope it proved helpful fr u

The geometry of a molecule can be given as either the electron geometry or the molecular geometry. The VSEPR theory (Valence Shell Electron Pair Repulsion theory) can be used to determine the geometries of molecules. Electron geometry includes the lone electron pairs present in a molecule.
Example:
In NH3 ; same Molecular and Electron structure ; but in case of NH2. molecular and electron geometry different. Because CH4 stable structure..so it have same molecular and electron structure.
Hope it would help you


text:
How to make oil more heavy?
I would like to increase the density of a given oil (or anything that repels water) so that it is slightly heavier than water; to be able to precisely control how slow the oil would fall in the water.
What can I add to the oil (or the water) to accurately set its density?
I am looking for a cheap solution, that I can easily do at home.
Edit
Another way to create slow and tiny drops would be to drop water into oil. Yet, the oil surface tension prevents water from falling down ; I asked another question for this problem.

Something similar to lava lamps?

How are you planning on using these dense hydrophobic liquids?

@andselisk yes, my goal is to make something like lava lamps ;-)

how about bromonated vegtable oil?

Bromonated vegtable oil seems nice; how can I make it? Where can I buy this? Should I mix brome and vegetable oil until I get the right density?

Halogen-substituted aliphatic hydrocarbons are usually miscible with oils. Pick the compound by its density, price, safety, boiling point and other characteristics.
For example, trichloroethylene (trico) has a density of ~1.6 g/cm3, and a Material Safety Data Sheet shows a vapor pressure of ~60mm Hg at room temperature and a time weighted average (TWA) toxicity limit of 50 ppm (269 mg/m3) from ACGIH and OSHA. Though it's readily available, trico might be too toxic and volatile for your use. A similar chemical is tetrachloroethylene, sold as "perc", with the same toxicity issue.
In general, going down the halogen group increases density, so bromine or iodine substituted compounds would be denser than those with fluorine or chlorine.

I was going to suggest using mothballs (apparently 1,4-dichlorobenzene nowadays). Halocarbons are simple solutions, but toxicity is indeed an issue.

Thanks for your answer! Sorry I am not active on this thread right now (but I am still interested! ). I am not a chemist at all, so I don't know where to buy "Halogen-substituted aliphatic hydrocarbons", nor which one to buy. I am looking for something safe, relatively cheap, and easy to manipulate. Once I get this, I just have to mix it with oil until I obtain the proper density?

@arthur.sw, see the example above: "trico" is commonly sold for removing grease spots, as is "perchloroethylene" AKA "perc". As for your use, assuming you want to make "lava lamps", yes, it would take a bit of experimenting, mixing light oil and denser additive to get something that just floats or sinks in water with a small temperature change. You would also need to experiment with dyes to see which ones will not mix into both liquids.

Unfortunately, trichloroethylene and perchloroethylene are both forbidden to sale in Europe : / They sale substitutes though but I guess it does not have the same density...
I looked for a list of "Halogen-substituted aliphatic hydrocarbons" but could not find any. Where can I find it?

Of course trico is available in Europe! See http://www.bannerchemicals.com/trichloroethylene-triklone/ for example. That said, you might need a license to obtain it, because the ignorant, malicious or careless could harm themselves and the environment.

It might be available for professionals, but I could not find any place / site to buy a small amount of triclo for a reasonable price. Another way to create slow and tiny drops would be to drop water into oil ; but the oil surface tension prevents water from falling down ; I asked another question for this problem.