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Balancing Redox Reactions (oxidation number method)
I'm confused about balancing redox reactions using the oxidation number method. I know that in order to get my coefficients for the two reactants I need to find the simplest whole number. However, when I come across certain reactions I'm not getting the same answer as the answer key.
For example, $\ce{Cu(NH3)4^2+ + S2O4^2- + SO3^2- + Cu + NH3}$
I know that $\ce{S2}$ is oxidized: $+3 \rightarrow +4$; Change= -1 but since there are two sulfur I know that I need to multiply -1 by 2, so I get -2 electrons for the reactant $\ce{S2O4^2-}$.
$\ce{Cu}$ is reduced: +2 => 0; Change= +2
My total change would be -2 and 2. If I multiply the two numbers I get 4 and the simplest common factor for 4 is 2 and 2. So my coefficient for $\ce{Cu(NH3)4^2+}$ should be a 2 and my coefficient for $\ce{S2O4^2-}$ should also be a 2. Right? However, on my answer key both $\ce{Cu(NH3)4^2+}$ and $\ce{S2O4^2-}$ have a coefficient of 1. I'm not sure how they got that answer.

Simplest whole number is not a thing at all. You need to find the least common multiple of 2 and 2; what can it be? Not 4, mind you.

Correction the least common multiple.
I think I made a mistake, the total change should be -4 since I had to multiply the change for copper (which was 2) and the change for sulfur (which was -2). So shouldn’t the coefficients have been 2 for Cu(NH3)4^2+ and 2 for S2O4^2-?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{ }$ for chemical formulas/equations, $ $ for math term/equations, and $\pu{ }$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies.

You don't have to multiply 2 and 2. You have to find their least common multiple.

Oh gosh. I’ve been doing that for some of the other reactions and have gotten the correct answer. No wonder I was confused. Thank you Ivan.

Thank you for welcoming me A.K. I’ll make sure to correctly format equations next time.

It seems that you have the answer, but I'll work it out anyways.
Given the reaction:
$$\ce{Cu(NH3)4^2+ + S2O4^2- -> SO3^2- + Cu + NH3}\tag{1}$$
You can separate it into two half reactions so that you don't have to solve the whole equation at once.
$$\ce{Cu(NH3)4^2+  ->  Cu + NH3}\tag{2}$$
$$\ce{S2O4^2- -> SO3^2-}\tag{3}$$
Balance equation (2)
First just balance the stoichiometry... 4 molecules of $\ce{NH3}$ are needed
$$\ce{Cu(NH3)4^2+  ->  Cu + 4NH3}\tag{4}$$
Now notice that copper is going from $\ce{Cu2+ -> Cu}$ so we have to add two electrons on the left to get the charges to balance.
$$\ce{2e- + Cu(NH3)4^2+  ->  Cu + 4NH3}\tag{5}$$
That gives the copper half-cell reaction.
Balance equation (3)
First we need to double the $\ce{SO3^2-}$ on the right
$$\ce{S2O4^2- -> 2SO3^2-}\tag{6}$$
Notice that copper starts of as the $\ce{Cu(NH3)4^2+}$ complex, so the solution is basic. We have more oxygen atoms on the right, so in water you can use the mythical reaction $\ce{2OH- -> H2O + O}$ to balance. We need to add $\ce{4OH-}$ to the left, and $\ce{2H2O}$ on the right to get the stoichiometry right.
$$\ce{S2O4^2- + 4OH- -> 2SO3^2- + 2H2O}\tag{7}$$
Now the atoms balance, but the left side has a -6 charge, but the right has a -4 charge. So we need to add two electrons on the right to balance charges for the equation.
$$\ce{S2O4^2- + 4OH- -> 2SO3^2- + 2H2O + 2e-}\tag{8}$$
That gives the $\ce{S2O4^2-}$ half-cell reaction.
Finish
Equation (5) has two electrons on the left, and equation(8) has two electrons on the right. So we can just add those equations and cancel out the common electrons from both sides to get:
$$\ce{Cu(NH4)^2+ + S2O4^2- + 4OH- -> Cu + 4NH3 + 2SO3^2- + 2H2O}\tag{9}$$


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how can a covalent molecule with a permanent dipole still have temporary dipoles?
As I have read online, temporary dipoles occur due to the random fluctuations of the electron cloud. The electrons move randomly to one side of the molecule creating a negative partial charge while the nucleus is "left on its own" on the other side - creating a partial positive charge.
My question is :
If there is a permanent dipole , won't that stop the electrons from moving wherever they want ? They would be fixed to a certain region, right?
The only solution I see is that the "permanent dipole" isn't really permanent. Could it be that it fluctuates as well? That would allow the electrons to move around and create temporary dipoles.

Why would anybody care about temporary dipoles when a permanent one is around? But then again, no, a permanent dipole won't stop anything. Electrons in a polar molecule are no less and no more "free" than in any other molecule.

I read online that dispersion forces play a greater role in the boiling temperatures of molecules than dipole-dipole forces, so according to that one would care about temporary dipoles...

IT isn't that temporary dipoles don't happen when a molecule has permanent dipoles, it is that they are far weaker and therefore less important.

Both temporary and permanent dipoles will both play a part in what the boiling point of the compound is.
Permanent dipole-dipole forces acting between molecules will mean the majority of molecule will end up aligning themselves to minimise repulsion from partial charges. but this still doesn't stop van der Waal's forces acting between any two molecules. If the idea of electrons continually moving is thought about then it makes sense that just because a bonding pair may spend more time closer to one atom than the other, it doesn't mean they can still move about - so temporary charges can still arise as they move.
So the boiling point will be a result of the permanent and temporary dipole forces acting between molecules. But the contribution from temporary dipoles will be insignificant compared to permanent dipoles.

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Why do bubbles form on the side of a plastic bottle?
I have a plastic bottle filled with drinking water where only one side of the bottle has air bubbles. I know that water is not pure and there are dissolved gases, and my theory is that there are small microscopic holes on one side of the plastic bottle that make bubble formation possible. Or may be its a temperature difference between the two sides (which shouldn't be the case - the bottle was kept in room temperature, but it was closed for a day or more). What could be the cause?

Does this answer your question? Where do the bubbles in a glass of water come from?

Your reasoning seems good. A number of things can affect nucleation of bubbles, such as surface defects, which you state, as well as temperature (was one side of the bottle facing the sun, perhaps?), and even sound.
A fun demonstration is to put a cup of carbonated beverage in an ultrasonic bath (e.g. parts cleaner or humidifier). Here, cavitation serves to nucleate bubbles.

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Balancing the equation for the preparation of dichromate from chromite ore

$\ce{FeCr_2O4 + Na2CO3 + O2 -> Na2CrO4 + Fe2O3 + CO2}$

I have tried balancing it by writing the reduction and oxidation half equations viz. :
$\ce{2Fe^{+II}Cr_2^{+III}O4 -> 4Na2Cr^{+VI}O4 + Fe2^{+III}O3 + 14e^-}$
And:
$\ce{3O2^{0} + 12 e^- -> 2Fe2O3^{-II} } $
Then I multiplied the first equation by 6 and second by 7 and added them and though the electrons got cancelled, it turned out to be really hard to make any progress.
Could someone please provide the technique for balancing such an equation?

$\ce{4FeCr_2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$ This is a simple balance.

I have the answer @soumikdas. I wish to know the method to arrive at it.

Balancing is a case of trial and error most of the time and just spotting what works and thinking logically about the equation

As of someone mentioned here, balancing most chemical equations is a a case of trial and error. Your equation is easy to balance by this way than go to extremes.
1) A relatively complicated formula within this equation is $\ce{FeCr2O4}$ on the reactants' side (LHS) of the equation. Since $\ce{Fe}$ atom in it is converted to the resultant product, $\ce{Fe2O3}$ and no other $\ce{Fe}$ bearing products on products' side (RHS), you should start with putting coefficient $2$ in front of $\ce{FeCr2O4}$ to balance $\ce{Fe}$ in LHS.
2) It would make four chromium atoms in LHS. Thus, you need to balance that by putting coefficient $4$ in front of $\ce{Na2CrO4}$ on RHS because it is the only $\ce{Cr}$ bearing molecule among products.
3) That action would make eight sodium atoms in RHS. Thus, you need to balance that extra $\ce{Na}$ by putting a coefficient $4$ in front of $\ce{Na2CO3}$ on LHS because it is the only $\ce{Na}$ bearing molecule among reactants.
3) Now, you would find four carbon atoms in RHS. To balance that extra $\ce{C}$ you may need to put a coefficient $4$ in front of $\ce{CO2}$ in LHS because it is the only $\ce{C}$ bearing molecule among products.
4) Now you have balance all atoms except for oxygen atoms. Thus, count the number of oxygen atoms in RHS and LHS, separately. If you count correctly there are 27 $\ce{O}$ atoms in RHS and only 20 of them in LHS. To balance that extra 7 $\ce{O}$ atoms, you should put a coefficient $\frac{7}{2}$ in front of $\ce{O2}$ in LHS. This would complete the balancing your equation:
$$\ce{2FeCr2O4 + 4Na2CO3 + 7/2 O2 -> 4Na2CrO4 + Fe2O3 + 4CO2}$$
In addition, some people do not like to keep fractions as coefficients in their balance equations. So, you may multify whole equation by 2 to get integers as a coefficients:
$$\ce{4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$$


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chemical equilibrium and exothermic reaction
here is an interesting question about the chemical equilibrium and Le Chatelier's principle.
the following pressure show the partial that produced in the equilibrium reaction at time (t1 and t2)
$$\ce{N2 + 3H2 <=> 2NH3 \Delta H =-92 KJ}$$
at point t1 hydrogen is added to the previous equilibrium system at this point on the curve and after a period a new equilibrium state took place at point t2
so
A = ...
B = ...
C = ...

what i say is the answer is A=H2 ,B=N2 and C=NH3
becuase by adding more hydtogen thus increasing the conc. the reaction will move forward so the temp will increase increasing the pressure according to The Pressure Temperature Law so the the pressure of N2 increases but again also the pressure of NH3 should increase because the no of moles will increase as the reaction will shift forward so I am lost here

Since, H2 is added at the equilibrium state, it's initial pressure will be more and starts decreasing as it goes into the reaction with N2. This is clearly depicted by graph A. While the pressure of N2 starts decreasing from the level it was in equilibrium state, represented by graph C. Also, the reaction shifts forwards in order to gain equilibrium, pressure of NH3 will increase. So, the correct answer would be,

A = H2, B = NH3, C = N2


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Using a significantly large metallic sphere, could we draw out the electrons from the negative terminal of a battery?
I've pored over many articles on electrochemical cells, and I understand that on the oxidation side of a zinc-copper Daniell cell, the anode is highly negatively charged because zinc participates in oxidation and loses electrons as the ions go into the zinc sulfate.
So since the anode is rich in free electrons, is it possible to take an electrically neutral, large conductor (a metallic sphere for example), touch it to the negative terminal of the battery so that the free electrons will distribute themselves over its surface, discharge the sphere, and repeat this until the negative terminal is neutralized?

