text:
Slater Matrix of Methane Molecule
I am going through Hartree-Fock calculations in detail, but the textbooks I am referring all considered homonuclear molecules as examples. So I am trying to come up with my own the Slater determinant for Methane. Kindly let me know if my application is correct!
For Methane, we have in total 10 electrons (Carbon 6e, and four Hydrogen with 1e each). We also know that valence shell electrons of C and H engage in bonding. The $1s$ shell of Carbon does not engage in bonding. I am also going to denote $+$ for spin-up and nothing for spin-down. Please note that $1s$ for first two columns in the following matrix describes the $1s$ orbital of carbon which does not involve in bonding.
$$\begin{pmatrix}1s(1)^+ & 1s(1) & S_1(1) & S_1^+(1) & S_2^+(1) & S_2(1) & S_3^+(1) & S_3(1) & S_4(1) & S_4^+(1) &\\\ 1s(2)^+ & 1s(2) & S_1(2) & S_1^+(2) & S_2^+(2) & S_2(2) & S_3^+(2) & S_3(2) & S_4(2) & S_4^+(2) \end{pmatrix}$$
For brevity, I am not writing out the rest of the eight rows in matrix. Now  $S_n$ (spatial part of the wavefunction $S_n$), above in the matrix, is defined as the following:
$S_n=c_{n,1}1s + c_{n,2}2s + c_{n,3}2p_x + c_{n,4}2p_y + c_{n,5}2p_z$, where $n$ runs from $1$ to $5$. Why? because we know that these electrons engage in bonding, hence we have to take a superposition of orbitals (mixing). So in total we have 5 equations e.g. $S_1, S_2,...,S_5$.
Now each such superimposed MO (the $S_n$) can hold 2e each for a total of 10 electrons. Now the carbon $1s$ took away 2 electrons, so 8 electrons remained. And I have hit redundancy, because theoretically each $S_n$ can take 2e, so I have 10 electrons which can be accommodated, but I have only 8. Moreover in the Slater matrix above, I can't find a way to fit in $S_5$.
I am missing something. I would be glad if someone can throw some light on this.

+1 If you don't get an answer here within a few days, you might consider migrating the question to matter modelling SE. (This is not to say in any way that the question is off-topic here)

Whenever some says they are just starting out with electronic structure theory, I always recommend that they read Szabo and Ostlund's Modern Quantum Chemistry. The book is great introduction to HF and post-HF methods (some of the discussion of coupled cluster is a bit outdated) and its very affordable. It is nice to have around as a reference.
Getting to your actual question, I think you are confusing yourself by trying to hybridize the orbitals. During the HF procedure, you will solve for a set of MOs that is are linear combinations of the AOs, so there isn't a need to form these $S_n$ functions beforehand.
There is also the issue of actually forming the Slater determinant; while the solution to HF will be a Slater determinant, you generally don't explicitly write it out this way. Generally, you assume that each MO in the Slater determinant is a linear combination of the AOs and you solve for the coefficients of each AO, such that the energy is minimized. So the Slater determinant for methane will have 10 (initially unknown) MOs in it to account for the 10 electrons and you will then need to build these MOs using the Roothaan-Hall form of the Hartree-Fock equations.


text:
Can anyone explain, why there are coefficients 2 in the equation for rate of change [A]?

The reaction mechanism for the decomposition of $\ce{A2}$ is thought to be:
\begin{align}
\ce{A2 &<=>[$k_1$][$k_1'$] A + A} & &\text{ (fast)} \\[0.2cm]
\ce{A + B &<=>[$k_2$] P} & &\text{ (slow)}
\end{align}
The rate of formation of intermediate $\ce{A}$ is given by:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{A2}] - \color{red}{2}k_1'[\ce{A}]^2 - k_2[\ce{A}][\ce{B}] \approx 0
\end{align}

In the equation above, why is the coefficient of $2$ present for both the forward and reverse reactions (highlighted in red)?

Welcome to Chemistry! A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

([])/=_ [ ]−_(−) []^−_ [][] for reaction mechanism 1. →+         _ and 2.  +→        _(−)
and 3. +→         _

It would be really great if you could [edit] your question. On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here.

I did it for you this time, but please do take a look at the guide which Martin linked.

Thank you, this is my first question, so I leran.

There is an answer to a v. similar question here https://chemistry.stackexchange.com/questions/151970/steady-state-approach-how-to-find-rate-equation-for-intermediate/151989#151989

$$\ce{A_n ->[$k$] nA}$$
For the reaction given above, rate of formation of $\ce{A}$ is defined as:
$$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$
In this case, we have three different reactions that involve A. They are:
\begin{align}
\ce{A_n &->[$k_1$] nA} \tag{1}\label{1}\\
\ce{nA &->[$k_1'$] A_n} \tag{2}\label{2}\\
\ce{A + B &->[$k_2$] P} \tag{3}\label{3}\\
\end{align}
Writing the rate equations for $\eqref{1}$, $\eqref{2}$, $\eqref{3}$ and summing them up to get total rate, we get:
\begin{align}
\frac{1}{n}\frac{\mathrm d\ce{[A]1}}{\mathrm dt} &= k_1\ce{[A_n]} \tag{4}\label{4}\\
-\frac{1}{n}\frac{\mathrm d\ce{[A]2}}{\mathrm dt} &= k'_1 [\ce{A}]^n\tag{5}\label{5}\\
-\frac{\mathrm d\ce{[A]3}}{\mathrm dt} &= k_2\ce{[A][B]} \tag{6}\label{6}\\
\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} &= \frac{\mathrm d\ce{[A]1}}{\mathrm dt} + \frac{\mathrm d\ce{[A]2}}{\mathrm dt}+\frac{\mathrm d\ce{[A]3}}{\mathrm dt} \tag{7}\label{7} \\
\end{align}
Substituting values from $\eqref{4}$, $\eqref{5}$, $\eqref{6}$ into $\eqref{7}$, we get:
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{n}k_1\ce{[A_n]} - \color{red}{n}k'_1\ce{[A]^n} - k_2\ce{[A][B]} $$
Here, $n = 2$, therefore
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{2}k_1\ce{[A_n]} - \color{red}{2}k'_1\ce{[A]^2} - k_2\ce{[A][B]} $$

Thank you very much! But I still a little be confused, shouldn't it be [A] to the power of n in the equation (5)? That gives quadratic equation for the intermediate for the given mechanism..

I tried this way to solve next problem. The suggested reaction mechanism is the same, but first reaction is switched. In the textbook,(which is Atkins PhysChem), coefficient 2 appears only for forward reaction. Is it mistake or the I need to use different approach?

@Tara Please use the answer field only for actual answers. If you are new to the site, I suggest you take the [tour] and learn more about it. I have converted your answer to a comment here; if you have more follow up questions, it might be better to ask again and link here for context, see also [ask].

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text:
Calculating the ionic strength for an aqueous solution containing potassium ferricyanide and sodium sulfate
I have the $100$ microM of K$_3$Fe(CN)$_6$ is dissolved in water and the following data:
$$\begin{array}{c|c}
\text{Experiment} & \ce{[Na_2SO_4]}/\pu{M} \\ \hline
\mathrm{A} & 0 \\
\mathrm{B} & 1\cdot10^{-3} \\
\mathrm{C} & 3\cdot10^{-3} \\
\mathrm{D} & 8\cdot10^{-3} \\
\mathrm{E} & 1\cdot10^{-2} \\
\hline
\end{array}$$
I want to calculate the ionic strength for each experiment and I know that I should use the equation:
$$I = \frac {1}{2} Σ c_i z_i^2$$
and that it for this case will look like:
$$I = \frac {1}{2} ((1)^2*[\ce{K+}]+(-3)^2\cdot[\ce{Fe(CN)6^3-}] + (1)^2*[\ce{Na+}]+(-2)^2\cdot[\ce{SO4^2-}] )$$
However, when I calculate this for the different experiments, I don't get the number which is in our answer key. There is no calculations in the answers and my answers differ quite much. This is the way I calculate the different ionic strengths:
$$I_{\mathrm{A}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot0+(-2)^2\cdot0 ) = 5\cdot 10^{-4}$$
$$I_{\mathrm{B}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(1\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) ) = 3\cdot10^{-3}$$
$$I_{\mathrm{C}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(3\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) ) = 8\cdot10^{-3}$$
and so on.
In the answer key $I_{\mathrm{A}}=0.6\cdot10^{-3}$, $I_{\mathrm{B}}=3.6\cdot10^{-3}$ and $I_{\mathrm{C}}=9.6\cdot10^{-3}$. As you can see the difference increases and I don't understand what I am doing wrong. I would truly appreciate if someone could point out my mistakes. Thank you!
EDIT
I am also given this but I don't think they matter for the ionic strength:

When dissolving $\ce{K4Fe(CN)_6}$, it produces $4$ ions $\ce{K+}$ and $1$ ion $\ce{[Fe{CN}–6]^{4-}}$ . Why do you admit that it produces a negative ion containing some potassium like $\ce{[KFe(CN)_6]^{3-}}$ ?

@confused - I have edited your title, removing MathJax (the use of which makes it harder to locate posts via search engines). Also note Maurice's comment. You have specified potassium ferricyanide as one of your salts, but you have the ferrocyanide anion later on. I might be off, but I encourage you to double-check.

That is the way the problem was asked. It states that  Fe(CN)$_6$$^{3-}$ reacts with an organic ion to form  Fe(CN)$_6$$^{4-}$.

In aqueous solutions, $\ce{K3Fe(CN)6}$ and $\ce{Na2SO4}$ will completely dissociate:
$$\ce{K3Fe(CN)6 (aq) -> 3 K+ (aq) + Fe(CN)6^3- (aq)} \tag1$$
$$\ce{Na2SO4 (aq) -> 2 Na+ (aq) + SO4^2- (aq)} \tag2$$
Thus, from the equation $(1)$, $c_\ce{K+} = 3c_\ce{Fe(CN)6^3-} = 3c_\ce{K3Fe(CN)6}$ and from the equation $(2)$, $c_\ce{Na+} = 2c_\ce{SO4^2-} = 2c_\ce{Na2SO4}$
Hence for condition A $([\ce{Na2SO4}] = 0)$:
$$I_{\mathrm{A}} = \frac {1}{2} \left((1)^2 \cdot (3\cdot10^{-4}) + (-3)^2\cdot(1\cdot10^{-4}) + (1)^2 \cdot 0 + (-2)^2\cdot 0 \right) = 6\times 10^{-4} = 0.6\times 10^{-3}$$
Similarly, for condition B $([\ce{Na2SO4}] = 1\cdot10^{-3})$:
$$I_{\mathrm{B}} = \frac {1}{2} \left((1)^2\cdot(3\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(2\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) \right) = 3.6 \times 10^{-3}$$
And, for condition C $([\ce{Na2SO4}] = 3\cdot10^{-3})$:
$$I_{\mathrm{C}} = \frac {1}{2} \left((3)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(6\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) \right) = 9.6 \times  10^{-3}$$
So on for other conditions. Note that for all calculations, ionic strength of $\ce{K3Fe(CN)6}$ is always $0.6\times 10^{-3}$ since $[\ce{K3Fe(CN)6}] = \pu{1.0 \times  10^{-4} M}$ for all conditions.

