What does "the magnesium salt of EDTA" mean? I am attempting to prepare an ammonium chloride/ammonium hydroxide buffer solution ($\ce{pH}=10 \pm 0.1$) for titrating water hardness with calgamite and EDTA.

In the 17th Edition of the Standard Methods for the Examination of Water and Wastewater, the reagent preparation section for titrating water hardness (Section 2340 C.) states:

Mix the $\ce{NH4Cl+NH4OH}$

  Add $\pu{1.25 g}$ magnesium salt of EDTA (commercially available).

Dilute with distilled water.

  After researching, I have assumed it to mean Magnesium Disodium EDTA (hydrate), CAS# 14402-88-1.

In reading the source material and searching online for synonyms of "magnesium salt of EDTA," I can't seem to find an exact hit on the term. Searches included PubChem, ChemSpider, Sigma-Aldrich, and Cole-Parmer, among others. Each seemed to point me toward "Magnesium Disodium EDTA Hydrate." I was hoping that with the amount of knowledge on this site, someone might recognize the term "salt of" and be able to help me out and clarify the term, or confirm that I am using the correct EDTA salt.

The authors have some strange kind of humor... it's a riddle.

Below you might see

  2) If the magnesium salt of EDTA is unavailable, dissolve 1.179 g disodium salt of
  ethylenediaminetetraacetic acid dihydrate (analytical reagent grade) and 780 mg magnesium
  sulfate (MgSO4⋅7H2O) or 644 mg magnesium chloride (MgCl2⋅6H2O) in 50 mL distilled water.
  Add this solution to 16.9 g NH4Cl and 143 mL conc NH4OH with mixing and dilute to 250 mL
  with distilled water.

From the molar mass of $\ce{MgSO4 * 7 H2O}$ ($\pu{246.47 g/mol}$) you can calculate the number of moles of magnesium the "magnesium salt of EDTA" should contain. With the help of the data sheet of your purchased magnesium salt of EDTA you can calculate the required quantity in grams.

You also could calculate possible stoichiometries of the imaginary "magnesium salt of EDTA" if you like. But be aware that not all commercially available magnesium salts of EDTA have a well defined stoichiometry.


How to filter out magnesium acetate (a soluble salt) quickly and cheaply from water I am conducting an experiment that involves reacting a weak acid and a weak base to form water and a salt with this overall equation

$$\ce{Mg(OH)2(s) + 2CH3COOH(l) <=> 2H2O(l) + Mg(CH3COO)2(aq)}$$
Currently, I am trying to filter out the salt to purify the water; however, there are specific restrictions.

The filtration process must be able to:

be done quickly. I can't wait a long time for the water to be distilled
be done at home. This filtration needs to be something that the average person can  could do with household ingredients or ingredients they can buy at a drug store (Assume that the $\ce{Mg(CH3COO)2}$ solution is given to them pre-made)

So far, I have tried to use Gelatin as a clarifying agent, which is typically used to clear beer. However, I couldn't find much info on using gelatin with salt water. Does anyone have any insight about using gelatin as a clarifying agent in this solution? Is there a better way to do such a thing that still meets the requirements?

Ion-exchange resin can be made with $\ce{H+}$ cations on one type of resin and $\ce{OH-}$ anions on another type of resin. These would leave the water without the any of the magnesium or acetate ions effectively without replacing the ions with other ions. This is how we deionize for general chemical use. This happens according to the reactions below:
$$\ce{2HSO3R + Mg^{2+} + 2C2H3O2- -> Mg(SO3R)2 + 2HC2H3O2}$$
$$\ce{R4N+OH- + HC2H3O2 -> [R4N+ +\ C2H3O2^{-}] + H2O}$$
In these examples, the sulfonates/sulfonic acids and ammonium cations are tethered to the resin so that they can't dissolve.

The downside to this is that if you are trying to measure something else in the system, if it is ionic, it will be removed as well.

I've never had to source ion-exchange resin myself, so I'm not sure of the easiest way to go about it. There's pleanty of deionization resin on Amazon. The mixed stuff is simple to use, just mix it with your water, agitate for a while and then filter it out and you'll have removed all the ions. That is expensive, though, because it's single use.

You can also buy regeneratable ion exchange resins. You don't want to mix the regeneratable stuff, and you have to do two separate ion-exchange steps, one with the cation-exchange and one with the anion-exchange. It's pretty easy to regenerate, though, as far as I know. You just use strong acid with the cation-exchange to redissolve the soluble cations and use a hydroxide solution to regenerate the anion exchange resin. It can't be done forever because there are some cations and anions that just won't want to redissolve, but if you're doing it a lot, might be worth it. I can't seem to find the anion stuff except in bulk.

I'm guessing the mixed type would be more appropriate for most applications. If you really need it quickly, I think you can buy the filters at Lowes/Home Depot and just run it through those or open them up and get the resin out.

Update: These filters are the only ones I saw that said they actually have the "dual ion-exchange system." Their website actually says it gets down to "000 total dissolved solids," which is pretty impressive relative to other water filters from what I've seen (depending on what "000 TDS" actually means).


Reaction between free radicals and water Free-radical substitution does not occur in aqueous solution as free radicals react with water (from school textbook). What does the reaction between free radicals, say Br, and water produce?

Bromine radical reacts with water

$$\ce{Br* + H2O -> HBr + *OH}$$

OH radical reacts with bromine

$$\ce{*OH + Br2 -> BrOH + Br*}$$

to give a new bromine radical, and so on, until the bromine is totally consumed.


Chemical time scales In calculating UV/VIS spectra with computational procedures (say, TD-DFT), it is commonly assumed that the geometry of the ground-state is still valid also for the excited state due to the short time scale (I more or less took this statement from here).

So I asked myself, are there good references on time scales for chemical processes?
Some phenomena may be obviously faster than others, but others (say, two molecular excitations) may be not that obvious to compare.

I am looking for something simple, like a table (or tables) with examples.

According to the Franck-Condon principle which you are implicitly referring to, absorption is indeed very fast compared to all other processes.

If you are looking for time scales of photophysical processes, I can suggest you for example B. Valeur's book "Molecular Fluorescence: Principles and Applications", chapter 3 (Characteristics of fluorescence emission). You'll find this list of characteristic times:

absorption: $10^{-15}$ s
vibrational relaxation: $10^{-12} - 10^{-10}$ s
$\ce{S_1}$ lifetime (fluorescence): $10^{-10} - 10^{-7}$ s
intersystem crossing: $10^{-10} - 10^{-8}$ s
internal conversion: $10^{-11} - 10^{-9}$ s
$\ce{T_1}$ lifetime (phosphorescence): $10^{-6} - 1$ s

as well as examples of molecules with their photophysical data.

For chemical processes, it depends on many things, such as concentration, diffusion rates... and you would need to precise your question (type of photochemical reaction, type of quenching, etc.).


Getting the pressure using the equation of states - will the answer be the same? When I get the pressure using the different equations of state (e.g. van der Waals, Redlich-Kwong, etc.), will the answer still be the same?

In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter.

For example, the van der Waals EOS gives pressure values at high temperature and low pressure for noble gases that are close to what you would obtain with the ideal gas law.

Since it is typically easy to calculate pressure from an EOS given temperature and volume, you can try this yourself and see how much of an impact the choice of EOS makes for your particular system at a particular point in phase-space.


Does radioactivity affect chemical reactions? Do compounds of radioactive elements show a different behaviour? What happens when the radioactive part of the given compound decays? e.g. Radium chloride; what happens to it when radium decays?

I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay can cause damage to crystalline solids.

Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the host crystal. The crystal can then be polished and etched to reveal these so called "fission tracks":

(source: Stanford Fission Track Thermochronology Laboratory)

Heating of a damaged crystal will cause the damage to "heal" and the tracks to disappear. Thus, radioactive decay can indirectly induce chemical reactions. This routinely occurs in the deep Earth. It also has some very useful applications in geochronology, which you can read about in the Wikipedia article about fission tracks.


What exactly is a spontaneous reaction? What I understand by a spontaneous reaction is that - A reaction whose G is negative and which occurs without the input of energy.

Gasoline reacts with oxygen only if a spark is provided. So how can it be a spontaneous reaction?

Do nonspontaneous reactions occur in nature? If so, are they always slow? If they are slow then what is the difference between a slow spontaneous reaction and a nonspontaneous reaction?

Are nonspontaneous reactions active energies more than that of spontaneous reactions?

Spontaneous reactions simply mean that the products will be lower in free energy than the reactants ($\Delta G$ is negative). If $\Delta G$ is negative, then it is energetically favourable for the reaction to occur - in other words, there will be a release of energy due to the reaction.

The magnitude of $\Delta G$ (how large it is) does not say anything about the reaction speed. For example, the reaction between gasoline and oxygen in the atmosphere at room temperature will be extremely slow, despite a large, negative $\Delta G$. This is because this reaction has a large activation energy, $E_A$.

Hold on. When two reactants collide, they interact to form an extremely unstable structure called the transition state. The transition state is high in energy and is NOT energetically favourable, and thus the transition state quickly collapses again and becomes either the reactants or the products. You can see this in the following picture.

Depending on the kinetic energy of the reactants, they might not reach the top of the curve, in which case they will fall back to being reactants. But if they collide fast enough and are positioned appropriately with respect to each other, the top of the curve is reached, and the products will form. The energy required to form the transition state is the activation energy.

With gasoline and oxygen, the activation energy is high. So even though a tremendous amount of energy is released during the reaction, it needs a spark to overcome its activation energy. From here on, the heat generated from the reaction supplies the activation energy.

If this reaction was not spontaneous, we might be able to force the reaction along, but the reaction would not continue on its own.

As you know, not all reactions run to completion. This is because Gibbs free energy is dependent on concentrations of reactants and products, so as the products accumulate, and the reactants are used, $\Delta G$ becomes closer to 0, eventually reaching equilibrium, where $\Delta G = 0$. Visualise this scenario on the above picture. If there is no difference in Gibbs free energy, the reactants will still get to the transition state and turn into products. But the energy required for products to reach the transition state and becomes reactants is equivalent, and thus the rate of the forward reaction is equal to the rate of the reverse reaction.

