Is there an example of a closed shell molecule with a permanent ground state magnetic dipole? It seems that molecules of point group C$_{n}$ and C$_{nh}$ can in principle support a permanent magnetic dipole moment (Barron, 2004). However, I cannot think of any closed shell molecules that physically have a magnetic dipole in the ground state.

Is this ever observed, and can you give an example of such a case?

Citation: Barron, Laurence D. Molecular light scattering and optical activity. Cambridge University Press, 2004.

Updated 4/6/2017: It turns out that singlet ($^1\Delta_g$) molecular oxygen is paramagnetic due to orbital angular momentum only, as all electrons are paired. It has a magnetic dipole moment of $1\hbar$ and can be determined experimentally. (This makes sense from a computational perspective, given that the lowest restricted Hartree Fock solution of singlet oxygen is always complex, and only complex wave functions yield non-zero expectation values of orbital angular momentum.)

For closed shell (even number of electrons) molecules the answer is no.

After a little more reading, I found the answer. The reason is that the magnetic dipole is odd under time reversal (Table 4.1, p. 218 in Barron, 2004).

Apparently, (Barron, p. 221) "the expectation value of a time-odd operator [e.g. magnetic dipole] vanishes for states invariant under time reversal, which can always be constructed for an even electron system and hence any nondegenerate state".

I do not believe this applies to transition properties, however, as we can see magnetic dipole allowed transitions.

Citation: Barron, Laurence D. Molecular light scattering and optical activity. Cambridge University Press, 2004.


Effectiveness of Grignard reagent So for my experiment, I am synthesizing grignard's reagent (phenylmagnesium bromide). I am trying to change the dryness of ether and see if there is a change in products. Is there a simple method to quantify the results? I need to figure out the effectiveness of the product. If water is introduced, the reaction will probably form alkanes and magnesium hydroxide rather than phenylmagnesium bromide. It has to be a simple method since I am working in a school lab.

For a grignard reaction to work, the ether must be free from water. This is usually achieved via distillation over sodium/potassium using benzophenone as an indicator. As you are in a school lab, the solvent provided to you is likely already dried/dry enough for purpose.

If the solvent is a little wet, two things will happen:

Firstly, it is more difficult to get the grignard formation to 'initiate' as the water effectively quenches the reagent as it forms. However, with persistence it is usually possible to get them to initiate when making simple grignards such as PhMgBr.

The other issue however, is that as the water is quenching any grignard reagent formed until all the water is used up, you will make less reagent than you think you are (in this sense the reaction is self-drying) meaning if you do think your ether is wet, you should aim to make a greater quantity of reagent than required.

It is possible to titrate a grignard reagent to identify the actual concentration of solution made, although in an undergraduate practical it is likely that you will use the reagent in excess anyway.

In terms of identifying your product, you should carry out the reaction, work it up, purify, and use a technique such as NMR to see if your product has formed. If this is unavailable, simple IR spectroscopy can help you determine if the functional groups in your starting materials have changed (e.g. Carbonyl to alcohol).

Good luck with your reactions


lindane extraction from aqueous solution using hexane I am studying the degradation of lindane using iron nanoparticles. I have prepared lindane solution with an acetone and water mixture (1:9 v/v). I tried to extract lindane after reaction with nanoparticles using hexane as a solvent. a viscous third layer has formed between the two layers. I have no idea why this happened. Can anyone please explain why this has happened?

According to this publication, lindane has the following properties:

Practically insoluble in water
Specific gravity = 1.85
Melting point = 113 $\mathrm{^oC}$
Solubility in acetone > 200 g/L
Solubility in n-heptane* = 10-14 g/L
* Used as surrogate for hexane solubility for this argument.

Notes: lindane is by far the most soluble in acetone, though only a small amount could have been used in the 1:9 acetone:water solvent.  If we assume that the reaction mixture were extracted with an equal volume of hexane (conservative I think, probably used more hexane), then the solution would be about 1:10:10 acetone:water:hexane. Given the properties of lindane, I'd expect it to make for a gooey, separate, small layer between the water and hexane.

Prediction:

  Bottom layer - Water, undissolved iron, dissolved iron chlorides.
  Middle layer - Acetone, lindane, partially reacted lindane, some iron chlorides.
  Top layer - hexane,
  Small amounts of lindane and organic byproducts.


In the DNA backbone, which bonds exactly are considered ester bonds? Following are the three different answers to this question that I reached by googling. Which of them is right?

The covalent bond between the 3' C and the O, and the covalent bond between the 5' C and the O:

2. The covalent bond between the P and the O that is bonded to the 5' C, and the covalent bond between the P and the O that is bonded to the 3' C (deduced from the following image):

3. The -O- (two covalent bonds with an O) that link between the P and the 5' C, and the -O- that link between the P and the 3' C (deduced from the following image):

Maybe my confusion stems from a more general misunderstanding:
When two alcohols are linked through an ester linkage (i.e., both OH groups are replaced with an -O-), we get R1-O-R2. In this case, do we say that each of the covalent bonds with the O is an ester bond? Or is the whole -O- considered a single ester bond?

(Note that I am not asking what exactly a phosphodiester bond is, which is (if I understand correctly) everything that connects the 5' C and the 3' C of adjacent nucleotides.)

All of them are right, regarding ester bonds themselves. But only one of them is right, regarding DNA context. AMP/ADP/ATP/GMP are based on ribose ( see RNA ), not on deoxyribose ( DNA ).
Phosphoric acid has 2 ester bonds in DNA.
You have somehow confused an ester and an ether bond.
$\ce{R_1-O-R_2}$ is an ether bond,
$\ce{R-O-X}$ is an ester bond, where $\ce{X}$ is a carbonyl ( from carboxylic acids or carbonic acid ) or some nonmetal like $\ce{S}$ or $\ce{P}$ for esters of inorganic acids like $\ce{(CH3O)2SO2}$.
If we are to consider it as a bond or a group depends on the point of view. It is both.
It is a group, if we focus on the $\ce{CO-O}$ atom group of esters of organic acids or $\ce{O}$ for mineral acid esters.
It is a bond ( or a bridge ), if we focus on what the group connects.


