Can naphthenic acid be ionized in n-dodecane? I am not very clear about the difference between ionization and dissolution. I know that naphthenic acid can dissolve in n-dodecane, but I do not know if it is ionized in n-dodecane.
Thank you!

It would not ionize (though the correct term is that it won't dissociate, ie. the O-H bond in the acid won't cleave giving carboxylate anion).

It's because n-dodecane has no lone electron pairs, so it can't complex the hydrogen cation. Unless we're talking about extreme vacuum or plasma, there are no free $H^+$, it always exist in a form of a complex cation with a molecule of solvent or the like. But such solvent must have some electron donor: lone pair, double bond, etc. to form the complex.

The only other possible way here would be self-dissociation, where one of the substance molecules accept the hydrogen carion from the other one, but here it is also borderline impossible as the additionally protonated acid molecule would be very unstable and immidetely such auto-dissotiation would reverse.

The fact that the naphtalenic acid dissolves in n-dodecane is due to low polarity of both, that is acid molecules don't bind strongly to each other and n-dodecane molecules don't bind strongly to each other.


What do these labels for molecular electronic states mean? What do these symbols mean in excited states?

$${}^2\!A_2,{}^2\!B_1,{}^4\!A''$$

I am confused with these representations, found in the abstract of this paper. I think it is kind of a representation of electronic excited states. The left superscript means spin multiplicity ($2S+1$). What do the other parts of these symbols mean, and how can I write these symbols with other molecules?

As stated in this answer, these are irrep (irreducible representation) labels for molecular symmetry point groups. In the context of chemistry, point groups are usually introduced when learning about structural symmetry (atoms and bonds). This is a broad topic with many technical points, too many for a single answer, so I won't cover the basics but hopefully the few things needed to make the connection between what you read in a book and what's present in this paper.

The key is recognizing that the structural symmetry of molecules leads to symmetry in their molecular orbitals. Take the water molecule as an example. Water's molecular structure can be described at best by the $C_{2v}$ point group, so we say it has $C_{2v}$ symmetry. Composing its molecular orbitals from atomic orbitals is actually an exercise in symmetry, as only atomic orbitals of identical symmetry may combine. Here is the LCAO-MO diagram for water, with the oxygen 1s orbital left out due to its extremely low energy; it won't mix with the hydrogen AOs.

Here is the character table for the $C_{2v}$ point group.

AOs of identical symmetry (they transform identically for any given symmetry operation) are shown in the same color. A few things are worth pointing out:

s orbitals are perfectly symmetric (a sphere), so they will always transform as the totally symmetric irrep of a molecular point group. In $C_{2v}$, that's $A_1$. Once you learn what the operations $E$, $C_{2}$, $\sigma_{v}(xz)$, and $\sigma_{v}(yz)$ do, you can convince yourself that the $\sigma$ combination of $\ce{H2}$ AOs is also completely symmetric under those operations, so it also belongs to the $A_1$ irrep. The p orbitals are figured out from the 3rd column, d from the 4th.
The 2s and $\sigma$ AOs combine to give the $3a_1$ MO. The "3" is because it's the 3rd such MO with $a_1$ symmetry. The symmetry label is in lowercase because it for a 1-electron state. Symmetry labels are in uppercase for many-electron states, such as the total wavefunction, as you see in the linked paper. This is a matter of convention, nothing more.

Neutral water is a closed-shell molecule; there aren't any unpaired electrons, so it has singlet spin multiplicity. How do you determine the symmetry of the total wavefunction? A Slater determinant can help, or even just the Hartree product part of it. Take a fictitious system with 4 electrons in 2 orbitals. The Hartree product will look something like

$$
\Psi_{\text{Hartree}} = [\phi_{1}(1)\alpha(1)][\phi_{1}(2)\beta(2)][\phi_{2}(3)\alpha(3)][\phi_{2}(4)\beta(4)]
$$

This is a product wavefunction that might be made from an electron configuration such as $1s^{2}2s^{2}$, $1s^{2}2p_{x}^{2}$, $4p_{x}^{2}4p_{z}^{2}$. Each of these MOs has a certain symmetry; the total symmetry is given by the product of the individual symmetries. For this step one needs the point group product table (again for $C_{2v}$):

$$
\begin{array}{|l|l|l|l|l|}
\hline
            & \mathbf{A_1} & \mathbf{A_2} & \mathbf{B_1} & \mathbf{B_2} \\ \hline
\mathbf{A_1} & A_1          & A_2          & B_1          & B_2          \\ \hline
\mathbf{A_2} & A_2          & A_1          & B_2          & A_1          \\ \hline
\mathbf{B_1} & B_1          & B_2          & A_1          & A_2          \\ \hline
\mathbf{B_2} & B_2          & B_1          & A_2          & A_1          \\ \hline
\end{array}
$$

If the electron configuration of the system is indeed $4p_{x}^{2}4p_{z}^{2}$, this results in a total wavefunction symmetry of

$$
B_{1} \times B_{1} \times A_{1} \times A_{1} = A_{1}
$$

The spin components don't contribute, only the spatial components. Note that in any product table, the product of any irrep with itself is always the totally symmetric representation in that group, hence the diagonal being all $A_{1}$. Product tables are also symmetric, since irreps are commutative under multiplication. Since this is a singlet, the ground-state wavefunction can be specified as $^{1}\!A_{1}$. For a closed-shell system (RHF), due to the pairs of electrons always leading to $A_{1}$ symmetry, it will always have a total state symmetry of $^{1}\!A_{1}$. If you have an open-shell system of any kind in a ground or excited state, with either an even or odd number of electrons, the total wavefunction symmetry will be determined by electrons in those orbitals alone, since the product between fully-occupied orbitals will always be of $A_{1}$. In the simplest case of one unpaired electron, the total symmetry of the ground state will be the symmetry of the SOMO.

Now, on to the paper. The most important part for our understanding of the state label assignments is Table 3:

All the systems considered have either $C_{2v}$ or $C_{s}$ symmetry. Note how going from a point group of higher symmetry ($C_{2v}$, 4 irreps) to one of lower symmetry ($C_{s}$, 2 irreps) leads to states collapsing into fewer irreps due to fewer symmetry operations that can distinguish them. For the plots (here the last row of Figure 6),

there is the isosurface of the molecular orbital (really a natural orbital), the orbital number and symmetry, and the natural orbital occupation number (NOON) for that NO in that state.

There are a number of good Dover publications (1, 2, 3) for learning about group theory in chemistry and spectroscopy that are highly recommended, since the primary peer-reviewed literature isn't usually a good resource for learning this. An excellent resource for character tables is WebQC.


Why isn't ΔG equal to zero for this equilibrium reaction? In the reaction $\ce{E(g) <=> P(g)}$ at 25 °C, equilibrium is reached when the pressure of the product is 0.100 that of the reactant. What is $\Delta G$ in joules?

Using $\Delta G = RT\ln K$, my answer is $-8.31 \times 298 \times \ln (1.00) \ \text{J}$.
However, the answer is $-8.31 \times 298 \times \ln (0.100) \ \text{J}$. Wouldn't this mean that $\Delta G > 0$, therefore favoring the reactants? (As an aside, why would the reaction even be shifted to the left since it has a higher pressure?)

You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium.
And this quantity can very well be different from zero.
For non-standard conditions the Gibbs free energy of reaction is

\begin{equation}
 \Delta G = \Delta G^{0} + RT \log K
\end{equation}

and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.

For a derivation see this answer of mine.


What is the energy of the London dispersion force holding together two atoms of the same element? I read somewhere that the energy is around $U=-\frac{3\alpha I}{4r^6}$ where $\alpha$ is the polarizability and $I$ is the ionization energy but I'm obtaining way too large energies for a lithium crystal.

The shortest distance between two lithium atoms in the crystal is $304 pm$

The first ionization energy of lithium is $520.2 kJ/mol$

I've found various polarizability constants online in papers but I can't really figure them out or the units they are using.

Also, I suspect that I want the dispersion force between $\ce{Li^+}$ ions and not neutral $\ce{Li}$ atoms.

Polarisability as used here has dimensions of volume. The London formula is $$U \approx -\frac{3}2\frac{I_aI_b}{I_a+I_b}\frac{\alpha _a \alpha _b}{r^6}$$  for two species a and b so you just need to square $\alpha$.


Are there ethanol and methanol analogues with more than one hydroxyl group? Ethanol and methanol are ethane and methane derivatives, respectively, with one hydrogen replaced by an hydroxyl group. Are there molecules based on either where more than one hydrogen atom has been replaced by an hydroxyl group? If so, what are they called?

Generally speaking, they're called diols when bearing two hydroxyl groups. If they're on the same C, this is called a geminal diol, if they are on neighboring Cs, they're called vicinal diols. The C on geminal diols has a higher oxidation state than that of an alcohol, i.e. which is why they could be seen as relatives of an aldehydes (or ketones).
In a diluted aqueous solution, formaldehyde $\ce{H2C=O}$ actually predominantly exists in its geminal diol form $\ce{H2C(OH)2}$. This is no longer true for 1,1-ethanediol (the hydrated form of acetaldehyde).

The simplest vicinal diol 1,2-ethanediol is better known under its trivial name ethylene glycol (a common ingredient of anti-freeze formulations).

Triols with the hydroxyl groups on the same C (with the oxidation state of carboxylic acids) are generally unstable.


How to extract wavefunction from GAMESS output? Is there a way to extract the molecular wavefunction from GAMESS output?

I know there are packages that can extract the eigenvalues or the exponents and show them as a table but that is not of much help as anyone can easily find those values in the output file by looking at the numbers.

What I am looking for is the complete wavefunction equation. Is there such a package or something available. Something like this simple extractor I have written for STO-3G (this is just a very primitive one that only works for small STO-3G outputs ) :

I'm not too familiar with the input structure of GAMESS. But, I found and input in which you can find how to call the AIMPAC keyword GAMESS input with AIMPAC keyword and this is how wfn file looks like. In principle, when you indicate the AIMPAC keyword, the GAMESS itself print the wave function in a file with wfn extension at the same folder of the input/output.