The difference between reactions that will reach an equilibrium and the ones that run to completion is the magnitude of the $\Delta G$ at standard conditions (denoted $\Delta G°$). This is a measure of how energetically favourable the reaction "inherently" is. If $\Delta G°$ is large and negative, the reaction will run to completion. If it is smaller, the reaction will establish an equilibrium at some point.

That's at least generally true for reactions, where all of the species stay in the same phase I think. If you envision a reaction where one of the products is a gas, that escapes the solution of reactants, then the products will be unable to reform into reactants.


Odorless smoke from 1812 Overture cannon fire I went to an outdoor performance of the 1812 Overture last night. At the finale, a sequence of 'cannon fire' occurred.

In past years the smoke has had the expected sulfur odor but this year the dense cloud of white smoke that drifted over the crowd had no (to my nose) detectable odor at all. In all other ways the detonations were normal- extremely loud, jet of flame, dense white smoke. Any ideas about what explosives were used?

Most likely they replaced the black powder with nitrocellulose also known as smokeless powder (although it does generate some smoke especially when used in artillery) which does not contain sulfur and hence the lack of smell.


How can water catalyse a reaction between iodine and aluminium? A few drops of water can initiate a reaction between iodine and aluminium.

https://www.youtube.com/watch?v=SKSU72-1ERc

How can this happen, since iodine is only slightly soluble in water?

A naked aluminum surface (freshly sanded) in air immediately grows about 4 nm of adherent aluminum oxide that passivates the surface.  Iodine slightly dissolves in water and disproportionates

$$\ce{I2_{(s)} + H2O_{(l)} -> HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$

Iodine is strongly solubilized by forming triiodide.

$$\ce{I2 + I- -> [I3]-}$$

The acids chew through the alumina surface barrier.  When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air.

http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm
$$\begin{alignat}{2}
&\ce{AlI3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -300.8\ \mathrm{kJ/mol}\\
&\ce{Al2O3_{(s)}}\quad &&\Delta G_\text{f}^\circ = -1582.4\ \mathrm{kJ/mol}\\
\end{alignat}$$
Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.


Why are carbons of Inositol chiral centers? I was asked to draw all possible stereoisomers of inositol (1,2,3,4,5,6-cyclohexanehexol). To obtain the answer I had to assume that all six carbons of the molecule are asymmetric, which (bearing in mind possible molecular symmetry) results on the following nine stereoisomers:[1]

The problem is that I fail to see why these carbons can be considered chiral centers. Given:[2]

  The most common cause of chirality in an organic molecule, although not the only one, is the presence of a carbon atom bonded to four different groups [...]. These carbons are now named chiral centers [...]

The chirality of the molecule is not the reason of my doubts, I can see that the only chiral isomers are D- and L- chiro-inositol. Since all carbons in the ring possess a hydroxyl group, can they not be considered to be bonded to equivalent substituents, making them symmetric? Is the possibility of the -OH groups to be above or below the plane of the ring what makes them asymmetric?

From Wikipedia
McMurry, John. Organic Chemistry. CENGAGE Learning, 2008. 7th edition.

Consider 2,3-butanediol.

This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.

A carbon center is asymmetric if it has 4 different substituents regardless of whether or not the whole molecule is chiral. The ring substitution in inositol is such that the centers around the ring are not identical (including when you look at stereochemistry--for example, (R) has higher priority than (S)). This of course isn't enough to help you easily make stereochemical assignments for all of the centers of inositol, but hopefully, this allows you to understand why the centers are still chiral centers.

As an aside, there are a few centers that cannot be chiral because they lie on a mirror plane. Interestingly though, changing the configuration of the center produces a different molecule. These centers are called pseudochiral, and we label them with lower case r and s.


How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced? I have the formula $\ce{Co(C2O4)*H2O}$ in a reaction that forms cobalt oxide $\ce{Co3O4}$
My data here says that my oxalate hydrate weighed $\pu{0.3283 g}$ and my product oxide weighed $\pu{0.1158 g}$ which I have calculated as:
$$0.1158 \times \frac{\pu{176.79 g}\, \mathrm{(cobalt\,oxalate)}} {\pu{240.79 g}\, \mathrm{(cobalt\,oxide)}} =\pu{0.08502 g\,\mathrm{(cobalt)}}$$
Then $\pu{0.3283 g (reactant)} − \pu{0.1158 g (product)}$ gives me $\pu{0.2125 g}$ water removed divided by $\pu{18.016 g/mol}$ to get $\pu{0.01179 mol}$ $\ce{H2O}$ in the sample.
How do I use this data to determine the empirical formula? I am not sure how to determine the molar ratios from this point.

I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.
Some parts of your calculation are correct, you do however forget that $\ce{CO2}$ is released. This would be my approach:
$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$
This means that there was three times as much $\ce{CoC2O4}$ in your sample.
$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$
That equals a weight of $\pu{212.1 mg}$. Your original sample weighed $\pu{328.3 mg}$, so there is $\pu{116.2 mg}$ left that seem to be $\ce{H2O}$ ($\pu{18.015 g/mol}$), giving you.
$$n_\ce{H2O} = \pu{6.45 mmol}$$
The ratio of cobalt oxalate to water then is $\left(\frac{6.450}{1.443} = 4.5\right)$ which implies the empirical formula is $\ce{2CoC2O4.9H2O}$


Understanding Avogrado's hypothesis I have read that gases under identical conditions of temperature, pressure, and volume contain the same number of molecules, but every gas has different types of atoms so their force of attraction is also different then how can this be true?

If we consider **Ideal gas equation **
and ignore the type of interaction which are present in the real gases then ;
$PV=nRT $
wherevP= pressure ; v=vol. ; n= moles; R = universal gas constant ; T= temp.

$PV/RT=constant $
so $n=constant$

so if$ n$ is constant then if we multiply it with avogadro's constant $N_A $ it would give you number of molecules .
Again that's when ideal gas .


Shouldn't the pH at the equivalence point always be 7? I learned in class that the equivalence point in an acid-base titration is reached when the solution contains an equal amount of substance of $\ce{OH-}$ and $\ce{H+}$ ions. However, in a weak acid and strong alkali titration, the pH at the equivalence point (when I added the right amount of acid and alkali to make it a neutral solution) is 9. Also, in the titration of a strong acid and weak alkali, the pH of the solution when I have added the correct amounts is 5.

My question:

If I add 1 mole of ethanoic acid to 1 mole of sodium hydroxide, I should get a neutral solution and the pH should be 7. However, why is the pH 9? This does not make sense to me as a pH 9 suggest that the concentration of $\ce{OH-}$ is higher than the concentration of $\ce{H+}$. But how is that, I reacted the correct amounts?

Two kinds of reactions have to be considered:

Neutralisation of the weak acid with the strong base
Secondary reactions from the products

The neutralisation is fairly straightforward:
$$\ce{HA + NaOH -> NaA + H2O} $$

However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated base of the acid can also react (as a base):
$$\ce{A- + H2O <=> OH- + HA}\; ,$$

thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.

So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.


How many bonds can nitrogen form? As known, nitrogen could form 3 bonds based on octet rule, because it has 5 valence electrons. That means it needs 3 bonds.

On the other hand, why sometimes nitrogen forms 4 bonds?

I'd like to try answer the question from the title regarding the maximum number of atoms nitrogen is capable of bonding with, and also slightly expand my comment.

Metal nitrido complexes are commonly known to have up to 6 metal centers bound with a single bridging $\ce{N3−}$ ion, located in an octahedral cavity. An interstitial nitrogen can contribute 5 electrons, and the rest is provided by the group 9 and 10 metals which are electron-rich (typically, $\ce{Rh}$, $\ce{Ir}$).

There are few more exceptions where formal C.N. for nitrogen is 7: complexes of lithium amides based on $\ce{\{Li14N10\}^{6-}}$ cluster framework [1,2] and an inclusion nitrido-cluster $\ce{\{Co2RhN2\}^{3-}}$ [3]. Unfortunately in both crystal structures [1,2] with 6- and 7-fold coordinated nitrogens of $\ce{N-Ar}$ groups those are heavily disordered. Structure [3] is more suitable for the representation.

tris(Tetramethylammonium)($\mu_7$-nitrido)-($\mu_6$-nitrido)-decakis($\mu_2$-carbonyl)-undecacarbonyl-deca-cobalt-rhodium(I) $\ce{[Co10RhN2(CO)21]^3-}$ [3] contains two non-equivalent 6- and 7-fold coordinated nitrogen atoms ($\mathrm{N2}$ and $\mathrm{N1}$, respectively), sharing a triangular face:

$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$;
$\color{#0A7D8C}{\Large\bullet}~\ce{Rh}$;

Cluster core wireframe model without carbonyl ligands:

Atom $\mathrm{N1}$ with C.N. 7 is coordinated with 6 cobalts and 1 rhodium, forming a capped trigonal prism. Interestingly enough, $\mathrm{Co1}$ is a capping atom, not rhodium:

     N1   SYMM Co5   Co4   Co6   Co3   Co2  Rh1  Co1
Co5  1.90 I    -     -     -     -     -    -    -
Co4  1.91 I    135.0 -     -     -     -    -    -
Co6  1.92 I    79.5  80.2  -     -     -    -    -
Co3  1.98 I    129.7 85.1  80.2  -     -    -    -
Co2  2.00 I    82.3  140.2 128.4 75.4  -    -    -
Rh1  2.18 I    80.7  81.1  128.3 144.9 95.3 -    -
Co1  2.43 I    143.5 70.3  136.5 66.4  70.1 78.6 -

Both interstitial nitrogens play the role of internal ligands, which provide
cluster valence electrons (CVE), but don't contribute to steric hindrance between external ligands such as carbonyls, making the cluster more stable [4, ch. 1.18]

Bibliography

Armstrong, D. R.; Barr, D.; Clegg, W.; Drake, S. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wright, D. S. Angew. Chem. Int. Ed. Engl. 1991, 30 (12), 1707–1709. DOI 10.1002/anie.199117071.
Armstrong, D. R.; Ball, S. C.; Barr, D.; Clegg, W.; Linton, D. J.; Kerr, L. C.; Moncrieff, D.; Raithby, P. R.; Singer, R. J.; Snaith, R.; Stalke, D.; Wheatley, A. E. H.; Wright, D. S. J. Chem. Soc., Dalton Trans. 2002, 0 (12), 2505–2511. DOI 10.1039/B107970K.
Costa, M.; Della Pergola, R.; Fumagalli, A.; Laschi, F.; Losi, S.; Macchi, P.; Sironi, A.; Zanello, P. Inorg. Chem. 2007, 46 (2), 552–560. DOI 10.1021/ic0608288.
Metal clusters in chemistry; Oro, L. A., Braunstein, P., Raithby, P. R., Eds.; Wiley-VCH: Weinheim; New York, 1999. ISBN 978-3-527-29549-4.