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text:
Limitations of the simple harmonic oscillator?
I understand that the simple harmonic oscillator can be used to model the behaviour of molecules at low vibrational states. But why does this not work for molecules that are highly vibrationally excited?

Good question, I await to see what deep analysis the down voters are preparing as a criticism. But essentially it's the theory of small oscillations (e.g.http://www.nhn.ou.edu/~gut/notes/cm/lect_10.pdf) - for any potential energy curve near enough to a minimum it looks like a parabola, and motion in a parabolic PE curve is SHM. However in real systems the PE curve is not truly a parabola, and the larger the amplitude that the oscillations are, the more the motion will deviate from SHM - and this is what you are referring to above. A proper answer needs maths, hence a comment at the moment.

I second @IanBush's opinion. In the absence of a full answer, I'll add that in the meantime have a Google for a Morse oscillator, which is more illustrative of the excited state PE curve.

Real molecules dissociate given enough vibrational and rotational energy, a 'harmonic oscillator' molecule can never dissociate as the potential goes to infinity. A Morse oscillator give a more realistic model of a molecule and approximates the HO at low energy. Also experiment shows that energy gaps in real molecules decrease as energy increases, this can never happen in the HO.

BTW, a simple analogy of deviation from a simple harmonic oscillator is a pendulum. For small amplitude swings, there is a definite period, but as amplitude increases, the period changes -- and once "over the top", there is no definite period.


text:
Which of the given compound has the highest rate of dehydration?
I came across this question recently:

For which compound acid catalyzed dehydration rate is highest?

Now, they all are secondary alcohols so their dehydration should follow E1 mechanism and the stability of the carbocation formed in the RDS should be considered while comparing their rates.
Following this rule, I think, the answer should be (B) as ketone group shows (-)R effect in (C) and (-)I effect in (A) and (B), all which destabilize the carbocation formed.
But the actual answer is (A). How can that be? Is there some other mechanism to follow? I can’t understand what am I doing wrong.

Basically, B and C have no difference when come to dehydration. So, according to your argument, B and D have similar rates.

Driving force here to getting faster rate of A is gaining conjugation in the product.

Notice, that the first compound (A)can show tautomerism, after drawing the enol form we find that the carbocation lies at the allylic position, which means that it is stabilised through resonance forms, hence the compound A has highest rate of reaction.
Hope it helps.

Oh. I see it now. So due to allyl alcohol formation, the enol form is more stable than keto form, which is quite rare. And thus the enol form undergoes dehydration and not keto form. Is that the correct deduction from your answer?

@ApoorvSharma, in this particular  compound the most stable is the keto form only, however the stable form is always the least reactive one under normal conditions, hence in this case enol form being less stable reacts faster as compared to that  with keto form.


text:
why did a mixture of water, sugar, starch effervesce after adding a little bit of glycerol as humectant? cooking at 130 degrees celsius
I am trying a new formulation of lokum (turkish delight) which is a gel composed of starch, water and sugar. After adding 0.5% of glycerol to a 170 Kg of a boiling mixture, it became fugitive and effervesced. I need to know at which stage I should add the glycerol to the mixture. before cooking? or after I turn off the heat?

How can a mixture of water plus something bear a temperature of $130°C$ without being evaporated ?

You may want to ask this at https://cooking.stackexchange.com/

Whenever you add anything to a boiling mixture, you risk boiling over. Two possibilities present themselves: 1) the glycerol mixed with the formulation sufficiently to increase its viscosity and the boiling continued without the bubbles breaking.
But perhaps 2) the glycerol did not mix well, but spread over the surface of the hot formulation and cooled it, making this surface layer more viscous from the cooling rather than glycerol thickening. The lower layers continue to produce bubbles, but they don't break the surface.
Perhaps removing the formulation from the heat, allowing the bubbling to cease, and then adding the glycerol would work better.
Boiling after adding the glycerol could be a problem too. Don't they use glycerol in bubble-blowing formulations for children?
Anyway, use a big pot.


text:
Effect of +I on the alpha hydrogen in aldol reaction
We have $\ce{(CH3)2CHCHO}$. Is the alpha-hydrogen acidic enough to react with a base? There is +I effect of two methyl groups, however there the carbanion is getting stabilised through resonance. And, as we know that inductive is a weak effect, so will it proceed through aldol reaction or will Cannizzaro reaction occur? Moreover, if we are considering  the inductive effect to be significant,  won't it reduce the electrophilicity of carbon of the carbonyl group, hence reducing the chances of Cannizzaro reaction? What could be the products if the reaction is done separately by using conc. base in one time and dil. base for other?

Looks like a duplicate: https://chemistry.stackexchange.com/questions/65089/why-can-enolizable-aldehydes-undergo-cannizzaro-reactions/92477#92477

Does this answer your question? Why can enolizable aldehydes undergo Cannizzaro reactions?

Is the alpha hydrogen acidic enough to react with base?

The answer is yes. As you said +I effect of the methyl group stabilizes the carbocation. It is weakly acidic but not weak enough to not be taken by any base.

Will it proceed through aldol reaction or will Cannizzaro reaction occur?

It will proceed through Cannizzaro mechanism even though it has an alpha hydrogen. That is because if you take it through aldol, the attacking nucleophile in the second step is $\ce{-C(CH3)2(CHO)}$ which is a very bulky nucleophile and so it isn't able to attack on $\ce{CH(CH3)2(CHO)}$. Think, if this reaction undergoes than the product will have $\ce{CH3}$, $\ce{CH3}$, $\ce{CHO}$ and $\ce{-C(OH)(CH(CH3)2)}$ groups all on one carbon. The steric crowding will be so immense that the reaction becomes infeasible. That is why it undergoes Cannizzaro reaction which has $\ce{H-}$ as the attacking nucleophile (in the second step).
I think your other questions won’t arise if you think this way.

Thanks for the answer, so what's the major thing which is preventing the aldol reaction, steric or the reason of alpha hydrogen not being acidic enough, for example if we add formaldehyde , considering only the cross aldol, will it happen, since the steric  hinderance  is not very much?

@green_32: You might want to look at this. https://chemistry.stackexchange.com/questions/65089/why-can-enolizable-aldehydes-undergo-cannizzaro-reactions/92477#92477


text:
Steady state approximation for three-step cycle

How to write the steady state for $\ce{A}?$
$$\ce{A ->[$k_1$] B ->[$k_2$] C ->[$k_3$] A}$$

Species $\ce{A}$ seems to appear on both side of equation as it uses itself to produce some of itself. I have checked some online tutorials but I found none cover this scenario.

One approximation could be $\ce{B <=> C}$ and at steady state, $\ce{[B]}$ and $\ce{[C]}$ do not change.

A, B and C remain constant in this scheme.

They come to equilibrium with amounts proportional to the ratio of rate constants.  If all k's are equal then the population is 1/3 each.

Related.


text:
What are the ozonolysis products of phenanthrene?
What is / are the product(s) of ozonolysis of phenanthrene?

I am not sure I understand how to proceed. Which of the $\ce{C=C}$ double bonds is most susceptible to this type of reaction?
I am confused given the resonance and aromatic character of the compound which doesn't make me feel comfortable doing ozonolysis the usual way.

In anthracene (an isomer of your compound), the central ring is less aromatic (and susceptible e.g., to the Diels-Alder reaction) than the two outer ones.  Similar here, in phenanthrene, the double bond at position 9,10 is easier attacked by ozonolysis, than the other positions (relatively speaking).  So start from there.  You might infer from other resources as well, e.g. OrganicSynthesis, too. And it is much easier to keep track of the number of electrons with explicit dashes for single and double bonds, than with Thiele rings.

@Ashish In case of anthracene, Clar's rule is helpful.  On occasion, Wikipedia sub-section's tagged "chemistry" (example) note general pattern.  An identification of these may be helped by the structure explorer, too.

Phenanthrene reacts quickly with one mole of ozone at the 9,10-double bond according to this paper here and the OrgSyn procedure here referred to above by @Buttonwood. In chloroform a polymeric mono-ozonide is obtained. In MeOH the reaction takes a different course resulting in the formation of a cyclic methoxy-hydroxy-peroxide and a cyclic dimethoxy-peroxide. Reductive work up (Zn/AcOH or Me2S) of these products gives the di-aldehyde.


text:
Steady-state approach - how to find rate equation for intermediate?

The reaction mechanism for the formation of $\ce{NO2}$ is:
\begin{align}
\ce{NO + NO &<=>[$k_1$][$k_1'$] N2O2} & &\text{ (slow)} \\[0.2cm]
\ce{N2O2 + O2 &->[$k_2$] NO2 +NO2} & &\text{ (fast)}
\end{align}

What is the equation for rate of change of the intermediate $\ce{N2O2}$?
My solution is as follows:
The rate of formation of intermediate $\ce{N2O2}$ is given by:

\begin{align}
\frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - \color{red}{2}k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0
\end{align}
But the answer in the text book says:
\begin{align}
\frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0
\end{align}

Can not figure out why my answer is wrong.

Write the rate law for the rate of reaction of creating $\ce{NO}$ from $\ce{N2O2}$ (reverse of slow reaction). The is the rate of consuming $\ce{N2O2}$ for that component reaction.

Split the scheme into three parts and write down each step. Start by definition with
$$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$
then the first step $\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=k_1[\ce{NO}]^2\quad$ or $\quad\displaystyle \frac{d[\ce{NO}]}{dt}=-2k_1[\ce{NO}]^2$
and so $\ce{N_2O_2}$ is produced as
$$\displaystyle \frac{d[\ce{N2O2}]}{dt} = k_1[\ce{NO}]^2$$
In the reverse equilibrium step $\ce{N2O2}$ is lost (by dissociatiing) so $\displaystyle \frac{d[\ce{N2O2}]}{dt} = -k_1^{'}[\ce{N2O2}]$
and in the last reaction it is also lost by reaction with oxygen therefore $\displaystyle \frac{d[\ce{N2O2}]}{dt}=-k_2[\ce{N2O2}][\ce{O2}]$.
Adding up all three rates for $\ce{N_2O_2}$ gives
$$\displaystyle \frac{d[\ce{N2O2}]}{dt}=k_1[\ce{NO}]^2-k_1^{'}[\ce{N2O2}]-k_2[\ce{N2O2}][\ce{O2}]$$
You can now find the steady-state conditions.
(In your last equation there is $2k_1$ and it is quite common to see it written this way, all it does is to change $k_1$.)