After you get the wfn file, you have to integrate it with (I recommend) 1. AIMPAC or 2. MULTIWFN (Both are free).


Why don't ions have one equilibrium potential? (Nernst equation) I know that equilibrium potentials are dependent on the ratio of ion concentrations inside and outside of the cell and temperature. I also know that the equilibrium potential is reached when there is no net movement of ions across the membrane, i.e. the chemical and electrical gradients are equal. Why don't the ions move until they reach the same ratio of inside and outside concentration every time? For example, why doesn't ion "x" always diffuse until the ratio of inside to outside concentration is 15 to 1? Thanks in advance!

Good question.  The answer (as I understand it) is that if there were only a single type of permeable ion, then there would be only one equilibrium ratio, and it would depend on the transmembrane potential.

However, in real life, there (a) is usually more than one membrane permeable ion, and more importantly (b) the transmembrane potential is not a fully independent variable but is affected by the transmembrane permeation of all those ions.

Imagine sovling a single instance of the Nernst equation when you know the electrical potential $\Delta \psi$:

$\frac{\Delta G}{F}=\text{ion motive force}=0=-m \Delta \psi+ RT\ln{\frac{[X_{out}]}{[X_{in}]}}$

If you know $\Delta \psi$ (and $R$ and $T$) and you are at equilibrium (so $\Delta G$ is 0), then the ion concentration ratio will always go to the same value, as you suggested in your question.

But now suppose there are many equilibrating ions, say sodium, potassium, protons, and chloride.  And suppose you don't know $\Delta \psi$.  Now, there is a Nernst equation for each species, as well as Gauss's law that $\nabla \centerdot \bf{E} = \frac{\rho}{\epsilon}$.  And probably also total material balances like $V_{out}[X_{out}] + V_{in}[X_{in}] = N_{tot}$ for each ion.   To fully solve the system to determine equilibrium, all equations have to be solved at once!  I hope you can see that (a) doing this is complicated and (b) if parameters like the volume of the cell $V_{in}$ or even more likely the total amount $N_{tot}$ of ions in the system changes for one ion, it could completely change the ratio $\frac{[X_{out}]}{[X_{in}]}$ for other ions.  That's why in general, the ratio is not the same every time.


Is there a systematic naming system to name carbonyl derivatives? I had asked a question earlier about how the “$\ce{-COOCl}$” functional group would be named.

I had received a satisfactory answer that helped me completely understand how that compound was named.

However, I have a much broader question in mind now.

Consider the "$\ce{-COOH}$" functional group. We know that it's called a carboxylic acid group, however I used to think, why wasn't ethanoic acid simply called as a 1-hydroxyethanal? I now know that "1-hydroxyethanal" has properties vastly different from both aldehydes and alcohols, and as a result, was named differently to acknowledge this difference. Many acid derivatives have been named (and behave) differently, as a result of the $\ce{C=O}$ group nearby.

However, these may not be the only molecules that are affected by the nearby $\ce{C=O}$ group. My earlier question had dealt with the $\ce{-COOCl}$ functional group (not to be confused with acyl chloride $\ce{-COCl}$). However, there are a lot more:

Of course, I did try to name these compounds myself.

Cyanide is sometimes treated as a pseudohalide, so my guess is ethanoyl cyanide.
It may be an anhydride of hydrogen sulfide and ethanoic acid, but I'm not very sure.
Loong gave me an interesting way to name certain "anhydride" compounds in my previous question. This looks like an anhydride of hydroxylamine and ethanoic acid. So according what Loong told me, my guess is ethanoic hydroxylamine anhydride.
This one look's rather crazy to me. Had this been posessing a $\ce{C-O-S}$ instead of a $\ce{C-S}$, this would have been an anhydride of sulfuric acid and ethanoic acid. But it's not, so I have no clue on how to name it.
This one looks very familiar to an amide, infact a phosphorus analog of amide. For some reason, I feel like calling it ethanephosphide, although I don't know why.
Like cyanide, even azide is considered a pseudohalide, so my guess is ethanoyl azide.

Please keep in mind, I am not asking for the explanations to the naming of all the six compounds I showed above, although it would be great if I could get the names alone. Rather, what I mean to ask is is there any simplified nomenclature that can deal with all these types of compounds in a simple way? Although I only showed six, I can go on as long as my imagination permits.

The names (2 of 6) are rather correct (but note that acetyl and  acetic names are preferred over ethanoyl and ethanoic), except following:

$\ce{Ac-SH}$: ethanethioic acid, more specifically ethanethioc S-acid (since there are two possibilities of “monothiation”)

  P-15.5.3.4.1

  (…)

  $\ce{CH3-CO-SH}$
  thioacetic S-acid
  ethanethioic S-acid (PIN)

  $\ce{CH3-CH2-CS-OH}$
  thiopropionic O-acid
  propanethioic O-acid (PIN)

$\ce{Ac-O-NH2}$: Anhydride name cannot be used as hydroxylamine is not a name of acid. It's also not a recognized acid functional group, so the name probably should be regarded a substituted ketone, with preselected prefix aminooxy (P-68.3.1.1.1.2), 1-(aminooxy)ethan-1-one. (Although pubchem entry lists wrong name: amino acetate; O-acetyl hydroxylamine is acceptable I think.)
$\ce{Ac-SO3H}$: “α-oxosulfonic acid“, 1-oxoethane-1-sulfonic acid
$\ce{Ac-PH2}$: phosphide as alleged analogy to amide (with different meanings in inorg. and org. chem.) is not used, the name should be 1-phosphanylethan-1-one (or acetylphosphane).

  P-64.1.2.1 Pseudoketones

  (…)

  $\ce{H2P-\overset{1}{C}O-\overset{2}{C}H2-\overset{1}{C}H3}$
  1-phosphanylpropan-1-one (PIN)
  propanoylphosphane


The reactions of grignard reagent Question

  Find y in this question :

The $\ce{ -CH2-Cl}$ below according to me will not take part the reaction (or please correct me) thus my answer is y=2 but actually it's y=3

The third Grignard reagent molecule could extract a proton from the methylene group which is in an electron withdrawing environment.  This type of enhanced acididity on carbon is referred to by the term "carbon acid" .

While this reference deals with carbonyl groups, conjugation with a phenyl ring can do the same thing to the methylene group.

We do not see any change of the methylene group in the final product because it's just deprotonated, and the acid workup brings the proton back.  We would, however, see the third Grignard reagent being protonated to its parent hydrocarbon, in this case methane which bubbles off.


Does sanding the oxide layer of a magnesium sample break bonds? We all know that if we keep a piece of magnesium metal exposed in open air, an oxide layer forms on it. This layer prevents further oxidation of the bulk metal. If we then rub it with sand paper, the magnesium metal can react again because the oxide layer is removed. So my question is:

Does that mean abrasion with sandpaper can break the oxygen bonds with magnesium?

I upvoted the question because it is actually very insightful. We may think of simply removing the MgO layer, and voila, there is bare magnesium. I think not.

If you took bare Mg (no oxide at all) in a vacuum, and dropped powdered MgO onto the Mg, then exposed the metal to an argon atmosphere, I think you could blow off the oxide powder because there would be no significant adhesion.

In other words, the unreactivity of the oxide-coated magnesium arises because the oxide coat is strongly bonded to the metal. This transition from metal to metal oxide occurs over a few atoms and is probably very irregular. After all, even the surface of the metal won't even be flat to within 1 or 2 atoms.

Visible scratches on the metal after sandpapering prove that some mechanical effect has removed metal. If you remove metal, you must have broken some metal-metal bonds; and in the turmoil, at least some metal - oxygen bonds must also have been broken - even tho they might soon reform with O2 in air.

A similar, but chemical, film removal occurs when stainless steel, which is likewise protected by a very thin oxide coat, is treated with hydrochloric acid. The film is removed, and the steel corrodes. Because the reaction is so thorough, removing the whole oxide coat and a few atomic layers of metal as well, we don't stop to think about the transition layer. Different thicknesses can produce different colors as well as different degrees of protection.


Photometric titration If we have  $\ce{A + T -> P}$
where $\ce{A}$ is the analyte, $\ce{T}$ is the titrant, and $\ce{P}$ is the product,
photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

When stable complexes formed such as $\ce{Bi}$-EDTA, they will not absorb the radiation (or in small amounts), but when unstable complexes formed such as $\ce{Cu}$-EDTA, they will absorb highly. The same goes also for the acids. Strong acids will absorb much less the electromagnetic radiation than the weak acids.
The reason for this is that strong acids are completely dissociated.

Why does that indicate the photometric titration ?

photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

Almost ;-)

Regardless whether you monitor a complexometric titration by eye or with the help of a spectrometer, it is a good idea to use an indicator, such as Eriochrome Black T. Then, it doesn't matter, if or how  strong the EDTA complex of the metal absorbs!

Instead, you monitor the absorption of the indicator, that is liberated from the metal-indicator complex upon addition of EDTA.


Validity of rationale for balancing chemical equations The reason given for balancing chemical equations is the law of conservation of mass. They say that we balance equations with keeping in mind that atoms of every element included in the reaction must be equal on both sides of equations. And they just add some coefficients for this purpose.

But I apprehend why is the amount of substance not brought under consideration while doing this?

For example when we write,

$$\ce{H2 +O2->H2O}$$

That is; until we don't know how much of $\ce{H2}$ and $\ce{O2}$ was included in the reaction, how can start to balance it? How can we decide that atoms are not equal on both sides?  However, the molecular formulas given above are only representing the substances.

When you write the equation

$$\ce{H2 + O2 -> H2O}$$

you are implicitly saying that the stoichiometric coefficient of every substance is $1$. It is like writing the equation $a + b = c$. By not having coefficients in front of $a$, $b$, or $c$, you are implying that $1a + 1b = 1c$ and not $2a + 3b = 4c$ or anything like that.

Obviously that cannot be true for the reaction above because if you react one $\ce{H2}$ molecule and one $\ce{O2}$ molecule, you get one $\ce{H2O}$ molecule and one leftover oxygen atom that doesn't know what to do with itself. The only balanced equation that does not lead to stray atoms is

$$\ce{2H2 + O2 -> 2H2O}$$

or any integer multiple of that, i.e.

$$\ce{$(2n)$H2 + $(n)$O2 -> $(2n)$H2O}$$

Humans prefer to work with small integers or half-integers, so $n$ is usually either chosen to be $1$ or $1/2$.