When does a solute form ions? If I have $\ce{HCl}$ and I put it in water, it will form ions. But, if I take $\ce{HCl}$ and put in benzene, it won't. How do you know that? If I had other substances, how would I know if it forms ions or not?

In water, HCl reacts to form $\ce{H3O+}$ and Cl-.

Then, water molecules surround the respective ions in a stabilizing orientation.
The partially negatively charged O atoms of water interact with the $\ce{H3O+}$.
The partially positively charged H atoms of water interact with the Cl-.

None of the above can occur in benzene because benzene is very reluctant to accept a proton and because benzene is non-polar.

Generally speaking, polar solvents have an ability to stabilize ions that nonpolar solvents do not.


Is the Hartree-Fock procedure the same thing as the self-consistent-field (SCF) theory? Is it right to think that the Hartree procedure is called the SCF theory, as the notion of self-consistent field comes from approximating multi-electron wavefunction as multiple one-electron wavefunctions. Since we don't know the final charge density, we must iterate until it is self-consistent?

No.

The Hartree-Fock method encompasses a set of assumptions used to formulate and solve a series of equations embodying a mean-field, time-independent Schrödinger equation.

The self-consistent field method is a computational technique used in myriad applications where the equations to be solved are implicit—that is, where it is impossible to isolate an analytically solvable expression for a variable and its derivatives.  To the best of my knowledge, any solution method involving a "guess-and-check" approach can be considered a 'self-consistent solution method'.

The Hartree-Fock method is an example of a use-case for a self-consistent field procedure, but they are not synonymous.


How does a d4 high-spin complex have a spin-allowed transition without violating the selection rule for total angular momentum? 1) I am confused by the fact that the spin-allowed transition for a high-spin d4 complex is 5E→T2

Doesn't this violate the spin selection rule for total angular momentum?

2) Also, for a d6 complex, some sources I've consulted with mention a one spin-allowed transition for a high-spin complex, and no spin-allowed transitions for a low-spin complex.

From my understanding of the spin selection rules and the use of the Tanabe-Sugano diagram, I found no spin-allowed transitions for a high-spin complex and 4 spin-allowed transitions for a low-spin complex (i.e. 1A1→1T1; 1A1→1T2; 1A1→1A2; 1A1→1A2)

The photon has one unit of angular momentum and as angular momentum is conserved this must be taken up by the molecule; in terms of symmetry species this means that the product of the symmetry species of the two states involved and of the transition dipole must be worked out. The electron spin in not affected by the photon angular momentum so the spin selection rule applies $\Delta S=0$ unless there is also some spin - orbit coupling then otherwise spin-forbidden transitions can occur but with generally low probability compared to a spin allowed one.

To work out the symmetry species involved in a transition the value of the integral

$$\int\psi_i\mu\psi_fd\tau$$

between initial $i$ and final state $f$ has to be evaluated.; $\mu$ is the transition dipole. This integral is very hard to evaluate but fortunately symmetry can help.

Using symmetry the integral can be determined to be zero (transition forbidden) or not zero. The equation to use is

$$\Gamma\psi_i \Gamma\mu \Gamma\psi_f = \;'A_1\,'$$

where $\Gamma$ means the symmetry species, $A,\; E$ etc and $'A_1\,'$ means the product must contain the totally symmetric representation in the point group used. This is the label given to the top row of characters in the table ($A_1$ or $A_{1g}$).

The transition dipole symmetry is found in the column of the point group just to the right of the characters and are the symmetry species with x, y, and z. In the $T_d$ and $O$ types of point groups the transition dipoles are either the $T_2$ or $T_1$ symmetry species respectively.

Thus in your $A_1\to T_1$ example the symmetry product is

$$A_1 \otimes T_1 \otimes T_1 = A_1$$

in an $O$ point group. The symbol $\otimes$ means multiply characters in each symmetry species column by column.  Hence you can work out the product  by multiplying the characters together or more easily by looking up the values in direct product tables.

In this case this is easy to do as anything multiplied by $A_1$ is unchanged (as its totally symmetric) and any symmetry species multiplied by itself always contains the totally symmetric representation $A_1$. So the transition
$A_1\to T_1$ is allowed in octadedral point groups.

In the tetrahedral point group $T_d$ where the transition dipoles each transform as $T_2$ the product is

$$A_1 \otimes T_2 \otimes T_1 =  T_2 \otimes T_1$$

and now the characters have to be multiplied and the the irreducible representation found (see link to answer below). Instead looking up the product in direct product tables gives

$$T_2 \otimes T_1 = A_2+E+T_1+T_2$$

and in this case as the totally symmetric representation $A_1$ is absent the transition would not be allowed. This means that any transition will be weak, i.e. has a very low but finite extinction coefficient. There are effects in molecules that 'break' the symmetry, such as vibrations distorting the molecules 'ideal' symmetry and these can make forbidden transitions slightly allowed.

This answer gives some details on point groups.
Understanding group theory easily and quickly


Reaction of anisole and succinic anhydride in presence of aluminium chloride Predict the structures of the intermediates/products in the following reaction sequence:

In the first step, I know that aluminium chloride  is playing some role,  as otherwise hydronium ion would have reacted with succinic anhydride  to give succinic acid.
I thought of the following two possibilities:

Friedel Crafts: But till date, I have only seen Friedel Crafts type reactions and have not seen aluminium  chloride complex with either acid or anhydride. I could also not find it on Google.

Chloride formation: But I have also never seen aluminium chloride forming a chloride in reaction with anhydride. I am not sure if it does that, does it?

So, what is the reaction mechanism for the formation of (A)?
Source: Joint Entrance Exam (JEE) 1996 India

Aluminium chloride ($\ce{AlCl_3}$) is mainly used for this Friedel-Crafts alkylation or acylation, where this compound acts as catalyst, and also controls the production of the desired compound.

For example, in this case, the reactants are anisole ($\ce{Ph -OMe}$) and succinic anhydride. So, the main reaction which drives these reactants towards product is the nucleophilic attack by the para carbon (w.r.t the $\ce{OMe}$ group) of the phenyl ring at the carbonyl carbon of the succinic anhydride, and an addition-elimination of the attached oxygen of the ester part in succinic anhydride.

For this, we need the state of the reactants in a proper way. We have to make sure that, the oxygen of $\ce{OMe}$ does not attack prior to the phenyl ring, and also, the electrophilicity of the carbonyl carbon needs to be increased to much extent, which is generally not so much due to present resonance in the compound.

$\ce{AlCl_3}$ plays a special role over controlling these two phenomena. Being a Lewis acid, it forms a Lewis acid-base interaction with the oxygen of $\ce{OMe}$ and also with the oxygen atom of $\ce{C=O}$ group in succinic anyhdride. Due to this interaction, a δ+ charge is introduced on both oxygen atoms described before. This increased formal charge helps to increase the electrophilicity of the carbon due to increased -I effect of the attached oxygen and also reduces the chances of attacking by $\ce{OMe}$ group. This catalyses to form our desired product with the proper reaction between the reactants.


What metabolite intermediate will accumulate in a mammalian cell-free extracts capable of glycolysis if lactate dehydrogenase is inhibited? I know the reaction proceeds as follows:

Pyruvate accepts the electrons from NADH to form lactate. What would be the intermediate in this reaction? Would it be NADH or NAD+?

As you probably know, lactate dehydrogenase catalyzes the interconversion of pyruvate and lactate, with electron transfer via NADH/NAD⁺. If lactate dehydrogenase is inhibited, the fate of pyruvate in the cell extract will depend on what cell structures as wells as enzymes and other compounds are available. In other words, the answer to your question will likely be strongly dependent on the organism, as well as the particular circumstances of the experiment you are performing.

To give you a speculative answer: Assuming that the metabolic pathways of the mammalian cell still function and that oxygen is available to accept electrons in respiration, pyruvate could be metabolized through the tricarboxylic acid cycle, forming energy-rich compounds and carbon dioxide. If other fermentative pathways are available, pyruvate could be funnelled into those.

In any case, predicting what compunds would accumulate would require more information than that which is given in your question.


Reusing boiling chips I was wondering why cannot boiling chips be reused.

As far as I know, they are microporous materials, often coated with PTFE.

They work by, I suppose, giving the heated substance a small cavity, in which a phase transition is favored.

Yet, the golden rule of boiling chips is: they cannot be reused.

I was wondering why they cannot be treated the way dessiccant materials (for instance, molecular sieves) are used, for instance by washing them after use and driying them in oven (especially when non-dangerous materials are involved).

According to this post at www.reference.com/science

  Boiling chips are generally porous, trapping heating bubbles and
  allowing solvent vapors to form in the air cavities. They should
  always be added to a solvent prior to the beginning of a heating
  cycle; adding to a hot liquid may cause it to violently boil over.
  Boiling chips and stones are only used once, as the absorptive
  material becomes impregnated with the first solvent used in.

And Wikipedia confirms the basic idea of why they shouldn't be used more than once:

  The structure of a boiling chip traps liquid while in use, meaning
  that they cannot be re-used in laboratory setups.