Reactivities: interhalogen compounds versus halogens My textbook says that interhalogen compounds are more reactive than halogens, because the $\ce{X-X}$ bond in halogens (for example, dichlorine or $\ce{Cl2}$) is stronger than the $\ce{X-X'}$ bond in interhalogens (for example, chlorine monofluoride or $\ce{ClF}$).

(In the symbols that I have used, $\ce{X}$ and $\ce{X'}$ have been used to denote dissimilar halogen atoms.)

So my question is: why is the bond between dissimilar halogens weaker than the bond between atoms of the same halogen?

I am aware that the bond between two identical atoms will be a completely covalent bond due to no difference in the electronegativities of the participant atoms. So, the $\ce{X-X}$, bond as in $\ce{Cl2}$ will have no ionic character.

In case of an interhalogen compound, however, the difference in the elecronegativities of the (dissimilar) participant atoms will cause the bond to become polar and thereby acquire ionic character.

So is it the ionic character of the interhalogen compounds that causes them to be more reactive than halogens? But then, this means that ionic bonds are weaker than covalent bonds.

But, the answers here as well as here say that an ionic bond is stronger.

But this link says the very opposite. Okay, the people writing the answers there have written it to suit an eight grader's understanding, but even so, they have stated that covalent bonds are stronger than ionic bonds.

So which alternative am I to believe? I know that most bonds are neither purely ionic not purely covalent, but intermediate between these two extreme characters, but this only serves to increase my bewilderment.

Can somebody please resolve this, or tell me where I have gone wrong in my reasoning as regards the reactivities of the interhalogen compounds in comparison with halogens?

And finally, please tell me the reason why interhalogens are more reactive. Thanks!

Here is the answer for which bond (ionic or covalent) is stronger.
There is no direct answer to this question as it depends on atoms that take part in bond formation.For example you might be knowing that a Fluorine- Fluorine bond is weaker than a chlorine-chlorine bond due to high inter electronic repulsion.In another example iodine-iodine bond is weaker than chlorine-chlorine bond.Similarly there is certainly this sort of variation in ionic bond as well.
So if you compare the weakest ionic bond and strongest covalent bond then covalent bond would be stronger.And if you compare stongest ionic bond and weakest covalent bond, then of course ionic bond would be stronger.Also in many substances its not only the bond strength that matters but also the crystal structure.
An example where covalent bond is stronger:Carbon-Carbon bond is stronger than NaCl bond.
Whereas NaCl ionic bond is more stronger than Br-Br covalent bond.

So it depends on the atoms that participate in bond formation.

visit https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf for bond enthalpy.


How 'heavy' should an element be, to be a "Heavy Metal"? I've come across the term "Heavy metals" innumerable times in articles, mostly pertaining to environmental issues.

Is there a weight range (of sorts) against which an element (metal) is classified as a "Heavy metal"?

Often I see "Heavy metals" refer to Cadmium (Atomic number 48), Mercury (Atomic number 80) and Lead (Atomic number 82), but does this term also apply to metals of intermediate atomic masses [such as Tungsten (Atomic number 50), Gold (Atomic number 79) and Tin (Atomic number 81)]?

I can't seem to find any chemical literature that deals with this, so I suspect that the term "Heavy metal" has no real serious scientific definition.

There is no true, accepted definition of heavy metal. I was taught to apply the option a metal that has density equal to or over $5.0\ \mathrm{g/cm^3}$.

Other variants include a different density range, specific gravity over density, environmental impact, atomic number, toxicity, or atomic mass, even chemical properties. See here$^{[1]}$ for further information and references.

$[1]$ John H. Duffus. '"Heavy metals"—a meaningless term?'. IUPAC technical report. Pure and Applied Chemistry, $(2002)$, 74(5), pp 793$-$807.


Reaction of aqueous sodium carbonate with aluminum foil Aqueous sodium hydroxide is known to react with aluminum foil in an exothermic, hydrogen-gas releasing reaction like $$\ce{2NaOH(aq) + 2Al(s) + 2H2O(\ell) → 2NaAlO2(aq) + 3H2(g)}$$

However, on youtube, it looks like someone was able to substitute NaOH with Na2CO3 and get a vigorous similar-looking reaction as well: https://www.youtube.com/watch?v=ggPWdwcsyqM

Which makes me wonder — did they also form $\ce{NaAl(CO3)2}$ as well? E.g., via
$$\ce{2 Na2CO3(aq) + 4Al(s) + 6H2O(\ell) -> NaAl(CO3)2(aq) + 3NaAlO2(aq) + 6H2(g)}$$

Or was sodium carbonate merely used as a source of NaOH, from $$\ce{Na2CO3(aq) + H2O(\ell) \rightleftharpoons NaHCO3(aq) + NaOH(aq) }$$

It is rather:

$$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis.

$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$

The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in.