How to convert from w/w% to molarity? After reading a number of explanations about it after googling, I'm a bit confused on this. Some explanations are doing it using a solution density, others aren't, though most are. On one site it's showing different densities for a solution of a compound at different w/w% values. I'm now very lost as to how you convert from a w/w% to molarity. Can someone clarify?

Weight percent is essentially the same as (I arbitrarily picked grams as the mass unit):

$$
\%~\mathrm{w/w~X} = \mathrm{g~X\over g~solution}
$$

In order to get to the $\mathrm{mol\over L}$ units of molarity, you have to convert by multiplying by the solution density, $\rho$, and dividing by the molecular weight of $\mathrm X$, $M_\mathrm{X}$:

$$
\mathrm{g~X\over g~solution} \times \stackrel{\left(1\over M_\mathrm X\right)}{\mathrm{mol~X\over g~X}}
 \times \stackrel{\left(\rho\right)}{\mathrm{g~solution\over L~solution}}
 = \mathrm{mol~X\over L~solution}
$$

One key thing to remember is that $\rho$ is the density of the solution, not of the pure solvent. If you're working with dilute solutions (below $\sim\!100~\mathrm{mM}$ for aqueous) these two densities will be similar, but in concentrated solutions they will diverge.


On the molybdenum(III) oxide-catalysed oxidation of alkanes to carbonyls The following reaction is specified in my notes;
$$\ce{CH4 ->[Mo2O3, \Delta]HCHO}$$ The intermediate steps involve the dehydration of an alcohol to yield the final product.
Wouldn't this only work if there's acid in the reaction mixture? Or is it that water functions as a proton donor, inducing the protonation of the hydroxyl and causing the dehydration?

The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow.  Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.
On industrial scale, formaldehyde is prepared from methanol by an  heterogeneous redox-reaction.  The oxygen needed may come from oxygen gas, or simply air.  Thus, the last step known as Formox process is
$$ \ce{2 CH3OH + O2 -> 2 CH2O + 2 H2O} $$
(from wikipedia)
$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.


Hydrogen peroxide: catalyst changes colour during peroxide decomposition When adding iron(III) chloride to hydrogen peroxide, the hydrogen peroxide decomposes.

Iron(III) chloride is a catalyst.

But when it is first added, it changes its colour to dark brown, then near the end it turns to orange.

  Why does the colour change?

I heard that it has to do something with reaction intermediate. Correct?

It is seen on: iron(III) chloride

Also, the colour changes when different catalyst is used like iron(III) nitrate and potassium iodide potassium iodide, iron(III) nitrate

Recently, I answered your question and relevant information can be found from that answer.

At first, I thought that the brown colour is due to ferrous chloride that formed during the course of reaction but since the reaction is occuring in a aqueous solution, it should be green in colour. Ferrous chloride is brown in anhydrous form. So I thought to be either due to formation of a complex, iron(II, III) oxide($\ce{Fe2O3}$), or iron(III) oxide-hydroxide($\ce{FeOOH.nH2O}$) or a mixed salt of oxide-chloride i.e. iron oxychloride($\ce{FeOCl}$).

The following information is from the paper 2 of that answer:

  According to the mechanism proposed by KREMER and STEIN, an
  intermediate oxygen complex of iron with oxidation number +V is
  primarily formed by the reaction of $\ce{Fe^3+}$ with $\ce{H2O2}$.
  This complex reacts with another $\ce{H2O2}$ molecule to water and
  oxygen thereby reforming $\ce{Fe^3+}$.

  $$\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> [Fe^{V}O]^3+ + H2O ->[H2O2] Fe^3+ + 2H2O + O2}$$

  According to the mechanism proposed by HABER and WEISS the $\ce{Fe^3+}$ ions
  initiate a radical reaction, after which the chain reaction consumes
  the hydrogen peroxide. This mechanism can explain the high reaction
  rate very well.

  Chain initiation:  $\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> Fe^2+ + HOO. + H+}$

  Chain propagation: $\ce{Fe^2+ + H2O2 -> Fe^3+ + 2OH.}$
  $\ce{Fe^3+ + H2O2 + OH. -> Fe^3+ + HOO. + H2O -> Fe^2+ + H+ + O2 + H2O }$

That intermediate oxygen complex of iron with oxidation number +V maybe responsible for the brown colour of solution.

As for the potassium iodide, the color you are seeing is basically food coloring. Various food coloring  is used to make the foam produced colorful. If food-color is not used, still color is seen. That color is basically color of pure soap foam. This elephant toothpaste experiment not uses food-color but still it has color that is color of soap-foam.


Do carboxyl groups behave like bases and amino groups behave like acids when peptide bonds are created? Peptide bonds are synthesized when the carboxyl group of one amino acid molecule reacts with the amino group of the other amino acid molecule, causing the release of a molecule of water.

Water molecules consist of a hydroxyl from a carboxyl group and a hydrogen from an amino group.

And that seems a bit strange to me, since carboxyl is an acid and should give a hydrogen (instead of hydroxyl) and amino groups are bases and should get proton (and not give a hydrogen).

So for me it looks like in creating this bond, the carboxyl group behaves like a base and amino group behaves like an acid. Is that correct?

Yes, an amino acid consists of a carboxyl group, commonly called carboxylic acid, plus the amino group. And yes, a carboxyl group can act as an acid, i.e. release its proton to give a carboxylate. Conversely, an amino group can act as base, its free electron pair accepting a hydrogen to give an ammonium derivate. That is in fact what happens in aquaeous media or in the solid state of amino acids: the carboxylate protonates the amino group creating a zwitterion.

But these are only one type of reactions both groups are capable of performing. In the case of the amino group, as long as it is not protonated (i.e. still has an unbonded electron pair) it can attack nucleophilicly. Maybe you have heard of $\mathrm{S_N}2$ type reactions, that are rather similar.

The nucleophilic attack of the nitrogen could happen on the carboxylate’s proton — but then we wouldn’t be gaining anything, since it would just be the same acid-base reaction. (And this wouldn’t be considered a nucleophilic attack, usually.) Instead, the nitrogen needs to go to the carbonyl group — which happens to be the most positive carbon in an amino acid, so should be a great target for a nucleophile.

The initial result, as I tried to show in the scheme below, is a tetrahedral carbon with a negative charge on the ex-carbonyl oxygen and a positive charge on the nitrogen. If one of the nitrogen’s protons jumps across to the ex-hydroxy oxygen, that one gets positively charged, and then electron pairs move, we release water and gain our peptide bond.

(The picture is really only the basic mock-up and no real effort put into it. It neither exactly matches the biological peptide bond mechanism, nor does it match the mechanism that a chemist would use in the lab. But it conveys the basic picture.)

So while you did technically use the basicity (or postivie-charge loving) of the nitrogen, you see that the acidity of the carboxyl group plays no role in the reaction itself.


Is the compound CM-7 in Red Comet fire grenades azeotropic dichloromethane? Red Comet Manufacturing changed the liquid in their fire grenades from carbon tetrachloride to something they referred to as CM-7 in 1955.

Is CM-7 the same thing as azeotropic dichloromethane? If not, what is the composition of CM-7? What are the gases given off by CM-7 when it is exposed to heat?

Passive fire suppression systems by Red Comet included a glass bulb with colorless transparent liquid, and had a sticker:

  MC-7 This fire extinguisher contains azeotropic chlormethanes®

Registered trademark symbol already signifies that marketing already prevailed at that time. Indeed, among chlormethanes $\ce{CH_xCl_{4-x}}$ chloromethane $\ce{CH3Cl}$ is flammable. Dichloromethane $\ce{CH2Cl2}$ is not combustible itself, but supports burning and flashes. Both chloroform $\ce{CHCl3}$ and $\ce{CCl4}$ are not flammable, but don't form azeotrope and still would produce phosgene $\ce{COCl2}$ when react with atmospheric oxygen. Also, the company denied presence of $\ce{CCl4}$ in MC-7 (Kiplinger's Personal Finance, Oct 1956, p. 48):

  CM-7, which has replaced carbon tetrachloride in all our grenade extinguishers. CM-7 does not produce toxic gases. It converts toxic gases created by the fire itself into additional fire-extinguishing chemicals.

As it seems like Red Comet doesn't hold any patents for inventions in chemistry, most likely it's a hydrochlorofluorocarbon that's been known at that time.

These days MC-7 extinguishers are sold as vintage items, and uniformly it's claimed (e.g. here and there) that the liquid component is 1,1,2-Trichloro-1,2,2-trifluoroethane, which at that time indeed has been studied as a very unreactive (even upon heating) vaporizable liquid agent (e.g. Fire Research Abstracts and Reviews, Vol. 3
1961, p. 148).

Starting 1960s, several acts in Montana, Chicago, and later across US no longer permitted usage of both tetrachloromethane and MC-7 as fire-extinguishing chemicals. Halons took over.


How many bonds can Hydrogen make? Recently I saw a video which told me that Hydrogen has a valency of 1, i.e, Hydrogen can only bond with 1 other atom. But since hydrogen wants a complete shell, it can have 2 covalent bonds with 2 electrons. Please explain.

Apart from some Group 13 weirdness, hydrogen can only make one bond. Covalent bonds require pairs of electrons and hydrogen can only have two electrons bound in one covalent bond.


How to stop a lye + oil + water reaction in aluminium container? I was making soap using a guide for the first time. I accidentally used an aluminium container. It has lye and olive oil in it and was kept on the stove; now it is bubbling and not ceasing (I've turned off the gas).

What should I do to stop it?

Slowly add vinegar. DO NOT TURN THE FLAME BACK ON.

Most likely, the alkali concentration is high enough that you've dissolved away the passivating oxide on the aluminum, and the exposed aluminum metal is reacting with water to form $\ce{Al(OH)4-}$ and hydrogen gas.  Adding vinegar should bring the $\mathrm{pH}$ back down into a range where the aluminum can passivate again, stopping the reaction.