From personal experience however, I have to disagree with these absolute "cannot be done" sort of statements.  I used to rinse the used chips in an appropriate solvent and then heat them at some temperature well above the solvent's boiling point overnight and don't remember ever having problems.

I realize that the boiling point isn't completely relevant when you can have porous, high surface area material with adsorbed solvent.  Also, as I stated in the comments, I don't remember using PTFE coated chips.


Use of osmosis to measure molecular weight of macromolecules I found a question in a biochemistry book.

How can osmosis be used to measure molecular weight of macromolecules? Is it applied practically?

A general answer without the usage of many technical terms would be most suitable for me.

It has to do with the fact, that osmotic pressure of any solution is dependent only on the relative number of particles in the solution (in case of dilute solutions), irrespective of their nature. Hence osmotic pressure is called a colligative property.
Therefore, you take a certain known weight of the substance you want to calculate the molecular mass of, prepare a solution and measure the osmotic pressure. Then you compare this to the known osmotic pressure of standard solution which tells you the relative number of particles required to  create such a pressure. From this and the knowledge of the weight you used to prepare the solution, you can get the required molecular weight (weight of Avogadro's number of particles). In practice, certain complications, like dissociation and association might complicate the matter a bit.
This method is widely used to calculate the molecular weight (especially) of polymers. See Berkeley and Hartley method.


What is the term for passing an eluate back through an adsorption column? In my experiment I am passing a solution through an adsorption column, collecting the eluate, then passing the eluate back through the same column.

Is there a concise term which describes this process?

This process is similar to re-cycling chromatography or recycling HPLC. In a more strict sense, it does not do any good, at least theoretically, if you keep on recycling all the eluate and transfer it back to the column.

What makes more sense is that you collect fractions and choose certain fractions for recycling.


What is the relation between Rate of reaction and rate of consumption of reactants? The rate of the following reaction is 0.300M/s. What is the relative rate of change of each species in the reaction?$$\ce{A + 3B -> 2C}$$

We are supposed to find A and B and I understand that we have to use $\frac{ [\delta]A}{[\delta]t}$ and then use a negative for the coefficient. But what I don't understand is, why for B we had to multiply -3 but not -1/3? And is it the same for A?
Also, I am confused on how I can differentiate between using these two equations of rates. The Delta of reactants/delta of time, versus the rate $v(nu)=k[A]m[B]n$
equation.

For your first part you're supposed to multiply it by -3 instead if $\frac{-1}{3}$  as the rate of reaction is given for one mole. That is if $R$ is the rate of reaction it is defined as .
$$\frac{-d[A]}{dt} = \frac{-d[B]}{3dt} = \frac{d[C]}{2dt} = R$$
So the rate of consumption of B is $-3R$.

As for your second part it's a tad unclear for me if you could further edit your question.I could take a shot at answering your second part as well


Why does neopentane have a higher melting point than n-pentane? Why does neopentane have a higher melting point than n-pentane?

I know that we can say that, due to stacking, it has a higher melting point as it freezes easily. But then, why doesn't this happen in the case of n-hexane too?

n-Hexane has a melting point of −95 °C whereas 2,2-dimethylbutane has a melting point of −98 °C. Instead, if we go by stacking, 2,2-dimethylbutane should have a higher melting point.

image source

Many symmetrical molecules undergo a phase change before they melt.  This is usually interpreted as the ability to rotate without disrupting the lattice structure.  This has a much greater effect on the entropy of the solid than on its energy.  This reduces the entropy of fusion more than it reduces the enthalpy of fusion, thereby raising the melting point which equals delta-H/delta-S.  This is very dramatic in the comparison of melting points of octane, cyclo-octane, and bicyclo-octane.


Preparation of ammonium acetate buffer for HPLC Buffer solutions are typically prepared by dissolving an appropriate salt in water and then adjusting the $\mathrm{pH}$ to the desired position by addition of the conjugate acid or base.
For example, to prepare a $\pu{100 mM}$ ammonium acetate solution at $\mathrm{pH}~9,$ you’d first prepare a $\pu{100 mM}$ ammonium acetate, and then add ammonium hydroxide dropwise until the desired $\mathrm{pH}$ was achieved (a titration, using a pH meter to determine endpoint).
I’d like to work out (roughly—I understand that what I’m about to describe is nothing more than an approximation, but it’s a reproducible approximation at least which is good enough for our needs) directly how much ammonium hydroxide and acetic acid I would need to make up the same solution.
I understand the fundamental Henderson–Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A}]}{[\ce{HA}]}$$
But as I see $\ce{A}$ is ammonium and $\ce{HA}$ is ammonium acetate. Thus, in my example I don’t really simplify things that much.
Additionally the majority of examples I’ve found look at acidic buffers rather than alkali ones. Does my equation therefore need to be rearranged further to take account of this?

to prepare a 100 mM ammonium acetate solution at pH 9, you’d first prepare a 100 mM ammonium acetate, and then add ammonium hydroxide dropwise until the desired pH was achieved.

Well, this procedure is good enough for a rough screening solution- call it a quick and dirty method. The main issue is that one would not prepare a solution of 100 mM ammonium acetate beforehand and add drops of ammonia to that. Whenever you need to prepare a buffer solution, you should know what volume is desired and what concentration is needed.
For HPLC, let us say you need 1000 mL of buffer containing 100 mM ammonium acetate. One would dissolve the required amount of ammonium acetate in about 900 mL water in a 1 L beaker, add ammonium hydroxide dropwise till you reach the desired pH, add more water till you are close to 950 mL. Check the pH again, and adjust as desired, and then quantitatively transfer the solution to a 1 L volumetric flask and then make up the volume to 1 L. This will be your analytically correct 100 mM ammonium acetate buffer.
Now, coming to the buffer equation. Again, for that you need to define the concentration and volume. What would ask yourself what should be concentration of ammonia should be present in 1 L of 0.1 M ammonium acetate solution such that the final pH is 9 solution.
You would use the pK$_a$ of ammonium ion and from the equation you would figure out the concentration of ammonia.
See example no. 2 here: Buffer calculation
And also study example 3. Ammonia buffers
Now the problem is that conc. ammonia is sold as w/w % solution. Using the density and percentage, you need to determine the molarity of ammonia solution as well to actually prepare a buffer.


Why are the halogens good disinfectants? I've been searching around the internet for a while and I know that Chlorine, Bromine and Iodine are used as disinfectants.

My question is, what is the property of the halogens that make them suitable for killing microbes? Is it just their toxicity?

The halogens, particularly in their diatomic free states and within various oxoacids, are strong oxidizing agents by virtue of their high electronegativities, electron affinities, and reduction potentials. The polarizability of the heavier halogens also makes them almost uniquely versatile as both good leaving groups and strong nucleophiles, depending on conditions. Additionally, the relatively low bond dissociation energies of the halogens contribute to their aptitude to react by free-radical mechanisms (coupled with the exothermicity that usually accompanies the propagation of those reactions).

In their diatomic forms, they readily add across the $\pi$-bonds of alkenes and alkynes (which are ubiquitous and essential structural features of myriad types of biomolecules), and they can oxidize numerous other functional groups under the right conditions and in combination with other substances. Various oxoacids that incorporate halogens are often extremely strong oxidizers as well (household bleach, for example, is an aqueous solution of sodium hypochlorite, which is mild by comparison to certain other halogen-containing oxoacids). In those reactions, they typically serve to add oxygen to molecules in place of, e.g., hydrogen, while themselves serving as leaving groups (or parts of leaving groups).

The antiseptic properties of halogens (and halogen-containing oxoacids) are directly consequent to those chemical properties. At least one source suggests that elemental iodine, for example, reacts by mechanisms similar to the ones described above to disrupt key structures of bacterial enzymes, proteins, and lipid membranes: it oxidizes the thiol groups of cysteine or methionine to disulfides, interfering with protein synthesis; it reacts with unsaturated lipids to alter their structure; etc. These are biological equivalents of the very same reactions that are routinely carried out in organic chemistry labs. If you want more details/examples, they should be easy to find with a web search.

In short, the very same awesome chemical reactivity that makes them so indispensable in organic synthesis, also means that they are highly reactive towards (and hence, usually harmful to) all manner of organisms.


Why are hydrogen ions always associated with another molecule? I was reading N.C.E.R.T. class XI part 2 book and chapter Hydrogen,when I came across this statement:

  Loss of the electron from hydrogen atom
  results in nucleus ($\ce{H+}$) of ~1.510–3 pm size.
  This is extremely small as compared to normal
  atomic and ionic sizes of 50 to 200pm. As a
  consequence, H+ does not exist freely and is
  always associated with other atoms or
  molecules.

So a question arose in my mind that why it is a consequence of small size of hydrogen nucleus that $\ce{H+}$ does not exist freely and is always associated with other atoms and molecules?

The way I understand it is (and my understanding is by no means perfect, or complete), as you pointed out correctly: a hydrogen ion is in fact a proton. The proton is a "bare charge" and as you rightly said, "tiny". a
This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.

For instance, in an aqueous solution it forms $\ce{H_3O+}$ (hydronium ion, which are further solvated by water molecules and form a manner of aggregates).

An extremely interesting way of looking at it was pointed out in the comments (just putting it here, so no one misses it)

  An interesting way to thinking about the reactivity of a bare proton is to calculate its charge density and compare with other monocations, such as Li+. This will make it clear why free protons will abstract electrons from just about anything. –  Nicolau Saker Neto

Do note (as far as I know, someone might correct me if I am wrong) free protons are fairly stable (by free I mean not bound to electrons or other nucleons) and usually don't decay into other particles. They exist in several naturally occurring situations, high energy plasmas come to mind.


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

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

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

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

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

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

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

References:

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


Why ginger reacts with sparkling water? I think that the ginger ale is born this way.
Anyway, if you put the ginger in a glass and add sparkling water, small drops splash out of the glass, the water is fizzier, more effervescent.
Why? what makes the two of them react?