That's possible, but you will need awfully many large metallic spheres.

"the anode is highly negatively charged" This statement is false. Even a small build of excess charge would be heavily disfavored due to magnitude of Columbic forces. The negative sign is not the charge. It is the sign of the relative potential.

The anode is highly negatively charged; more negative than the cathode which has less negative charge.

Neither electrode can be "highly" charged. They will accumulate charge until the resulting voltage is sufficient to oppose the redox reactions, at which point no further charge is produced. Thus, only a tiny amount of charge can be stored in this manner and only a minute amount of the cell will be depleted in each cycle.

Yes, this is possible, you have effectively described using a battery to charge a very weak capacitor, then discharging the capacitor.  The charge stored will be minimal (possibly immeasurable), pico-coulombs or less and the number of cycles to fully discharge then battery may be on the order of billions or trillions, but in theory the process described is possible (more properly: plausible).


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Angular Momentum Quantum Number and Orbitals
In our lecture I was told the angular momentum QN is equal to  $n - 1$ of the principle QN, which corresponded to the orbital shape.  How is this case, for example in Barium the principle QN is 6 with, so it should have at least an f orbital because the angular momentum QN contains 3, which corresponds to the f orbital, but in it's electron configuration it does not have an f orbital. Why is this?

"In our l"? what was this supposed to be?

It has the f orbitals, they're just empty, and so they get left out of the (ground state) electron configurations.
F orbitals are high energy orbitals for their energy level.  The 4f orbitals (in n=4) don't get filled until after the 6s orbital is filled (and kinda sorta one 5d gets filled - f orbitals don't really follow nice patterns when they get filled).
For a good read on the filling pattern of orbitals (which, btw, is very much a generalization - there are plenty of exceptions), here's a good link.


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Why does the neutralisation of any strong acid in an aqueous solution by any strong base
Why does the neutralisation of any strong acid in an aqueous solution by any strong base always result in a heat of reaction of approximately –57 kJ mol−1?

Definition of enthalpy of neutralization: CHANGE IN ENTHALPY WHEN ACIDS AND BASES REACT TOGETHER AND FORM 1 MOLE OF H2O
Strong acids and bases have 100% dissociation in water, meaning that 1mol of water is formed from the reaction of any two (1mol quantities of) strong acids and bases
Notice that the definition of the neutralization of enthalpy is based on moles of water formed; we don't care about the salts
Thus, enthalpy of neutralization is constant regardless of the strong acids or based used


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Can a solid have a vapor pressure without the ability to sublimate?
Does sublimation occur when a solid turns into a gas and THEN vapor pressure is produced or does the solid first have a vapor pressue and then  when the vapor pressure of the solid equals the atmospheric pressure then it sublimates? Please explain.

Vapor pressure is, by definition, the pressure that a gaseous phase exerts after its reaction (some phase -> gas) rate  equalled the rate of the inverse reaction (gas -> some phase) of phase transition.
Therefore, the solid is all the time sublimating off the substrate, as well as the vapour is sublimating back to solid phase (though in grains, not back to the substrate). IUPAC mentions the vapour pressure at equilibrium as 'saturation vapour pressure', indicating that the terminology is also used out of equilibrium.

Vapor pressure does not require a reaction

@iad22agp , as a state function, vapor pressure does not require a reaction per se, but by defining it on phase transition equilibrium, it is properly defined in terms of at least two reactions, in comparison to a non-equilibrium state.


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What is the solubility of quinine suphate in 98% ethanol?
I have a result from an experiment that really bothers me, yet I found no sources online that can explain the situation.
While quinine is relatively soluble in ethanol because they are both non-polar. However, the result was not homogeneous. Even worse, the solution freezes when we left it sitting for a moment (it turns into solid mass), certainly not within our expectations. what we did was preparing 300 mg of quinine and dissolve it into 10 ml of ethanol and spin it in vortex spinner for 2 minutes.

The solubility of quinine sulfate in ethanol is temperature dependent. This reference 1 describes it as slightly soluble in EtOH at 25C, soluble at 80C. What you describe sounds like the material crystallised out of solution when you let it stand.
Note: Ethanol is generally regarded as a polar solvent, and while the freebase of quinine is non-polar, any salt of it is highly polar.

so i have gotten the wrong idea about the polarity. though i still believe that we can dissolve non polar molecule in ethanol? but why did our solution crystallised when we let it stand? is there any relation towards the freezing point then? more importantly, what is the evaluation towards the solubility since it has crystallised? are we going to say that quinine is insoluble in EtOH? could you please explain further on this?

My speculation is that spinning it also heated it, when you let it stand it cooled and the material that had dissolved crystallised out.  Quinine freebase is EtOH soluble (1g/0.8ml). Do you need the solution to be of the salt?

if you can provide, please do so, i need as many informations available apparently

Solubilty taken was taken from here: https://www.mpbio.com/product.php?pid=02150152&country=222

I'd like to know what you mean by telling "..the solution freezes.." If I'm correct about your observation, I think it underwent organogelation of the solvent. Modified quinines are known to supergelators: https://pubs.rsc.org/-/content/articlelanding/2016/sm/c5sm02723c/unauth#!divAbstract.

it basically turn into a solid mass when we spin the solution in vortex spinner and let it sit for a moment.

Thanks to @mathew-mahindaratne This may be relevent https://en.wikipedia.org/wiki/Organogels


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Is benzenesulfonic acid or cyclohexanesulfonic acid more acidic?

Which of the two sulfonic acids above is a stronger acid?
I think that cyclohexanesulfonic acid 2 should be more acidic as there is no competitive resonance as in 1. Is that correct?

Draw out the full line-bond or Lewis structures for each and you will see that there is no way for the benzene ring to contribute to resonance structures that involve the sulfonic acid group, or the sulfonate conjugate base. Instead, the important contribution is the inductive effect. Benzene rings are more electron-withdrawing than alkyl groups, so benzenesulfonic acid should be more acidic than cyclohexanesulfonic acid. A comparison of pKa values for $\ce{PhSO3H}$ ($-2.8$) vs $\ce{MeSO3H}$ ($-1.9$) supports this assertion (Can. J. Chem. 1978, 56 (17), 2342–2354).

Doesn't the reason why Ortho substituted benzoic acids are more acidic than benzoic acids due to SIR, support that 2 is more acidic?

@user137644 No, ortho effect is much different then inhibition of resonance and has nothing to do with this question. https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases


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What is the magnetic moment of potassium hexacyanochromate(II)?
I calculated the magnetic moment of $\ce{K4[Cr(CN)6]}$? in the following way:
$\ce{Cr^{+2}}$ has $4$ d electrons. And since $\ce{CN-}$ is a strong field ligand, the electrons will pair up in the $\mathrm{t_{2g}}$. The electronic configuration would be $\mathrm{t_{2g}^4}\;\mathrm{e_g^0}$. Or more specifically, $\mathrm{d}_x^2\; \mathrm{d}_y^1\; \mathrm{d}_z^1\; \mathrm{d}_{z^2}^0\; \mathrm{d}_{x^2-y^2}^0$.
The magnetic moment will be a contribution of both orbital and spin magnetic moments.
$$\mu = \root\of {n(n+2)+L(L+1)},$$
where $n$ will be $2$ as the unpaired electrons are $2$.
$L$ will be $2\times 2 + 1\times 1 +0\times 1 = 5$ corresponding to $\Sigma m_l$.
Solving I get $\mu = 3.74$. But the experimental value is $3.2$.
Why is there this deviation?

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

For transition metal complexes, to a first approximation you should not include any orbital angular momentum, so the magnetic moment is described by the spin-only formula: $\mu = \sqrt{n(n+1)}\mu_\mathrm B = 2.83 \mu_\mathrm B$. In general, this is a very rough approximation (it holds well in some cases), and bearing that in mind, $2.83$ is pretty close to the experimental value of $3.2$ you quote. To go beyond the spin-only formula, however, is more difficult and involved than one might think.

Note also that the "real" d-orbitals $\mathrm d_{xy}$, $\mathrm d_{yz}$, $\mathrm d_{zx}$, $\mathrm d_{x^2-y^2}$ do not possess well-defined values of $m_l$ (they are linear combinations of $m_l = \pm 1$ and $\pm 2$), so your summation of $m_l$ values is inappropriate anyway.

@orthocresol I am aware that the orbital contribution does not generally hold good. But since the system in this case is triply generate, (with 3 possible arrangements in the $t2g$ orbital), hence I expect the orbital contribution to be prominent. (This scenario also fulfills the 3 requirements of orbital contribution that are stated in your link)

@orthocresol Could you elaborate a little on the $m_l$ values. (not being well defined). I am having a little trouble with those.

Atomic orbitals are solutions to the Schrodinger equation (eigenfunctions of the Hamiltonian), but the choice of solutions is not unique, as any linear combination $a\chi_1 + b\chi_2$ of degenerate eigenfunctions $\chi_1, \chi_2$ is still an eigenfunction itself. So, when you have degenerate orbitals (like the five d-orbitals), you can add and subtract them as you wish to make a "new" set of d-orbitals. The five d-orbitals that possess specific values of $m_l$ are not the same as the five d-orbitals that are usually drawn ($\mathrm d_{xz}, \cdots$) - they are linear combinations of each other.

Back on the topic of the magnetic moment - you are correct that the ground state (which has a $T$ term symbol) possesses orbital angular momentum. I don't want to try to explain how this leads to deviations in the magnetic moment from the spin-only formula - it is really quite involved, and although I learnt about it I don't feel comfortable writing an answer - so maybe someone else will answer, but in the meantime, consider looking for a book on physical inorganic chemistry (it's too specialised for the general inorganic texts).


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NMR sweeping magnetic field
I am currently doing a experiment in the lab which is using a continuous wave NMR, by sweeping the magnetic field, and I am having trouble understanding how the theory works for this type of experiment.
So I currently understand if a proton is not placed in a magnetic field it will spin on it axis of rotation due to the magnetic moment.
My first question is on what causes this magnetic moment, from what I have found so far that the proton can be modelled a sphere with infinitely  many of  tiny circular current loops which run along the axis, but what I can understand is that for a dipole moment to be formed a current going in a circular loop generate the dipole moment.

So dose the dipole cause the proton to spin which generates the magnetic moment and if so what causes the tiny dipoles along axis? It seem to me there a rotation being caused by a rotation.