The batch of soap is probably already ruined, so I wouldn't worry about that.

In the future, I would recommend using a cast iron or mild steel (not stainless) pan for this procedure.  Plain iron is remarkable for its resistance to alkali, in my experience.


Specific gas constant for two gases I have a mixture with 1 part of gas A, and 7 parts gas B. This ratio is in terms of mass.

Gas   Molar Mass (kg/mol)
A     0.17034
B     0.034

I'm trying to find the specific gas constant of the mixture.

$$
R_{\text{sp}} = \frac{R}{M}
$$

I thought it might be
$$
M = 0.17034\frac{1}{8} + 0.034\frac{7}{8}
$$
but I can't work out if that makes sense given the units, or whether I should be using atomic mass or something.

The molar mass $(M)$ of a mixture with known mass fractions $(w_{i})$ is calculated from:

$$
\frac{1}{M} = \sum_{i}\frac{w_{i}}{M_{i}}
$$

So in this case
$$
\frac{1}{M} = \frac{1/8}{0.17034}+\frac{7/8}{0.034} = 26.469\ldots\\
M = \frac{1}{26.469\ldots} = 0.037779872\ \mathrm{kg}/\mathrm{mol}
$$


Air ionization, positive ions Sharp air cleaners has a technology called PlasmaCluster. They say that it produces negative ions $\ce{O-}$ and positive ions $\ce{H+}$ (the difference with classical ionization is that ionization produces only $\ce{O-}$, PlasmaCluster also produces $\ce{H+}$)

The air contains oxygen molecules so oxygen can be ionized. But the question is how H+ is created ? I know that the air is $5.5 \times 10 ^{-5}\%$ Hydrogen gas. Is this small quantity of hydrogen in the air ionized, or is water in the air is split via electrolysis, creating the $\ce{H+}$ ions?

The plasmacluster supposedly works by ionizing water molecules in the air into $\ce{H+}$ and either $\ce{O^2-}$ or $\ce{HO-}$ ions. These ions then, supposedly, actively attach themselves to and break down pollutants, neutralizing their charges by pulling them apart. After they achieve this, the ions come back together to reform water.

At least, this is what their website says. They are correct that water can be split into positive and negative ions as well as the fact that these ions can come back together to reform water. I am unsure that the ions will stay separated from each other long enough to break up pollutants, and it is not very clear how positive and negative ions break up "pollutants." I find it unlikely that these ions will kill bacteria, which is one of the claimed benefits of this type of ionizer (I am not a biologist, so I could be wrong about this). Even if these ions are fatal for bacteria, odds are the plasmacluster cannot produce concentrations sufficient to do this. Also, if it could produce such concentrations, I might be somewhat concerned about what those concentrations of ions could do to humans.


What decides the order of priority of functional groups in organic compounds? Is the order of defined by IUPAC or we follow some special conventions to know the order of priority?

It’s an arbitrary agreement by IUPAC, although note
that there is some correlation between the
oxidation state of the carbon and the priority
(more oxidized groups tend to be higher
priority). Generally, the functional group which shows its characteristics more dominantly even when present with other group gets a higher priority.


the perfect campfire As far as chemistry goes, there are laws or observations that can be useful to determine the perfect shape and disposition of the wood in a campfire ?

For example what chemistry says about the 2 most popular "shapes"

teepee
log cabin

or even other variations such as the swedish stove ( 1 log with the top splitted in multiple segments )

The properties that I'm looking for:
 - easy to start
 - long lasting

The properties that I would like to have but I can discard:
 - significant heat generation

From many years as a trained firefighter, I can tell you that there is certainly science involved. There are a number of very heavy calculations for calculating things like solid combustible burn time and heat release rates for combustible materials, which might possibly be useful to predict the perfect shape, fuel size and arrangement for a campfire, but are probably well beyond the scope of producing the perfect sausage sizzle. However, a number of key factors that influence fire behaviour, and which must be considered in building the best campfire include:

Size of fuels. For a given fuel type, the larger the fuel piece, the greater the resident burn time, but the greater the energy input required to obtain ignition. So small fuels will ignite and burn first, and provide energy through preheating to ignite larger fuels.
Availability of fine fuels (typically those under 5mm diameter). For ignition to occur, you need to have fine fuels, especially in the absence of other freely burning ignition agents (fire starters, a cup full of petroleum etc). As heat rises, the fine fuels need to be placed at/near the base to preheat larger available fuels. Plenty of fine fuels is critical - you can't have too much.
Available air capacity. Compact fuels that do not have sufficient air flow are poor propagators of fire. Oxygen is one of the critical elements of the fire triangle (or quadrilateral if you choose), and if not being positively injected (bellows etc), there must be sufficient air flow between fuels to allow drawing in of fresh oxygen supplies (As fire burns, the hot gases rise, and fresh oxygen is drawn in at the base, and must be able to freely disperse.
Exposure to radiant and convection heat energy. Heat energy passes via radiant, convection and conductance pathways. Solid fuels need to be preheated to release the combustible gases, and this requires sufficent exposure to radiant and convection energy pathways, which is usually above the burning fuel source.
Fuel moisture. Critical factor. Many fuels with greater than about 15% fuel moisture will have difficulty igniting; around 25% most fuels have difficulty in sustaining burn. Fine fuels exchange moisture with the atmosphere far more readily, and so weather conditions must form a real factor in your calculations (temperature, relative humidity, wind). On hotter, drier days, larger diameter fine fuels can be used for ignition. Drier fuels also burn hotter and provide greater energy output. Moist fuels release lower energy flames, taking longer to ignite other fuel.
Resident burn time. Different materials (including different types of wood) have different resident burn times, and so your choice of fuels will be significant. Iron bark, for instance, has a greater resident burn time, and greater energy output than radiata pine. This derives from the density and composition of the fuels themselves. Fine fuels, of course, have a very short resident burn time, so it is important to have enough to ensure sufficient preheating of larger fuels. How much will depend on fuel moisture!
Surface area of fuels. Smooth, polished wood will take longer to ignite than rough sawn timber

In addition, you may have to stipulate some other limitations:
Availability of renewable fuels. Is you campfire fuels source static, or are you able to add more fuels as the campfire burns. This clearly influences how long it lasts, and heat generation
Stability of design. As the finer fuels burn, there is a point of critical collapse, where the top fuels collapse on the sub-layers. This will influence a number of factors, including availability of air and exposure to preheating.

My design for my favourite campfire is somewhat similar to your log cabin design. It is a crib design, which means it is layered, quite often with the bottom layer of fine fuels (tinder) sitting on a spaced grid of larger sticks to allow air flow underneath. The mix of large fuel and smaller fuels (kindling) is staggered to give stability and access to preheating. The overall shape gets smaller as the layers go up, so the final shape is somewhat square-pyramidal. Separation of fuels needs to be close, while still ensuring maximum air flow.


Is there a commercial liquid with refractive index greater than n=2? For an application I need a commercial, transparent liquid with refractive index $n>2$.

Is this kind of liquid available?
What's the chemical composition of this?

On the web, only $n=1.81$ is routine.

Definition of refractive index:

$n=\frac {c}{v}$

If n>2 then this means that the velocity of light in the material is less than half than in vacuum so there must be a strong interaction with the material.

 One of the highest refractory index in liquids is 1.79(20°C) possessed by a solution of sulfur in methylene iodide($CH_2I_2$). Liquids are less dense than solids and this explains why it's difficult to have a refractory index greater than 2.
I found a nature article that says:

  Selenium monobromide, Se2Br2, has a higher
  refractive index than that of any pure liquid hitherto
  recorded. Prepared by direct combination, the value
  for $\nu$ is 1.96±0.01 rising to 2.02 on exposure to the
  atmosphere, owing to decomposition of the bromide,
  with separation and reabsorption of selenium.

https://www.nature.com/articles/133066b0

EDIT: From a more recent paper(2016), selenium monobromide is confirmed as one of the liquid with the highest refraction index(>2). However, diiodomethane is more easily accessible commercially.

  A promising candidate solvent is the organic liquid diiodomethane (CH2I2), which is one of the liquids with the highest
  known refractive index values (n = 1.74). While other high
  refractive index liquids exist (phenyldi-iodoarsine (C6H5AsI2)
  with n = 1.85 and selenium monobromide (Se2Br2) with n=
  2.1), diiodomethane has the key advantage of being commercially available. In addition, diiodomethane is an excellent
  solvent, and many liquid formulations using salts dissolved in
  diiodomethane are reported to increase the refractive index
  and are even available commercially.

https://www.osapublishing.org/ao/abstract.cfm?uri=ao-55-12-3165


Friedel Crafts reaction of anisole Does anisole undergo friedel-crafts reaction or does the oxygen atom complex with Lewis acid?

We know that yields of this reaction are reduced in case of phenol but does the same happen in anisole?

It does indeeed.
This reference describes the F-C reaction of anisole under AlCl3 or BF3 catalysis in good yields with the expected o/p direction.
J. Am. Chem. Soc. 106(1984) 5284

There are numerous other references

It should be noted that Anisole + AlCl3 is a method for demethylation to the phenol.


Why is the orbital angular momentum of a pi electron along the axis of two atoms' molecule one? I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not know why (I'm a student studying physics)

I currently have a preliminary understanding of this:

In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+>-|->)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.

Is my understanding correct?

Let the molecular axis be the $z$ axis, A and B the two atoms forming the bond. The $2p_z$ AO has atomic quantum number $m_l = 0$ and so has $L_z =0$. The MOs formed from
2pz AOs are therefore $\sigma$ MOs, bonding and antibonding.
The $p_x$ AO is a linear combination of $m =+ 1$ and $m = -1$ AOs and has $|m| =1$.

$2p_x \equiv \frac{(2p_1 + 2p_{-1}) } {\sqrt 2}$

$2p_y \equiv \frac{(2p_1 - 2p_{-1}) } {i\sqrt 2}$

Therefore the MOs made from the $2p_x$ and $2p_y$ AOs have $|m| = 1$  and are $\pi$ MOs.


Why is the oxime tautomer more stable than its corresponding nitroso isomer? We know, generally, that the keto tautomer is more stable than the corresponding enol tautomer. I would appreciate an explanation that goes beyond the thermodynamic reasoning, too.