Initial $\mathrm{pH}$ is ( see notes below line ):

$$\mathrm{pH_{init}}=14 - 0.5 ( \mathrm{p} K_\mathrm{b2} - \log{ [  \ce{Na2CO3} ]} )\\=  7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [  \ce{Na2CO3} ]} ) $$

because $\mathrm{p} K_\mathrm{a2} +
  \mathrm{p} K_\mathrm{b2} =   \mathrm{p} K_\mathrm{w} = 14 $ ( at $\mathrm{ 25\ ^{°}C}$ )

As $ \mathrm{p} K_\mathrm{a2} = 10.329$:

$$\mathrm{pH_{init}}=12.17 + 0.5 \cdot  \log{ [  \ce{Na2CO3} ]} $$

..and when some $\ce{OH-}$ is spent and $\ce{HCO3^-}$ is formed:

$$\mathrm{pH}=10.329 + \log \frac {[  \ce{CO3^2-} ]} {[  \ce{HCO3^-} ]}$$

From Henderson–Hasselbalch equation

$$\ce{pH} = \ce{p}K_\ce{a} + \log \left( \frac{[\ce{A^-}]}{[\ce{HA}]} \right)  $$

If there is supposed all $[\ce{H+}]$ comes from acid dissociation and majority of acid is not dissociated, then  :

$$\ce{pH}   \overset{cca}=
\ce{p}K_\ce{a} -   \ce{pH} - \log   {[\ce{HA}]_\ce{total}}
$$

$$\ce{pH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{a} - \log   {[\ce{HA}]_\ce{total}} )$$

Alternatively

$$\ce{pOH} = \ce{p}K_\ce{b} + \log \left( \frac{[\ce{BH+}]}{[\ce{B}]} \right) \\  \overset{cca}=
\ce{p}K_\ce{b} -   \ce{pOH} - \log   {[\ce{B}]_\ce{total}}
$$

Then

$$\ce{pOH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{b} - \log   {[\ce{B}]_\ce{total}} )$$

$$14 - \ce{pH} \overset{cca}=  \frac12 ( 14 -
\ce{p}K_\ce{a} - \log   {[\ce{B}]_\ce{total}} )$$

$$\ce{pH} \overset{cca}=  7 + \frac12 ( \ce{p}K_\ce{a} + \log   {[\ce{B}]_\ce{total}} )$$


How to calculate the pH of a buffer after HCl was added? A buffer is made from $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ benzoic acid, $K_a = 6.3\cdot10^{-5}$, and $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ sodium benzoate.
a) Calculate the pH of this buffer

For this do I simply use $-\log(6.3\cdot 10^{-5})$?

b)Calculate the new pH when $0.010~\mathrm{M}$ $\ce{HCl}$ is added to $100~\mathrm{mL}$ of this buffer.

This one I do not understand.

For (a), the Henderson-Hasselbalch equation, $$\ce{pH} = \mathrm{p}K_a + \log([\ce{A^{-}}]/[\ce{HA}]),$$ comes in handy. Because your molarities and volumes of the acid and its conjugate base are equal, this indeed reduces to simply $\ce{pH} = -\log(6.3 \cdot 10^{-5})$.

For (b), the volume of $\ce{HCl}$ added is required, as the concentration of the solution alone is not sufficient information. The standard practice is to assume that $\ce{HCl}$ (being a strong acid) reacts fully with the conjugate base in your buffer solution to produce an equal amount of the conjugate acid (i.e., if $x$ moles of $\ce{A^{-}}$ are consumed by $\ce{HCl}$, $x$ moles of conjugate acid $\ce{HA}$ are produced). Therefore, you can use the Henderson-Hasselbalch equation to recalculate the $\ce{pH}$, subtracting the moles of $\ce{HCl}$ added from your conjugate base, and adding that some number of moles to your conjugate acid.


No phenolphthalein mono-anion? This is a follow-up to the question here.  There is no mono-anionic stage shown, does it "disproportionate" to the neutral molecule and the di-anion?

This is an answer to a question of mine, which I posted a few days ago, Deprotonated form of phenolphthalein.

I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deprotonated; the monoanionic form exists, but is colourless.


What does the "total energy" of a molecule mean for the Chem3D MM2 minimized energy calculation? I was fiddling around with Chem3D and the MM2 calculation for finding a molecules minimized energy.

Benzene was assigned a total energy of -0.5741 kcal/mol while biphenyl was assigned a total energy of 4.5397 kcal/mol.

What is the "total energy" relative to? In other words, why does benzene have a negative total energy while biphenyl has a positive total energy? What is the 0 energy state that they are being compared to?

General definition of energy

Usually, the total energy of a system is subdivided into kinetic energies of all particles and potential energies of the interactions between them, where the later terms are most often limited to the strongest Coulomb interactions only. In atomic units we get then:

$$
    E
    =
    - \sum\limits_{\alpha=1}^{ν} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{ν} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, the zero of energy corresponds to the case when all the terms are zero. Kinetic energy is zero when velocity is zero, while the Coulomb interaction energy is zero when particles are infinitely separated from each other. Thus, the zero of energy corresponds to the case when all the particles are infinitely far away from each other and not moving. In accordance with that choice of the zero of energy, the energy of any stable molecular system in the modeled environment (usually, an isolated molecule in the gas phase) has to be negative.

Molecular mechanics definition of energy

Molecular mechanics (MM) operates with a different kind of energy. To start from, usually, it is just the potential energy. Secondly, in MM we think about a molecule in terms of atoms and bonds (rather than nuclei and electrons) and the potential energy is divided into contributions from bond stretching, angle bending, torsional rotations around single bonds, non-bonded interactions, etc.
$$
    E = E_{\mathrm{stretch}} + E_{\mathrm{bend}} + E_{\mathrm{torsion}} + E_{\mathrm{nb}} + \dotsc \, ,
$$
where the first term, for instance, is typically proportional to the sum of the squares of extensions from the equilibrium length for all the bonds:
$$
    E_{\mathrm{stretch}} = \sum\limits_{\mathrm{bonds}} k (r - r_{\mathrm{e}})^2 \, .
$$
Now, we could take the energy corresponding to the equilibrium length $r_{\mathrm{e}}$ as the zero of energy and proceed similarly for the other bonded terms, which would mean that (ignoring usually small non-bonded interactions) the MM potential energy is zero when the molecular system is completely "unstrained" (i.e. when all geometrical parameters have their "natural" equilibrium values). However, different molecules are bonded differently, and thus, the zero of energy would, in general, be defined differently.  Each potential energy will thus be computed relative to a different zero of energy which makes the comparison of energies for any two different molecules meaningless.