So moving on when the proton is placed in the a uniform magnetic field it will either align it self with parallel or anti parallel. It would not align it self fully due to the rotation of the proton so the proton will actually precess around the axis in which is in the direction to the static magnetic field. Now when the proton is in a state parallel to the static magnetic field it is in a low energy state and when anti parallel it is in a high energy state so the energy difference is given by:
$$\Delta E=hf=g\mu _nB_0$$
Now applying this to a sample of high hydrogen 1 like glycerine that has been place in CW NMR. When the sampled is placed in the static filed according to the Boltzmann distribution there will be more protons parallel to the field that non-parallel so now in cw as I understand a constant RF signal is being applied and a varying magnetic field perpendicular to the static field.
So my second question is

Dose the varying magnetic field vary the energy levels until resonate occurs?

I believe it dose as this would explain the nmr of glycerine that was take in the lab, as show below.

Now the real trouble I am having with these graphs is that one it looks to me that there is a dampening effect occurring, which I believe to be the relaxation time of the sample, but what I cant visualise is how the resonance is occurring within the sample it self in terms of the physics with respect to all the protons precessing.

Are the individual proton within the sample at different depths become out of phase with each other and the varying magnetic field is decreasing over time, and why dose each resonance peak have a negative voltage associated with it? How dose this negative voltage tell me about what is occurring with in the sample?

If anyone could with any of the theory behind CW NMR and how the signals being receive can inform me on what occurring with in the sample it would, it would be appreciated. Thanks in advance.

I'd suggest not trying to understand the phenomenon of spin physically - the proton is not literally spinning, nor are there tiny wires running clockwise around it. It is quantum mechanical angular momentum...

What causes this magnetic moment?
  So does the dipole cause the proton to spin which generates the magnetic moment and if so what causes the tiny dipoles along axis? It seem to me there a rotation being caused by a rotation.

Spin is a property predicted when relativity theory is applied to electromagnetic theory, leading to the Dirac equation. It is an intrinsic property of subatomic particles, like charge. It explains magnetic phenomena such as NMR. It is a quantum property of NMR active nuclei (when measured it takes on discrete values). For spin-${\frac{1}{2}}$  nuclei, it is related by a multiplicative constant to the magnetic dipole moment of the nucleus. Thanks to spin, nuclei have an intrinsic and fixed magnetic moment in an external magnetic field. One component of this moment adopts a discrete orientation with respect to the applied field. Depending on the orientation, two states of different energy, usually called $\alpha$ and $\beta$, are possible for a spin-${\frac{1}{2}}$ nucleus.
We can define the resonance frequency in the presence of the external field using the equation you provide. For a spin-${\frac{1}{2}}$ nucleus it is proportional to the difference in energy between the $\alpha$ and $\beta$ states. It is the frequency at which the nuclear moment nutates about the applied field.

Does the varying magnetic field vary the energy levels until resonance occurs?

Yes. In the CW experiment, you measure the electrical signal induced by the sample in a pick-up coil while simultaneously irradiating the sample with a second orthogonal RF coil, and sweeping an applied magnetic field. The sweep alters the resonance frequency of the spins. When this frequency equals the irradiation frequency spins are excited and transition between the two states - this is resonance, the points at which the oscillating red curve (the field sweep) crosses the origin. The spins now oscillate together coherently back and forth between $\alpha$ and $\beta$ states, yielding a fluctuating magnetic field which induces an oscillating current in the pickup coil. As the field sweeps past the resonance point, the spins begin to lose energy - to relax. Relaxation causes the signal to eventually decay to the baseline.

Are the individual proton within the sample at different depths become out of phase with each other and the varying magnetic field is decreasing over time?

The NMR signal is an average, reflecting the behavior of a large ensemble of spins, acting coherently (nearly identically) during an experiment. The total signal is a sum over all excited spins in the sample. The applied field is as uniform across the sample as possible, so that all spins behave similarly. But in a way the spins act independently. Some oscillate faster or slower than the mean. The differences add up over time, leading to dephasing of the signal.
There is usually no spatial information in such an experiment unless special tricks are used.

and why dose each resonance peak have a negative voltage associated with it? How dose this negative voltage tell me about what is occurring with in the sample?

This question is not clear. The spins exist in states whose coherent oscillations induce a sinusoidal magnetic field in the RF pickup coil. The electric signal induced in the pickup coil is an AC voltage, alternating between negative and positive values until it decays completely.


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Link1 and multiple output files in Gaussian: doable?
I have recently started learning Gaussian in my Computational Chemistry class, and I have recently learned how to use the "Link1" option, in order to run multiple calculations from a single input file.
What I was wondering is: is there an option that allows the creation of multiple output files (one for each "Link1" section), thus allowing an easier interpretation of the results for each calculation?
I couldn't find anything about the problem on the manual or online.

Can't guarantee because I haven't tried this myself, but I believe what you are looking for is "punch=all". The one issue this might have is I think it only produces a single fort.7 file, so it probably requires an additional option or a separate program to ensure later link1's are sent to different outputs.

That's indeed an useful option (better than nothing, thank you!)
Apparently, as far as I see, there are no options to change the filename, but this might be a start :)


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Why the bond strength of HCl is more than HBr?
Most people say that it is because the 3p orbital of chlorine when overlapping with the s orbital of hydrogen covers more area than when 4p orbital of bromine overlaps with the s orbital of hydrogen which makes bond between H and Cl more stronger than the bond between H and Br. But how can we say that the area of overlapping is more for 3p and s orbital compared to overlapping of 4p and s orbital?

"Less reactive" is not well defined.

What do you mean?

I mean, when you use these words without clarification, you can and will get misunderstood.

I edited my question. I think it is more clear now.

Yes, that's better.

You can think of the overlapping volume between the electron clouds, but a simpler form to show why this happens is when you consider the electrical potential between the atoms. Even tough this isn't an ionic compound, it has many characteristics that are similar.
If I remember correctly, this is the expression for the potential energy between charged particles: $E_p = k\frac{Q_1 \cdot Q_2}{d}$
Since the radius of $\ce{Br}$ is greater than the radius of $\ce{Cl}$, the potential energy involved in $\ce{H-Br}$ bond will be smaller than $\ce{H-Cl}$ bond.
Another thing that came to my mind, and I'm not sure, is that as the 4p cloud of $\ce{Br}$ is bigger than that of $\ce{Cl}$, the relative amount of the volume that is overlapped with the 1s cloud of $\ce{H}$ gets smaller with $\ce{Br}$ than with $\ce{Cl}$, which means the connection is weaker.

How does potential energy of electrons in the outermost shell of an atom are even related to bond energy? Its just a part contributing in the bond energy. The interaction energy between electrons of one atom and the other atom and the interaction energy between electrons of one atom and protons of the other atom also play a role. By electrons I mean all the electrons, not just valence electrons. And for that you need to know the positions of electrons which is not possible because of the uncertainty principal.

I am a bit confused in the last part.

If you aren't convinced by the overlap area argument,  then think of it in terms of polarization.  The H+ cation will polarize Br- more as compared to Cl-, so HBr will possess a more covalent character as compared to HCl, or conversely, H-Cl bond will be more ionic than the H-Br bond. And as we know, ionic bonds are generally stronger than covalent bonds

No.  Ionic and covalent bonds are,made of the same stuff, electrostatic attraction between opposite charges with similar magnitude at similar distances.  They are roughly equal in strength.

@OscarLanzi Sorry,my bad. I guess I wanted to convey the idea that more electrostatic attraction will be present in H-Cl as compared to H-Br due to differences in charge/ radius ratio, and hence the charge from chlorine will be more than bromine in the formula of colombic force.

@OscarLanzi Also, I used the word generally, so it is kind of true. https://chemistry.stackexchange.com/q/8281/54655


text:
Gaseous States and Laws
What are the factors on which the value of the Boyle temperature depends? And if the Boyle temperature depends on the pressure, then why?
As at the Boyle temperature a real gas becomes ideal, it should be independent of pressure, but on reading an article on internet it confuses me so please help me understand why.

The energy of an ideal gas is independent of pressure, but what do you mean by "it should be independent of pressure".

The Boyle temperature only depends on the form of the interaction between a pair of gas atoms. It is not dependent on the pressure.
It is defined as the temperature at which the second virial coefficient $B_2(T)$ becomes zero. Near that temperature, over a wide range of pressures, it is true that the equation of state of the gas is quite close to the ideal gas equation, because typically the higher virial coefficients give a relatively small contribution. However, it is not quite right to say that the gas "becomes" an ideal gas. The behaviour is really a consequence of two cancelling contributions, at that particular temperature.
You can see a formula for the dependence of $B_2$ on both the temperature and the pair potential here. So, if you know the pair potential, and can do the integral (numerically, perhaps) you can calculate $B_2(T)$ for any temperature. If you plot it as a function of $T$, almost always it will go through zero somewhere: this defines the Boyle temperature.
The reason that there is a temperature at which $B_2$ becomes zero, for most gases, is that at very high temperatures $B_2$ is positive, being determined by the repulsive interactions between atoms (which reflect the fact that atoms have a nonzero size), while at low temperatures the attractive forces play the largest part in determining $B_2$, and these make it negative. So somewhere in between, it must go through zero. If you look on this page at the section on the (approximate) van der Waals equation of state, you can see these two contributions to $B_2$: $b$ (atomic size) and $-a/RT$ (attractive forces). So in this case, the Boyle temperature, where $B_2$ is zero, is $T_\text{B}=a/(bR)$.


text:
hydrogenation using H2 + Ni2B
I am having trouble understanding why hydrogenation with $\ce{Ni2B}$ gives anti-addition.
all my book says is this:

Could someone explain the mechanism.

Your book says wrong Ni2B (P-2 Catalyst) is used as a surface for the reaction to take place (as it contains free electrons).
The H2 is adsorbed by the surface of Ni2B. Similarly the molecules of alkyne are adsorbed by surface of Ni2B. Slowly stepwise the adsorbed hydrogen and the adsorbed alkyne forms bond and syn-addition takes place before the organic molecule leaves the catalyst surface. As a consequence, both hydrogen atoms usually add from the same side of molecule which results in formation of  cis-alkene (or Z-alkene).
Similar reaction is hydrogenation of alkene using Pt as catalyst, it is also syn addition.
See adsorption on wikipedia
Also Reaction on surfaces Wikipedia


text:
Mobility of Cations in the Environment
I have recently performed an experiment in which I tested the affects of phytotoxic cations on the rate of germination of the common navy bean. I did this out of  curiosity, as I have recently been researching arsenic (III) and arsenic (V) mobility in water systems, and I wanted to know weather or not it was the mobility of the ion or its general toxicity to a cell that is more influential on the overall phytotocicity. Generally, arsenic (III) is much more toxic due to the fact that it is so mobile while arsenic (V) is less toxic due to the fact that it is less mobile. After testing the oxygen usage in each of my different germination environments, I was quite surprised at the results. I had placed 100 navy beans in 1 molar solutions of copper, lead, iron, and cobalt chloride (the lead chloride may have been a little under 1 M, as it is not very soluble in water); in each of these solutions, I let these beans germinate for 12 hours and then tested the oxygen output of each sample. Here were my results:

I did not think the mobility of these ions would have such an effect, as I know lead is less mobile than other cations in water systems; however, my major problem is interpreting the top three results, all of which gave fairly similar results. I am having trouble figuring out where this result is coming from. I know that all of the ions are phytotocic to some degree, and I would expect all of the ions to have a detrimental affect on the germination of the seeds. Is it more likely that these findings are the result of the phytotocicity of the ions or the mobility of the cations across the cellular membrane?