According to this Wikipedia article:

  Carbonated water (also known as club soda, soda water, sparkling water, seltzer water, or fizzy water) is water into which carbon dioxide gas under pressure has been dissolved. Some of these have additives, such as sodium chloride, sodium bicarbonate or similar, but seltzer water is almost always composed of water and carbon dioxide with no other additives.

So the only possible candidates for a chemical reaction from the sparkling water are $\ce{CO2}$, and water itself. The other salts present at low levels would not have sufficient reactivity to produce the relatively high-energy results observed. And water can be ruled out as a reactant because ginger itself of course is a plant that contains water (or contained water previously if dried ginger was used). For the sake of thoroughness, I went to my kitchen and added tap water to dried, ground ginger and observed no effervescence.

This leaves $\ce{CO2}$ as the only candidate for causing the observed reaction upon the addition of ginger to sparkling water. A chemical reaction can be ruled out from a couple perspectives. First, as $\ce{CO2}$ is a very stable, completely oxidized compound, ginger would have to contain some strong reducing agent for it to react. We know this is not the case as the ginger is produced and stored in an oxidizing environment (our atmosphere of $\pu{21\%}$ oxygen). The atmosphere also contains on the order of $\pu{0.04\%}$ $\ce{CO2}$, which is less than that found in sparkling water but would still have quenched any $\ce{CO2}$ reactive component, albeit in slower less dramatic fashion.

At this point we've determined the effervescence must be a result of the $\ce{CO2}$ in the sparkling water, but we've eliminated the possibility of a chemical reaction. Because the carbonation process leaves the sparkling water supersaturated in $\ce{CO2}$, the only thing preventing the $\ce{CO2}$ from bursting out all at once is the lack of sites for $\ce{CO2}$ bubbles to nucleate. And this is what the ginger provides, particularly if it were in a ground, high-surface area form, but even in the case of raw ginger slices there are many more surface sites available than a glass container.

Bottom line, TL/DR: There is no chemical reaction, rather the ginger just provided a relatively high surface area for the rapid formation of $\ce{CO2}$ bubbles from the $\ce{CO2}$ supersaturated sparkling water.


Express the maximum work from a voltaic cell The net cell reaction of an electrochemical cell and its standard potential is given below:
$$\ce{ Mg + 2Ag+ ->Mg^{2+} + 2Ag} \ \ \ \ \ \ \ \ E^\circ=3.17\:\mathrm{V}$$
The question is to find the maximum work obtainable from this electrochemical cell if the initial concentrations of $\ce{Mg^{2+}}=0.1\ \mathrm{M}$ and of $\ce{Ag+}=1\ \mathrm{M}$.

The solution just uses the Nernst equation to find the potential at this concentrations and uses $\Delta G=-nFE$ to calculate the Gibbs free energy change which is then equated to the maximum work obtainable.

But this is just the work obtainable per mole at this concentration only. As soon as the reaction proceeds, the concentration change and so does the value of $\Delta G$ per mole and therefore the maximum work obtainable is different for different concentrations.

Therefore, to find the maximum work obtainable, shouldn't we first calculate the equilibrium constant of this reaction (I calculated it to be $K_c=1.89\times10^{107}$), then the equilibrium concentrations(which is probably $\ce{Mg^{2+}}=0.6\ \mathrm{M}, \ \ce{Ag+}=0.178\times10^{-53}\ \mathrm{M}$) and then use something like:
$$
\int_{\mathrm{in}}^{\mathrm{eq}}\Delta G* \mathrm{d\,M}
$$
where $\Delta G$ is per mole and the $dM$ is a small amount in moles that the reaction proceeds at that concentration.

Please help me verify my method and to proceed further.

There is a simpler approach to this problem.

First, since both initial concentrations are not 1 M, you need to determine $E$ for the initial conditions, where $Q$ is the reaction quotient $Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$:

$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$

The value of $n$ for this reaction is $n=2$, and assuming $T=298\ \mathrm{K}$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$

You can then go and convert this potential into free energy or maximum work or whatever else you want.

At equilbrium $E=0$ and $Q=K_c$, so we can calculate the equilibrium constant if we want to compare it to $Q$ and determine which direction is favored. We don't need to. Since $E_i>0$, this reaction procedures toward the products.

Now, as the reaction proceeds the concentrations of $\ce{Ag+}$ and $\ce{Mg^{2+}}$ will change. From the law of mass action:

$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$

After some time $\Delta \ce{[Ag+]} = 0.25 \text{ M}$, so $\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$ and $[\ce{Ag+}]=0.75 \text{ M}$ and $[\ce{Mg{2+}}]=0.225 \text{ M}$. The value of $Q$ has changed and so has the value of $E$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$

As the reaction proceeds, the value of $E$ decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of $E$ occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since $E$ decreases as the reaction proceeds, so too does the work the cell can do at any moment.

If you want the total work, then you need to integrate with respect to $Q$, not worrying about the actual concentrations:

$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$


Arrhenius equation in exothermic reactions According to the Arrhenius equation, the rate of reaction increases when the temperature increases. But according to Le Chatelier's  principle, the reaction rate decreases when the temperature increases.

So is Arrhenius equation applicable on exothermic reactions?

The Arrhenius equation is absolutely applicable to exothermic reactions and it will give results in accordance to Le Chatelier's principle.
Applying the Arrhenius equation for the forward reaction yields:
$$k_\mathrm{f} = A \exp\left(\frac{-E_\mathrm{f}}{RT}\right),$$
applying the Arrhenius equation to backward reaction yields
$$k_\mathrm{b} = A \exp\left(\frac{-E_\mathrm{b}}{RT}\right).$$

We know that $E_\mathrm{f} -E_\mathrm{b} = \Delta H$. Dividing $k_\mathrm{f}$ by $k_\mathrm{b}$ we obtain:
 $$\frac{k_\mathrm{f}}{k_b} = A \exp\left(\frac{-\Delta H}{RT}\right).$$
For an exothermic reaction $\Delta H < 0$. So $-\Delta H/(RT)$ is a positive number. As $T$ increases, the exponent decreases and thus $k_\mathrm{f}/k_\mathrm{b}$ decreases, which is nothing but the decrement of the equilibrium constant.
Thus the result is in perfect accordance with Le Chatelier's principle.

Also, you can apply the van't Hoff's equation which states:
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{\Delta H}{R}\cdot\left(\frac 1{T_1} - \frac 1{T_2}\right),$$
which also gives the decrease of rate of reaction with increase in temperature as $\Delta H < 0$.


Reaction of phosphorous acid (H3PO3) and potassium hydroxide (KOH) For the reaction of phosphorous acid ($\ce{H3PO3}$) and potassium
hydroxide ($\ce{KOH}$), write (a) the balanced molecular equation
and (b) the net ionic equation. Note that phosphorous acid is
a diprotic acid.
$$\ce{H3PO3(aq) + 2KOH (aq) -> 2 H2O(l) + K2HPO3 (aq)} \tag{a}$$
$$\ce{H3PO3(aq) + 2 OH-(aq) -> 2 H2O(l) + HPO3^2-} \tag{b}$$

The answers are above. Can somebody explain why it is $\ce{K2HPO3}$ in (a)? I am familiar with monoprotic acid, and this is my first encounter of diprotic acid. Thanks.

You have made the assumption that all the hydrogens are attached to oxygens. This is not a good assumption; one of the hydrogens is bonded directly to phosphorus and is not removed by OH-.
A better description of phosphorous acid is HPO(OH)2 with one oxygen doubly bonded to phosphorus  wikipedia


Which isomers of C6H4(OH)2 has a higher melting point?why? At first I thought that the first isomer must have the highest melting point since it's the most polarized isomer. So I looked them up on Pubchem to be sure and the answer was exactly the opposite;the third isomer has the highest melting point and it's not polar.

Any idea why?

Thanks!

Polarity rules the day if there is no other interactions involved. There is quite a lot of potential interactions in organics, that may overwhelm polarity by orders of magnitude.

To name a few.

There is a well known effect of halogen-halogen interactions, that may raise bp/mp dramatically. This is due to so-known dispersion forces, caused by correlation of electron movement in two atoms. Sometimes it is described as interaction of 'momentary dipoles'. To produce strong dispersion forces, atoms involved must be large and have a lot of electrons. Naturally, heavier atoms in a column are typically larger and have more electrons, thus we have strong increase of mp/bp for most haloalcones whith shift from chlorine to iodine.

Another well known effect is formation of intermolecular hydrogen bonds. Hydrogen bonds are common in compounds containg $\ce{O-H}$ or $\ce{N-H}$ bonds, but sometimes may pop-up in other, usually quite exotic settings. The most common form of hydrogen bonds is a bond of form $\ce{R-O-H\cdot\cdot  \cdot O(R)-R}$ . Such a bond may be stronger of weaker. It is stronger if the second oxygen atom has strong negative charge and the hydrogen involved has strong positive charge. Hydrogen bond is stronger if it is linear, i.e. the entier $\ce{O-H\cdot\cdot  \cdot O}$ fragment is linear. Intermolecular hydrogen bonds have competition : intramoleculear hydrogen bonds. Intramolecular hydrogen bonds prevents two potential sweet spots from interacting with other molecules, and thus a compound capable of formation of intramolecular bonds typically has lower mp/bp than compound that cannont, but have same amount of potential hydrogen bonding sites.

And, finally, come electrostatic interactions. They are not limited to just polarity, some compounds are ionic and this typically results in high bp/mp. Ionic bonding is common in inorganics, but may occur in organics as well, say, $\ce{N(CH3)4 PF6}$

Now, about example at hand. All three molecules are capable of formation of hydrogen bonds. So, the one that is better in it wins the contest of bp/mp.

The 1,2-disubstituted compound is capable of forming intramolecular bonds, and the oxygen atoms are a bit crouded. This makes bonding a bit weaker.

The 1,3-compound is better in this regard, but has $\ce{OH}$ groups in meta-position. It is known, that oxygen in such group is capable of donating some of its electron density into aromatic ring. However, if we put two oxygens in 1,3-pattern,  the two will suppress donation of each other thanks to resonance effects. The last compounds would have the most polar hydrogen bond. It also has a high symmetry, which helps. It is well known, that slight asymmetry mar lower mp considerably. Consider, for example, toluene. It has higher bp than benzene but significantly lower mp.