Determine balloon diameter needed to lift a 7 pound weight with various gases My question is regarding setting up this equation. The givens are as follows:

Gases are helium, methane and hydrogen. (Test all three)
Assume balloon is spherical.
Assume temperature is $25~\mathrm{^\circ C}$

So I need to find the diameter of the balloon needed to lift a 7lb weight. I setup my equation as follows. $$m=\frac{PVM}{RT}$$

Next, I subtract my first gas ($\ce{He}$) from ambient air and convert my $m$ to equal grams in the weight. so I set it up like so.

$$3730.15~\mathrm{g}=\frac{PVM}{RT}-\frac{PVM}{RT}$$

Next I factor out the pressure and volume and fill in my numbers.
$$3730.15~\mathrm{g}=PV\left(\frac{(28.9~\mathrm{g~mol^{-1}})}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}-\frac{8~\mathrm{g~mol^{-1}}}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}\right)$$

This is where I am stuck at as I don't know what makes sense to use for pressure so I can solve for volume? Or maybe this is all setup wrong and makes no sense?

For a real balloon, the balloon itself will have mass, and the balloon may have higher pressure inside than atmospheric pressure.

Absent additional information, for a homework problem, you would assume mass of the balloon is zero and pressure is atmospheric pressure.

Note: in the OP 8g/mol is wrong.


Does the addition or multiplication rule have precedence deciding the significant figures in 5×5.364? How many significant figures should be present in the answer of
  $5\times 5.364$?

Addition rule: The result cannot have more digits to the right of the decimal point than either of the original numbers.
Multiplication rule: The result must be reported with no more significant figures as there are in the measurement with the few significant figures.

Using rule 1,
$$5\times 5.364 = 5.364 + 5.364 + 5.364 + 5.364 + 5.364 = 26.82,$$
one would use 4 significant figures.

Using rule 2,
$$5 \times 5.364 = 26.82 = 30,$$
one would use 1 significant figure.

Which one is correct?

If this is the question in its entirety, then you should not try to 'guess' what the origin, and therefore the accuracy, of the measurement 5 is. You should take it on face value, and complete the calculation using the multiplication rule, as the equation is requiring a multiplication operation. In this case, you should have one significant figure only.

The only ambiguity you need to consider in calculating your significant figures is whether the '5' is an exact number or not. Exact numbers come about either by definition (1 hour = 3600 seconds) or by counting (5 measurements were made) and have an infinite number of significant figures. Hence, they do not affect the accuracy of your measurements. Exact numbers are usually integers (speed of light and number of centimetres in an inch are a couple of exceptions). My guess is that your question is simply a case of going through significant figures calculations, using the information provided. To avoid ambiguity in the case of a single significant figure measurement, one could consider writing 5 as 5E0, or 5x100.

If you know 5 to be an exact number (ie the question gives that information), then you would still complete the calculation using the multiplication rule, but assume infinite number of significant figures for 5. In this case you would have 4 significant figures.


Gibbs Free Energy Paradox From my experience so far with chemistry, it seems like every reaction that happens is to attain stability. So shouldn't a lower entropy be good? Since less disorder means more stability.
But by Gibbs Helmholtz equation, a negative entropy change value favors a positive $\Delta G$.... But shouldn't a reaction which leads to stability be spontaneous?

In chemistry it has often been explained that entropy increase had something to do with increasing disorder. The word disorder stems from Helmholtz (1882). But this is outdated, it was only a first attempt at theoretical interpretation.

Entropy = Spreading of energy or of particles of molecular (or sub-molecular) scale against constraints on their free motion. These constraints may be outer (volume) or inner (aggregative) constraints. This spreading leads to quantum mechanical stabilization effects (which may be explained with the “particle in a box”).[1] — This is the reason why entropy increase leads to more stability.


Why do aqueous solutions of sodium chloride and silver nitrate swap ions? So, we know that:
$$ \ce{NaCl(aq) + AgNO3(aq) -> NaNO3(aq) + AgCl(s)}$$
(In other words, the ions swap and a precipitate forms).
However, both of those compounds seem pretty stable, since $\ce{Ag}$ and $\ce{Na}$ cations both have a 1+ charge, and $\ce{Cl}$ and $\ce{NO3}$ both have a 1- charge. Since they are stable, I would assume no reaction would take place since both ions are happy with the ions they are currently bonded to (sodium and silver, respectively).
However, $\ce{Na}$ is higher on the activity series than $\ce{Ag}$, and we know that these aqueous solutions actually contain dissociated ions, so I assume this has something to do with the double-replacement. But why is this? Why does this reaction take place, even though the original compounds appear stable at first glance?
Edit
I read this question, and it says

Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium.

The above compounds of sodium chloride and silver nitrate appear to be at an initial equilibrium, do they not?
Edit 2
The question isn't why silver chloride is a solid, but rather, why does this reaction take place if the initial two compounds are stable to begin with?

While solutions of silver nitrate and sodium chloride are in stable equilibrium on their own, once mixed the equilibrium is changed. When the salts are dissolved, you no longer have silver and nitrate or sodium and chloride ions associated with each other, you have a mixture of individual ions in solution.