This is a very interesting question and has a strong physical chemistry component, but I have some trouble understanding the data. Is germination expected to exhaust a finite oxygen supply in a growth chamber, and this is what the "divergence" indicates? Basically the metals are expected to inhibit cellular respiration? Why does lead and DI show similar results?

Also, you are monitoring cumulative oxygen consumption post-germination over ~2.5 h? For all measurements you measure difference with respect to initial measurement at time 0 and call this divergence? What data do you have otherwise on the health of the germinated beans? The large difference suggests to me that the germination of beans in lead and DI has proceeded faster but is arrested in the presence of the other metals, as indicated by their low oxygen consumption...

At the start of my experminent, I placed the beans that had already germinated in their respective toxic metal cation solutions for 18 hours into 5 separate containers (including my control, DI water); I then put 5 oxygen sensors into my germination chambers and measured the current oxygen levels inside of the containers. For 2.5 hours I measured the current oxygen levels of each of the different containers. The divergence of oxygen concentration is simply referring the the the difference between...

...the starting oxygen concentration to the concentration at the time of measurement (this was intended to show the amount of cellular respiration that had occurred). I intended for all of the metal cations to inhibit the cellular respiration of the seeds to some degree, as all were considered phytotoxins (chemicals that are toxic to plants); however, I was surprised at how little the lead cations did to inhibit the rate of respiration, as the results showed the lead tainted seeds were almost identical in terms of health to the control. What I am very curious about thought...

...is if my results seem to demonstrate the initial effects of the phytotoxic ions and not the cumulative effects. I know lead most likely would have the worst cumulative effect, but I had only soaked the seeds in the water for 18 hours, so the lead was not likely mobile enough to have an effect. What my main question in all this really is is weather or not these results seem to indicate that the ions with the most detrimental effect in my experiment (cobalt) was the most mobile in the water system or the most toxic at a cumulative level.

My educated guess is that since the concentration of the heavy metals is so high (1 M seems very high), diffusion (distribution through diffusion of solvated ions) effects would be negligible. Discrimination through active uptake followed by specific cellular toxicity is probably most important in determining the effect.  But that's a guess. You need to check refs for comparative cytotoxicity, there are plenty I am sure. You should also run repeat experiments to see variation across probes, chambers etc.

The metal toxicity question is inarguably more the domain of chemical biology, more than pure p-chem.


text:
Reactivity and its meaning
When we say that “potassium” is reactive, are we referring to a potassium atom or bulk potassium as in the solid metal? In other words does “potassium” refer to a type of atom or an agglomeration of potassium atoms?

Typically it refers to the bulk, but you can never be sure. For this and other reasons, the word "reactive" is more confusing than meaningful, and hence is better avoided.

Reactivity is a vague concept in chemistry. It is rather confusing than useful. Reactivity involves both thermodynamic and kinetic factors, it predicts whether a chemical compound will undergo a chemical reaction (either by itself or with other compounds) or not and if it does how fast is the reaction going to completion i.e., rate of reaction.
Reactivity is also used to compare or study rates of two or more chemical reaction. For example when we say reactivity of HBr is more than reactivity of HCl - here we refer to how fast the compund will give acid-base reaction.
Reactivity is sometimes referred to bulk and sometimes to a single molecule but mostly it is referred as bulk, but it is not predictable.
In pure compounds, reactivity is regulated by the physical properties of the sample. For instance, grinding a sample to a higher specific surface area increases its reactivity. In impure compounds, the reactivity is also affected by the inclusion of contaminants. In crystalline compounds, the crystalline form can also affect reactivity. However, in all cases, reactivity is primarily due to the sub-atomic properties of the compound.

Hi there, so I am currently learning about redox reactions. So is the “oxidation state of a potassium atom” the same as the “oxidation state of potassium”?

The only oxidation state of potassium is +1. It cannot go to higher values, like +6 !!!

@Maurice Thank you for pointing out, I forgot why I did that there.

When we say that potassium is reactive, it is understood that it is more reactive than sodium. This can be easily demonstrated in classrooms by dropping a small piece (max 0.1 g) of those metals into a tub of cold water.
Sodium reacts in a relatively calm way by melting and moving quickly at the water surface, yielding Hydrogen gas H2 which can be set on fire with a match. The H2 flame is one or two millimeters high. It stays and move erratically around the sodium globule. The whole is not dangerous. It is worth the trouble showing it in a classroom. No smoke is produced.
$$\ce{2 Na + 2 H2O ->  2 NaOH + H2}$$
$$\ce{2 H2 + O2 ->  2 H2O}$$
Potassium reacts so violently with water that both metal and gas take fire in such a way that it is a dangerous reaction. Intense smoke is produced, made of potassium oxide, hydroxyde and superoxide KO2, which are corrosive and toxic. It should not be shown in an ordinary classroom.
$$\ce{2 K + 2 H2O ->  2 KOH + H2}$$
$$\ce{K + O2 ->  2 KO2}$$
Don't try to use more than 0.1 g of these metals. The reaction becomes explosive with higher amounts.


text:
Which one is the stronger base?
I'm studying Wolff-Kishner reduction and I don't understand why after this step

the leaving group is $\ce{OH-}$ instead of $\ce{^-HN-NH2}$. I guess it's because hydroxide ion is less strong than the other although it would be useful to know if i can calculate pkb of diazanide (namely $\ce{^-NH-NH2}$) starting from data on hydrazine. Also I'm considering pkb of $\ce{OH-}$ equal to $0$ because it's acting as a base.

It's rather called hydrazide. They're stronger bases then respective alkoxides (and OH-).

A hydrazide is a derivative of a carboxylic acid [RC(=O)NHNH2]. A hydrazone is a derivative of a ketone [R2C=NNH2] or aldehyde [RHC=NNH2]. ;)

Mirko: The mechanism you present for the early stages of the Wolff-Kishner Reduction includes hydroxide necessary for the decomposition of the yet to be formed hydrazone 6 of the ketone 1 (or aldehyde). One may preform the hydrazone and then do the actual reduction in the presence of hydroxide in a high boiling hydroxylic solvent (Huang-Minlon variation1).

The mechanism for hydrazone formation is a reversible reaction and is often conducted by removal of water to shift the equilibrium to the right.2 Realizing that most commercial hydrazine contains 15% water, it can act as the agent for proton exchange in adduct 3. One need only use the electron pair in structure 4 to eject hydroxide contrary to the elimination proposed in the last step of your mechanism. There is no need to be concerned about the anion of hydrazine as a leaving group. Hydroxide only has to deprotonate the acid 5.

1) Huang-Minlon, J. Am. Chem. Soc., 1946, 68, 2487.
2) Hydrazones and oximes are not as susceptible to hydrolysis as are imines owing to the resonance stabilization provided by the second heteroatom.


text:
Which reaction rate formula is suited for diatomic molecules in a plasma?
I am an experimental physicist, so I unfortunately do not know much about chemistry except the basics. For my current study, I am investigating diatomic molecule formation in plasma. These diatomic molecules are often not stable, but they get excited and emit light, which makes them interesting for spectroscopy.
What I want to know is: If I have atoms, ions and electrons in my plasma, and I have some temperature T, which model for the molecular reactions would best describe this system? What kind of reaction rate formula would I need?
To give a specific example and to explain my current assumptions:
In plasma of samples containing Ca and Cl (among other elements), I get emission from the diatomic molecule CaCl. I assume that the concentrations correlate with the stoichiometry of the sample. Now I want to know the amount of CaCl that forms in the plasma, and how it relates to the concentrations in the sample. I currently assume that the reaction can be described like this:
$$\ce{Ca + Cl -> CaCl}$$
So it should be very simple, as every element is already present either as an atom or as an ion. (Note that I do not really understand how the ions play into this, I'm assuming that the reaction is the same but that an electron is captured during the reaction - if someone could elaborate on this, I would also be grateful.)
As for the reaction rate, I have found the following rate law:
$$
r = k [A]^m [B]^n
$$
Where m and n are integers and [A] and [B] are the concentrations of two reacting species A and B. What I don't understand is the physical interpretation of m and n - I have a feeling that for single-element reactants like the ones in my case, they should both be 1. Is that correct, or is there a way these could be wildly different, and maybe even non-integers? Because my experimental results can best be described by this formula if m and n are non-integers (m = 0.6 and n = 1.3, for example). But that confuses me a lot and makes me think that maybe this type of reaction rate formula might not be suited for reactions of atoms/ions in a plasma.
I have found also another formula for the bimolecular reaction on a surface (called the Langmuir–Hinshelwood mechanism):
$$
r = k \frac{K_1[A]K_2[B]}{(1+K_1[A]+K_2[B])^2}
$$
While I have no idea how this would relate to my situation, the results I get from this formula are quite nice, and I do not have to change anything from integers to non-integers. So maybe this could be a suitable explanation instead?
I should note that, if I say "it fits my experiments", I mean that I have done several experiments with different concentrations of Ca and Cl in my sample, and I try to fit them all by saying that the intensity of my CaCl emission should be proportional to the reaction rate, and the reaction rate parameters have to be the same in each case. There are a lot of assumptions here, but since the plasma itself is very complex, this is the best I can do for now.
Thank you very much for your help!
EDIT: After reading some of Paul's suggestions, my understanding is now that I have to think about every possible reaction that might take place, and that I have to distinguish between the excited states CaCl(A) and CaCl(B) and the ground state CaCl(X):
$$
\ce{Cl + e- <=> Cl-} \\
\ce{Cl <=> Cl+ + e-} \\
\ce{Ca <=> Ca+ + e-} \\
\ce{Ca + Cl <=> CaCl(A)} \\
\ce{Ca + Cl <=> CaCl(B)} \\
\ce{Ca+ + Cl- <=> CaCl(A)} \\
\ce{Ca+ + Cl- <=> CaCl(B)} \\
\ce{CaCl(A) + M <=> CaCl(X) + M} \\
\ce{CaCl(B) + M <=> CaCl(A) + M} \\
\ce{CaCl(B) + M <=> CaCl(X) + M} \\
\ce{Ca+ + Cl + M <=> CaCl+ + M} \\
\ce{Ca + Cl+ + M <=> CaCl+ + M} \\
\ce{CaCl+ + e- <=> CaCl(A)} \\
\ce{CaCl+ + e- <=> CaCl(B)} \\
\ce{CaCl(A) -> CaCl(X) + h\nu} \\
\ce{CaCl(B) -> CaCl(X) + h\nu} \\
$$
This just seems ridiculously complicated, if you ask me. (And I have not even started taking into account that Ca also reacts with O to CaO.) Surely this can be simplified to some extent? Some of these reactions are probably very unlikely?
But so for each species, I would then set up a reaction rate based on the equations above:
$$
\frac{d[\ce{Cl}]}{dt} = - k_1[\ce{Cl}][\ce{e-}] + k_1'[\ce{Cl-}] - k_2[Cl] + k_2'[\ce{Cl+}][\ce{e-}] + \dots \\
\text{etc.}
$$
Then I can solve this set of equations either numerically or by simplifying something.
Am I missing something here? Are the equations correct? Do you think they can be simplified in some way?