In the nitroso-oxime tautomerism, the nitroso form is analogous to keto isomer and the oxime is analogous to the enol form.
Then why is the oxime more stable than the nitroso isomer?

In keto-enol tautomerism generally the keto tautomer is more stable than its enol counterpart because in the former a significant ionic character exists which stabilises it. However, there are certain special cases. For instance if the enol form is aromatic(phenol), or very bulky mesityl groups are attached, etc.

In oxime-nitroso tautomerism however, the electronegativity difference comes into play.

In oxime, nitrogen is attached to carbon by a double bond;

and in nitroso, nitrogen is attached to oxygen by a double bond.

Now it is known that greater the difference in the electronegativity values, stronger is the bond. This property makes the nitrogen carbon double bond stronger hence making oximes more stable than nitroso.

(electronegativity values: N=3.04, O=3.44, C=2.55)


How can I generally determine which bonds are longest/shortest in a molecule? My task is to arrange the bonds from longest to shortest. Although I have the answer (C,D,B,A), I'm having a very hard time understanding how to derive this answer. Any clues would be helpful, especially those that help me in other related problems.

Because atomic radii of elements tend to decrease across a period (source), and bond length increases with atomic radius, the bond lengths are expected to decrease in the following order: C (C-C bond) > B, D (C-N) > A (C-F).

The difference between B and D is that in D, the nitrogen is $sp^3$-hybridized, while in B, it is $sp^2$-hybridized due to conjugation of its lone pair with the aromatic ring 2. This results in partial double bond character of the C(ring 2)-N bond, similar to aniline:

More $s$ character leads to shorter bonds, and therefore the C-N bond B is shorter than D.

In summary, the bond lengths decrease in the following order: C > D > B > A.


How to separate platinum from steel? I'm currently extracting some platinum from an electrode. One part seems stuck in some way (possibly welded) to a piece of steel. I'm guessing it's steel because it's magnetic. I did a heat test on the piece. Results:

Left side is platinum, right side suspected steel, and the platinum is not oxidized, as expected. What would be the best/easiest way to

determine if this is actually steel/iron;
dissolve or otherwise separate it?

My resources are limited, I've got a propane torch and household items, and can of course order chemicals if needed, but no acetylene torch or anything.

Cut off as much of the "steel" as possible (hacksaw or angle grinder will work - use gloves & eye protection) then immerse the stub in 33% hydrochloric acid until the "steel" remnants are fully dissolved. Neutralise the acid with bicarbonate before disposal.


Cram's rule for nucleophilic addition to carbonyl groups I recently heard about Cram's rule for predicting the stereoselectivity in nucleophilic additions to carbonyl groups. Can anybody explain what this is and whether it is an accurate rule?

Cram's rule, proposed by Donald Cram in 1952,[1] was the earliest model proposed for rationalising the stereoselectivity of nucleophilic additions to carbonyl groups with α-stereocentres. The model involved assigning each α-substituent a relative "size" (small, medium, and large), then placing the carbonyl oxygen antiperiplanar to the largest of these three groups. The nucleophile then attacks the carbonyl group opposite the larger of the two remaining groups (i.e., the medium group). This is best explained with a diagram:

In the reduction of (S)-3-phenylbutan-2-one with L-Selectride (a bulky source of hydride ion), the anti alcohol is formed as the major diastereomer. Cram's rule rationalises this by placing the phenyl group (RL) anti to the carbonyl oxygen. The hydride nucleophile, drawn here as H− for simplicity, then preferentially attacks over the proton (RS) instead of the methyl group (RM).

Although Cram was able to demonstrate that his rule worked for a number of nucleophilic additions, there are some conceptual issues with it. Most obviously, in the reactive conformation, the phenyl group is eclipsing the methyl group on the ketone; this torsional strain is worsened when the nucleophile adds into the carbonyl, as the resultant product is initially in a staggered conformation.

In 1968, Hugh Felkin pointed out two observations which the Cram model could not explain.[2] One is to do with the reduction of cyclohexanones, which we will not go into here, as nowadays a slightly different model is used to explain reduction of cyclic ketones. The other was about how the stereoselectivity of C=O reduction varied when the ketone was made progressively  more bulky.

According to Cram's model, as the size of R increased, the steric repulsions between R and the phenyl group (RL) should increase, leading to a decrease in stereoselectivity of reduction. However, Felkin showed that the opposite was true: the anti product was formed with higher stereoselectivity when R increased in size from Me to t-Bu.

To cut a long story short, Cram's rule was eventually replaced by the Felkin–Anh model. Here, the large α-substituent RL was instead placed perpendicular to the C=O bond. This leads to a choice of two possible conformations. The nucleophile attacks the carbonyl group along the Bürgi–Dunitz angle, and the attack over RS instead of over RM is favoured:

Like Cram's rule, the Felkin–Anh model correctly predicts the diastereoselectivity observed in this reduction. However, it also allows us to rationalise why the stereoselectivity increases as the size of R increases. A larger R makes the gauche interactions between R and Ph more destabilising, which biases the conformational equilibrium towards the reactive conformer (on the right).

For more detail about the Felkin–Anh rule and other modifications which are necessary (depending on the exact substrate and conditions), see: Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones and the answers therein.

References

Cram, D. J.; Elhafez, F. A. A. Studies in Stereochemistry. X. The Rule of “Steric Control of Asymmetric Induction” in the Syntheses of Acyclic Systems. J. Am. Chem. Soc. 1952, 74 (23), 5828–5835. DOI: 10.1021/ja01143a007.
Chérest, M.; Felkin, H.; Prudent, N. Torsional strain involving partial bonds. The stereochemistry of the lithium aluminium hydride reduction of some simple open-chain ketones. Tetrahedron Lett. 1968, 9 (18), 2199–2204. DOI: 10.1016/S0040-4039(00)89719-1.


What is the relative size of the (M+2) peak? The $(M+1)$ peak is often considered in the high-resolution mass spectra of organic molecules as it reveals the number of carbon atoms in the sample. In general, it is known that the ratio of the size of the $M$ to $(M+1)$ peaks is $98.9 : 1.1 \times n $ since the relative abundance in nature of $^{13}$C is $ 1.1$% for the mass spectrum of an organic molecule containing $n$ carbon atoms and no heteroatoms. It is mentioned that the $(M+2)$ peak is statistically insignificant on this site. However, I believe that only applies for organic molecules with a relatively small number of carbon atoms and this peak would become significant when considering the mass spectra of larger organics. Using simple mathematics, I derived that the ratio of the $M$ to $(M+2)$ peak is $98.9^{2} : 1.1^{2} \times _nC _2 $. I would like to verify if this is correct. If it is not, could someone then correct it by posting an answer?

Let’s assume your compound is $\ce{C_nH_xO_y}$. Thankfully, both hydrogen and oxygen are elements that only have one significant naturally occuring isotope. Therefore, we can treat the entire contribution of $\ce{H_xO_y}$ as a constant $c$. All we need to answer is how large the $M+1$ and $M+2$ peaks are given the number of carbons, $n$. (This treatment is not entirely correct as highlighted by orthocresol, but it is close enough for my purposes.)

Assuming you do not have any isotope enrichments, each carbon will independently have a chance of either being $\ce{^12C}$ or $\ce{^13C}$ (again, we will ignore all other isotopes such as $\ce{^14C}$). The independence is the big key word here. We can use general principles of stochastics to calculate the result.

The probability of all carbon atoms of one molecule of your compound are $\ce{^12C}$ is:
$$P(\ce{^12C_nH_xO_y}) = 0.989^n$$
The $M+1$ peak is represented by one single atom being $\ce{^13C}$. Again, the principles of stochastics apply:
$$P(\ce{^12C_{n-1}^13C1H_xO_y}) = \left ( n\atop 1\right) \times 0.989^{n-1} \times 0.011$$
And finally for the $M+2$ peak:
$$P(\ce{^12C_{n-2}^13C2H_xO_y}) = \left ( n \atop 2\right) \times 0.989^{n-2} \times 0.011^2$$

Using this, we can plot what the probability of each peak is – and use that as their relative heights.

We see that from $n=90$ the $M+1$ peak is actually larger than the $M$ peak. From $n=128$, even the $M+2$ peak will be larger than the $M$ peak. And from $n=181$ the $M+2$ peak becomes the largest of these three.

Of course, further peaks will also start appearing, meaning that the mass spectra of very large molecules will become difficult to analyse.

Now the numbers from which $M$ is no longer te principal peak are quite alrge. Unless you are synthesising maitotoxin, you will probably not need to resort to any analysis that does not only involve $M+1$.


What would the chemical name be for C13H8Cl3NO Formula

  C13H8Cl3NO

SMILES

  C1=C(C(=CC(=C1)Cl)Cl)N(C(C2=CC=CC=C2)=O)Cl

I found the diagram on the left in a book and drew the one on the right using

https://pubchem.ncbi.nlm.nih.gov/edit2/index.html

And got the SMILES description from that.

Any clues as to what might be an IUPAC name and the formal way to write it out?  Have I have even drawn it correctly? I could not do it exactly because the two rings kept connecting if I followed the same orientation.  I don't know enough about double ring compounds to even hazard a guess.

Is there a automatic naming engine out there?

A compound called British Impregnite found in The Scientific method by Louis F. Fieser p.137

Well, let's reconstruct that starting from the very right side, where it says a $\ce{C_6H_5}$. The ring and the $\ce{CO}$ group would be a benzaldehyde if it had an $\ce{H}$ instead of an $\ce{N}$, right? Or a benzoic acid if it was $\ce{OH}$ instead of $\ce{N}$. So what would it be if it had an $\ce{NH_2}$-group? It would be a benzamide. If the $\ce{N}$ is substituted with for example a chloride we call that N-chlorobenzamide. And now we have another $\ce{N}$-centered ligand, the second phenyl ring. The ring has three positions, the one where it's connected to the rest of the molecule would be 1, so that makes a 2,4-dichlorophenyl.