But silver chloride has very low solubility, about 1 mg can dissolve in 1 L of water. So when a silver ion and chloride ion meet, they quickly fall out of solution. So if we had an equilibrium between the dissolved ions and the small amount of "soluble" silver chloride, that equilibrium would be driven toward the silver chloride by the precipitation. When compounds escape the solution by precipitating or evaporating, they are removed from the equilibrium, so trying to replace them to restore equilibrium requires converting more free ions into silver chloride and pushes the reaction to completion.


Do molecules like NH2 cation exist? I saw that the book said that nitrogen cations must have four bonds and no lone pairs. I wasn't sure why you couldn't have a lone pair and 2 bonds.

As @Mithoron said, $\ce{NH2+}$ ion is called parent nitrenium ion. The following text is an abstract from a paper1 which discussed its generation and various trapping pathways:

[...] (It) has been generated in various solutions through photolysis of 1-amino-2,4,6-triphenylpyridinium perchlorate. It was
found that the hydrocarbons, toluene and triphenylmethane trap
$\ce{NH2+}$ through two independent pathways. (1) free radical pathway
which leads to bibenzyl (in the case of toluene) or the trityl radical
(in the case of triphenylmethane). (2) hydride transfer
pathway which leads to benzyltoluene (in the case of toluene) or
triphenylmethanol (in the case of triphenylmethane). Results from dilution, triplet sensitization, and EPR experiments lead to the assignment of the hydride transfer pathway  to the singlet state of $\ce{NH2+}$ and the radical pathway to the triplet state of $\ce{NH2+}$ [...]
$$\ce{PhCH2+ + NH3 <-[PhCH3] ^{1}NH2+ -> ^{3}NH2+ ->[PhCH3] NH3+^{.} + PhCH2. }$$

There is also another paper2 which discuss about the reaction of $\ce{NH+, NH2+}$, and $\ce{NH3+}$ ions with $\ce{H2}$ at low temperatures which leads to formation of ammonium ions (the general pathway of $\ce{NH4+}$ generation in interstellar environment):
$$\ce{N+ ->[H2] NH+ ->[H2] NH2+ ->[H2] NH3+ ->[H2] NH4+}$$
This reaction scheme is very much temperature dependent as it can lead to different pathways. For instance, reaction between $\ce{NH+}$ and $\ce{H2}$ has two pathways:
$$
\begin{align}
\ce{NH+ + H2 &->[k^{a}_{NH^{+}}] NH2+ + H} &\quad \Delta H^{\circ} = \pu{-1.9 eV}  & \tag{1}\\
\ce{NH+ + H2 &->[k^{b}_{NH^{+}}] H3+ + N} &\quad \Delta H^{\circ} = \pu{-0.8 eV} & \tag{2}\\
\end{align}
$$
The former pathway occurs at 300 K while the latter one occurs at 15 K.
Reference

On the Solution Chemistry of Parent Nitrenium Ion NH2+:  The Role of
the Singlet and Triplet States in Its Reactions with Water,
Methanol, and Hydrocarbons Sanjay Srivastava, Michael Kercher, and
Daniel E. Falvey The Journal of Organic Chemistry 1999 64 (16),
5853-5857 DOI: 10.1021/jo990324+
Reaction of NH+, NH2+, and NH3+ ions with H2 at low temperatures
The pathway to ammonia production in the interstellar medium
S. Rednyk, Š. Roučka, A. Kovalenko, T. D. Tran, P. Dohnal, R. Plašil and J. Glosík, A&A
Volume 625, 2019, DOI: 10.1051/0004-6361/201834149


How can electrons manage to be at the same distance from the nucleus at all positions in p, d and f orbitals? In quantum mechanics electrons are arranged in certain orbitals namely s, p, d and f orbitals.The position of the electrons are determined by 4 sets of quantum numbers n, l, m and s. Electrons can only be at a fixed distance from a nucleus which is determined by the quantum number n, but in p, d and f orbitals how  does it manage to be at a fixed distance at every point in the region of these orbitals?(here is an example of p orbital in image)

I think you're confusing what the orbital shapes represent. The shapes are probability distributions-which means that you're most likely to find the electron somewhere within the shape.

  Electrons can only be at a fixed distance from a nucleus

This is false

  which is determined by n quantum number

The n quantum number tells you how big the orbital is-not precisely how far away the electron is.

In reality, the electron is moving around the atom, and is probably somewhere within the orbital shape.


How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine? I was given the first structure, and then drew the other 5 resonance structures:

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity.

My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There is charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule.

Hope I'm correct so far; here's my question:

In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised?

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.

Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).

  In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$-hybridised?

If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$-hybridized so that the the lone pair is in a $\mathrm{p}$-like orbital to ensure appropriate symmetry for $\pi$-overlap. Put another way, if an atom is $\mathrm{sp^2}$-hybridized in one resonance structure, then it is $\mathrm{sp^2}$-hybridized in all of them. Atoms that are $\mathrm{sp^2}$-hybridized and $\mathrm{sp^3}$-hybridized have differing geometries, which is not permitted in the resonance phenomenon.

In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$-hybridized (not the other way around).

In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems, which is not very good but will give you the idea.

  My reasoning is that the first structure contributes more.

The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list.