You oversimplify things greatly. It is very unlikely that only one reaction will suffice to describe the kinetics in your plasma. Also the reaction you mention must involve a third scattering body to take away the binding energy or requires a very fast radiative recombination rate. It is also likely that the electrons in your plasma are responsible for exciting (part of) your molecules. You'll have to consider a more complex mechanism to predict the production of the excited states of CaCl and observed fluorescence.

The points you mention (third scattering body, fast radiative recombination) are exactly the kind of thing that I wanted to know more about. This is why I am asking the question in the first place. As for the excitation, I would have thought that this can be regarded as a separate step. First, I need to know how many molecules even form before I can look at their emission rate.

The point about kinetics is that - in general - it involves coupled differential equations, so you should look at all relevant processes simultaneously. I think you're best of consulting a text book on the topic such as Laidler's Chemical Kinetics. Articles that might be of interest are J. Photochem. 25, 389 (1984) or Phys. Rev. E 57, 4684 (1998).

Thank you for the literature recommendations. In the beginning of the first article, three reaction equations are presented. Then they say that based on these, the intensity of an O2 emission should be I = k1[O]^2[M]/(1+k2/k3 [N2]). This is very interesting, can you maybe help me understand what they are doing there? I mean, this seems to be the type of derivation that I have in mind - start with something complex, but end up at something simple.

They assume equilibrium conditions so that the reaction rate equals zero. Taking the expression for the rate of formation of excited oxygen molecules and setting the rate to zero allows to write the equilibrium concentration of O2 in terms of the other species.

Alright, now that I understand that step, there's no real magic there... I will try to edit my question today in order to reflect my new understanding of the topic. While I am grateful for your help, the question still stands: Which reactions have to be taken into account and how do they have to be formulated when dealing with a plasma? I understand now that it might be more complex than I thought, even though this confuses me because I thought it would be one of the easier cases in chemistry.

After some further research, I have decided on the following solution. First, let us define the reactions that we assume are involved in the formation of diatomic heteronuclear molecule AB:
$$
\ce{A + B ->[k1] AB^{*}} \\
\ce{AB^{*} ->[k2] A + B} \\
\ce{AB^{*} + M ->[k3] AB + M} \\
\ce{AB^{*} + M ->[\gamma k3] AB(A) + M} \\
\ce{AB(A) + M ->[k4] AB(X) + M} \\
\ce{AB(A) ->[k5] AB(x) + h\nu} \\
$$
Here $\ce{AB^{*}}$ is a precursor of molecule AB that dissociates easily and needs to relax by collision with M to some form of AB, which may be excited or not. A fraction $\gamma$ of these collisions will produce AB(A), which is molecule AB in the electronic state A. This can either collide with another species M again, or it can emit a photon. In both cases it transitions to the ground state AB(X). Only the last transition is the one that produces my molecular emission, so that is the one that interests me. Using the defined reaction constants, I can describe the formation of $\ce{AB^{*}}$ and AB(A) this way:
$$
\frac{d[AB^*]}{dt} = k_1 [A] [B] - k_2 [AB^*] - k_3[AB^*]\\
\frac{d[AB(A)]}{dt} = \gamma k_3 [AB^*] [M] - k_4 [AB(A)] [M] - k_5 [AB(A)]
$$
(Note that [M] represents different concentrations depending on which species is most dominant in each of these terms, i.e. [M] cannot be factored out.) Finally, I can make the assumption that these reactions are very fast in comparison to the changes in my plasma, so that I have a quasistatic process. This needs to be validated by actual experiments, but let's assume it for now. Then we can say that the concentrations do not change, i.e. $\frac{d[AB^*]}{dt} = \frac{d[AB(A)]}{dt} = 0$, and we get:
$$
[AB^*] = \frac{k_1 [A] [B]}{k_2+k_3} \\
[AB(A)] = \frac{\gamma k_3 [AB^*] [M]}{k_4 [M] + k_5} = \frac{\gamma k_1 k_3 [M]}{(k_2 + k_3)(k_4 [M] + k_5)} [A] [B]
$$
Now, since my intensity is proportional to $k_5 [AB(A)]$, I can write my intensity as:
$$
I \propto [A] [B]
$$
Lo and behold, it's the formula I started with. The difference is that I now know that it's not a single reaction rate that is involved here, but multiple ones, which intertwine with one another.
What this exercise also taught me is that the reaction orders m and n are really based on the stoichiometry of the reactions, i.e. if I need 1 Ca and 1 Cl then m and n will be 1. This has been something that has been very opaque in the literature I've read (in which it was often emphasized that m and n can be fractions as well, but it was not mentioned under which circumstances).
Other than that, I guess I'm a bit bummed out that nobody answered my question.


text:
questions about the synthesis of iodobenzene from aniline
I am conducting an experiment about synthesis of iodobenzene from aniline, referring to this site(and various sites) (http://www.prepchem.com/synthesis-of-iodobenzene/). Below is my experimental process.

Put $\pu{9.5mL}$ of conc. $\ce{HCl}$, $\pu{9.5mL}$ of distilled water, $\pu{2mL}$ ($\pu{0.0215mol}$) of aniline in a $\pu{100mL}$ round flask with ice bath, and wait until the temperature of mixture drops below $\pu{5℃}$.
Dissolve $\pu{1.56g}$ ($\pu{0.0226mol}$) of $\ce{NaNO2}$ in $\pu{50mL}$ of distilled water, and cool it with ice bath.
Once the temperature of mixture dropped below $\pu{5℃}$, put the solution of $\ce{NaNO2}$. the solution must be poured very slowly not to raise temperature above $\pu{10℃}$.
Stir the mixture for $\pu{10 min}$.
Dissolve $\pu{ 3.58g}$ ($\pu{0.0216mol}$) of $\ce{KI}$ in distilled water. I used the water in very small quantities because $\ce{KI}$ is very soluble in water.
Put solution of $\ce{KI}$ to the mixture, then color of the mixture turns to be black, and gas bubbles (may be nitrogen gas) rise from the mixture.
After $\pu{1-2 h}$, gently heat the mixture (I set the temperature about $\pu{50-70 ℃}$).
Again after $\pu{1-2 h}$, it seemed that there are no more gas bubbles, so I stopped stirring and heating.
Layer separation occurred, (I guess) because iodobenzene is insoluble in water. so I moved it to separatory funnel and took the black oily organic layer presumed to be iodobenzene.
Here is the stage I am stuck on. according to the reference site, it says "use $\ce{NaOH}$ to decolorize the iodobenzene" (iodobenzene is colorless liquid). I put $\ce{NaOH}$ and made the pH of mixture 7, but its color is still black.

Question 1. Do I have to make the mixture basic to decolorize it? (actually, I tried making it basic, but it was not decolorised. Should I wait a moment after making it basic?)
Question 2. According to the reference site, iodobenzene have to be distilled using steam-distillation, but should I distill it with steam-distillation, not just ordinary distillation?
Question 3. Are there any problems with my experimental process?

This procedure looks basically OK. The key thing here is keeping the reaction mixture below 5C. It is good practice to cool your KI soln to 5C, and add it slowly so that the temperature of the reaction mixture stays below 10C until addition is complete. Using very concentrated KI soln may be an issue as the reaction is exothermic, better to have it more dilute and cold.
I would use an organic solvent (EtOAc or Et2O) at stage 9 to extract the product simply because you are doing this on a relatively small scale. On this scale chromatography is a better option than distillation for final purification.

Thank you for your answer. If I have to use distillation, Can I use ordinary distillation, not steam-distillation? (question 2), and can I ask you about question 1, too?
+) If I use organic solvent, can I separate iodobenzene by separatory funnel and rotatory evaporator? (and some anhydrous magnesium)

Yes to both questions, ordinary distillation will do it.  The organic phase containing the iodobenzene will most likely still require purification after drying. I don't see what you are removing by base washing except excess HCl, but that is not a reason to omit the step. The procedure says that the dark colour is due to iodine, so a wash with sodium metabisufite will remove that. If you are happy with the answer please accept it.

@Waylander: I think you mean  extraction of the product using an organic solvent at stage 9.

The boiling point of iodobenzene is about $\mathrm{188~^oC}$, thus, it's better to purify using fractional didtillation under reduced pressure (if you are reluctant to purify it by chromatography).

@MathewMahindaratne You are correct, text edited to correct this


text:
Is there absolotely no relation between R/S configurations and d/l optical rotations?
Is there such a relation between R/S configurations and d/l optical rotation that, if we experimentally know that R configuration of a fixed compound gives 'l' rotation then can we infer that the S configuration(or the mirror image) of the same compund must give 'd' rotation?

Well, yes, that much is obvious. But beyond that there is no relation.

Note that R/S notation is mostly for tetrahedral chiral centers. You can have an inherently chiral structure, with chiral substituent that does not cause big optical rotation change.
E.g.
6I-O-[(2R)-butan-2-yl]-α-cyclodextrin and
6I-O-[(2S)-butan-2-yl]-α-cyclodextrin

will probably have very similar optical rotations, with the same sign.
However these are diastereomers, not enantiomers, but you did explicitly require them. It's just to mention that there can be a catch.
Also note that d/l descriptor are now strongly discouraged ans (+) or (−) should be used, instead.

There is no simple relationship that links the R,S configuration to the direction of optical rotation
The rotation of polarised light by optically active compounds is caused because linearly polarised light can be though of as a combination of (mathematically speaking the two are equivalent) two opposite circularly polarised light beams and they are themselves chiral (think of the fields as forming a spiral in space: spirals have handedness). The chiral molecule reacts differently with light of different chirality (as chiral molecules do with their enantiomers). This is one way to explain the interaction of light and chiral molecules.
But that interaction is complicated to explain (you need a fair amount of quantum mechanical math to talk about it at all). Moreover there is no simple way to tell from a know chirality in a molecule which direction it will rotate light.
But we do know that, whichever direction one chiral molecules rotates light, its enantiomer will rotate light by the same amount but in the opposite direction. Beyond that we have no simple models to tell us the magnitude or the direction of rotation even when we know the absolute configuration of the chirality of the molecule.


text:
Do acid derivatives (anhydrides/amides/acyl halides) give positive Iodoform test?
Technically, having a methyl carbonyl group should enable them to give positive Iodoform test.