Consequently, compound 3 (hydroquinone) wins the prize.


2,2,6,6-Tetramethylpiperidine - Brønsted Basicity I understand that this molecule can only act as a Brønsted base because it is sterically hindered by its multiple methyl groups.

However, isn't the proton bound up in solution? Doesn't that increase the effective size of the proton? How then exactly is a proton expected to reach the site of basicity if it's tied up with multiple water molecules? Or is the size of the hydrated proton still quite small? Or does the proton separate from the water molecules once the proton is sufficiently close to the site of basicity (the lone pair on the nitrogen)?

I don't see much hinderance[note that they all have been optimised to stable configurations]
Blue: nitrogen,White: Hydrogen,Grey: Carbon

Side View:

Top View:


What does under-stoichiometric or stoichiometric mean? When people talk about oxides like $\ce{SiO2}$ or $\ce{Al2O3}$, they use expressions like stoichiometric or under-stoichiometric. I understand that this refers to the relative composition of materials, but I don't really know what it exactly means.

Say for $\ce{SiO2}$, does under-stoichiometric mean that it isn't exactly $\ce{SiO2}$, but more like $\ce{SiO_{1.8}}$? Does the term usually refer to the oxygen composition?

Yes, the convention is to refer to the oxygen atoms when talking about over- and under-stoichiometric in metaloxides. Normally they also indicate this explicitly by writing $\ce {SiO}_x$ (like in the article here for example) or $\ce {Al_2O}_x$. Being explicit about it is a good habit because it avoids ambiguity.


C-4 oxidation of octanoic acid to 4-oxooctanoic acid I am leading a project in which I am attempting to synthesize 4-oxooctanoic acid in a (mostly) biological pathway.
Currently, I have reached a dead-end. I have a way to create octanoic acid through bacterial growth, I could not seem to find a way to proceed to the 4-oxo step with enzymes so I turned to chemistry.

Although this seems to be impossible due to the structure of octanoic acid. There appears to be no way to specifically target the fourth carbon to obtain the 4-oxooctanoic acid.

My question is: Am I correct in assuming that converting octanoic acid to 4-oxooctanoic acid is impossible (at least via simple oxidation)? Or would there be a better way to go about doing this?

Thanks in advance

PS: Here are pubchem links to the two structures involved.

https://pubchem.ncbi.nlm.nih.gov/compound/octanoic_acid#section=2D-Structure

https://pubchem.ncbi.nlm.nih.gov/compound/4-oxooctanoic_acid#section=Top

If I was tasked with making 4-oxooctanoic acid I would not start from octanoic acid! There is no easy way of making 4-oxooctanoic acid directly from octanoic acid. It is however a relatively trivial exercise from other starting materials.

My suggestion is to react the commercially available ethyl succinyl chloride (ethyl 4-chloro-4-oxobutyrate) with n-butylmagnesium bromide (also commercially available) modified with manganese(II)chloride/lithium chloride as described in Tetrahedron Lett. 1992, 33 (31), 4439–4442. This will give the ethyl ester of the required 4-oxooctanoic acid. A simple base hydrolysis gives 4-oxooctanoic acid. You could probably use the widely available n-butyllithium in place of the Grignard.


Why are acids corrosive? I'm in the second year of my biochem major, taking organic chemistry, and I feel like I've learned a lot about acids. The most popularly recognizable property of acids is their corrosive power. There are pop-culture depictions of acids everywhere, and all they seem to do is corrode stuff. Yet, with all I've learned about them, it's never been explained to me why they have this property, or how and when it works.

Strong acids disassociate completely, weak acids do less so. When they do disassociate in water, they form $\ce{H3O+}$. This pulls the electrons from the thing it is reacting with. Lets take this to be Fe. ($\ce{H3O+}$ rips the electrons from the substance). So the reaction would go something like: $\ce{Fe + 2H3O+ -> Fe^2+ + 2H2O + H2}$. The $\ce{Fe^2+}$ formed is soluble in water. This also reacts with the other ion from the acid. This is what makes the acids corrosive.


Nitric acid acting as oxidising agent In the following mechanism, nitric acid acts as an oxidising agent to remove 2 H atoms from the intermediate (in the blue square)

My first instinct if given only the reagents would be to protonate something, such as the amine group for example, rather than removing hydrogens.

So my question is how can I identify if $\ce{HNO3}$ will act as an acid or an oxidising agent, given only the reagents? Do I just have to try out both and see which one leads to a reaction? Or is there a property of nitric acid that I'm missing?

The reactions of nitric acid could be due to 1) its acid nature, 2) its oxidizing properties, or both.

Orthocresol was correct that the substrate could predetermine which course was more likely: acid-susceptible substances (like alkalies, or alkaline materials) would respond to the acid nature; oxidizable substances would respond to the oxidizing power (like organic alcohols and sugars).

An interesting case is pure iron: in dilute nitric acid (less than 10%), it dissolves. In concentrated nitric acid, it is oxidized (passivated) and does not dissolve.

Increased temperature would also increase the likelihood of an oxidation reaction compared to an acidic reaction, because the activation energies for oxidations are usually higher.


In an octahedral complex, what happens to the electrons donated by the ligand? I am trying to understand how complexes are coloured. After some reading, I found out this was due to the d-d splitting induced by the coordinate bonds of ligands to the central metal ion. The electrons in the d-orbitals then transition between each other, and absorb light in the process.

However, I do not understand what happens to the electrons donated by the ligand. Don't they form a sigma bond with the d-orbitals of the metal ion? In which case, wouldn't they go into the d-orbitals and 'fill it up', thus disallowing transitions?

Or is it the case that the d-orbitals of the central metal ion are not involved in the bonding at all?

This is a very good question, I must say. It requires the understanding of the very fundamentals.

You're right, if the electrons from the ligand pair up with the electrons of the metal, the electrons cannot undergo $d$-$d$ transitions, and no color would be seen.

But do they pair up? Think carefully. Recall one of the most basic assumptions of the Crystal Field Theory: The attraction between the ligand and the metal is assumed to be purely electrostatic in nature. This very important assumption is the sole reason for the explanation of colors of these complexes. As the ligands approach the metal atom, they create a crystal field. These fields tend to repel the electrons of the metal and increases their energy. A perfectly symmetric spherical  crystal field would increase the energy level of all the orbitals equally and to the same level. But this isn't the case in actual coordination complexes. An octahedral crystal field would approach along the axes of the $d_{x^2+y^2}$ and $d_{z^2}$ and tend to repel them to a greater extent than the other orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$). This difference in energies of the orbitals is known as crystal field splitting. We can now group these orbitals into two groups, three low energy $t_{2g}$, and two high energy $e_g$ orbitals.

Keep in mind that the ligands do not pair with the electrons of the metal. They simply repel the orbitals electrostatically and increase their potential energy in the process.

So, now how do the complexes actually get their color then? When the orbitals split, the difference in their energies is called Crystal Field Stabilization Energy (CFSE) and is denoted by $\Delta_{\text{o}}$. When photons of light are incident on the complex, it absorbs the photons which possess the energy equal to that of the value of $\Delta _{\text{o}}$. From quantum theory of electromagnetic waves, it's known that the energy of a photon is given by:

$$U = \frac{hc}{\lambda}$$

Where $h$ is known as the Planck's Constant, with a value of $\pu{6.626×10^{-34}Js}$, and $\lambda$ is the wavelength of light. If the value of the wavelength lies in the visible light spectrum, you can find the color of the light absorbed by the complex.

How do you find the color emitted out? There's a very easy and fun way to find that out. Take a look at this color wheel:

Find where the color of absorption lies. Then the color of the complex lies on the opposite side of the color of absorption.

This is a brief discussion about the colors of complexes. Let me know if you need more clarification.


Are these different neutralization pathways? When sodium oxide is added into water, it reacts with water to form $\ce{NaOH}$.
$$\ce{Na2O(s) + H2O(l) -> 2 NaOH(aq)}$$

This means that if we were to add an acid, it would be neutralized by the sodium hydroxide.
$$\ce{Na2O(s) + H2O(l) + 2 HCl(aq) -> 2  NaCl(aq) + 2 H2O}$$

Is the following equation a totally different reaction pathway, or is it simplified from the reaction pathways above?
$$\ce{Na2O(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l)}$$

Both fast reactions happen in parallel, acting as "first comes, first served".

Concentration of water molecules ( $\pu{55 mol/L}$ in water) is much higher than $\ce{[H3O+]}$  for diluted acids, so oxide anions meet water molecules much more frequently.

If the oxide anion meets water molecule, it will not wait for hydronium.
Additionally, hydronium ions would be rather depleted by reaction with hydroxide ions next to the oxide surface.

Be aware that $\ce{Na+}$ ions are just bystander/spectator ions, so it is more illustrative to write equations as ionic equations:

The main reaction:
$$\ce{O^2− + H2O -> 2 OH−}\tag{1}$$

The secondary alternative reaction:
$$\ce{O^2− + H3O+ -> OH− + H2O}\tag{2}$$

The final reaction:
$$\ce{H3O+ + OH− -> 2 H2O}\tag{3}$$


Cyclisation reaction through hydrazine The following reaction confuses me - I don't understand how and why it happened (couldn't figure out the mechanism):

I thought this is some Wolff Kishner sort of reduction, but the reagents don't match (no alkaline medium). I am not sure how hydrazine lead to cyclisation - for starters, I guess it added to the carbonyl oxygen (aldehyde, as it is more reactive) as an ammonia derivative to form a hydrazone. What's next?

It's essentially just the same reaction once and then over again. The basic medium of the Wolf-Kishner reduction is to remove the hydrazone hydrogen, releasing the $\ce{C=N}$ electrons onto the carbon. In this reaction, the hydrazone instead further reacts with the other carbonyl to form an aromatic 6-membered ring.