Thank you! My final solution is:$$\displaystyle \frac{1}{2}\frac{d[NO_2]}{dt}=-\frac{d[N_2O_2]}{dt}$$
$$\displaystyle \frac{1}{2}\frac{d[NO_2]}{dt}=k_2[NO_2]^2\quad$$
That means:
$$\displaystyle +\frac{d[N_2O_2]}{dt}={2}k_2[N_2O_2][O_2]$$
So, the overall rate law:
$$\displaystyle \frac{\color{red}{2}k_1k_2[O_2][NO]^2\quad}{k’_{-1} + k_2[O_2]}$$
And we can simplify to:
$$\displaystyle \frac{\color{red}{2}k_1k_2[O_2][NO]^2\quad}{k’_{-1}}$$

If you apply Goldberg & Waage's Law of Mass Action to the rate determining (slowest) step, we will get:
$$\frac  {d[\ce{N2O2}]}{dt}=-k_2 [\ce{N2O2}][\ce{O2}]$$
Moreover, we have:
$$\frac {k_1}{k'_1}=\frac {[\ce{N2O2}]}{[\ce{NO}]^2}$$
If this is substituted to the rate equation obtained before, the rate expression involves concentration of products and reactions, rather than intermediates:
$$\frac {d[\ce{N2O2}]}{dt}=-\frac {k_1 k_2 [\ce{NO}]^2[\ce{O2}]}{k'_1}$$
This should be the final expression. Your expression is correct too, as $2k_1[\ce{NO}]^2-2k'_1[\ce{N2O2}]=0$, hence my expression and yours are equivalent.
I would love to hear feedback.

Thank you! You are right for given conditions, pre-equilibrium approximation with fast first step. Problem in textbook call for steady-state approach.  Sorry for misleading!


text:
Stability of meta ethyl and meta methyl benzene carbocations respectively
It is known that Methyl has a greater +I effect than Ethyl, but here, in this case, the condition is satisfied in the para benzene carbocations whereas, in the meta forms, it is the opposite. Shouldn't meta methyl benzene carbocation be more stable than meta ethyl benzene carbocation? Please, Can anyone explain?

I would like to point out that the species you are talking about are not benzene carbocations but phenylic carbocations.

And yes, no resonance or hyperconjugation effects are observed from the meta position and thus the higher inductive effect of ethyl group should make it a stabler structure.

Sorry for the mistake.

Does this mean methyl has a greater +M effect than ethyl. How?

Ethyl and methyl don't have an M effect they have a H effect.

We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. So, IV>I.
Now note that both of these structures must be less stable than II and III, because these are stabilized by both hyperconjugation and inductive effect, and hyperconjugation is a more stabilizing phenomenon than inductive effect. Since II has more $\alpha$-H than III(3 and 2, respectively), there would be more hyperconjugative structures for II, hence more stability. Hence, finally, the order is I<IV<III<II. This is the first option.


text:
Comparing the rates of hydrolysis of different haloalkanes
The title may not completely explain my exact doubt so I'll explain it here:
My teacher gave me the following problem during a lecture about SN1 reactions and I had to determine the order of the rate of reaction ($k$):

As in the rate determining step of an SN1 reaction, the halogen (leaving group) would leave to form the carbocation. I simply looked at which $\ce{C-X}$ bond would easily break and concluded that:
$k_1 > k_2 > k_3$ where $k_i$ is the rate of the given reactions respectively.
I had another approach to this by looking at acid-conjugate base relations.
As $\ce{HI}$ is a strong acid, it's conjugate base, $\ce{I-}$ would be a weak base hence making it a good leaving group (Weak bases are generally good leaving groups).
Which was correct as my teacher further explained. But he mentioned some more points, which I think, weren't suitable to be talked about but it still confused me.

He mentioned the solvation of the respective halide ions ($\ce{X-}$) in a polar protic solvent (here, $\ce{H2O}$):
$$\ce{Cl-} > \ce{Br-} > \ce{I-}$$
as the iodide ion is larger in size, the water molecules don't surround and 'trap' it as much as they surround the chloride or bromide ion.
But here's where things got a bit confusing. Technically, if the chloride ion is solvated the most, it is trapped by the solvent and isn't a naked nucleophile anymore while the iodide is. So Iodide would have more tendency to attack the substrate back.
Looking at it this way, the chloride ion was seeming to be a better leaving group which it obviously isn't supposed to be.
So changing the way of looking at this is changing the conclusions as well. The former approach leads to the correct answer, which is:
$$k_1 > k_2 > k_3$$
Yet, the latter seems to indicate:
$$k_3 > k_2 > k_1$$

What's going on here? I would really appreciate if someone could clear out this confusion.

This confusion arise because OP did not look at the energy profile of $\mathrm{S_N}$1 mechanism. The first order kinetics of $\mathrm{S_N}$1 reactions (e.g., reactions given in the question) suggests a two-step mechanism in which the rate-determining step consists of the ionization of the substrate (here, it is an alkyl halide), as shown in the following diagram:

The first step, $\ce{R3C-X <=> R3C+ + X-}$, which is the ionization of alkyl halide, is the slowest and the rate determining step. It is true that this step is reversible and is in equilibrium.
However, the next step(s), $\ce{R3C+ + H2O <=> R3C-O+H2 -> R3C-OH + H+}$, is fast and gives the intended product(s). As shown in the energy profile of this mechanism, a carbocation is formed as a high-energy intermediate in first step, and this intermediate bonds immediately to nearby nucleophiles (either $\ce{X-}$ or $\ce{H2O}$ in this case). The rate of this latter reactions would be depends on the values of $\mathrm{\Delta E_\ce{X}}$ (reversible reaction of first step) and $\mathrm{\Delta E_\ce{2}}$ (the second step, $\ce{R3C+ + H2O <=> R3C-O+H2}$), respectively. The value of $\mathrm{\Delta E_\ce{X}}$ may depend on how fast $\ce{X-}$ could take away from the electrophile, meaning the rate of solvation of the respective halide ions come to play here. However, since this is an example of hydrolysis reaction, protic solvent molecules are much more around the electrophile than halide ions, and this competitive reverse reaction is minimal. It is also worth noting that final step, $\ce{R3C-O+H2 ->[X-] R3C-OH + HX}$ is irreversible (not shown in the profile) and would have gone towards to give products eventually.
Therefore, the rate is only depend on the first step. Such reactions would be considered unimolecular and would follow a first-order kinetics. That is the reason to call it $\mathrm{S_N}$1 reactions and having undergo with relevant mechanism. The first step depends on the leaving group ability of the halide. As you learn in the subject, the leaving group ability increase in halogens when going down in the periodic table: $\ce{I- \gt Br- \gt Cl- \gt F-}$ similar to $\ce{C-X}$ bond dissociation energy. Thus, rate of hydrolysis of alkyl halides are:
$$\ce{R3C-I \gt R3C-Br \gt R3C-Cl \gt R3C-F}$$
Hence, $k_1 \gt k_2 \gt k_3$.

Note: Your teacher is correct of the hydration rates of halide ions: $\ce{I- \lt Br- \lt Cl- \lt F-}$ (Ref.1)

References:

Kenzo Hiraoka, Susumu Mizuse, and Shinichi Yamabe, "Solvation of halide ions with $\ce{H2O}$ and $\ce{CH3CN}$ in the gas phase," J. Phys. Chem. 1988, 92(13), 3943–3952 (https://doi.org/10.1021/j100324a051).

One way to interpret the solvent effect in this particular problem is that, as you mentioned, the iodide could attack the carbocation if it weren't trapped as much as the other halides were. But, that doesn't necessarily occur because the step-2 of $\mathrm{S_N}$1 mechanism is the fast step where the $\ce{H2O}$/ any other nucleophile attacks the carbocation immediately. Hence, the second solution that you arrived at: $K_3 \gt K_2 \gt K_1$, stands invalid.
But, it is to be remembered that the action of polar protic solvents, does increase nucleophilicity as we go down the periodic table, which doesn't happen otherwise.
Also, though the trapping of iodide is less compared to the other halides, it still remains 'trapped' for it to attack the carbocation back.
So, taking into account the above reasonings, I think the answer, $K_1 \gt K_2 \gt K_3$ can be validated.

Update: I asked my teacher the same question and he told me that, the iodine could attack some amount(%) of the carbocation back and it will become a reversible reaction, but this won't be considered because the starting material remains as is. The remaining (in %) carbocation, there is nucleophile attack, hence will be considered the major product.

I think you should include your update as it is in the answer. Also, the attacking nucleophile here is not $\ce{OH-}$, but water. I'd like you to correct that as well.

The rate law for the SN1 reaction is rate = k[substrate] as the rate-limiting step is formation of the intermediate carbocation.
In terms of leaving group ability, I- > Br- > Cl-. This is due to the polarizability of the leaving group and the ability to "spread out" the negative charge over a larger electrostatic area.
As such, I would say k1 > k2 > k3.
Intimate ion pairs may happen to a small extent but do not play a large role in determining the kinetic here; this would be more useful to explain inversion versus retention of configuration here, but with a trimethyl-substituted carbon, the point is moot.


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Practical to measure topography of a large area (mm-scale) with scanning probe microscopy (SPM)?
I'm currently very new to nanotechnology. I do understand that it is possible to achieve an excellent topography image of a sample by SPM. However, is it possible to do SPM over a large area (for instance like in SEM, which is possible to image over 1x1 mm)? Theoretically speaking, I do not see any problem except that measurement time might be very high.
So, is it possible to scan over a large area (mm-scale)? If not, what can the reason be if one excluded measurement time. If yes, which technique will give an advantage for such measurement?
I would appreciate if anyone could explain the theory/mechanism behind this or point me to a place where I can obtain a better understanding.

Cross posted in Physics SE

You mean, self-study of entries like wikipedia with its crosslinks at the end of the entry, chemistry libre texts (there are multiple entries there) still left you thirsty to learn more?