They don't give iodoform test. The reason is the decreased electrophilicity of the carbonyl carbon can't lead to the end of the reaction.

related https://chemistry.stackexchange.com/questions/74315/does-acetamide-respond-positively-to-iodoform-test

Possible duplicate of Does acetic acid give a positive result with the iodoform test?

Possible duplicate of Does acetamide respond positively to iodoform test?


text:
Why is the current draw dropping in water electrolysis
We've done an experiment regarding water electrolysis. Basically, we just separated the oxygen and hydrogen gas in a beaker that contains a magnesium sulfate solution. We set the power supply voltage to 12.46V with an initial .09A and final .01A over 45 mins. I was wondering as to what causes the drop in current draw. I have my suspicion that the concentration of the solution is decreased since we're using copper wires, and there was a displacement reaction between copper and magnesium sulfate.

If you look at the copper wires after the electrolysis, the cathode (connected to the negative pole of the current source) will be white. This electrode is coated with Mg(OH)2.
The low activity of copper on the electrochemical scale makes it unlikely (impossible) for copper to displace magnesium from an aqueous salt.
However, the initial voltage is high enough to theoretically evolve metallic magnesium, which would instantly react with water to produce H2 and Mg(OH)2, which will precipitate because of its low solubility. The Mg(OH)2 adheres, at least to some degree, on the copper electrode and interferes with current flow. The resistance to the electrolysis at the same voltage leads to a lower resulting current.
Even at electrode potentials below that needed to evolve metallic Mg, current carried by the aquo Mg++ ion will result in an increased pH and deposition of Mg(OH)2 around the cathode. An interesting question is whether the deposit from a low voltage electrolysis has the same adhesion as from a high voltage experiment. Maybe low voltage electrolysis would deposit Mg(OH)2 farther out in the solution and not be so blocked off by adhesion to the electrode.

Along with my observed result of the electrolysis process, there was a lump of white stuff which was as you've mentioned was precipitate and how about the bluish stuff? The copper electrode we used might have had caused that. But, I was wondering as to how it has been stripped off in a solid form and adhere with assuming sulfate from the magnesium sulfate?

The copper anode is subjected to acidification and oxygenation. This would tend to favor some dissolution of copper, but not stoichiometrically - i.e., not one Cu into solution for each Mg out. The white lump was Mg(OH)2 that started adhering to the cathode, then got blasted off by bubbles of H2 forming under it. Continued electrolysis is going to be a battle between H2 formation and Mg(OH)2 adhesion and blockage of the cathode surface. A better anode would be stainless steel (or Pt).


text:
How is it that earth's inner core is still solid?
Earth's inner core is made up of an alloy of Iron and Nickel.
The important note here is that it is solid.
The temperature at the inner core is about 5,763 K = 5,490 °C.
Now the melting point of both solid iron and nickel is about 1,500 °C, and even though they are combined, I still think that 5,490 is much greater than the melting point of the alloy.
Regardless, how is it that the inner core still stays solid? I would expect that at such higher temperatures, a phase change to liquid, or maybe even gas would occur.
Why does it not?

Because pressure.

There's a typo, a digit missing in "5,77 K".

Wikipedia's article on the inner core links to Science 2013, 340, 464-466, which states that "[...] we conclude that the melting temperature of iron at the inner core boundary is 6230 ± 500 kelvin."

The melting point of the vast majority of materials increases as more pressure is applied. At https://www.researchgate.net/figure/312054885_fig1_Figure-231-Experimental-pressure-temperature-phase-diagram-of-pure-iron-a is a phase diagram showing how this variation occurs for pure iron:

From https://physics.stackexchange.com/questions/184032/what-is-the-pressure-at-the-center-of-the-earth-or-a-neutron-star the pressure at the centre of the Earth is 365GPa. At this pressure we can see from the graph that the melting point of pure iron is well over 5000k. Thus, given the experimental uncertainties and the that the core is not pure iron, it is easily conceivable that the core of the Earth is solid, or at least a large proportion of it is.

A liquid's boiling point is defined at a standard
pressure, so also a solid's melting point is dependent on the applied pressure.
Liquid iron, under pressures at the center of the earth, is compressed to
similar density to solid iron, and our best theories say that it has
to become solid, even at those temperatures, at that pressure.
Iron's temperature-versus-spacing parameter (thermal expansivity) is
about 12 * 10^(-6) per Kelvin; its pressure-versus-volume parameter (bulk
modulus) is 170 GPa.   So, 10,000 degrees heating has
an atomic-spacing expansion factor of 1.12, while a center-of-planet
pressure, 320 GPa at 5000km, gives a volume compression factor of
2, or a distance compression factor of 1.3, so it makes some
sense that the pressure and temperatures in the Earth's core
will not let iron expand from a solid to liquid: the pressure
is the winning factor here, not the temperature.
Detailed observations DO find a solid center in the planet, but
our knowledge of the exact nature of that solid is mainly
theoretical, based on heavy computation of the quantum mechanics
of atomic ensembles held apart by their electron shells and
pushed together by gravity.


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Are there amide versions of lactides and lactones?
In my book, I read that Lactides and lactones are cyclic esters which contain two and one ester group in them, respectively. They are formed when $\alpha$-hydroxy; and $\gamma$- or $\delta$- hydroxy carboxylic acids; respectively, are esterified.
Now, there exists a certain similarity between amides and esters (both being carboxylic acid-derivatives), such as the similarity between ester bonds in polyesters and amide bonds in polyamides.
In this context, I have a question: Do amides have compounds similar to lactides and lactones? What are they commonly known as?
Please let me know a good source that I could refer to in addition to any answers posted, or please cite any sources used.

By the way, there are also lactols.

Amide analogue of lactons are lactams, their tautomeric forms are lactims. Citing from the IUPAC Nomenclature of Organic Chemistry (Preferred names 2013):

P-66.1.5.1 Lactams and lactims
Intramolecular amides of amino carboxylic acids, $\ce{-CO-NH\bond{-}}$, are
  called ‘lactams’ and their tautomers, $\ce{-C(OH)=N\bond{-}}$, are ‘lactims’. Lactams named in two ways:
  (1) as heterocyclic pseudoketones;
  (2) by substituting ‘lactam’ for the ‘ic acid’ ending of the systematic ‘oic acid’ name for the parent acid without the amino substituent, and inserting a locant designating the position of the amino group between the ‘o’ and the ‘lactam’. Method ‘lactams’. Lactims are named in the same way, using ‘lactim’ in place of ‘lactam’.
(1) generates preferred IUPAC names.
Examples:

pyrrolidin-2-one (PIN)
  butano-4-lactam
(…)

3,4,5,6-tetrahydropyridin-2-ol (PIN)
  pentano-5-lactim

(Note that the numberings depicted are for the preferred names (PINs) based on nitrogen heterocyclics)
There's no (at least IUPAC) term for lactides analogue (‘lactide’ is not used in IUPAC names themselves anyway), but they exist.
In traditional or general names, the Greek letter numbering is used, e.g. ε-caprolactam, or β-lactam four-membered ring part in bicyclic penicillin skeleton.

Lactide is a cyclic lactone di-ester, derived from lactic acid (https://en.wikipedia.org/wiki/Lactide). Thus, there is no name for lactam version. However, it could be a series of derivatives of piperazine-2,5-diones. Simplest member is piperazine-2,5-dione, derived from glycine.

@MathewMahindaratne Can you please add an answer? It seems comments eventually get deleted... This is good enough for an answer.

@AbhigyanC They need to obsolete/not useful etc. and flagged (or deleted by author). I'd expect comments that add important stuff to answer to last.

Cyclic amides are known as lactams.  They can be a β-lactam which is a 4-membered ring, a γ-lactam which is a 5-membered ring, a δ-lactam which is a 6- membered ring, an ε-lactam which is 7-membered ring, and so on.
An example of a drug containing a β-lactam ring is ezetimibe.  Antibiotics that contain a β-lactam ring are collectively known as β-lactam antibiotics which include penicillins such as amoxicillin, cephalosporins such as cephalexin, cephamycins such as cefoxitin, carbacephems such as loracarbef, carbapenems such as doripenem, monobactams such as aztreonam, oxacephems such as moxalactam.
Examples of drugs containing a γ-lactam ring include levetiracetam, doxapram, glimepiride.
Examples of drugs containing a δ-lactam ring include perampanel, milrinone, pirfenidone.
Example of a drug containing an ε-lactam ring fused to a benzene ring is benazepril.
There are also groups that are half ester and half amide.  They are known as carbamates (or urethanes).  Some examples of cyclic carbamates are rivaroxaban, metaxalone, zolmitriptan, furazolidone, linezolid  etc.


text:
NMR analysis regarding cis-1-isopropyl-3-phenylcyclohexane
I have come across a question in my textbook:

The textbook gives this answer:

I understand the explanation for the first compound. However, for the second structure, I am not understanding why the 2 methyl groups are diastereotopic. Is there not free rotation about the $\ce{C-C}$ bond, connecting the $\ce{CH(CH3)2}$ group to the ring such that the 2 methyl groups become equivalent? The answer key makes it seem as though there is no rotation and the methyl groups are fixed in place. I feel as if I am misunderstanding the question. Please help me clear up the confusion.

Substitution is frequently the easiest approach. Replace a hydrogen on each of those two methyl groups with a deuterium to create two new compounds? What is the relationship between those two compounds? Identical? Hydrogen atoms are homotopic. Enantiomers? Enantiopic. Diastereomers? Diastereotopic.

The two methyl groups do not see the same average environment
The issue here is not whether the two methyl groups are freely rotating about the bond to the CH group: the issue is whether they see the same environment on average when the bond between the CH and the cyclohexyl ring rotates.
The easiest way to see that they don't is to build a model and play with it.
If you do you will see that their average environment can never be the same no matter how much rotation you do. Imagine rotating them so that the CH is facing the same way as the Ph with respect to the ring. The two methyls are different: one will be closer to the Ph than the other. Now rotate the CH-ring bond 180 degrees. The CH now points away to the opposite side of the ring to the Ph. But the environments of the methyl groups are not reversed: the other methyl is now closer to the Ph but is now on the opposite side of the ring so sees a different environment. So that rotation does not equalise the average environment for each methyl.
This is generally true in chiral compounds.

Your explanation makes sense. I made a model and tried rotating the CH-ring bond and noticed that at no particular instance are the methyl groups the same distance away from the phenyl group AND the cyclohexane ring. I was wondering, however, if we were to remove the phenyl group, then would the methyl groups become equivalent? (I also noticed an instance of symmetry in my new structure without the phenyl. Is that at all relevant?)