How to distinguish two molecules with same formula but different geometry with vibrational spectroscopy There is a sample that we know is all made up by only one of these two molecule:

how you can see they have the same formula but two different geometries. The question is if it is possible to know by which of these two molecules the sample is made with an experiment involving vibrational spectroscopy (IR or Raman). The following is a quite wide answer I gave to myself but I'm not sure it is right, so please correct me or confirm what I have written.
We know that if we are considering small oscillations, the vibrational part of the wave function of a molecule with $N$ atoms is the product of $3N-6$ quantum armonic oscillators eigenfunctions, and we also know that in a classic way to proceed the nuclei can move in $3N-6$ differents normal modes. Fixed one of these normal modes, if it is for example IR active, we know that all the $3N-6$ oscillators moving in that normal mode can absorb photons from the external radiation. So if the external radiation has a frequency band enough large to contain all the proper frequencies of the oscillators, we'll notice $3N-6$ absorbtion lines. While if we are considering a not IR active normal mode of course we won't see any absorbtion lines. In general the nuclei of a molecule move on a linear combination of all the normal modes. So we can think that each IR active normal mode of the combination gives the same $3N-6$ lines while each not IR active mode gives zero lines. So I would expect to observe for my real sample 12 absorbtion lines (I am not considering overtones and the noise of the media) which correspond to the proper frequencies of the oscillators. So since the two molecule have exactly the same oscillators (it only changes their position), I should observe the same 12 lines on both cases, so I'd say that it is not possible to distinguish between the two. The same argument can be used for the Raman. However I'm almost sure that I'm wrong and a way there must exists.

The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $x, \,y, \,z$ and $xy, \,xz$ etc operators. The $x, \,y, \,z$ operators transform as does a dipole (so IR transitions) and  act as an alias for the symmetry species.

For example in the $C_{2v}$ point group each  IR transition must be one of $A_1,\,B_1 ,\, B_2$ species and as there are 12 transitions  there must be several of the same symmetry but different frequency.

The Raman transitions correspond to product terms , $x^2,\, xy$ etc. (because Raman spectroscopy depends of the polarisability which is proportional volume and so area in cross-section ) and these have also $A_1,\,B_1 ,\, B_2$ symmetry species as well as $A_2$ species (in $C_{2v}$) so we do expect to see some different transition frequencies between IR and Raman spectra. We do not necessarily know what the symmetry species are for these transitions, just that there is a difference.

In the case of $D_{2h}$ point group as there is a centre of inversion, there are no common transitions between IR and Raman spectra. Look at the point group and you will see this.

In more detailed analysis the polarisation of Raman signals can be used to determine which transition is $A_1$ etc. Spectroscopy books give details of this.

look at Understanding group theory easily and quickly for an explanation of point groups.


Covalency of metal oxides in high oxidation states So, metal oxides are generally basic, right? But in higher oxidation states their covalent character is predominant. For example in
Mn2O7 the Mn - O bond is covalent. Why is that?

Metal oxides are not always basic.  In many cases where the bonding is strongly ionic, the oxide ions do act as a base.  However, if the metal is in a high oxidation state an ionic bond would require a prohibitively high charge on the metal ion; the ionization energy required to put seven positive charges on one manganese atom far exceeds what can be reclaimed from the electron affinity of oxygen and electrostatic attraction alone.  The energetically preferred option for $\ce{Mn2O7}$ is to form polar covalent bonds instead.  Such bonds are similar to those in $\ce{CO2, SO3, P4O10,}$, etc., and thus $\ce{Mn2O7}$ will act similarly to those oxides -- as an acid anhydride.


What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation? In the first propagation step of the radical chain mechanism of alkane halogenation, why does the halogen actually abstract hydrogen from the alkane? My organic chemistry textbook so far has went into details on MOs, hybridized orbitals, and general thermodynamics of reactions but has been lacking in the by and large why of reactions. Is the $\ce{C-H}$ bond broken by sufficient collisions from halogen free atoms and the hydrogen goes to the halogen as a result of $\ce{F}$/$\ce{Cl}$/$\ce{Br}$ being more electronegative than carbon, or what should cause this to happen?

why does the halogen actually abstract hydrogen from the alkane?

Look at the two tables of bond energies on this webpage.  The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds.  In the top table we see that the H-Cl bond strength is 103 kcal/m.

If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be

$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$

Look at the top row in the bottom table again.  As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.

Edit

  As I understand, the first propagation step in the free-radical
  monochlorination of methane is slightly endothermic but is driven to
  completion by the exothermic nature of the second propagation step

The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.

$$\ce{CH3-H + Cl{.} -> CH3{.} + H-Cl}$$

$$\ce{CH3{.} + Cl-Cl -> CH3-Cl + Cl{.}}$$

The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.

These steps are consistent with my earlier argument.  You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate "back of the envelope" calculation and view of what is going on.

As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species.  It is reactive because it doesn't have an octet of electrons.  By undergoing the above-reactions it can regain its octet and become a stable entity.

Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram).  Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products.  So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.


Tollen's Test and Werner's Theory The production of silver ions in Tollen's test violates Werner's theory of
non - ionizable secondary valency since we are using the diamminesilver(I) complex as the reagent. How does one reason this??

Ionizable and non-ionizable in this context refers to the observation that the counterions of a complex ionize in solution and can thus be precipitated and the ligands of the complex cannot and are thus non-ionizable.

There are two things at work here:

All complexes exist in an equilibrium with the free ligands. In the case of diamminesilver(I), $$\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]^+}$$ the equilibrium is overwhelmingly toward the complex and no AgCl will be precipitated if chloride is added. However, if we add iodide instead, we will actually observe an AgI precipitate because AgI is far less soluble than AgCl and there is sufficient $\ce{Ag+}$ that we see some precipitate. Therefore, there is actually always a little bit of $\ce{Ag+}$ free in solution.
The Tollen's test is a redox reaction. Unlike the case above where we're precipitating an ionic compound and thus need to form an ionic bond, electron transfer to the silver is quite easy and once reduced, it is unfavourable for the silver to remain in solution. Even more extensive ligand coverage in something like $\ce{[Fe(CN)6]^{4-}}$ doesn't prevent redox reactions from occurring. Even some metalloporphyrins undergo redox reactions at the metal centre.