The question you have to ask yourself is what is the length scale of the features you want to study as this will inform you of the correct technique to use.
The most common type of SPM is the atomic force microscope (AFM) which has an astoundingly high resolution (atomic scale in the right conditions), but as a result is limited to relatively small scan sizes with the largest avaliable being about 150 x 150 µm. There are AFMs that can stitch together multiple images and give you massive images, but as you said collection time will be very very long (if this is what you want to do, look for an AFM in a clean room that is primarily used for metrology in silicon wafer processing as these generally have the software to do the stitching).
If you only care about microscale features, then a stylus profilimeter would do the trick. It is basicaly a large AFM so it has most of the same advantages/problems. I haven't really used these so I am not sure how well they image or if they are mostly used for single line profiles.
Better than a stylus profilimeter, in my opinion, are the suite of 3D optical profilimeters. There are lot of options here from different manufacturers using different technologies, but they all have similar characteristics: xy-resolution is limited by diffraction (depends on objective used, ~200 nm) and z-resolution is sub nm; scans are very quick (can be less than second); can stitch many scans together to form very large images (I have seen an example of a whole coin); they struggle with transparent materials; they can't see reentrant geometry.
If you require nm resolution over a mm scale, I would go with AFM and just take images of representive areas as data analysis would be really difficult if you had to analyse AFM data over millimeters.
Finally, the last piece of advice I'd give you is something all students hate: go read the literature. This is the best way to figure out what technique to use. What I found helpful was to make a table of all the different techniques used in similar pieces of research and make a list of their pros and cons, how easy are they to learn, do I have access, what does data analysis look like etc. Even if you end up using the first technique in the table, you have learned a lot about where and when other techniques are useful.


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Reacting Grignard reagent with amides

Propene in the presence of $\ce{HBr}$ and $\ce{H2O2}$ gives A.
A on reaction with $\ce{Mg}$ in dry ether gives B.
B on reaction with acetamide, $\ce{CH3C(O)NH2}$, gives C.

In the above question, I got A as 1-bromopropane and B as propylmagnesium bromide. As for C, I got
$$\ce { CH3-CH2-CH3 + CH3C(=O)NHMgBr}$$
The ratio of moles of the Grignard reagent vs. acetamide is not specified. Should I add an additional propyl Grignard and arrive at $\ce{CH3-CH2-CH3}$, $\ce{CH4}$, and $\ce{ CH3C(OH)N(MgBr)2}$ ?
I need help with figuring out the product in the reaction of B and acetamide.

The pKa of acetamide is ~25 - easily acidic enough to be deprotonated by Grignard, and that is where the reaction stops in my opimion.

https://chemistry.stackexchange.com/questions/21931/do-grignard-reagents-react-with-amides https://chemistry.stackexchange.com/questions/88361/leaving-group-stability-when-treated-with-grignard

In the reaction that you have mentioned above, the reaction essentially stops after the reaction has taken place in a 1:1 ratio. That is because the electrophilicity of the polar bond of the acid derivative has reduced sharply so it cannot undergo further reaction.
However, had the nucleophile been $\ce{RLi}$ instead of $\ce{RMgX}$ ($\ce{RLi}$ is more ionic and hence a better nucleophile), further reaction would've been possible, although it provides the product in low yield. R attacks C=O resulting in the formation of $\ce{R-C-O- Li+}$ (The other groups remain as they were- the only difference being that it is $\ce{NHLi}$ instead of $\ce{NHMgX}$)

Here is information on how you can format (no coding involved!) your answer.


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Is the spin of the transferred 4s electron always opposite to the spin of 3d electron in chromium?
I am studying atomic structure and found out that Chromium has an exception in electronic configuration. One electron from 4s jumps to 3d to achieve stability.
let's say the spin of transferred electron is +1/2. Then the other 4s electron should have -1/2. Then, are all the 3d electrons of +1/2 spin or some may have -1/2 spin too?

Hund's rule still applies here.

@Zhee,but the multiplicity changes,so what would be correct

Your question is not particularly clear on when you want to make this observation. Immediately? After a long time?

whenever it becomes stable

Define "stable".

I mean until it reaches its ground state electronic configuration

Didn't you just answer your own question?

but I have asked that,does the spin cobversion happens when electron from 4s goes into 3d,as to attain max. possible stability?if yes,then how does the spin conversion happens,if no,how can the electron which before had +1/2 in 4s can have -1/2 when it goes to 3d?

I think spins between different atoms don't necessarily correlate. You'd have to express the new eigenstates as a function of the old eigen states. That already means you may not know what the spin is after transferring the electron... You can learn more about spin flips here.

@zee,ohhk,but why then the multiplicity of chromium defined,as if we dont know the spin of the electron that jumped,so it may be two different values ?

It might be in multiple states before collapse of the wave function.

@Zhe,so,is there still uncertainity regarding multiplicity of chromium?


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How does a spectrophotometer measure absorbance based on the intensity of detected light?
We know that $\displaystyle A = -\log\frac{I_t}{I_0}$, where $I_0$ is the intensity of light without a sample absorbing, and $I_t$ with the sample. For determining the absorption $A$, we need both $I_t$ and $I_0$, but how does a spectrophometer detector measure $A$ knowing only $I_t$ ? (The detector is placed after the sample in the optical path, so it does not measure $I_0$ in front of it.)
Does the detector measure $I_t$ or the ratio $I_t/I_0$ ?

I'm not sure I understand the question because you haven't defined any variables and it's not clear what the difference between "It" and "If" is, but I guess you are missing the fact that the sample solution and the reference (aka "blank") are measured separately (or there is a scheme allowing for automatic subtraction of a "blank" signal).

There are two possibilities:

If there is only one optical pathway, you record i) the blank sample for all wavelengths of interest and store the detector's intensity information in function of $\lambda$ in a working memory.  Light passing the blank is then assumed to equate $I_0$. Then you record ii) the wavelength dependent intensities $I_\mathrm{abs}$ with your sample of interest, and report the subsequently corrected data as spectrum $\mathrm{Abs} = f(\lambda{})$.

.or.

Your instrument permits the simultaneous recording of sample and blank to record both intensity data and reporting the already corrected absorption in one run.  Be sure, though, to assert which of the two paths (= two sample holders) is about your sample, and the blank (which may depend on spectrometer and software).


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Spectroscopy: What is an instrument response function?
Many papers involving ultrafast spectroscopy measure the instrument response function and model it as a Gaussian, but the sources on any of this are never cited and I have no intuition for what this could mean. What is the instrument response function intuitively and why is it represented as a convolution?

I Googled "What is the instrument response function" and got this hit. You might want to do some perfunctory research and amend your query to include what your efforts yielded.

@ToddMinehardt Good hit! This query is really better suited to the signal processing stack exchange: they are all about linear, time invariant (LTI) systems, impulse responses, frequency domain transfer functions, and on and on. TLDR: in the time domain, the output of a LTI system is the convolution of the input with the impulse response function of the LTI system. In the frequency domain, the Fourier transform of the output is the product of the transfer function and the Fourier transform of the input.

More generally, you want to look for an "impulse response" function, which dictates how the spectrometer (or any LTI system) responds to input. It's a bit long, but the Signals and Systems course on MIT OCW is a good place to learn these things. Of course, it raises the question of whether your spectrometer is truly LTI :-)

Usually this a measured via autocorrelation of the laser pulse and there are various ways of doing this. Then you have to assume a pulse shape and fit to the autocorrelation. Actually what you measure is the convolution of the actual pump and probe pulses with the molecular response. If the response is very long, say nanoseconds then you can measure the rise time profile, perhaps  as a few tens of femtoseconds depending on you laser system, and use this to obtain quite a good estimate of the pulse duration by  fitting assumed pulse shape to the data. This ignores any chirp in the pulses.

There are great answers, I wish I could select both of them. Thank you!

Suppose that the lifetime of electronically excited atoms or molecules is to be measured by exciting them with a pulse of light and their fluorescence measured as it decays with time. This fluorescence could be observed with a photodiode or photomultiplier, whose output voltage is measured with an oscilloscope. Before doing this experiment, two questions have to be answered;

Is the laser used to excite the molecules of short enough duration that the molecules or atoms can be excited quickly enough before any significant number can decay back to the ground state?

Is the detection equipment (photodiode, oscilloscope) used able to respond quickly enough to measure the decaying fluorescence properly?

If either one or both of these conditions cannot be met, then the data will be distorted by the relatively slow response of the instrument. The convolution curve in the figure shows how this distortion affects data. In this figure, the top curve is the ideal decay of the excited state, but it could represent any ideal response. This behaviour would be observed if the molecules could be excited with an infinitesimally narrow laser pulse and measured with a photo-detector with an unlimited time response. The second curve is the actual shape of the laser pulse, and/or detector response, and is the 'instrument response' drawn on the same timescale. Clearly, this has a width and a rise and decay time that is not so very different to that of the ideal response. The lower curve is the convolution of the ideal response with the instrument response, and is what would be measured experimentally and clearly has characteristics of both curves. A log plot of the data would show that only at long times does the convoluted response have the same slope as the ideal one. It makes no difference if the instrument response consists of a slow 'driving force' for the experiment, in this case a long-lived light-pulse, or a slowly responding detector or both, because the effect producing the convolution is the same.
To understand how convolution works, suppose that the overall instrument response is made up of a series of $\delta$-function impulses. Suppose these impulses are made at ever shorter time intervals, then the effect is that of smoothly exciting the object. After a while these impulses stop. Each of the impulses elicits an ideal response but because there are many of them, their responses must be added together. The result is the convolution; the effect is shown below. It is always assumed in the convolution that the response is linear with the impulse, which simply means that doubling the impulse doubles the response and so forth.
Written as a summation, the convolution at point $k$ is
$$\displaystyle C(k) = \sum_{i=0}^k f(i)w(k - i )   \tag{32}$$
This sum evaluates just one point; to calculate the whole convolution, the index $k$ must now be varied from 1 to $n$, which is the number of data points, making a double summation. One reason Fourier transforms are used to calculate convolutions is that the fast Fourier transform algorithm, FFT, is far quicker on the computer than calculating the convolution as a double summation, particularly for a large number of data points. (However, FFT should not be used if the signal at beginning and end of data is not of the same magnitude. This caused severe distortion in the convoluted data and is a real drawback when doing experiments, i.e. it is safer to use summation, even though it is slower to calculate. )

Suppose that the pulse exciting the sample has a shape given by some function $f$, the ideal experimental response $w$, and the convolution $C$. The terms can be written down at each time if it is assumed, for the present, that the impulses are discrete and the data is represented as a series of points at times $1, 2, 3$, and so forth; $f$(6), for example, represents the value of $f$ at the sixth time position. The first point of the impulse is $f$(1) and this produces the response
$$\displaystyle f (1)[w(1) + w(2) + w(3) + \cdots]$$
The second and third impulses produce
$\displaystyle f(2)[w(1) + w(2) + w(3) + \cdots]$ and $f(3)[w(1) + w(2) + w(3) + \cdots]$.
The convolution is the sum of these terms at times 1, 2, 3, and so on therefore;
$$\displaystyle\begin{align}
C(1)& = f (1)w(1)\\
C(2)& = f (1)w(2) + f (2)w(1)\\
C(3) &= f (1)w(3) + f (2)w(2) + f (3)w(1)\\
C(4)& = f (1)w(4) + f (2)w(3) + f (3)w(2) + f (4)w(1)\\
\end{align}$$
You can see how the series folds back on itself, hence the name convolution.

(+1) Excellent example and explanation!