Why did they provide the value for the enthalpy change of fusion? The metal gallium melts when held in the hand; its melting point is $\pu{29.76 ^\circ C}$. How much energy as heat is removed from the hand when $\pu{5.00 grams}$ of gallium initially at $\pu{20.0^\circ C}$ melts? The value of $\Delta H_{fusion}$ is $5.576~\mathrm{kJ~mol^{-1}}$ and the specific heat of gallium is $0.374~\mathrm{J~g^{-1}K^{-1}}$. Take the final temperature to be $\pu{29.76 ^\circ C}$.

I got:

$$q = (\pu{5.00 g})(0.374~\mathrm{J~g^{-1}K^{-1}}) (\pu{9.76^\circ C})$$

$$q=\pu{18.3 J}$$

So I reason that since $\pu{18.3 J}$ of energy in the form of heat is required to melt gallium from at an initial temperature of $\pu{20.0 ^\circ C}$ to a final of $\pu{29.76 ^\circ C}$ then this is all the energy required to do this but why did they give $\Delta H_\text{fusion}$?

You calculated the energy required to change solid gallium at 20°C to solid gallium at 29.76°C.  Once solid gallium reaches 29.76°C, additional energy is required to melt it.  The energy required is given by the $\Delta H_{\text{fusion}}$.  So to complete the question, transform the $\Delta H$ units to a grams basis instead of a mole basis, multiply it by the weight of gallium in the problem, and then add that result to the one you already calculated.  That will be the energy required to change solid gallium at 20°C to liquid gallium at 29.76°C.


Why is moist hydrogen made to react with chlorine to give hydrogen chloride? Moist hydrogen reacts with chlorine in diffused sunlight to give out hydrogen chloride. Why is moist hydrogen used? Why not dry hydrogen?

H2O has h ions so it adds more + ions so that the cl- gets reacted faster towards the H+ ions so water is a catalysts or booster
I hope u understnd


How was Avogadro's law discovered? What was the experimental setup and/or reasoning that Amedeo Avogadro, in 1811, used to arrive at the conclusion that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules"?

This law bears Avogadro’s name because he implicitly provided circumstantial evidence that it is true, by successfully applying it as explained below. But, strictly speaking, Avogadro’s law has not been discovered neither by experiments nor by reasoning since it had already been in the air if one had accepted the permanent chaotic movement of particles in gases as suggested by Daniel Bernoulli in 1728.

Also Dalton reflected about the equal volumes−equal numbers hypothesis before Avogadro, but discarded it as impracticable because he did not factor in the idea that gaseous elements might be polyatomic molecules; in consequence he thought that the equal volumes−equal numbers hypothesis were not useful to explain the known physical and chemical facts about gases. (According to Avogadro, Dalton had additionally some strange caloric objections against this view, see p. 59 of the link in the next but one paragraph.)

Hence, the idea that elements might be polyatomic molecules was not accepted at that time (1811). It was even declined thereafter for a long time (by Berzelius who dominated chemistry then).

Being a physicist and mathematician, Avogadro did not himself accomplish chemical reactions with gases, but he could built on existing gas density measurements and the following two results of Gay-Lussac (1808), which he discussed in his article in the Journale de physique, de chimie …, 1811 (all the following page numbers refer to this text) with the equal volumes−equal numbers hypothesis as starting point (!):

Gay-Lussac's combining law: “Not only do gases combine in very simple proportions” by volume, “but the apparent contraction of volume, which they experience on combination, has also a simple relation to the volume of the gases, or at least to that of one of them” (p. 58), and
his experimental combining volume ratios for different gas reaction, e.g. of the formation of water (1 + 1 = 2), ammonia (1 + 3 = 2) from the elements (p. 60).

When comparing all these data, Avogadro proceeded in two steps:

he discussed that, as a consequence of the equal volumes−equal numbers hypothesis, one could easily bring into line the densities of elementary gases with the relative molecular weights of the gases; for instance the density of oxygen being about 15.074 times larger than that of hydrogen (p. 59).
Then he had (p. 60) the idea that gaseous elements could consist of polyatomic molecules (“molecules composées”). Subsequently he discussed extensively the known combining volume experiments of gases in this light and showed that so all hitherto contradictions thereof could be solved.

In this way, Avogadro had simultaneously also proved by circumstantial evidence that the equal volumes−equal numbers hypothesis, from which he started his eventually successful reflections, must be true because he had demonstrated that it is one of the important factors for the causal explanation of all those gas data by particle theory.

Hence, the so called “discovery” of Avogadro's law consists merely in the success to bring all these gas data into accord, which vice versa substantiates that it is a law, whereas before this application, it was unclear how the equal volumes−equal numbers hypothesis could be generally made use of.


Is buckminsterfullerene aromatic? According to Wikipedia,

  The $\ce{C60}$ molecule is extremely stable,[26] withstanding high temperatures and high pressures. The exposed surface of the structure can selectively react with other species while maintaining the spherical geometry.[27] Atoms and small molecules can be trapped within the molecule without reacting.

Smaller fullerenes than $\ce{C60}$ have been distorted so heavily they're not stable, even though $\ce{M@C28}$ is stable where $\ce{M\,=\,Ti, Zr, U}$.

Some of us have heard and learned about the "rules" of aromaticity: The molecule needs to be cyclic, conjugated, planar and obey Huckel's rule (i.e. the number of the electrons in $\pi$-system must be $4n+2$ where $n$ is an integer).

However, I'm now very skeptical to these so-called rules:

The cyclic rule is violated due to a proposed expansion of aromaticity. (See what is Y-aromaticity?)
The must-obey-Huckel rule is known to fail in polycyclic compounds. Coronenefigure 1 and pyrene figure 2 are good examples with 24 and 16 $\pi$ electrons, respectively.
Again, Huckel fails in sydnone. The rule tells you that it's aromatic, while it's not.

The planar rule is not a rule at all. We're talking about "2D" aromaticity when we're trying to figure out the $n$ in $4n+2$. The "3D" rule is as following:

  In 2011, Jordi Poater and Miquel Solà, expended the rule to determine when a fullerene species would be aromatic. They found that if there were $2n^2+2n+1$ π-electrons, then the fullerene would display aromatic properties. - Wikipedia

This would mean $\ce{C60}$ is not aromatic, since there is no integer $n$ for which $2n^2+2n+1 = 60$.

On the other hand, $\ce{C60-}$ is ($n = 5$). But then this rule strikes me as peculiar because then no neutral or evenly-charged fullerene would be aromatic. Furthermore, outside the page for the rule, Wikipedia never explicitly states that fullerene is not aromatic, just that fullerene is not superaromatic. And any info on superaromaticity is unavailable or unhelpful to me; including the Wikipedia "article" on that topic.

So, is $\ce{C60}$ aromatic? Why, or why not?

Aromaticity is not binary, but rather there are degrees of aromaticity.  The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene.

Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be involved) overlap in a pi-type system.  As the examples above indicate, the stabilization can be large or small.

Let's consider $\ce{C_{60}}$:

Bond alternation is often taken as a sign of non-aromatic systems.  In $\ce{C_{60}}$ there are different bond lengths, ~1.4 and 1.45 angstroms. However, this variation is on the same order as that found in polycyclic aromatic hydrocarbons, and less than that observed in linear polyenes.

Conclusion: aromatic, but less so than benzene.

Magnetic properties are related to electron delocalization and are often used to assess aromaticity.  Both experiment and calculations suggest the existence of ring currents (diamagnetic and paramagnetic) in $\ce{C_{60}}$.

Conclusion: Although analysis is complex, analysis is consistent with at least some degree of aromaticity.

Reactivity - Substitution reactions are not possible as no hydrogens are present in $\ce{C_{60}}$.  When an anion or radical is added to $\ce{C_{60}}$ the electron(s) are not delocalized over the entire fullerene structure. However, most addition reactions are reversible suggesting that there is some extra stability or aromaticity associated with $\ce{C_{60}}$.

Conclusion: Not as aromatic as benzene

Resonance energy calculations have been performed and give conflicting results, although most suggest a small stabilization. Theoretical analysis of the following isodesmic reaction

$$\ce{C_{60} + 120 CH4 -> 30 C2H4 + 60 C2H6}$$

suggested that it only took half as much energy to break all of the bonds in $\ce{C60}$ compared to the same bond-breaking reaction with the appropriate number of benzenes.

Conclusion: Some aromatic stabilization, but significantly less than benzene.

This brief overview suggests that $\ce{C_{60}}$ does display properties that are consistent with some degree of aromatic stabilization, albeit less than that found with benzene.


Pyramid Structure Name I would like to know what the name is for the structure drawn below. This looks like a pyramid with 4 atoms at each vertex. Does the name of the structure depend on whether the atoms are all the same or if they are a combination of different ones?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) and the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) include polyhedra and corresponding polyhedral symbols that describe the geometrical configuration.

When an atom is attached to three atoms or groups (e.g. coordination compounds with a coordination number of three), the geometry may be trigonal planar or trigonal pyramidal. For the structure given in the question, the geometry is obviously trigonal pyramidal, not planar. Thus, the coordination polyhedron is a trigonal pyramid; the corresponding polyhedral symbol is TPY-3.

This coordination polyhedron is not to be confused with a tetrahedron (polyhedral symbol: T-4), which applies to a coordination number of four, or an atom that is attached to four atoms or groups.


HPLC: Peak height calculation I am trying to implement a small python script to detect peaks and calculate the corresponding area. The y-values are in mAU and x-values in minutes but there are some unit conversions in most HPLC softwares. For example in this image (found here):

The peak at 7.87 min has a height of 329451 but only approximately 20 V in the chromatogram. How can I calculate this height?