Summarizing we get N-chloro-N-(2,4-dichlorophenyl)benzamide


If volatility depends on intermolecular forces, then why in Raoult's law, two substances have different volatility? I have a doubt, Ideal solutions are solutions where intermolecular forces between solute-solute and solvent-solvent are nearly equal to solute-solvent. Since intermolecular forces between A-A = B-B, doesn't that mean, they both have same volatility? But in Raoult's law, the graph shows that one of the vapour pressure is lower than the other resulting in two substances  having different volatility. I'm confused, where did I go wrong?

Assume surface molecules are restrained by a QM harmonic potential, with energy levels described as $$E_n = \hbar \omega(n + \frac12)$$
where $$\omega=\sqrt{\frac{k}{\mu}}$$
is the frequency of the harmonic oscillator
and $$\mu=\frac{m_1m_2}{m_1 + m_2}$$
is its reduced mass.
The actual shape of the potential is determined by $k$, the force constant, ie classically we would write $$U_{pot}=\frac12 k x^2$$
Therefore we might have two particles with identical restraining potentials, but due to differences in reduced mass they might differ in the nature of the available QM energy levels and their thermal populations. This means you can in fact have solute and solvent interacting via identical potentials and yet having different volatility.

It can be readily shown that a condition satisfied by an ideal mixture that follows Raoult's law is that the heat of mixing

$$\Delta_{mix} H=H_{12}-\chi_1 H_{11} - (1-\chi_1)H_{22}$$

be equal to zero.  Note that $H_{12}$ depends on the composition (on $\chi_1$). However, identical potentials (force constants) does not guarantee that the condition is satisfied. Consider now a simple model that assumes that the oscillator consists of a dangling surface molecule and the rest of the bulk, such that $\mu \approx m_i$ (where $m_i$ is the mass of the dangling molecule). Then it can be shown that the ideality condition represented by the equation $\Delta_{mix} H=0$ holds for statistical mixing if the force constants are independent of the nature of the bulk (but may depend on the dangling molecule). This is probably not an unreasonable description of van der Waals bonding, which depends on the area of the interaction surface of the solute more than on the composition of the solvent (assuming chemically similar compounds eg hydrocarbons).


What is entropy change in electrochemistry? I keep seeing entropy change (or reversible effects) and irreversible effects (Joule heating) when reading about charging and discharging of Li-ion batteries. I understand the Joule heating part (the irreversible), but I have troubles understanding the reversible effects. What's going on in the structure of the electrodes that the process is reversible? As far as I know, reversiblities come from the fact that there is no friction or other losses. If one isolates a single ion, the charging or discharging, in my interpretation, would look roughly like in the figure below:

What happens inside the battery during this charge and discharge that would make one think of reversibility? And why is entropy important in this case? And why would a reversible process release heat?

I think your diagram seems plausible; the physical process happening here is when charging an applied potential difference (voltage) causes lithium ions to leave the crystal lattice and migrate to the cathode. When the battery discharges, the reverse happens, and lithium ions rejoin the anode.

The definition of reversibility is the infinitesimal change in some thermodynamic quantity in the system with respect to its surroundings without increasing the entropy. The electrode is crystalline, so Li+ ions can rejoin the lattice without increasing the configurational entropy, whilst infinitesimally changing the chemical potential.

If the temperature is low enough for crystallisation to be entropically and energetically favourable with respect to dissolving, then the free energy of the system will be reduced by adding an ion to the crystal lattice. The chemical potential induced by charging causes the lithium ions to return to the anode when discharging.


Is the following rate law possible for the given reaction? A researcher is investigating the following overall reaction:
$$\ce{2C + D -> E}$$
The researcher claims the rate law for the reaction is written as follows: $\text{Rate} = k[\ce{C}][\ce{D}]$.

Is the rate law equation possible for the given reaction?
If so, suggest a mechanism that would match the rate law. If not, explain why not.

The answer is yes. I don't know how to arrive at this answer. From what I  have learned, the following must be true for a) to be true:

The rate law for the slow step agrees with the experimental rate law
The elementary steps of add up to the overall reaction.

I would assume the following elementary steps from the rate the researcher has confirmed:
$$
\begin{align}
\ce{C + D &-> CD}\\
\ce{C + CD &-> E}
\end{align}
$$
But since $\ce{CD}$ has a larger size than $\ce{D},$ wouldn't it be less likely to have the proper alignment in a collision to enter the active state? Thus, wouldn't it be a slower reaction than $\ce{C + D}?$
Please inform me where my logic is wrong and why the answer is yes.

You should use steady state approximation which leads to the mentioned answer:

$$\frac{\mathrm{d}[\ce{CD}]}{\mathrm{d}t} = 0$$

$$
\begin{align}
\ce{C + D &->[$k_1$] CD}\\
\ce{C + CD &->[$k_2$] E}
\end{align}
$$

$$\ce{->[QSSA]}\quad k_1[\ce{C}][\ce{D}] = k_2[\ce{C}][\ce{CD}] \quad\to\quad k_1[\ce{D}] = k_2[\ce{CD}]$$

$$\frac{\mathrm{d}[\ce{E}]}{\mathrm{d}t} = k_2[\ce{C}][\ce{CD}] = k_1[\ce{C}][\ce{D}]$$


Removing wine stains from a decanter I have a decanter which I use fairly often to pour out wines. I typically rinse it with hot water after the last wine has come out, but I cannot prevent it from forming 'residual' wine stains over time.

The last time I wanted to clean it, I looked on the internet for advice on cleaning. There are tons of it (Stained wine decanter), but I noticed that there are quite a lot of things around that don't work (or don't work well) and I figured that the question is basically plain chemistry.

Therefore, my question is, can somebody explain how I can best clean the wine stains from my decanter? Please, also explain the chemistry behind the method you propose so I know why it should work.

It's a bit of a blunt instrument, but I'd tend to just go straight for hydrogen peroxide if it's available where you are. It breaks up all sorts of organic dyes via oxidation, and it slowly decomposes by itself, especially in sunlight, so you can just rinse (thoroughly) and leave it to dry in the sun.

(NB: I have never owned a wine decanter, and take no responsibility if your decanter dissolves, explodes, or spontaneously gains the ability to sing operatically.)


Should you work in metres-cubed or litres in chemistry? Generally, is it better to work in metres-cubed or litres in chemistry?

My undergraduate degree was in civil engineering and we always worked in $\mathrm{m^3}$.

However, I notice the gas constant $R$ is used in a few different forms:

$$R=0.0821\mathrm{\frac{L\cdot atm}{mol\cdot K}}$$
$$R=8.3145\mathrm{\frac{J}{mol\cdot K}}$$
$$R=8.2057\mathrm{\frac{m^3\cdot atm}{mol\cdot K}}$$

among others.

I guess it depends on the application but is it generally easier to work in litres or metres-cubed?

Values for the gas constant are tabulated for various different units as a matter of convenience, since when carrying out calculations, the units of the gas constant need to match out with the other values being used (which in turn depends on the expression being used in the calculation) and its often easier to look up a value of the gas constant than to convert everything else into a specific set of units.

Generally, chemists aim for SI units, however there is a great deal of discrepancy between, for example, chemists in the USA vs. chemists in Europe. This is especially apparent in physical chemistry.

For example, the SI unit for energy (+work, +heat) is joules, such that the 'correct' way would be to express energy is terms of kJ/mol or J/mol, however, for historical reasons, many american chemists still work in calories (ending up with kcal/mol etc).

In terms of the volume question, we have equally confusion.

For small quantities, volumes are measured in units like millilitres, micro litres etc, though cubic centimetres (CC) do occasionally get used. For calculations involving moles/concentrations however, we default to decimetres and metres cubed. (you should note that 1 decimetre cubed == 1 litre)

In conclusion, in an ideal world we'd all use SI units, but various quirks have meant that we don't. If in doubt, look at what other people in the literature are doing, and, whatever units you choose, make sure that the units in your calculation match up.


How does dimethyl sulfoxide transport other molecules through skin? Wikipedia says:

Use of DMSO in medicine dates from around 1963, when an Oregon Health & Science University Medical School team, headed by Stanley Jacob, discovered it could penetrate the skin and other membranes without damaging them and could carry other compounds into a biological system.

How does this work? Is there a term for this property? Can DMSO transport molecules through other barriers? What are the limitations on which compounds DMSO can transport? Which other compounds share this property with DMSO?

DMSO interacts with the outer membrane of skin, the stratum corneum.  The stratum corneum is a typical lipid bilayer and DMSO can solvate the exposed hydrophilic lipid head.  This results in a weakening of the lateral forces between the individual phospholipid molecules.  The weakened lateral forces result in a widening of the channel between the phospholipid molecules and the widened channel allows small molecules to pass through the lipid layer.  After the DMSO is removed the lipid bilayer restores itself back to normal.

Although it is agreed that small molecules can pass through the DMSO-affected bilayer, there is some evidence that larger molecules such as viral agents might also pass through.  There is also some controversy over the effect of DMSO on the blood-brain barrier.  Here is a link to a nice review of DMSO's pharmacological effects, including transport of larger molecules and transport across the blood-brain barrier.


Validation of _atom_site_frac_*: Range outside of [0,1)? I have the following cif file

loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
...
Al2 Al -0.35936(1) -0.08768(9) -0.24188(8)
Al3 Al -0.27044(1) -0.35380(9) -0.13044(8)
F1 F -0.5320(2) -0.2015(2) -0.3794(2)
F2 F -0.3800(2) -0.2437(2) -0.1891(2)
F3 F -0.0350(2) 0.0944(2) 0.1528(2)
...
N3 N 1.2782(3) 0.7260(3) 0.3726(3)
...
C2 C 1.0165(4) 0.6764(4) 0.2267(3)
C3 C 1.1053(4) 0.6806(4) 0.3576(3)
...
H1 H 1.015(4) 0.603(4) 0.147(3)
...
H6 H 1.068(6) 0.773(6) 0.220(5)
H7 H 1.100(5) 0.761(5) 0.441(4)
...
H10 H 1.303(5) 0.653(4) 0.319(4)
...
H12 H 1.070(5) 0.574(5) 0.364(4)
...
H16 H 1.328(6) 0.749(6) 0.471(5)
H17 H 1.313(5) 0.801(4) 0.348(4)

Of interest is the fact that many of these "fractions" of the cell length are greater than 1, and many are negative. To my sensibilities these atoms lie outside the unit cell.