@Porphyrin, Nice illustration. You can add that DFT can produce the same linear linear convolution if we make sure the length of the DFT is N1+N2-1 (by zero padding). There are nice tricks to get rid of circular convolution effects.

@M. Farooq , thanks I had forgotten to mention that.

Let me provide the intuitive idea so that you take the lead from here and understand the notion behind "instrument response" function. Any analytical/physical chemist, who uses an intrument, should have some basic idea of the instrument response. After that you should look up formal concepts of convolution on Youtube.
Whenever you collect experimental data using an instrument, you would like to have no distortions induced by the instrument in your data. For example, if your eyesight is weak, and you are trying to read a road sign, but it appears blurred to you, your eyes and brain have "convoluted" the true image of the road sign. Therefore, your eye/brain response (=instrument response) is to blurr the image and this is what you do not want. At least, one should be aware of this fact that the true image is not a blurred picture.
Hence people who wish to study lineshapes, linewidths in a spectrum they would like to know how their spectrometer set-up is distorting the true experimental result. In an ideal case there should be no distortion but a real instrument can introduce its own effects which we call as convolution.
Let us see a example to illustrate what Gaussian broadening does to a "true" signal shape, using an light absorbance detector (of a chromatograph). I did this just to understand the response function of that instrument.
Imagine that a light source is turning on and off periodically. It should look like a square wave to a light detector (signal is up when the light is on and it is down when the light of off). However, if you have increasing Gaussian broadening done by the instrument, you begin to loose the true shape of the square wave. If you have too much of Gaussian broadening, the square wave, appears to be completely distorted. Look at the top and the bottom figure. The top shows the relatively true signal shape, and the bottom figure shows the effect of Gaussian broadening on the same signal. All true information is lost.
Hope that highlights the importance of instrument response function in analytical chemistry/instrumental analysis.


text:
Stereochemistry & mechanism of β-cyano enone methylation
I am helping a friend with some upper-level mechanism problems. We came across this one. This is my proposal. Do you agree/disagree? Thank you for any input.

Source question:

In your mechanism, you have an anion next to a nitrile: that is good, but you should not draw it as a chair anymore, that carbon with a negative charge should be sp2, precisely because of the resonance stabilisation.

Fixing again... realizing it after you said that!

No problem. Now, the interesting part of the question IMO is the stereochemistry. The six-membered-ring-with-one-sp2-carbon should probably be drawn like cyclohexanone, and there are stereochemical models for nucleophilic attack on cyclohexanones; problem is I don't know what extent this is relevant for electrophilic attack of MeI on the carbanion...

@orthocresol - my initial thought was the C=O was going to be most reactive to the Grignard, then anion would be better resonance-stabilized next to the nitrile than adding beta to that group.

Why would the Methyl Grignard attack the carbon on the double bond next to the nitrile in your 2nd step? This seems highly unlikely to me. It would be more likely to add to the other end of the double bond in a Michael attack on the nitrile

It should attack the carbon beta to the nitrile to produce the resonance-stabilized anion as I have shown. How can we rationalize this stereochemistry is where I am slightly perplexed.

In the source question, I understand why the beta 1,3 carbon would be attacked (most electropositive from ketone/nitrile), but it isn't mechanistically productive. If the ketone is attacked first, it isolates the beta position and allows the setup of the resonance-stabilized carbanion.

You shows in your second structure, Grignard attacking the $\alpha$ carbon to carbonyl (not carbonyl anymore though), yet you shows methyl on  the $\beta$ carbon to carbonyl  with negative charge making $\beta$ carbon pentavalent (with negative charge). This should be fixed ASAP.

@mathewmahindaratne - fixed; apologies for the error - didn't catch it previously

The first two structures in the third line are not related to the starting cyanoketone. The carbonyl  group is in the wrong position in the first structure.

The details are here: https://pubs.acs.org/doi/pdf/10.1021/jo052102j

@user55119: You are correct about the error. Also, the carbonyl group is in the wrong position not only in the first structure, but also in the second structure. Rest is okay.

Thank you for the excellent reference. I started to consider magnesium coordination  but debated it.

Mechanism I developed from the paper @user55119 graciously provided. Appreciate it! Any refinements I appreciate. :)

One suggestion. Although the relative stereochemistry is what is important, try not to switch absolute stereochemistry as you have done in line 2. For the right hand structure in line 2, draw the same chair cyclohexane but shift all substituents one carbon to the left where axial becomes equatorial and equatorial becomes axial.


text:
Is there a public database for enthalpy of solution values?
I'm looking for a public database for enthalpy of solution values. The CRC handbook only has selected compounds available. The NIST webbook doesn't appear to have any enthalpy of solution values. Any help would be greatly appreciated!

Here?

It's not that easy. I found that same site through a quick google search, but the sites linked on that page are all subscription based, not public.

Your question still is a bit broad.  If possible, set a focus on the specific compound / compound class and the solvent intended.  Is it about dissolving salts in water, or organic compounds in each other?


text:
Converting a CIF file to a .com or .gjf file
I would like to perform DFT calculations and extract atom coordinates for calculations in Gaussian.
I am using Mercury and have been able to isolate the atoms that I am interested in simulating. However, when I save the CIF files and open them again i see something very strange.
Atoms that have fractional occupancy from the XRD are shown to be fully occupied, this gives me nonphysical molecules that contain to many atoms. How can I get rid of these fractional occupied atom positions?
I am attaching an image to illustrate what I mean.

When i retrieve the data from CCDC's webpage, the fractional occupied positions are marked with crosses, is there a way to clean the file?:

Technically, you are not allowed to "clean" the file since these positions are also occupied by atoms. And it's not "strange", most of structures contain atoms with fractional occupancies. Unless you are doing geometry optimization, you need to account for this. Alternatively, pick atoms from the same dominant site. I don't know about Mercury, but in Olex2 you can use showp <site #s> to hide minor occupancies.

@andselisk In principle one may use CCDC Mercury to edit an orientationally disordered .cif into two (or more than two) ordered .cif (Mercury handbook, section 12.1 / Manually Editing a Structure).  Compared to Olex2, it demands more attention.  Depending on the later intended computation, it might not be needed, though (e.g., CrystalExplorer lets you choose how to deal with disorder in .cif  when e.g., computing Hirshfeld  surfaces.)

@HanDen You state that the model model was published in a journal or / and in the CCDC database.  If you add a reference to the primary publication, the CCDC deposition number (sometimes in the experimental section or mentioned in the SI of a paper) or the CCDC structure code (six Latin characters, on occasion two additional digits), one could have a look at this special entry, because of CCDC's special site.

In general, OpenBabel may convert some of the information of a .cif into a Gaussian .gjf.  If your are fine with using the CLI, the command would follow the syntax pattern of
obabel -icif input.cif -ogjf -O output.gjf
The parameters -icif and -ogjf hereby instruct OpenBabel about the expected input / output format of the files, while -O actually triggers the program to write the output into a permanent record.  Further information may be obtained here.
Depending on the computation intended, the crosses Mercury displays may be very valuable information.  It is Mercury's way to tell the user that the structure is disordered, i.e. there is more than one orientation of a molecule in the unit cell.  For example, that a conformational flexible $\ce{-CH3}$ or $\ce{-CF3}$ group has been refined into more than one position (think staggered / eclipsed conformation of butane, there are more than one energetic minima, and one is the absolute one.)  But this orientational disorder may extend over larger portions and complete molecules, too (e.g., small solvent molecules).  It would be better if you could display more of the structure, and tell a bit in addition about the intended computation.
Do you know to read a .cif file about site occupancy factors?  E.g., for molecules with an up / down or ($\uparrow$ / $\downarrow$) orientational disorder and only two realizations, these often are refined to add up to $p(\uparrow{}) + p(\downarrow) = 1.00$.


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Why are zinc shot and mossy zinc not considered toxic but zinc granules are?
In our lab we have three forms of zinc. We were initially experimenting with these three forms to optimize our application. We have "zinc shot", "mossy zinc", and "zinc granules". The granules are ~20 mesh. I am tasked with creating chemical standard operating procedures (SOPs) for all reagents that are considered "particularly hazardous". Zinc granules are considered particularly hazardous by our University and is an "acute toxin to aquatic life with lasting effects" in a recent safety data sheet (SDS) while zinc shot and mossy zinc are not considered hazardous by our University or on comparable recent SDSs. I see zinc powder is also considered toxic. To me this suggests particle size dictates toxicity, but why? I understand surface area influences the rate of reactions but if the effects are long lasting, then why are all forms of zinc not considered to be an aquatic life toxin?

Fisher's SDS from 2015 for mossy zinc states it is "toxic to aquatic life;" and their one for zinc shot, 2018 states "very toxic to aquatic life with long lasting effects." Something's amiss.

Also, see the dated but still-relevant EPA publication on Ambient Water Quality Criteria for Zinc for more details.

It seems unusual because for one thing a great many tons of zinc rich paints have been, and are being used . I have been retired for awhile but i remember seeing no caution signs on the zinc powder ( typically -400 mesh/screen) used to mix these coatings. These zinc rich primers are a mainstay in many industries  with apparently no problems.

@blacksmith37 They are problems. Here's just one MSDS for a zinc-rich primer. It's "aquatic acute" and the rest of the (M)SDS is hardly encouraging. I suspect that back in the day we didn't know and/or didn't care about handling things that later turned out to be "problematic," such as DDT, leaded gasoline, benzene, tobacco, etc.

So older people who lived with galvanized plumbing are all going to die ?

@blacksmith37 Yes. We all do, someday.

The key to toxicity is concentration, which is related to how rapidly the material disperses (or concentrates) in the environment. Which is also related to water type (fresh or salt) and temperature. MSDS sheets are a bit of a joke. They are worded in a way that fends off lawyers rather than gives detailed useful toxicity data. Consider this MSDS for Sodium Chloride
Bear in mind this useful advice when using it at home: "Eye Protection Wear safety glasses with side shields (or goggles) (European standard - EN 166). Hand Protection Protective gloves"

NaCl SDS quote - “Keep away from food”. So perhaps you allude to a more appropriate question. How can students and labs inform themselves of the true toxicity or lack there of for any given chemical.