Height in HPLC simply refers to the height of the peak (= maximum value of the peak) above the baseline. The point by z1273 is right in the sense that this voltage to height conversion factor is instrument specific but this height cannot be in milliabsorbance units- it is too large for absorbance. This height is apparently "intensity". Anyway, you do not need to worry about the "true height" for peak detection. It will be fine to use the height of 20 V for the first peak at 7.867 min. In older instruments, the detector output was displayed as Volts or mV and the manufacturer provided a conversion factor from volts to light absorbance or sometimes as intensity in the manual. See this discussion here (https://www.chromforum.org/viewtopic.php?t=6442). You should really look at modern HPLC data. The one you are showing here is from an ancient instrument.
I am assuming you are using derivatives to detect peaks? For any chromatography program, I would suggest that you do a simulation first, because all variables will be in your hand. Make a chromatogram as sum of large and small exponentially modified Gaussians or simple Gaussians of various heights and areas and add some noise to test your program.


How can chlorine have 10 electrons around it in chlorine trifluoride? I'm relatively new to chemistry, and while doing some exercises about $\ce{ClF_3}$ I tried to draw where the electrons are. I checked Google and got this:

How is this structure possible? From what I've learned from high school every element wants 8 valence electrons. From the figure, $\ce{F}$ has 8, but $\ce{Cl}$ has 10.

Interhalogens consist of one $\ce{X-F}$ pair whith a regular two electron bond, and pairs of flourine atoms taking over one free electron pair of X (chlorine, bromine, iodine).

What forms are three-centre four-electron bonds (one pair of electrons from the chlorine, two single electrons from two fluorine atoms). That's why interhalogens always have an uneven number of fluorine atoms (except for the dimer $\ce{Cl_2F_6}$).

https://en.wikipedia.org/wiki/Three-center_four-electron_bond

If this is still octet rule is debatable, or at least not directly obvious (see the wp article). Let's just say that fluorine is a bad, bad guy who has things his way.


Galvanic corrosion between nickel brass nut and stainless bolt I am planning to install something on a boat that will be in salt water, and my question is, will the nickel plated brass fastener not corrode if I connect it to stainless steel?

I looked it up on the chart and it says that there should be less then 0.15 V difference in the "Anodic Index" for harsh environments and while brass and stainless is within the 0.15, nickel is just outside.

The environment is not entirely harsh as well, as these will be used to fasten just the windows, so while there could be water there, mostly it will be fresh water from rain and the occasional spray from salt water.

While I cannot comment in a technical level, I can speak from a background in aerospace experience (in my family) and lifetime of building things that live outdoors.

In time, dissimilar metal corrosion always sets in. Most stainless steels are inhibiting, not a 'way out' of the problem.

On top of that, I think the ambient salts will kills you on this in the marine environment.  If you have a regular maintenance plan, then perhaps you can re-service the bolts yearly, as needed.


What happens when you put a battery in saltwater? If somebody puts a battery in a salt solution, what path do the electrons take?

Salt dissolves in water to left behind positively charged sodium ion and negatively charged chloride ion. once you put battery in salt water, the sodium ion migrate towards the "negative tank" and chloride ion migrate towards the "positive tank". Salt water is much more conducive than regular water, so the battery would discharge more quickly and the current will break down the salt by electrolysis, producing hydrogen and chlorine gas.


Basic strength of nitrogen What is the order of basic strength of nitrogens in the compound given in the figure?

$$\ce{
H2\overset{.\!\!.}{\underset{(γ)}{N}}-\overset{.\!\!.}{\underset{(β)}{N}}H-\overset{\overset{\huge O}{||}}{C}-\overset{.\!\!.}{\underset{(α)}{N}}H2
}$$

The most basic is the γ-N since its lone pair can't go in conjugation. My doubt is between α and β nitrogens, whose lone pair can go in conjugation.

Tang Yong Boon certainly is in the correct direction (please refer to his comment).

I'll just wrap things up:
$$\gamma > \beta > \alpha $$
$$\mathrm{(Basic \ strength)}$$

Reasoning:

The $\gamma$ nitrogen is unquestionably the strongest basic site.
$\beta$-nitrogen follows, as its tendancy of lone pair donation is greatly enhanced due to the presence of an adjacent (ie. alpha wrt. itself) electron donating group ($\ce{-NH2}$), otherwise known as the alpha effect. Some explainations include destabilization and elevated energy of the ground state, resulting in lower activation barrier.
The $\alpha$-nitrogen is best called an acidic site.


Reaction of dihalides with excess of NaNH2 Gem-dihalides react with $\ce{NaNH2}$ to give terminal alkynes. Why is a terminal alkyne prefered?

For example,

I had thought that 2-pentyne would be preferred over 1-pentyne as it would be more stable due to hyperconjugation, but it says the final product is terminal alkyne.

Why?

this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.

Trusted source: Jerry March Organic Chemistry 6th edition.

  When the base is NaNH2 1-alkynes predominate (where possible), because this base is strong enough to form the salt of the alkyne, shifting any equilibrium between 1- and 2-alkynes. When the base is $\ce{OH-}$ or $\ce{OR-}$, the equilibrium tends to be shifted to the internal alkyne, which is thermodynamically more stable.


CI without Hartree-Fock SCF Usually textbooks on CI takes the SCF Hartree-Fock orbitals as a starting point. Can one use the configuration interaction method starting directly from the single-particle Slater determinants avoiding the iterative procedure. This question concerns mainly small molecules like the hydrogen molecule. Do I get a high inaccuracy in this case? If yes, why? Could this inaccuracy be decreased taking into account more configurations?

If you were to perform Full CI (within a finite basis) then I don't think that it matters whether you start with SCF orbitals or some other choice.  This solution will have the lowest conceivable energy no matter what.

But if you are doing truncated (less than full) CI, then you are very much at the mercy of your starting orbitals. In general, one would want to start with orbitals that resemble those of the actual state that you are trying to describe via CI.

Otherwise, your precious excitations (singles and doubles) may be consumed correcting the "slop" that you provided, rather than moving toward the full CI limit.


Derivation of the relation between temperature and pressure for an irreversible adiabatic expansion I have a formula for finding the final temperature in an irreversible adiabatic expansion:
$$T_2=\left[\frac{C_V+\frac{P_2}{P_1}}{C_p}\right]T_1,$$
where $T_1$ is the initial temperature.
$C_V$ and $C_p$ are the molar heat capacities at constant volume and constant pressure, respectively.

I am interested in how this formula is derived.

When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:

As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be

\begin{align}
  T_2=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_{p}}\right]T_1
\end{align}

Now, I will come to the derivation:
You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.

\begin{align}
  \delta Q = 0 \ .
\end{align}

This means that from the first law of thermodynamics you get:

\begin{align}
  \mathrm{d} U = \delta Q + \delta W = \delta W
\end{align}

where $U$ is the internal energy and $W$ is the work.
Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is $p_2$ in your equation, then this work is given by:

\begin{align}
  \delta W = - p_{2} \mathrm{d} V \qquad \text{where:} \qquad p_{2} = \text{const.} \neq f(V)
\end{align}

So, together with the first law of thermodynamics this yields:

\begin{align}
  \mathrm{d} U = - p_{2} \mathrm{d} V \ .  \tag{1}
\end{align}

Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that

\begin{align}
  \mathrm{d} U &= \underbrace{\frac{\partial U}{\partial T}}_{= \, C_{V}} \mathrm{d} T + \underbrace{\frac{\partial U}{\partial V}}_{= \, \Pi \, = \, 0} \mathrm{d} V \\
  \mathrm{d} U &= C_{V} \mathrm{d} T \tag{2}
\end{align}

where $T$ is the temperature, $C_{V}$ is the heat capacity at constant volume, and $\Pi$ is the internal pressure, which is equal to zero for an ideal gas.
Other relations that hold for ideal gases are the ideal gas law

\begin{align}
  p V = n R T
\end{align}

where $n$ is the amount of the gas and $R$ is the universal gas constant,
and

\begin{align}
  c_{p} - c_{V} = R  \tag{3}
\end{align}

where $c_{p} = \frac{C_{p}}{n}$ and $c_{V} = \frac{C_{V}}{n}$ are the molar heat capacities at constant pressure and constant volume, respectively.

Now, you just have to combine equations $(1)$ and $(2)$:

\begin{align}
  - p_{2} \mathrm{d} V = C_{V} \mathrm{d} T \ ,
\end{align}

integrate the result (under the assumption that $C_{V}$ is temperature independent)

\begin{align}
  - p_{2} \int^{V_2}_{V_1} \mathrm{d} V &= C_{V} \int^{T_2}_{T_1} \mathrm{d} T \\
  - p_{2} \left( V_2 - V_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

use the ideal gas law $V = \frac{n R T}{p}$ to substitute $p$ for $V$

\begin{align}
  - n R p_{2} \left( \frac{T_2}{p_2} - \frac{T_1}{p_1} \right) &= C_{V} \left( T_2 - T_1 \right) \\
  - n R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

introduce the molar heat capacity via $C_{V} = n c_{V}$

\begin{align}
  - R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= c_{V} \left( T_2 - T_1 \right) \ ,
\end{align}

and separate $T_1$ and $T_2$

\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \left( c_{V} + R \right) \ .
\end{align}

Finally, you can use equation $(3)$ to get the desired result:

\begin{align}
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \bigl(\underbrace{c_{V} + R}_{= \, c_{p}} \bigr) \\
  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 c_{p} \\
  \Rightarrow \ T_2 &=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_p}\right]T_1 \ .
\end{align}


How can I relate the reactivity series to electronegativity and ionization energy? I am trying to figure out how the reactivity series comes about. My understanding is that elements with a higher electronegativity will be more reactive than elements with a lower electronegativity, and that elements with a low ionization energy will be more reactive than elements with a high ionization energy.

Here is a chart of electronegativity (from Wikipedia):

This shows electronegativity decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.

Here is a chart of ionization energy:

This shows ionization energy decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.

Neither electronegativity nor ionization energy can predict both reactivity trends, so what is going on?

With less electronegativity, there is also less ionization energy. This means for metals, it is easier to lose the electrons. When it is easier to lose the electrons, the metal becomes more reactive because it uses less energy to fulfill its goal; to gain a full outer shell.

For nonmetals, the main goal is to gain electrons and not lose them. Thus, an increase in electronegativity and ionization energy means an increase in reactivity because the nonmetals want to hold on to their electrons and gain more vigorously to achieve the full outer shell.