Should the IUCr definition form CIF dictionary of _atom_site_fract_x state that the permitted range is 0.0 -> 1.0, but forgot?

Should these atoms be discarded, should they be mod'd into the unit cell, or should they be interpreted in some other manner?

I completely agree with @porphyrin comment. This is an example of crystal data which formatting does not entirely meet IUCr requirements, that's all. This happens a lot, and in most cases it's no big deal. This particular set of coordinates is just not uniform within a single cell. Different crystallographic software uses different means to "reassemble" atoms "correctly" and write changes to *.cif, e.g. Olex2 uses Standardise function for that:


Calorimetry, finding final temperature I was looking for some help on understanding this practice question from my text book, I know how to figure out enthalpy, but this is my first temperature change question and I am stumped. Any guidance would be greatly appreciated.

Nitric acid is neutralized with potassium hydroxide in the following reaction:
$$\ce{NHO3(aq) + KOH3(aq) -> KNO3(aq) + H2O(l)}$$
$$\Delta_\mathrm rH= -53.4\ \mathrm{kJ/mol}\ \ce{HNO3}$$.
$55.0\ \mathrm{mL}$ of a $1.30\ \mathrm{mol/L}$ solutions of both reactants, at $21.40\ \mathrm{^\circ C}$, are mixed in a calorimeter. What is the final temperature of the mixture? Assume that the density of both solutions is $1.00\ \mathrm{g/mL}$. Also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. No heat is lost to the calorimeter itself.

I know the answer is 29.7 degrees Celsius, I'm just not sure how to get there. I tried juggling around my equation "delta t r H= m c delta t" but since I don't know the final temperature i don't think the delta t works in this case. I've tried omitting the delta t but I get no where near 29.7 degrees Celsius.
So I took my formula Q=mc delta t and rearranged it, I now have delta t= Q / mc. Delta t= -53.4 kJ/mol / (55.0g)(4.19 J/g C) I changed the 55mL to g as the question says to assume the density is 1.00 g/mL. I also used the heat capacity as water given in my course data booklet). And I get -0.2317. I tried doubling the mass as there is two solutions but that doesn't work either


What does the shaded area represent? I'm not sure what the answer is, all of my friends and my chemistry teacher said it is A, I think it is obvious that the graph is talking about 2 catalysts so it could be D. But I have no idea what the E_a thing is, I've tried googling it to no avail. I got this question on an advanced Chemistry test. One thing for sure, it is not B because the word "additional" isn't there. And another thing, we can't say that one is the graph of the molecule without a catalyst and the other is its graph with a catalyst because according to the graph that would mean that sometimes the molecule without a catalyst is more active than without a catalyst which is obviously wrong.

Short answer: The two curves are the distribution of kinetic energy of molecules at two different temperatures. A catalyst won't change the distribution; only temperature change will change the energy distribution. The line marked Ea is the activation energy; a molecule has to have more than this to react. A catalyst could change the activation energy; since there is only one Ea, we can eliminate catalytic activity as an issue. All the molecules with less energy than this just bounce around, not reacting.

The curve with the higher peak represents the lower temperature. The one with the lower peak has a longer and higher tail out to higher energies, so the shaded area represents the greater number of high energy (reactive) molecules at the higher temperature. Answer: A.


Which elements exhibit allotropy? I was just reading this article about allotropes which relates allotropy with some specific elements and also this wiki article. Also some other articles some of which relate allotropy with catenation some with electronegativity and some with electronic configuration. I was understanding the relation of allotropy with all these factors but I am unable to get that how allotropy is dependent on them exactly. Can we predict the allotropy in elements (and also the extent to which they exhibit allotropy) just by concerning their chemical properties?

Given proper conditions, any element could have multiple allotropes For example, under sufficient pressure, hydrogen has a metallic phase, which may have been observed at Harvard recently.

The extreme example is degenerate matter, formed under the pressure of a collapsed star, though it could be argued that the elements are no longer the same, since nucleons themselves are eventually pooled in a "sea".

If you mean by your question "having allotropes at standard temperature and pressure (STP)", then the list is shorter, of course. Since this depends on the electronic configuration of an element and on the minimal-energy configuration of atomic arrangement, calculation is difficult, but not impossible, and there is software available for crystal structure prediction.

Now if I could only find a more convenient way to change graphite to diamond, without Superman's help...


offset in measuring decomposition potentials as far as I've understood to measure the decomposition potential of an electrolyte using a galvanic cell we just put two electrodes, connect them to a battery, and vary the external potential till when we observe a current. The potential at which the current starts to grow linearly is the decomposition potential of the electrolyte. There is not appreciable current before the decompositon potential because the electrodes are supposed to be inert and do not provide the ionic charges needed to "close the circuit". Is it right?
An other question. How the role of the electrodes is taken into consideration in the experimental setup? Is it considered an offset due to an eventual nonzero cell potential? I made this question because I'd like to be sure that I'm understanding the experimental setup.

Tirrel

Ok I will answer your  last question first. If you have two identical electrodes(i.e. same electrode material), the cell potential will be zero and there won't be any issues there.

You have to use inert electrodes so that you are not measuring the dissolution of the electrode itself.

The one part I am not sure you understand properly is the current growing. You want to vary the potential until the current raises exponentially when continuing to sweep. This will be a very sharp rise that will be hard to miss.


Assay of stainless steel I frequently need to discriminate between 304 and 316 stainless steel which cannot be done by eye.

304 SS has 18% chromium and 8% nickel

316 SS has 16% chromium, 10% nickel and 2% molybdenum

How can I do a chemical assay to discriminate between them and ideally identify their precise composition to 1% or better accuracy? I assume the basic procedure is to create a solution, then do a titration of some kind.

If the first step in your answer is... "dissolve the sample in acid," be aware that stainless steel is pretty resistant to most acids and I need a procedure that will work within 24 hours.

Also, please no answers like "use an xrf machine" or some surface detection technology. I need a chemical process that assays the whole sample, not just its surface.

Digest a sample of the metal in aqua regia, I have found that the gauze in a kitchen sieve can be digested in a very short time with a mixture of 16 M nitric acid and concentrated hydrochloric acid.

Next you should dilute it with water and filter it to remove any particles.

I would use either flame AAS or ICPOES to measure the metals after dilution to a suitable concentration in plastic ware. I would want to measure the Fe, Ni, Cr and Mo in the sample. The Fe to Ni and Fe to Cr ratio may help you distingish between a lot of grades of stainless steel.

One alternative for iron is to use 2,2'-bipy and hydroxylamine to form the deep red iron(II) tris bipy complex.

This can be measured with a UV / vis machine in a 1 cm cuvettes, you will have to make a series of solutions in volumetric flasks. You make 0, 1, 2, 5, 10 and 20 ppm iron by adding a 1000 ppm standard to the flasks, then add a solution of 2,2'-bipy in water, the hydroxylamine and then dilute to the line. Mix them well and then measure the red colour with the UV / vis machine.

I am sure that there are colourmetric methods for Mo and Cr but sadly I do not know them.


Are the elements lanthanum and actinium considered to be in the d block or the f block of the periodic table? Many periodic tables place lanthanum and actinium in the f-block of elements, for example, this periodic table from Los Alamos National Laboratory.

However, this table from the Royal Society of Chemsitry places lanthanum and actinium in the d-block.

Which is the most appropriate placement of these two elements on the periodic table? What is the most recent scientific consensus regarding this matter?

Lanthanum and Actinium are generally both accepted as f-block elements. You will notice in the Royal Society's periodic table, La and Ac are both color coded to match the other f-block elements. They do this to show the beginning of both the Lanthanide and Actinide series, while maintaining a reasonable symmetry for their periodic table (that is, without "holes" in their table, similar to the first link you provided).


How energy is released when chemical bond is formed? In books it is commonly written that whenever a chemical bond is formed the energy of the molecules/atoms gets lowered and hence energy is released. This is generally explained by diagrams like this

But I'm really unable to understand this process, I accept that before bonding the potential energy of the molecules were more and after bonding the potential is less and hence this difference of energy must be released but how does this lowering of potential energy happens? Let's just take an example of electrostatic situation (and as far as I know its the electric potential energy only that get's lowered and hence some energy gets released), say I have an electron at the origin and a proton at $x=x_1$ and then somehow it is moved to $x=x_2$ like this

   We know from equation of electrostatics that potential energy of proton at $x=x_1$ is $$ U_1 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_1} \\
\\
\textrm{and at} ~x=x_2 \\
U_2 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_2}$$
Since, $x_1$ is greater than $x_2$ (from figure) therefore $U_1 \gt U_2$. So, we energy at $x_2$ is less therefore there must be some release of energy, but how? If we let the field to do the work then no energy will be released because our proton gonna get the kinetic energy, if we do the work to move it from $x_1$ to $x_2$ then according the to the laws of physics I will do the negative work and hence energy will transfer from the proton to me. So, how does energy ever get released?

How in chemistry the energy gets released by the formation of bonds? I know that we can say without intricacies that potential energy gets lowered and hence the difference of energy is the energy that we get, but how does this happens? If we have some cation and then we put some anion in the reaction mixture then, of course, cations will get attracted towards the anion and hence will gain kinetic energy and according to me that's the only possible way to lower the potential energy. Is that kinetic energy that we measure as heat and say that the reaction is exothermic?

Any help or clarification will be much appreciated.

Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))

In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy $E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at $r_e$, the release of a photon with energy $h\nu=D_0+E_\text{kin}$ will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)

This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore $\text{H}_2$ molecules are unlikely to be formed via radiative association.

Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision.

When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.


Hydrolysing a compound by repeated methyl and hydride shifts The question is to predict the major product in the following reaction.

First of all the proton will abstract the bromide ion resulting in the formation of a primary carbocation which being unstable would rearrange.I have the following idea about how the reaction should proceed.

However I have doubts if such repeated methyl and hydride shifts are actually possible.Is the same product possible through a different energetically favourable mechanism?Thanks.