@ajschauer Yes. I am having the same problem working with Tungsten Carbide powder. Useless MSDS - and toxicology reports focus on tool grinding and composites with Nickel/Cobalt composites. The best I can assume is "don't eat it, don't breath it, but otherwise mostly harmless"


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Enthalpy of Reaction and Efficiency of Electrochemical Energy Devices
For electrochemical energy devices such as batteries or fuel cells working reversibly, efficiency is defined as: $$ \eta = \frac {\Delta_rG^⦵}  {\Delta_rH^⦵} $$ Since cell operates reversibly Gibbs energy change of the system is equal to electrical work done by the cell, so we can instead just use standard Gibbs energy of reaction as energy output. What I am not clear about is why is enthalpy of reaction used as energy input? Since electrochemical energy devices transform chemical energy of electroactive materials to electrical energy, energy input should be chemical energy, but I am not sure why enthalpy of reaction represents chemical energy? Enthalpy of reaction is given by this equation: $$ \Delta_rH^⦵ = \Delta_rG^⦵ + T\Delta_rS^⦵$$
Since cell operates reversibly, $\Delta_rG^⦵$ represents reversible electrical work (maximum non-PV work done by the cell) and $T\Delta_rS^⦵$ represents heat exchanged with surroundings to keep cell at constant T in reversible operation since change in Gibbs energy as measure of non-PV work is only applicable for closed systems at constant p and T.
Why is chemical energy sum of these two quantities physically speaking since I do know how to derive this equation from 1st and 2nd law of thermodynamics, but I am not sure in its physical interpretation in context of electrochemical cells?

I'm a bit unclear about what you are asking, is it why is $G=H-TS$ or why the efficiency is defined as it is ?

Why is efficiency defined as it is.

The efficiency is  $\eta=\displaystyle \frac{w}{\Delta H}$ where $w$ is the useful work output, and $\Delta H$ the reaction enthalpy. This can be written as $\displaystyle \eta=\frac{(\Delta H-Q_{in})}{\Delta H}$ and $Q_{in}$ is the heat added where $Q_{in}=T\Delta S$ when operated reversibly giving the efficiency equation you quote. In effect the efficiency accounts for heat losses or gains relative to the reaction energy available. Gibbs was first to realise that because heat from the surroundings can be added to or removed in a reaction the work done can be greater (or less than) than $\Delta H$. In your definition the efficiency can therefore be $\gt 100$%.

Yes, this is equation I've written since deltaG = deltaH - TdeltaS. Only, it didn't really answer my question why is enthalpy of reaction taken as energy input? It seems that chemical energy equals enthalpy of reaction since chemical energy is converted to electrical energy. What you said about efficiency being able to be greater than 100% is something I suspected, but I wasn't sure if that is true. Now you said it in your answer.

What I am also not sure is that if entropy change of reaction is positive than heat comes from surroundings to system and in that case energy output or change in Gibbs energy of the system is less than in case if no heat is exchanged which means that efficiency is lower than in other case. I am not sure why this is? Why does adding heat to the system reduce its efficiency?

The enthalpy would be the maximum amount of energy that could be recovered to do work in a chemical reaction if it were not for the fact that energy can be taken from or given to the surroundings. In other words some amount of heat, not from the chemical reaction, can be used to do work, ( or removed from doing work) and this amount under reversible conditions  is $T\Delta S$.

the efficiency is $\eta = 1-T\Delta S/\Delta H$ so if $T\Delta S$ is positive the efficiency is reduced for a given $\Delta H$

Yes, but why is it that if heat is given to the system A.K.A entropy of reaction is positive that efficiency is reduced? I can see that from equation as you stated, but I am not sure I understand physical significance/picture of it.

The heat is determined only by the entropy , which is a property of the chemical reaction, not the other way round.

Well, this is also something which I did ask before here, but I didn't get a good answer. When reaction happens entropy 9f the system changes because reactants and products are different compounds with different structure and aggregate state and because of that have different entropy. What does such entropy change have to do with heat exchanged since entropy change isn't due to heat exchanged, but due to reaction? This is that TdeltaS term which I am not clear about.

You are at constant temperature and reversible  conditions so $T\Delta S$ is determined by the reaction, so $q$  is not an arbitrary amount.

I agree with that, but that didn't answer my question. As I asked entropy change in the system is due to reaction not due to heat transfer. This is what I don't fully understand and I described it in my last answer.


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Particle tracking in colloids and uniformity in fractional part function
I don't know if its right place to ask this question since chemistry and programming both are involved.
I was working on particle tracking of series of pictures (.tiff) of colloids. And I was Using trackpy in python.
The code gives me the position of particle to suppixel accuracy.

There is another step involved is of checking the subpixel accuracy. In this method we check the uniformity in the decimal part of the position

a quick way to check for subpixel accuracy is to check that the decimal part of the x and/or y positions are evenly distributed. Trackpy provides a convenience plotting function for this called subpx_bias

Mask size is more or less the diameter of particle.
I am not getting how does evenly distribution of decimal part ensures that we are on the right track?
And how a dip shows that we are wrong somewhere?
You can also refer to Eric weeks website. on this website it is briefly mentioned that

One failure mode is if the length scale in feature is made too small, then all the x and y coords get 'rounded off' to the nearest pixel value. The above command plots a histogram of the fractional part of the x-coords of the image. The physically distributed positions should be random -- giving a flat histogram. If the histogram has two peaks (near 0 and 1), set the size parameter in 'feature' a little bigger, determine a new masscut, and repeat until everything is happy.

Suppose you generate uniformly distributed random real numbers between 2 and 3, having 3 decimal places. Then subtracting 2 from each one will give uniformly distributed (roughly flat histogram) decimal fractions between 0 and 1. But if you had simply rounded off the original numbers to integers, you would only get 2 and 3. So that test is just to make sure the “length scale” is not corrupting the data. Make sense?

But , then how does increasing of diameter (size parameter of feature) makes the histogram flat

If the mask size is too small, you are biasing toward integers: the particles are being sorted, as it were, into integer sizes. This is a quantization error or discretization error, With a larger mask size, you effectively “dither” the size estimation and reduce the quantization error. Kind of like adding a little white noise to dither the pixels and avoid the chunky quantization.

can you please write a answer, I am just a beginner and not so clear about mask size too

This is not really a chemistry question: it is an instrument behavior or statistics question. I have not done any particle tracking experiments. Maybe this question should be migrated (not cross-posted) at the signal processing stack exchange or the CV stack exchange.

This reads a little bit like the automated blob and cell counting in microbiology.  Maybe there already is a library about this in ImageJ / Fiji (example).  Because you aim to tackle the problem with Python, and it is kind of image processing, I infer from OpenCV's advertised use in ML OpenCV SimpleBlobDetector could be helpful.


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Does the critical temperature of a gas depend on the amount of the gas taken?
Does the critical temperature of a gas depend on the amount of the gas taken? Or is it independent of it? My book does not specify this point.
I think according to the gas laws, the critical temperature should depend on the amount of gas taken. Am I correct?

Critical parameters are intensive properties, independent on the system scaling.

I don’t think you're correct. The critical values (pressure, volume and temperature) are the properties of a gas and they don’t depend on how many moles of the gas you have. It's also evident from the formula of critical temperature:
$$T_\mathrm{c} = \frac{8a}{27Rb},$$
where $a,$ $b$ and $R$ are constant for the gas. There is no dependency on the amount of substance.
Further, you don’t find any data on the amount of a gas when you look for the value of it’s critical temperature. For example in table 3.10 here.

Critical volume is defined for one mole of the gas at critical temperature, but it's formula is : $3b$. So I am not sure we can base our arguments on these formulae.

I’m not completely sure but I think critical volume is defined for 1 mol because volume increases with an increase in no. of moles of the gas whereas temperature doesn’t. So we need to define critical volume for a certain quantity, i.e. 1 mol, but the same doesn’t apply on critical temperature. But again, this is my hypothesis which I haven’t cross checked anywhere.


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Energy Balance for a reactor

There's wastewater flowing into the reactor at $\pu{20 ^\circ C}$ and the reactor is maintained at $\pu{36 ^\circ C}$. The effluent of the reactor flows into the stripper, which is at $\pu{65 ^\circ C}$. How can you maintain the temperature at $\pu{36 ^\circ C}$ in the left reactor with fresh wastewater flowing in at $\pu{20 ^\circ C}$ and stripped wastewater flowing in at $\pu{65 ^\circ C}$?

$$\underset{\mathrm{input}}{\pu{20 ^\circ C}}\ce{->[$Q_i$]} \underset{\mathrm{reactor}}{\pu{36^\circ C}} \ce{<=>[Q_e][Q_r]} \underset{\mathrm{stripper}}{\pu{65 ^\circ C}} \ce{->} \mathrm{output}$$
When I did the mass balance, it's $Q_i = Q_s$. I don't know how you can get the $60\%$ of energy change during inflow using the energy balance.  How do you find that $Q_r = 60\%$ of $Q_i$?

What is the actual question here? I fail to understand, As an answer to your first question, we get that from the question itself. It is a statement and not something you need to prove.

Qr is supposed to be 60% of Qi. This is the answer. However, I am trying to figure out how to derive this using energy balance.

Is this the complete question? Could you share the exact phrasing of the question?

The question is: how can you maintain the temperature at 36 deg Celsius in the left reactor with fresh wastewater flowing in at 20 deg Celsius and stripped wastewater flowing at 65 deg Celsius?

Oh, so the question is what is Qr in terms of Qi?


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What is the correct reaction mechanism?
I came across this question:

I propose two things to happen:

It can undergo Aldol condensation

or

it can undergo SN by replacing $\ce{-Cl}$ with $\ce{-OH}$.

But apparently, both these mechanism give the wrong product. I’m unable to find out what is going on.

I see what went wrong. Such questions are often confused with aldol or cannizarro as they have a carbonyl group and a base is given.
Actually this is a completely different name reaction called Favorskii rearrangement. The key to identify it is that its applicable to α-halo ketones.
In Favorskii rearrangement, the base takes α-hydrogen from that side of ketone which is opposite to the side having the halogen. It then forms a 3 member cyclic ring as an intermediate product. The product formed is a carboxylic acid derivative, mostly ester.
You can learn more at organic-chemistry.org or wikipedia


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What is the LUMO of a molecule with an alkene and a carbonyl?
In a molecule that has both an alkene (C=C) and carbonyl (C=O) that are NOT conjugated, such as pent-4-en-2-one, how do you determine which π* anti-bonding orbital is the LUMO?
Essentially, I'm trying to reconcile two statements that I thought were true but apparently contradict here:

Increased electronegativity of an atom lowers the energy of its molecular orbitals. This lead me to think that π* C=O is the LUMO because the higher electronegativity of oxygen would lower the energy of the anti-bonding orbital (as well as the bonding orbital).
The lower the energy of the bonding MO, the higher the energy its anti-bonding MO is. This leads me to think that π* C=C would be the LUMO because π C=C is the highest energy bonding orbital, so π* C=C would be the lowest energy anti-bonding orbital.

Any help you could provide would be GREATLY appreciated!!

My guess is that the HOMO is the C=C pi bond whereas the LUMO is the C=O pi* bond. This would make sense as the electrophiles attack C=C first, and nucleophiles attack C=O first. Note that one molecule can have only one HOMO and only one LUMO.