Why or why not can hydroquinone and 2H-pyrrole act as ligands? I want to understand why hydroquinone and 3H-pyrrole can or cannot act as ligands in transition metal complexes. If they are ligands, then what is their denticity?

I'm trying to understand the overall co-ordination behaviour of these two molecules, so please consider all atoms and rings as possible electron donors.

I think the denticity for the second molecule should be 1, since there's only 1 donor atom N, but would it be a weak ligand as the nitrogen is double bonded to a C?

Zaworotko, M. J.; Stamps, R. J.; Ledet, M. T.; Zhang, H.; Atwood, J. L. Heterocyclophane complexes of transition metals. 1. Synthesis and crystal structure of both the (.eta.5-2.2pyrroloparacyclophane)tricarbonylchromium and the (.eta.6-2.2pyrroloparacyclophane)tricarbonylchromium. Organometallics 1985, 4 (10), 1697-1700. DOI: 10.1021/om00129a001

A chromium tricarbonyl complex of a pyrrole is reported, here the pyrrole coordinates with all four carbon atoms and the nitrogen atom. I think that the pyrrole is a six electron donor to the chromium.

Bentivegna, B.; Mariani, C. I.; Smith, J. R.; Ma, S.; Rheingold, A. L.; Brunker, T. J. Formation, Stability, and Structures of Borenium and Boronium Cations Derived from Pentamethylazaferrocene–Boranes by Hydride or Chloride Abstraction Reactions. Organometallics 2014, 33 (11), 2820-2830. DOI: 10.1021/om500348u

A pyrrole binds with all five non hydrogen atoms to an iron which is also bonding to a pentamethyl cyclopentadienyl anion. This is forming a protonated version of CpFe(C4N) which looks very much like ferrocene. In this case the bonding is the same as it was in the chromium compound.

Yamamoto, K.; Kimura, S.; Murahashi, T. σ-π Continuum in Indole-Palladium(II) Complexes. Angew. Chem. 2016, 128 (17), 5408-5412. DOI: 10.1002/ange.201601992

A pyrrole binds with two carbons as a pi ligand to palladium. These selected results suggest to me that pyrrole can bind as an alkene or as if it was a Cp ligand to transition metals.

Son, S. U.; Reingold, J. A.; Carpenter, G. B.; Czech, P. T.; Sweigart, D. A. Charge-Assisted Hydrogen Bonding and Other Noncovalent Interactions in the Self-Assembly of the Organometallic Building Block [(η6-hydroquinone)Rh(P(OPh)3)2]+with a Range of Counteranions. Organometallics 2006, 25 (22), 5276-5285. DOI: 10.1021/om0604425

It reports the binding of a rhodium to all six carbon atoms of hydroquinone. Here it acts as a six electron donor.

So there is nothing to stop the molecules in the question acting as pi ligands to transition metals.


What happened to mesoporous materials for drug delivery? A review on the topic can be found here: Wang, S. Ordered mesoporous materials for drug delivery. Microporous Mesoporous Mater. 2009, 117 (1–2), 1–9. DOI: 10.1016/j.micromeso.2008.07.002.

So the 21st century started with great hopes for these mesoporous materials - particularly for drug delivery. After initial success it seems the research in this field has fizzled out.  I'm just curious to why there hasn't been more progress?  Thanks so much!

I think that research of mesoporous materials for several uses is alive and well.  The link referenced within your question is for an article written in 2008.  Wikipedia has a good, though brief, article on its current research status without respect to medical use in particular:

  A procedure for producing mesoporous materials (silica) was patented
  around 1970, and methods based on the Stöber process from 1968 were
  still in use in 2015. It went almost unnoticed[8] and was reproduced
  in 1997. Mesoporous silica nanoparticles (MSNs) were independently
  synthesized in 1990 by researchers in Japan. They were later produced
  also at Mobil Corporation laboratories and named Mobil Crystalline
  Materials, or MCM-41.

This is an abstract available from PubMed published in 2007:

  Research on mesoporous materials for biomedical purposes has
  experienced an outstanding increase during recent years. Since 2001,
  when MCM-41 was first proposed as drug-delivery system, silica-based
  materials, such as SBA-15 or MCM-48, and some metal-organic frameworks
  have been discussed as drug carriers and controlled-release systems.
  Mesoporous materials are intended for both systemic-delivery systems
  and implantable local-delivery devices. The latter application
  provides very promising possibilities in the field of bone-tissue
  repair because of the excellent behavior of these materials as
  bioceramics. This Minireview deals with the advances in this field by
  the control of the textural parameters, surface functionalization, and
  the synthesis of sophisticated stimuli-response systems.

One possible competitor for sustained drug delivery is transdermal drug delivery.  It seems that this method, which has been used for centuries, is well established and thoroughly approved by agencies like the FDA in the US.  In the US anyway, approval of new medicine or medical devices happens very slowly, and mesoporous materials seem to still be more in the research phase and may not have the level of agency approval that competing systems have attained.


Equation for overall heat transfer coefficient The heat flow (in $[\pu{J/s}]$) through the wall of a tube of length $L$ and inner radius $R_0$ is
$$Q = (2πR_0L)U(T_{1\infty} - T_{2\infty}),$$
where $T_{1\infty}$ is the bulk temperature in the inside and $T_{2\infty}$ is the bulk temperature on the outside. The overall heat transfer coefficient reads as:
$$\frac{1}{UR_0} = \frac{1}{h_1R_0} + \frac{\ln{(R_a/R_0)}}{k} + \frac{1}{h_2R_a} \tag{1}$$
where $R_0$ and $R_a$ are the inner and outer radius of the tube, respectively, $h_1$ and $h_2$ are the inner and outer heat transfer coefficients, and $k$ is the thermal conductivity of the wall.
How does Eq. (1) simplify for the $T$-profile shown on the left?