The reaction will stop after first methyl rearrangement as it leads to tertiary allylic carbocation which is highly stable, so there will be no further rearrangement, as it does not lead to more stable carbocation intermediate. Thus one will get tertiary allylic alcohol.


Coordination compounds Which of the following complex is diamagnetic?

  $\ce{[Cr(Cl3)(NMe3)]}$
  $\ce{[Co(EDTA)]-}$
  $\ce{[Cu(CN)(NO2)(NH3)(Py)]}$
  $\ce{[Fe(NH3)6]^2+}$

For (4) I found that since $\ce{NH3}$ is a weak field ligand it cannot pair the electrons of $\ce{Fe^2+}$. Thus, (4) will remain paramagnetic.

But, for the rest of the options, I'm unable to clearly understand if they are paramagnetic or diamagnetic. I need help here.

1) Cr3+ has 3 electrons in 3d orbital. Obviously, it is a tetrahedral splitting, so configuration will be eg2 t2g1 which shows it is paramagnetic.

2) Here you need to know that for Co3+ all complexes are low spin except [CoF6]3- and [Co(H2O)3F3]. So, the complex given will be low spin with all 6 electrons paired & thus it will be diamagnetic.

3) Cu2+ has 3d9 configuration. So, it can never be diamagnetic as one electron will always be unpaired.

4) What you thought is right. But I would like to add that NH3 doesn't always behave like a weak field or strong field ligand. It depends on many factors which can't be explained using CFT.

Hope it helps!


Electrolysis of sodium chloride solution I am trying to understand electrolysis. My question  is: when electrolysis is done with a NaCl solution I have learnt that sodium metal is not formed. Why is this?

The electrolysis of molten NaCl indeed gives elemental sodium and chlorine gas:

$\ce {2NaCl (l) -> 2Na (l) + Cl2 (g)}$

However, in solution (namely aqueous solution such that you are doing the electrolysis on a solution of NaCl in water), the highly reactive Na produced reacts instantly with water to give sodium hydroxide, hence why no solid Na is ever observed.


How to model 98% ethanol solution concentration graph over time İ want to plot a graph of ethanol solution concentration over time .So we know that alcohol evaporates quickly in open air but is there a time when the rate of evaporation mathces that of water and concentration stops decreasing or increasing? İ know surface matters but lets assume surface is totally hydrophobic and doesnt interact with solution.Any ideas how to do it?

As Poutnik pointed out, if you boil a 98% ethanol solution (2% H$_2$O), then you will lose a few % of ethanol until you get to ~95.5% ethanol; then the rest of the solution will boil off as if it were a pure compound, for the composition will not change. Well, this is strictly true at the high temperature of boiling at 1 atm pressure. At about 0.1 atm, the azeotrope breaks, and so at that boiling temperature (about 35$^o$C, from a graph in https://wps.prenhall.com/wps/media/objects/3311/3391416/blb1105.html),

there is no azeotrope. I suspect that at lower temperatures than 35$^o$C, there may be no azeotrope. Nevertheless, if you start with a 98% ethanol solution, it won't change much until you have exported off most of the ethanol (like ~75% or so).
I would suggest cutting the concentration to about 50% ethanol, put it in a large beaker (~1 liter), stir with a small magnetic stirring bar at low speed to keep the contents somewhat homogeneous, put in a thermometer and a hydrometer with a range of 0.9 to 1.0 and engage a mild cross current of air across the beaker.
You will be able to determine the concentration of ethanol from a diagram like this:

from  https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html using the thermometer and hydrometer. Time is a variable, but is useless for determining any general scientific principle. The conditions of your experiment include air pressure (you can get that), air current (you can describe that - if you don't use any air current, it will take hours or days to do your experiment).
You can read off the volume of solution you have from the graduations on the beaker, and you can then plot the composition of the solution as its volume decreases by evaporation. The measurements are unfortunately not very accurate, but you will be able to make a graph and then spend several pages describing your lab work and potential errors and look up graphs and tables like the ones I have used. A likely error will be that the temperature will not be constant, the liquid level will drop so that the air current will have different strengths.
And as MaxW pointed out, the modeling may not be very "realistic". But I think people will be impressed with your use of scientific equipment and your analysis (which will be of the highest caliber, I'm sure).
While 50% ethanol is a lot less flammable than 98%, it still is a fire hazard, especially in quantities as big as a liter. Not only the liquid, but also the vapor is hazardous, so make sure that there are no flames or sparking devices within about 20 feet of your experiment. And have good overall ventilation: ethanol vapor can put you to sleep or make you dizzy. For extra security, you should plan to have a fire extinguisher handy (and know how to use it).


Implications of the excitation spectra The fluorescence excitation spectra show the change in fluorescence intensity as a function of the wavelength of the excitation light.
I'm interested in the certain physical implications of the above.
Does that mean, that at the wavelength of excitation maximum the most of the molecules get excited? Or do they just emit more photons per molecule per unit time?
Additionally, if the former is correct, is there any relationship between the percentage of excited molecules and the excitation spectra?

The fluorescence excitation spectra show the change in fluorescence intensity as a function of the wavelength of the excitation light. Does that mean, that at the wavelength of excitation maximum the most of the molecules get excited? Or do they just emit more photons per molecule per unit time?

You have to understand how the excitation spectrum is collected. You have to know (or at least have an estimate) of the emission maximum of a given molecule. For example, if we have quinine, I should know that it emits blue light and maximum wavelength of emission is ~ 450 nm.
Now we will fix the emission wavelength at 450 nm and scan the entire UV-VIS range for excitation. Say, we would excite quinine with 200 to 700 nm. One wavelength at a time, and count how many photons corresponding to 450 nm were emitted. This plot is called the excitation spectrum.
An excitation spectrum basically tells you which wavelength (from a given instrument) is going to show the highest intensity. I emphasized "a given instrument" because an excitation spectrum has to be corrected for the light source intensity. It will vary from instrument to instrument. Nobody corrects it because it is a long procedure as unfortunately, normalization cannot correct it.
Anyway, a corrected excitation spectrum should match in shape with the absorption spectrum provided that there is only one fluorophore in the molecule.
I wanted to connect the excitation spectrum with an absorption spectrum because we can now connect the molar absorptivity with the excitation. Consider molar absorptivity as a measure of probability of excitation. Higher the value of molar absorptivity, more molecules will absorb the light and hence more will emit. So your first interpretation is correct. It is a collective behavior of the all the molecules which are being illuminated in the cuvet.
Also remember that excitation emission is normally a one photon excitation and one photon emission in majority of cases until and unless you are using fancy lasers. So you cannot have one photon in, and multiple photons out in routine cases. In rarer cases, you can indeed have two photon in for excitation and one photon out. You need lasers again.


How the crystal structure varying for the same compound? i have synthesized the crystal Cyclohexylammonium acetate. Structure of this compound was already solved and it s under centro symmetric space group I2/a. But when i took single crystal analysis it comes under monoclinic C2/c space group. What is the reason for getting this kind of results?

a) Your structure analysis is wrong

or

b) the publication is wrong (your looked up the wrong substance, typo in the original work, ...)

or

c) You crystallised from another solvent, at different temperature, (pressure?) and got another modification.


What would be the state of water in interstellar space or vaccum? Consider the interstellar space: Far from radiations, heat, etc. You can imagine conditions equal to vacuum. At these conditions, what will be the state of water; Ice, liquid, vapours, or something else?

What I know is that in lab conditions in a vacuum flask if we do the same experiment with water:
First it starts to boil, as the pressure in vacuum is almost zero. As it takes no time for the water to reach pressure equal to surrounding pressure, vapour pressure is formed easily. Water starts to boil without any rise in temperature.
Second, since energy is spent in boiling all the kinetic energy is used up and the water molecules start to condense forming ice.

The question now is: What are the differences between lab conditions and interstellar space? Shouldn't the results be all the same?

tl;dr: ice

Let's have a look at the phase diagram of water:

In space, both the temperature and pressure are very low. That would put beyond the bottom left corner of the diagram, well within the ice field. So simply speaking, water would be ice.

However, there is the concept of vapour pressure. Any condensed material (just a fancy word for both solids and liquids) has that same material in the vapour around it. That's why you can smell alcohol, even though it's liquid as one example. That's why a puddle of water will eventually dry out even though it's below the boiling point of water.

The way vapour pressure works is that if you look on the phase diagram, for a certain temperature, the liquid-vapour and the solid-vapour lines have certain pressures associated with them. For example, if you look on 50 C, then the pressure on the curve is 10 kPa. Roughly speaking, this means that in atmospheric pressure, one tenth of the air (which is at 100 kPa) has to be water vapour. If it's any less, water will keep evaporating until it reaches that amount. On Earth, wind and diffusion will disperse the water so it will basically evaporate until there are no more liquid water left.

Now let's think what would happen in space. The temperature is very low, so the vapour pressure is very low. Thus, it should evaporate also in space. However, because the vapour pressures at such low temperatures are extremely low, you might end up with a situation where the gravitation of that ice blob is enough to overcome evaporation, so that will be an agonizingly slow process.


Why does the C-C bond break earlier than the C-F bond in the pyrolysis of perfluroethane? According to me $\ce{C-F}$ bond is highly polar and thus is a weaker bond as compared to $\ce{C-C}$. Hence, $\ce{C-F}$ should break first in the pyrolysis of perfluroethane. Instead, in J. D. Lee, it is stated that $\ce{C-C}$ bond will break first. Why?

I got this explanation from one of my seniors.

It is true that $\ce{C-F}$ bond is highly polar as fluorine is more electronegative than carbon. In fact it is this very difference in electronegativity that makes C-F bond very strong. This is because, in addition to the covalent bond strength between carbon and fluorine, there is an additional electrostatic attraction between positive carbon and negative fluorine.

We can think of it has the $\ce{C-F}$ bond has both covalent  and ionic character. This makes C-F bond extremely stable and hard to break.

In fact in organic chemistry we usually say that the $\ce{C-F}$ bond is not broken in reactions.

We can also verify this from their bond energies:

$\ce{C-F}$: $\pu{544 kJ mol-1}$
$\ce{C-C}$: $\pu{346 kJ mol-1}$