I think here you are missing the point on what exactly a HOMO and a LUMO are. They are descriptions of the energy splitting that occurs when quantum wave functions constructively, or destructively, interfere with eachother when in proximity. Because this system is not conjugated, the alkene and the carbonyl with both INDIVIDUALLY have their own HOMO and their own LUMO. In the case of both of these systems, the HOMO will be the pi bonding orbital and the LUMO will be the pi* anti-bonding orbital. Now, when it comes to which of these LUMOs will be lower in energy, it honestly will depend on the compound. Electronegative atoms, electron withdrawing groups, etc. narrow the energy gap between bonding and antibonding orbitals, making the LUMOs more reactive. So in this particular example, the carbonyl will probably have a larger pi* coefficient and thus be lower in energy. For more complex systems, reactivities need computer algorithms, clever thinking, or just experimentation to determine which energies are lower.

There is a contradiction in terms, here. Namely and literally a molecule can't have two highest (lowest) (un)occupied molecular orbitals. Beside this remark, what you say is pretty much useful to OP and stress that we should not be assertively pedantic when it comes to molecular orbitals.... Often this type of questions receive comments like "the orbital are those of the whole molecule" and stuff like that. But reality counts.


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Microfiltration for saturated solutions of highly hygroscopic compounds?
Is it possible to pass saturated solutions of compounds with very high solubility through filters with small pores (e.g. passing a saturated aqueous ZnCl2 solution at 90C through a 2mcm glassfiber filter with the help of partial vacuum)?
Will it work well? Are there preferred methods for this? If not, is there any (reasonable) solution to filter out microparticles from such solutions?

Filtration is a mechanical operation.  Thus, if you have a solution of $\ce{ZnCl2}$, which (assuming you use water as solvent) dissociated into ions of $\ce{Zn^{2+}}$ and $\ce{Cl^-}$, a typical filter will not retain these ions.  Instead of using some vacuum (potential danger to evaporate some solvent, thus altering the concentration of he solution), you may consider filtration under pressure (e.g., across a Schlenk fritt for larger scale, or a syringe filter at smaller scale [e.g., testing a spin coating]).

(credit)
If the solvent capacity for $\ce{ZnCl2}$ is not (this much) temperature dependent, you equally could perform the filtration near room temperature (e.g., $\pu{432.0 g}$ $\ce{ZnCl2}$ per $\pu{100 g}$ of water at $\pu{25 °C}$ listed here).

To be more precise: What I would like to do is to filter the saturated solution of highly hygroscopic compounds at a temperature at which as much as possible of the compound will cram into solution such as to maximize the common ion effect on crystalizing impurities in the solution (so in the case of ZnCl2 the idea would be to saturate it at 90C or so, then swirl slowly for a moment to allow for impurities to agglomerate into fewer/bigger crystals and then pass the solution through a 1-4mcm filter to, hopefully, get a solution of ZnCl2 of much higher purity on the other side of the filter).

I've never handled such highly saturated solutions so I'm not sure if visquosity or so would be a problem in passing them through a very small pore filter. One thing I had in mind was to be using some coarser sand above the filter such that larger impurity crystals don't get through to the small pore filter. One issue you mention is indeed that partial vacuum on the receiving side of the filter would risk clogging the filter by precipitation of the hygroscopic compound (by evaporating solvent).

Positive presssure is a possibility to explore (though I haven't thus far because my glassware won't hold much positive pressure at all). If I would have equipment able to hold say 2-3 atmospheres of positive pressure, do you think that would be enough (and that the pore size wouldn't be a problem if there's no vacuum on the receiving side of the filter)?

If the intend is to purify $\ce{ZnCl2}$, then -- beside the anticipated scale (g, kg, t) -- it must be clear in advance what the nagging impurities (and their concentration) currently are, and what (after purification) would be a level good enough. Otherwise the work will not find an end.  It could be that the undefined impurities are equally (or better) soluble in the solvent, too. Equally, filtration would proceed from coarse to fine filters just to prevent blockage, too.  A suitable pressure depends on the material, yet in addition e.g., shape and dimension of the container, too.

Small scale preparative column chromatography uses a gentle pressure (perhaps 1 bar) top push the eluent across the column, but column's preferred diameter is below 10 cm. What is the purpose of the ZnCl2 after purification? What did you find in Amareggo / Perrin «Purification of Laboratory Chemicals»? Given this entry I would not wonder if it is some kind of sublimation instead of a (directed) crystallization. Of course one would start with a coarse filter.

There's about 150ppm of other metal ions in there. Probably lead, copper, cadmium, iron, silver, mercury, arsenic. Maybe also thallium, selenium, gallium, germanium, antimony, tin. If I could cut that down a lot by such a simple process as this it would be great. I think it could help with silver, lead and thallium more particularly.

(it's for perhaps 300g of ZnCl2)

I've read somewhere it could be purified by sublimation in a stream of HCl but given the BP of 732C I believe this must apply to minute amounts and there's no way I can put up a setup for this. Zinc as a metal has a fairly high vapor pressure as does HCl but there's nothing to say ZnCl2 would too and I couldn't find any data on the vapor pressure of ZnCl2. Also with such a high BP sublimation would inedvitably carry along some mercury and cadmium.

I've thought about recrystalizing in ethanol but (a) I don't have any reliable solubility data and I can't trust Wikipedia on the single data point it gives, (b) I'm afraid zinc ions would catalyze the formation of aldehydes in the alcohol.

If sublimation, or high vacuum distillation (example, seems easier to perform than for magnesium) is not an option (physics group in an university?), then this demanding specifications of purity are better delegated to chemical supplier selling the material with an certificate of analysis.  E.g., sigma sells this with ≥99.995% purity  (#429430, max. 50.0 ppm trace metal), or at 99.999% purity (#229997, max 15 ppm trace metals). Crystallization from solution typically is not directed enough.

When you write "not directed enough" do you mean that it's not  directed at specific impurities? Would it be efficient at bringing the 150ppm trace metals down to say 5ppm trace metals after a few rounds?

"Directed" as in "zone melting" with a discrete layer of molten material passing the solid sample, e.g., to purify semiconducting metals in microelectronics (open access example).  While a melting point of $\pu{290 °C}$ could be attained with some resistor wiring around an horizontal glass tube, I do not know if after hours of contact normal lab glass eventually would be eaten away from a melt of $\ce{ZnCl2}$ ...


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Why is PbI₄ not composed of lead(II), iodide and triiodide?
$\ce{Pb^4+}$ is reduced by $\ce{I-},$ so $\ce{PbI4}$ cannot exist with $\ce{Pb}$ in $+4$ oxidation state. But if I replace three $\ce{I-}$ with $\ce{I3-},$ then we can have $\ce{Pb^2+}$ similar to $\ce{TlI3}$ which exists with $\ce{Tl+}.$ Why does this not happen?

$\ce{Pb^2+}$ prefers $\ce{[PbI4]^2-}$.

@Poutnik Then do we have any compounds where the triiodide and iodide ion exist simultaneously (a mixed salt)?

No idea, I cannot confirm nor reject it.

Partly because I have never even heard of a triiodide salt with a 2+ cation. Such compounds would likely decompose to give the normal iodide salt plus iodine. Even with a heavy alkali metal such as caesium, the triiodide ion is in effect partly decomposed, with one bond becoming longer than the other as if you had a complex of $\ce{I2}$ with a normal (mono)iodide ion.

Does this answer your question? Why triiodide ion does not form ionic bond with cation having +2 charge?

@Poutnik and sometimes $\ce{[PbI3]-}$

@nilayghosh Sure, depending on $\ce{[I-]}$ :-) One cannot get one without another.


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Reacting Grignard reagent with esters to excusively produce a ketone
Adding 2 moles of grignard reagent to an ester, gives out 2 alcohols. But, if 1 mole of grignard reagent is added, will the reaction exclusively produce a ketone? How to prepare a ketone in a reaction only with esters and grignard reagent as starting materials?
CH2=CH2-C(=O)OCH3 + CH3MgBr -> CH3-CH2-CH=C=O + CH3OH
Is the above reaction possible?

I would presume the answer to be $$\ce{CH2=CH-CO-CH3}$$ which is a ketone

No, you'll get a mixture. If you want a ketone then the standard way is to react the Grignard with a Weinreb amide RCONMe(OMe)

@Waylander So, is it impossible to isolate the ketone from reacting grignard and ester alone, irrespective of the no. of moles?

In $\ce{CH3-MgBr}$, magnesium increases the electron density toward the methyl group a lot. While it is not a salt like NaCl, you may assume that when it comes into contact with water (intentional during workup), the methyl group will react like a Lewis base to abstract (like a Bronsted base) a proton from water to yield $\ce{CH4}$, rather than to yield MeOH. This is one reason why, beside Grignard reagents, other organometallic reagents which are less basic (in terms of $pK_a$) are equally in use today.  Note, acidity (thermodynamics) is different from nucleophilicity (kinetics).

If you recall general reactivity scales of aldehyde vs. ketone vs. ester vs.  amide against a nucleophilic attack on the $\ce{C=O}$ group, you recognize that the ketone is much more susceptible to the reaction of the Grignard than the initially submitted ester.  Thus the advantage of the Weinreb amide even if it may require some extra work.

@Boson  It is impossible to get a decent yield of ketone from Grignard + ester. As @ Buttonwood states, any ketone you form is more reactive then any ester that would still be present.


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What caused a colour change when I mixed two different dish soaps?
I recently purchased "organic" dish soap refill made from saponified coconut oil and potassium soap (full ingredient list). Instead of waiting for the old dish soap (Dawn) to be completely finished, I just poured the refill into the bottle.
The refill was a translucent yellow, and Dawn is obviously translucent blue. However, as soon as they made contact with each other, some chemical reaction clearly happened and the whole thing turned into opaque dark brown. The resulting liquid still has cleaning properties but doesn't seem to be equivalent to either of the soaps alone.
I tried to figure out what could have possibly reacted, but couldn't figure it out. Thought I'd try here.

As every painter is aware, mixing two colors can result in another color. It is doubtful any chemical reaction took place.

Yellow and blue makes green (not dark brown). Plus the change in translucency, consistency and cleaning property of the liquid points to a chemical reaction.

@EdV, DAWN dishwashing liquid is not very stable with respect to strong alkalis and bleach. I have smelled ammonia like smell from decomposition.

@M.Farooq All the more reason not to mix chemicals at home! At least it was not bathroom cleaning products. So the solution is to flush it down the drain.

@EdV some things are learnt the hard way by follies. It is always a bad idea of mix cleaning products or even soaps.


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What does this representation of organic aromatic compounds mean?
I saw this structure in some problem. What does this mean? How to express this in form of a simple structure

This is the cyclooctatetraene dianion. One way to get to it is to start with cyclooctatriene and remove one hydrogen ion each from the two $\ce{- CH2 -}$ groups. This is related to the cyclopentadienyl anion, the more common aromatic anion.