@JamesBond I think that, were there no phenyl, the methyl groups would be equivalent as the average environment in the ring would now be the same from any direction and, perhaps more importantly, the molecule would no longer be chiral.


text:
Rate equation of a radical

Consider the following elementary reaction mechanism where both the forward and reverse reaction mechanism are important
  $$\ce{CO + O2<->[1] CO2 + O}$$
$$\ce{O + H2O<->[2] OH + OH}$$
$$\ce{CO + OH<->[3] CO2 + H}$$
$$\ce{H + O2<->[4] OH + O}$$
  How many chemical rate equations are need to determine the chemical evolution of a system defined by this mechanism? Write out the rate equation for the hydroxyl radical.

This question is from the book  Introduction to Combustion: Concepts and Applications by Stephen R. Turns.
My answer to the second part is
$$\frac{d[\ce{OH}]}{dt}=( k_{f_2}[\ce{O}][\ce{H2O}]+k_{r_3}[\ce{CO2}][\ce{H}]+ k_{f_4}[\ce{H}][\ce{O2}] ) - (k_{r_2}[\ce{OH}][\ce{OH}]+k_{f_3}[\ce{CO}][\ce{OH}] +k_{r_4}[\ce{OH}][\ce{O}])$$
Is this correct?
$k_f$ is the elementary forward rate coefficient
$k_r$ is the elementary reverse rate coefficient
[ ] refers to concentration

Second reaction stems from first one so it also should be present in equation, afaik.

Hei,  second reaction is included in the rate equation I wrote

I meant include first.  Gah, language is such a terrible tool.

Is there a difference between the rate equation and net reaction rate ? Because What I wrote is the net reaction rate and I have been told that my answer was wrong.

I would have a factor of 2 in the two terms for reaction 2 in the OH balance.


text:
Which is most stable resonance structure of 4-nitrophenoxide ion?
According to my coaching module the first structure is more stable. Why is this so? I feel both must be approximately equally stable since negative charge is on oxygen in both cases. Would someone please give me a hint?

There is no such thing as a most stable resonance structure. If anyone tells you otherwise, don't listen, it's wrong. See also: What is resonance, and are resonance structures real?

@Martin-マーチン I mean the major contributor to the final hybrid structure, why exactly is the first one the major one?

Consider aromaticity

@YUSUFHASAN would you please explain a bit more?

You were quite correct in saying that the negative charge is present on oxygen in both the cases, so the idea of comparing the negative charge being present on the most electronegative atom goes out of the window. All the octets are complete here, and there is no scope for minimizing charge separation. Now, I don't know whether you have studied aromaticity yet, but a quick glance through the Wikipedia page on the same will tell you that aromatic systems are tremendously stable. Hence, 1st option is the major contributor

@YUSUFHASAN thank you! Yes, I was actually still in the process of studying aromaticity, thanks!


text:
Methods of preparation of propane
My book States part C as the answer of the following question.
But according to me, part D, that is soda lime decarboxylation can also be a perfect answer besides the Corey House synthesis( that is the part C )the answer given by my book.
Please help in clearing my misconceptions.

Decarboxylation is usually assisted by the presence of an electron withdrawing group at the β position, so it may not be particularly effective here. Plus, the nucleophile generated by is not going to cause much elimination, so the Corey House synthesis is a better choice here

Can u please elaborate your second point.Moreover, then according to your first point,it implies that decarboxylation without a beta keto acid is surely not productive,can we say this way?

@yusufHasan Moreover,if there would have been a carbonyl group at the beta position it would lead to the formation of ketone,that we do not want because we want formation of alkane to occur...

Precisely, so if you look at the normal decarboxylation of the given compound, there is no additional factor favouring the decarboxylation or literally, 'removal of CO2'. What I said about the (c) option was that the substrate is a primary haloalkane, so it will have minimum chance to follow the E2 pathway, and also, CH3- will not be that strong a base to cause elimination in comparison to, say, EtO- so it is the best method here

Note that the question is asking for the BEST preparation of propane, not "a preparation that will give some propane". (c) is the best reaction.

I agree that Corey House synthesis will give good amount of yield of propane but at the same time there are sufficient conditions for decarboxylation to occur as well, presence of sodium salt of carboxylic acid as well as sodium hydroxide and calcium oxide support  the removal of co2.so it may have good amount of propane?what are the factors that distinguish the yield amount between the two and claiming Corey house will give propane in  excess…

Please follow this link( https://en.wikipedia.org/wiki/Corey–House_synthesis#Reaction_process_and_mechanism)

@chemophilic  The abstract of this paper https://pubs.acs.org/doi/abs/10.1021/ed040p212?journalCode=jceda8  states "Simple aliphatic acids (except for acetic) do not give good yields of the corresponding alkanes through decarboxylation, although many organic chemistry textbooks cite this as a general method for the preparation of alkanes". Perhaps you should read it.

Simple carboxylic acid or carboxylate ions (like CH3CH2CH2COO–) do not undergo decarboxylation to yield corresponding alkanes because in this case the leaving group would be a carbanion (in this case CH3CH2CH2–) which is very strong base and hence a very poor leaving group (the reason behind this is that strong base wants to give electrons, but leaving group has to take electrons and leave the compound, strong base is poor leaving group and weak base is good leaving group)
For decarboxylation to happen a keto group must exist to stabilise the negative charge, which is not present in salt of butanoic acid CH3CH2CH2COO–Na+

Yield of propane is very less if you chose decarboxylation, that is why Corey-House synthesis is best for preparation of propane.
I want to know why do you think that Option D is correct, mention in comment.

Is the negative charge itself not very stable in ch3ch2ch2coo-  because of equivalent resonating structure?when coo- posseses this kind of delocalization of e- then verily,it is a poor base and hence good leaving group?

@chemophilic yes negative charge stable on ch3ch2ch2coo- due to resonance but leaving group is not co2, leaving group is ch3ch2ch2- (I am mentioning again leaving group takes electron and leave the compound, that is why leaving group is ch3ch2ch2- which is strong base and very poor leaving group).

@chemophilic Also see that the first reaction is highly endergonic (meaning energy of products is greater than that of products) as product ch3ch2ch2- is very unstable compared to reactant ch3ch2ch2coo- which is stabilized by resonance. Therefore the reaction requires huge amount of energy which may decompose the reactant compound, and may not yield desired products, which is not the case with second reaction due to presence of carbonyl group the oxygen takes negative charge and is stable and energy required for reaction to proceed is low.


text:
Is phosphorine (C₅H₅P) aromatic?
Phophorine seems aromatic as it has 6 conjugated electrons. But the answer given is that it is not. This seems odd since pyridine has a similar structure and is also aromatic. Thus I ask is phospohrine aromatic or not?

Phosphinine is aromatic, but somewhat less than benzene.

Phosphorine (IUPAC: phosphinine) actually has aromatic character nearly as great (88%) as that of benzene.  According to the reference, phosphorine is sufficiently stable to be handled without air-free techniques; and it undergoes electrophilic substitution reactions similar to those of benzene.

With reference you mean the Wikipedia article? I have quite a bit of trouble believing this answer without knowing the context in which these 88% of aromatic character came about. Since there isn't even an agreed-upon rigorous definition of aromaticity, the number is without context equally as informative as a picked number from the phone book.

So ... you apparently think this is not aromatic?

No, that is not what I said. I just don't have any trust in that assessment of the 88% without any context of how this number came about.


text:
Is it possible to displace cerium from cerium sulphate using aluminium to get pure cerium metal?
I'm unable to find cerium any reactivity series, hence the question..

No
The std. electrode potential for Ce3+/Ce is -2.33V, the std. electrode potential for Al3+/Al is -1.66V. This shows that Al is not reactive enough to reduce Ce(III), it is reactive enough to take Ce(IV) to Ce(III) so I think you would get a mixed salt of Ce/Al sulfate.
This reference here states that calcium metal is used with cerium fluoride as a preparative method.
Note: you can't do this in aqueous solution. Your reactive metal will react with the water preferentially.

Is it possible to simply add a highly reactive metal like aluminium to a solution  of ceric sulphate? Forgive me is this question sounds silly but  I'm new to chemistry.

The electrode potentials show that Aluminium is not reactive enough to reduce Cerium. It requires metal with electrode potential below -2.33V.  Potassium,  Sodium and Lithium would  do it in addition to Ca. Ce has a reactivity close to Magnesium (-2.37V). Also note; you can't do this in aqueous solution. Your reactive metal will react with the water preferentially.

Not to mention that sulfates would get reduced to sulfides in the presence of such powerful reducing agents. That is wasteful of your reducing agent.

Thanks! May I know if it is possible to use electrolysis to extract cerium?

@mayur-sharmila yes it is. The reference I quoted in the answer mentions electrolysis of cerium chloride as a preparative method. If you are happy with the answer, please accept it.


text:
What governs the amount of smoke produced by an extinguished match?
First Chemistry question, so apologies if this isn't quite in the spirit of this SE.
I have noticed that when extinguishing a lit match by blowing it out, the amount of smoke produced can vary widely. Variables such as how long the match is allowed to burn before extinguishing, and whether it is blown out sharply or with a gentle prolonged draft, appear to influence the amount of smoke that results, but I have not yet been able to make enough observations to establish correlations.
Are there known factors that reliably contribute to the amount of smoke formed by an extinguished match? Is there an "ideal" technique of extinguishing a match that would maximise smoke production?
Thanks!


text:
Courses for Molecular Modeling?
I'm probably doing overkill trying to learn a bit of everything, but here's my story. I have a B.S. in Math with a minor in Microbiology, and I'm currently taking Computer Science courses online to meet the pre-requisites for a Master's in CS where I'll focus on machine learning. I want my thesis to be on the Molecular Modeling of a cellular process.
After receiving my Master's, because I want more research experience (and it will count as job experience for indistry), I want to work as a paid research assistant to a laboratory where I help code machine learning algorithms to aid biological modeling.
Let's say I do this for 3-4 years and I have permission to take up to three courses each fall/spring as a non-degree seeking student (realistic?). Would taking courses such as:
Physics: Modern Physics, Stachistical Mechanics, Thermodynamics, Quantum Mechanics, Computational Physics, Biophysics.
And
Chemistry: Physical Chemistry, Analytic Chemistry, Intermediate/advanced Organic Chemistry, Reaction Kinetics, Intro to drug discovery.
Be beneficial to me?
Sorry if this is too opinion based or off topic. However, what will probably benefit me the most when I get to a research assistantship, is that I'll speak to faculty and they'll  direct me on what courses to take. I'd still like your input though before the post is removed.

I think as you said it might be a bit too broad/opinion based for us to answer. Really any of these course could be beneficial or not depending on the particular focus of your thesis and the type of biological modeling you are going to do. Until you have more detail on that, we are kind of just guessing at what could be useful for you. Once you do, we could be some help, but again as you said your advisor will probably know better once you get to that point. @BryanAardvark

M.ghorabs answer is an example of this. I think it's a good answer that describes what would be useful for a particular program of biological research, but I could think of a program more focused on bioreactions and natural products that would require more emphasis on organic chemistry and kinetics and less on quantum mechanics. So the issue is that any advice we can give is going to depend heavily on the specifics of what you are looking to do, which you don't know yet.

Consider asking this question in chat.

I was a chemistry student and I would say Physical Chemistry helps, organic/analytical chemistry seems not quite useful. However, if you really wanna have a through understanding of physical chemistry, I guess courses in physics (possibly quantum mechanics and statistical mechanics, since we are dealing with a large number of molecules and atoms) help more; since you have a mathematical background, those courses may suit you more; because most of physical chemistry courses reduces mathematics part seriously (and for no good reason but they are difficult).

I’m not sure my answer is correct since I have not done modeling in chemistry and different fields in chemistry is of quite different flavor.

Someday I was wondering as you are doing but  I think that now my confusion has lessened considerably after talking with experts in the field and enrolling in a master program of pharmaceutical chemistry and drug design.
Different kinds of simulations can be called modelling of cellular processes. The first are the rules of systems biology and the second are the rules of chemistry or more fundamentally, quantum chemistry.
An example of systems biology is the simulation of the heart muscle fiber. a single cell, contractility as a function of the different ions concentrations, K, Ca, Cl, Na, etc.
The rules of quantum mechanics when applied to systems of atoms and molecules yields quantum chemistry. Quantum chemistry can calculate the tendency of a certain chemical reaction a function of the electron density and energy in certain regions. Molecular Mechanics and Molecular dynamics are more convenient when simulating bigger systems, larger number of atoms and molecules. With large number of molecules statistical mechanics must be taken into account to realistically simulate the relative population of molecules that have a certain energy.
In you question, I don't think that analytical chemistry has any thing to with computer simulation of cellular processes. From the the disciplines you outlined I thank that you are more interested in Quantum chemistry/Molecular mechanics/Molecular dynamics than that of systems biology. In my opinion you have to focus on: Physical chemistry, quantum chemistry, statistical thermodynamics,  Drug discovery and design. Also, In order to understand molecular dynamics you must have a background in motion laws of newton.
Occasionally you will need to gain a background in some specific points but, however, mastering the basics of the disciplines I mentioned will give you a comfortable platform from which you can move to your target easily.
Regards.


text:
Predict NMR spectra from the Mass Spectrometry
Imagine, you have a tissue and the Mass Spectrometry for it.
How would you get the corresponding NMR spectra knowing only the Mass Spec?
Is there is a special formula or a well-known technique?
I have an idea to build an ML model based on http://hmdb.ca metabolites data where you have those pairs (MS and NMR). But I worry it is completelly wrong.

Do you think is it a stupid idea?
If there is an easy way of converting MS -> NMR which everybody knows?

I'm sorry for a probably silly question, I came from mathematics.

Short of simply reverse-searching the mass spectrum in a database and finding out which molecule it corresponds to, I doubt it is possible to predict NMR data from MS data in any meaningful way. It's almost impossible to extract the structure of any random chemical from its MS.

NMR data typically has far more information about the structure of a molecule than MS data. The main datum from (hi res) MS is the mass of the molecular ion, which characterises the molecule globally by providing a molecular formula. Meanwhile, NMR provides some global information (e.g. H and C counts) but also directly provides a lot of local information. The only way to make up the difference would be to perform a lot of ion fragmentation with MS-MS, and then try to build the molecule back together from the pieces. There will still be things you're unlikely to figure out (e.g. chirality).

Sorry but I do think it "is a stupid idea" because those spectroscopies are based on entirely different aspects

@NicolauSakerNeto thank you for your comment! You said that NMR has far more information about the structure of a molecule than MS. Does it mean that the reverse transition NMR -> MS would be possible? At least some approximation?

@SteffX I understand that the principles are entirely different. But it is not obvious why necessary they don't have anything in common. Imagine, you have a camera and the painter - underlying principles of making a picture are very different, but it doesn't mean you will see the completely different output.

The reverse is more feasible, yes, especially since it is possible to perform multiple complementary kinds of NMR measurements. In many cases it is possible to determine the molecular formula (and full structural information) from these combined NMR data, making MS data completely "superfluous" (though corroborating data are very important). However, there are also plenty of cases (probably most) in which even full NMR characterisation is not quite enough, and some portion of the molecule may still be unknown. This becomes more likely as molecules grow large (e.g. biomolecules)

Sorry, but in my opinion this endeavour is unlikely to be successful.  The techniques are quite different by their underlying principles. This is, at least in part, why chemists working in (especially, but not limited to) organic sythesis use these and additional ones to get a sufficiently "rounded picture" about the substances analyzed, putting the information each technique provides together.

mass spectroscopy typically relies on fragmenting the molecules in question, and then separates these then charged fragments by their mass/charge ratio.  There are techniques keeping the molecules almost intact (softer ionization techniques, used for example in MS-mapping of tissues), and others (electon impact ionization) where the molecules are intentionally shattered into many differently sized fragments.  Depending on the ionization technique applied, the eventually intensity of these individual signals will vary. And if the molecules of interest are embedded in a matrix (e.g., a serum, a tissue), identification of this molecules by MS becomes even more complex.
NMR spectroscopy typically characterizes probes soluted, keeping the molecules of interest (smaller ones, or large proteins) basically intact, without fragmentation.  You basically probe your sample for specific atoms of a kind (isotopes) and their nearest chemical environment. Yet the NMR spectrum of a given sample -- say of an amino acid (building block for proteins) -- already looks differently for scrutinizing the either peripheral hydrogens, the backbone of carbon atoms, or nitrogens.

Addendum
I think a few pictures may provide a glimpse about the complexity here.  For a fairly simple molecule like acetophenone, given the molecular structure is known, a sketcher like Chemdoodle may compute within a blink of an eye an estimate how the NMR spectrum either about the hydrogens, or about the carbon atoms will look like.  This is possible because  the underlying physical principles are known.  In addition, there are very many similar compounds analyzed successfully by NMR, allowing (fine tuning) of details in the spectra simulated, e.g. where to expect the signals to appear.  Regarding MS, a conservative approach is to predict only the peak(s) that by theory should appear without change of the molecule in terms of its mass.  In other words, without fragmentation.  Hence Chemdoodle's estimation is this:

(accessed demo page)
There are equally projects to simulate MS spectra, like CFM-ID.  Under the menu Utilities -> Spectra Simulation, you have the possibilty to try out different modi of ionization (shortcut).  Acetophenone's structure, expressed in a machine-readable Smiles code is
C1=CC=C(C=C1)C(=O)C

Then, just choosing ESI mode about positive ions, without considering adducts, the three MS spectra simulated by CFM-ID of actephenone, in increasing hardness of ionization are these:

As you see, not only the presence / absence of the signals changes in function of the probability to record certain fragments of a specific mass/charge ratio (m/z), their relative intensity equally is affected by changing the parameters of ionization.
(Hovering the mouse over the streaks allows you to display the suggested fragments responsible for the signals in these simulations.  Because of the special conditions in a mass spectrometer, unstable specis you wouldn't be able to store in a shelf may be seen, too.)

Thank you for your answer! Now I understand, that the underlying principles are different. One question: you as an expert, would you ever "guess" how would NMR spectra look like given an MS? Some kind of an approximation or this is not possible too?

@GalinaAlperovich Indeed, there are MS fragments of certain mass/charge ratio typically observed again and again for certain key structures.  A signal at m/z = 91 is typical for benzyl derivates, a m/z = 149 of derivatives of phthalic anhydride (often found in polymers), for example.  But it were only a hint for the potential presence of such a sub-structure present. Other techniques (e.g., NMR, or IR, etc) then would be used to check if this is supported by their results, or not.  Especially, since the prediction of NMR or IR spectra about a given (known) structure is easier, than for MS.

@GalinaAlperovich An addendum in the answer illustrates the problem in further detail.

thank you for your time and explanation! I will accurately go through it to understand these new things better.


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Comparing the bond strength between a polar bond and non-polar bond
According to NCERT class 12 part 2, $\ce{ICl}$ has lesser bond enthalpy than $\ce{I2}$. I have read that non polar covalent bonds are weaker than polar covalent bonds as in polar covalent bonds partial opposite charges develop on the bonding elements which leads to extra electrostatic interaction between them which is absent in non polar bonds. So according to what I read the answer should have been opposite. Please let me know where this logic is wrong. Also using this logic I understood that ionic bond is stronger than covalent bond.

Do not generalize too much.At the end the polar character of a bond is an intrinsic result and not something that is formed after or on top of a preexisting bond. Size and shape of the bond matter, too. Note that I and Cl are both halogens so ICl isn't much polar.  The same is true for the final statement. Ionic bond are very strong and easily the covalent ones can be much weaker - at least in vacuum!

@Alchimista My book says the order of bond strength of O-Cl< O-Br <O-I so as to explain the oxidising power of hypohalous acid (as the ease of releasing oxygen) HOCl>HOBr>HOI but according to you should it not be like just the reverse as 2p-3p overlap is stronger than 2p-4p which is stronger than 2p-5p. Is wrong reason given in my book if yes then how will you relate oxidising power of hypohalous acid?

Now Jasmine I do not see the answer in a quick.  Still take what the book say. You take all given order or label as absolute... Do not do that. I didn't say what you claim.  It would be a mixture of electronegativity and geometry. You should post a specific qusr

In polar covalent bonds, say $A–X$, assume electronegativity of $X$ is greater than that of $A$ then their is polarisation of bond $A–X$, $X$ gains partial negative charge $\delta–$ and $A$ gains partial positive charge $\delta+$ but this charge is very less (as electrons of sigma bond are held tightly) to account for any significant increase in bond strength of $A–X$ bond but the $–X$ group in $A–X$ molecule tends to attract $A–$ group of another $A–X$ molecule in this way there is electrostatic interaction between two different $A–X$ molecules giving rise to intermolecular forces (Dipole-Dipole interaction, hydrogen bonding etc).
DO NOT CONFUSE
FORCES BETWEEN TWO DIFFERENT MOLECULES AND COVALENT BONDS WHICH EXIST IN A SINGLE MOLECULE. THEY ARE TWO COMPLETELY DIFFERENT THINGS
However, attraction or repulsion may exist between two parts of a single molecule (generally happens for organic  molecules) but doesn't significantly affects bond strength between them rather they tends to change orientation of those groups or may rotate about the sigma bond.
And for $ICl$ and $I_2$, bond of iodine and chlorine is weak due to polarisation. And bond between iodine and iodine is strong covalent bond. I hope you know that polarisation of electron cloud of sigma bond decrease overlapping between orbitals.