Image source: https://www.masterorganicchemistry.com/2017/05/17/frost-circles/

It represents a dianion structure that is, delocalisation of 2 negative charges around the ring. It's like extracting two $\ce{H+}$ from the structure of cyclooctatriene or adding two electrons to the structure of cyclooctatetraene.
That is how we represent the net resonance hybrid structure which is a combination of all the canonical structures we can make(which is what I assume you mean by 'simple' structures).


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In solid-phase peptide synthesis, why are the formation of aspartimides a problem but not "glutamides"?
I'm studying solid-phase peptide synthesis (SPPS) at the moment, and a common problem is the formation of aspartimides, which disrupt the synthetic process.
I understand how they form and why are they a problem. What I don't understand is why don't glutamic acid residues face the same problem? They have the same functional group, and a 6-membered ring is probably even more favoured than the 5-membered ring in an aspartimide.
Does anyone have any insight into this?

Since you understand how aspartimides (succinimides) form and why are they a problem in solid-phase peptide synthesis (SPPS), I'm not going to elaborate it further. Your main question is why don't glutamic acid residues face the same problem? Then, my question is who said they (glutamic acid residues) won't? Of course, they do. One example is discussed in Ref.1. You didn't recognized this cycle formation because the cyclic form is called Glutarimide Formation (contrast to aspartimides, which is actually a succinimide). The abstract states that:

[...] We found that the sequence of Glu(Gly)-OAll leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides. However, our results show that glutarimide formation is sequence dependent and can be inhibited by incorporating an amino acid like Lys(Boc) with steric hindrance from the protecting group. To prevent glutarimide formation, Ser(tBu) was used to replace the glycine in the GGG spacer adjacent to the residue of Glu, and a tailed cyclic RGD peptide, c[RGDfE(SGGKK-NH2)] was successfully obtained.

Initially, $\ce{c[RGDfE(GGGKK-NH2)]}$ was designed with $\ce{c(RGDfE)}$ linked through $\ce{Glu}$ to a tail consisting of a spacer of three $\ce{Gly}$ residues $\ce{(GGG)}$ and a linker of two $\ce{Lys}$ residues $\ce{(KK)}$. We found that the sequence of $\ce{Glu(Gly)-OAll}$ leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides (Note: $\ce{Glu(Gly)-OAll}$ is glutamic acid(glusine) $\alpha$-allyl ester):

The succinimide and glutarimide formations are not just common in SPPS. They can be formed under physiological conditions and can disrupt the biological processes (e.g., Ref.2,3).
References:

Junmin Zhu and Roger E. Marchant, "Solid-Phase Synthesis of Tailed Cyclic RGD Peptides Using Glutamic Acid: Unexpected Glutarimide Formation," J. Peptide Sci. 2008, 14(6), 690–696 (DOI: https://doi.org/10.1002/psc.975).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Spectroscopic Identification of Cyclic Imide b2-Ions from Peptides Containing Gln and Asn Residues,” J. Am. Soc. Mass Spectrom. 2013, 24(8), 1228–1241 (DOI: https://doi.org/10.1007/s13361-013-0661-6).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Effect of the Asn side chain on the dissociation of deprotonated peptides elucidated by IRMPD spectroscopy,” International Journal of Mass Spectrometry 2013, 354-355, 70-77 (DOI: https://doi.org/10.1016/j.ijms.2013.05.005).


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Entropy change and heat transfer in non-ideal heat engines
Please help me, a mathematician, to make sense of entropy. I know it's a topic with about a million questions already, so I understand if this gets ignored. Yet for the life of me, I cannot make sense of the previous answers.
So apparently the definition of entropy change in a system is $\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$ and this is a state variable. I also know that for irreversible paths, you have to find a reversible one that links the same thermodynamic states to calculate the entropy. Yet in my professor's notes it says
$$ \mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$
(is this even still an entropy difference? The notation $dS$ instead of $\Delta S$ confuses me) and elsewhere, talking about the Sterling process, it also says
$$ \oint_{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases } $$
so it really must be talking about the entropy change in the system (the machine) and not about the total entropy change of system + surroundings. But how can this be when for a system in a circular process the difference in any state variable is 0?

Pretty sure it should be $>0$, not $<$.

Clausius inequality. A bad fridge heats up the room more. https://coecs.ou.edu/Feng.Chyuan.Lai/thermoweb/Lecture13/lecture13.htm

@buckthorn this is just for the heat engine, not the reservoirs.

@Ritam_Dasgupta It should be < as in the Clausius inequality.

@Karsten Theis isn't entropy due to an irreversible process always supposed to increase? From the Clausius inequality itself?

@Ritam_Dasgupta For the entire process, this is true, it has to increase. For selected compartments, it may increase, decrease, or stay the same. In the heat engine case, it increases for the cold reservoir, decreases for the hot reservoir, and stays the same for the engine (after a complete cycle).

So apparently the definition of entropy change in a system is $\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$ and this is a state variable.

This definition is for the surrounding if it is boring, e.g. a temperature bath. The $_\mathrm{rev}$ subscript is just for the temperature bath, i.e. there could be irreversible processes elsewhere (e.g. in the system).

Yet in my professor's notes it says $$\mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$

The inequality is correct. Equating it to $\mathrm dS$ is not. The entropy change for a cyclic process is zero because entropy is a state function.

[OP...] elsewhere, talking about the Sterling process, it also says
$$ \oint_\mathrm{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases } $$

This is the Clausius inequality. If a heat engine is not ideal, it will not extract the maximal energy possible from the heat transfer from hot to cold reservoir. The claim about the entropy of the machine is incorrect. Indeed, the entropy of the machine stays the same (use your argument of taking a reversible path from one cycle to the next, i.e. do nothing). Overall, the entropy increases because the entropy loss in the hot reservoir is no longer matched by the entropy gain of the cold reservoir. Thus, we are not extracting the maximal work in the process.
I found this explanation helpful. They used $\delta Q$ instead of $\mathrm d Q$ to remind us that there is a path-dependency.


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Are all hydrolysis reactions non-redox reactions?
Are all hydrolysis reactions always non redox reactions? For example:
$$\ce{N2O3 + 2H2O -> 2HNO2}$$
The main reason I say that this statement is correct is because oxygen and hydrogen generally show oxidation states of −2 and +1 respectively when bonded to other atoms.
This means that any change in oxidation state needs to take place between the nitrogen atoms involved in the molecular structure and this is very rarely seen.
Hence most (if not all) hydrolysis reactions are non-redox in nature. Is there a mistake in my understanding?

Well, that would depend on what's you'll call hydrolysis - I wouldn't call reactions of oxides with water like that.

Hydration is sure different then hydrolysis, like, an opposite.

Exception: Hydrolysis of phosphorus pentachloride

Hydrolysis refers to any compound reacting with water, including cases where the water acts as an oxidizing or reducing agent.  Magnesium silicide, for instance, reduces water leaving both hydrogen and oxygen from the water in negative oxidation states:
$$\ce{Mg2\overset{-4}{Si} + 4\overset{+1}{H}_2O -> 2Mg(OH)2 + \overset{+4}{Si}\overset{-1}{H}_4}$$

Thanks @andselisk, although I still prefer Roman numerals for oxidation states in formulas to avoid confusion with ion charges.  Or is that not convention?

ONs are Roman when used in nomenclature, otherwise it's recommended to use Arabic numbers in order to avoid confusion with valency, iodine and vanadium symbols. Also, fractional ONs are quite common: how do you type in these with Roman numerals? You cannot confuse charge number with ON because ONs are placed above the element symbol and have preceding sign. Charges are placed in the upper right corner and are followed by the sign. For the IUPAC reference, see, for example, https://doi.org/10.1515/pac-2015-1204.

@andselisk Apologies for bothering you. Are you certain about your claim here? I ask for three reasons. 1) In the link (or PDF) you provide, the terms "arabic",  "above" give 0 hits. In the PDF itself, "roman" gives 1 result: "Similarly, OS as a nomenclature symbol is in roman numerals." But it does not say anything about writing OS in other situations. 2) In the IUPAC Green Book (p 50, 52), OS is explicitly written with Roman numerals in the upper right corner. 3) In IUPAC Red Book, recommendations for formulae (e.g., p 65–66), it is also put the upper right; also says to avoid fractional OS.

@andselisk References as pictures: see here.

No. One counter example for the proposed statement would be the case of hydrolysis of dinitrogen tetroxide:
$$\ce{N2O4 + H2O -> HNO2 + HNO3}$$
Here, the nitrogen goes from a $+4$ state to a $+3$ and $+5$ in nitrous acid and nitric acid, respectively.

Sucks to be whoever downvoted this.


text:
What is the IUPAC Name of [Cr(NH3)5Cl]Cl2.NH3?
What is the IUPAC Name for the complex $\ce{[Cr(NH3)5Cl]Cl2.NH3}$?
I can't even find it through the internet and also by employing a find in "IUPAC's Red Book".

You are not supposed to "find a compound" in the Red Book, you are supposed to apply nomenclature rules from this book to your compound.

@andselisk Fine but I actually searched for "monohydrate" as there's a similar complex like it(https://chemistry.stackexchange.com/q/128069/102878).

This answer is similar to this answer on $\ce{[Co(NH3)5(H2O)]Cl3}$, Thanks to Loong (Faded Giant) for the references.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore the systematic name of the co-ordination entity $\ce{[Co(NH3)5Cl]}$ would be "pentaamminechlorido" since the ammine ligands are named before the chlorido ligands according to alphabetical order.
The ionic composition of the complex salt $\ce{[Co(NH3)5Cl]Cl2}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

According to the above rules, the oxidation state of cobalt in the complex can be written in two ways - as (III) or (3+).
Similarly the chloride ions can be named in two different ways - chloride or dichloride.
Also, see:

IR-5.5 NAMES OF (FORMAL) ADDITION COMPOUNDS
[...]
The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space. The order of names of the individual components is, firstly, according to the increasing number of the components and, secondly, alphabetical. As the only exception, the component name ‘water’ is always cited last. (Note that this represents a change from the rule in Ref. 2 according to which the component names must follow the order given by the formula.) The numerals in the descriptor appear in the same order as the corresponding component names.
[...]

$\ce{CaCl2.8NH3} \equiv \quad$  calcium chloride—ammonia (1/8)

Therefore the systematic names for the compound $\ce{[Co(NH3)5Cl]Cl2.NH3}$ would be:

pentaamminechloridocobalt(III) chloride—ammonia (1/1)
pentaamminechloridocobalt(3+) chloride—ammonia (1/1)
pentaamminechloridocobalt(III) dichloride—ammonia (1/1)
pentaamminechloridocobalt(3+) dichloride—ammonia (1/1)