$U = k/R_0 \cdot 1/\ln(R_a/R_0)$
$U = h_1$
$1/U = 1/h_1 + 1/h_2 \cdot (R_0/R_a)$
$1/UR_o = (\ln(R_a/R_o)/ k ) + (1/h_2R_a)$

I had seen a similar $T$-profile in one of our exercises and there they used Eq. (1), however when selecting option 1, I get "incorrect answer". How can you tell from looking at the $T$-profle what the equation for the overall heat transfer coefficient will be?
All help and hints are appreciated!

This is the heat equation analogue to Ohm's law, in which resistances are additive (in series), and hence the inverse conductances are additive.

The way to reason is that the full equation (1) for the heat conductivity (1) contains three terms: one for the interior of the tube, one for the tube wall, and a third for the exterior. When you look at the profile, it becomes evident that, since temperatures immediately adjacent inside and outside the tube are equal, the tube has infinite heat conductance ($k=\infty$).

The right answer is therefore 3.


Comparing basic strength of nitrogen heterocycles Compare the basic strength of the following compounds:

  d > b > c > a
  c > b > a > d
  c > a > b > d
  a > d > c > b

I know that in compound (c), the π-bond between nitrogen and cyclopropenyl will polarize (due to aromatic stability) such that it increases electron density of the nitrogenous ring. Whereas, in compound (d), the π-bond will polarize reducing the electron density of nitrogenous ring. Hence, comparing compounds with similar nitrogenous rings, I get that the basic strength of (c)>(b)and that of (d)<(a). But I thought the ring containing more number of nitrogen atoms will be more basic as there will be more lone pair donors, so I thought the answer to be option (4), but the correct answer happens to be option (2). I would like to know that how is basic strength of the nitrogenous rings being compared?

It is likely that the electron pair (from the HOMO) being donated to the acid is from a π molecular orbital. Since the π bond is rather weak, the π electrons should be rather high in energy and is thus able to interact well with the LUMO of the acid. When nitrogen atoms are incorporated into the π system, the energy of the HOMO is lowered due to the high electronegativity of N. Thus, incorporation of N atoms into the system would decrease the basicity of the molecular ion. As such, A and D would be inferior bases, compared to B and C.

Choosing between A and D, and between B and C, we would have to base our criteria for the basicity on the extent of delocalisation. In D, the π system is larger, thus greater delocalisation was achieved, with a lowering of the energy of the HOMO. Thus, D is likely to be less basic than A. Similarly, the extent of delocalisation is also larger in C, thus it would have a lower-energy HOMO, compared to B. Thus, B is likely more basic than C.

Therefore, option (2) is likely the correct order of basicity.


Europium stability in +2 and +3 state I would like to ask a question about europium's stability in the $+2$ and $+3$ oxidation state.

The electronic configuration of europium in its neutral state is $\ce{[Xe] (4f)^7 (6s)^2}$.

Now, when in the $+2$ oxidation state, the electronic configuration is $\ce{[Xe] (4f)^7}$ and in the $+3$ oxidation state, it is $\ce{[Xe] (4f)^6}$.

Now, I thought the $+2$ oxidation state is more stable because it's a half-filled $\ce{f}$ sub-shell so there is less mutual repulsion between electrons in the same sub-shell.

However, an article from nature.com reads the following:

  Europium metal is now known to be highly reactive; the element's most stable oxidation state is +3, but the +2 state also occurs in solid-state compounds and water.

Now, I would like to ask, how is it possible for europium to be most stable in its $+3$ oxidation state, knowing what I have mentioned above?

This is going to be a rehash of Oscar Lanzi's answer, but this point needs to be driven home, so I make no apologies.

"Special" electron configurations - fully-filled or half-filled subshells - are only a relatively minor factor in determining the stability of oxidation states. I've written about this before in a slightly different context, but IMO it would be well worth reading this, as the exact same principles operate in this case: Cr(II) and Mn(III) - their oxidizing and reducing properties?

What determines the "most stable oxidation state"? We can consider what happens when we increase the oxidation state of a metal: firstly, we need to pay ionisation energies to remove electrons from the metal. On the other hand, though, metal ions in higher oxidation states can form stronger bonds (e.g. lattice energy in ionic compounds, solvation energy in solution, or more covalent bonds in molecular compounds). The balance between these two factors therefore leads to a "best" oxidation state.

For example, sodium could hypothetically go up to Na(II) and form $\ce{NaCl2}$. Theoretically, this compound would have a significantly larger lattice energy than $\ce{NaCl}$. However, sodium's second ionisation energy is prohibitively large, which means that it stays put in the +1 oxidation state. On the other hand, magnesium's second ionisation energy is not prohibitively large, and therefore Mg preferentially forms $\ce{MgCl2}$ over $\ce{MgCl}$.

For all lanthanides, the first three ionisation energies are all fairly comparable. There is a graph here which shows the variation in the ionisation energies of the lanthanides, which I reproduce below.

(source: Inorganic Chemistry 6ed, Weller et al., p 630)

For all lanthanides, these three ionisation energies are easily compensated for by the extra lattice energy / solvation energy derived from a more highly charged ion. The fact that Eu(II) has a $\mathrm{f^7}$ configuration only serves to make its third IE marginally larger than that of Gd. This difference is sufficient to make Eu(II) an accessible oxidation state (cf. different electronic behaviour of EuO and GdO), but not sufficient to make it the most stable oxidation state.

Note that "most stable oxidation state" depends on the conditions, too, but I assume we are talking about aqueous solution.


How does liquid mercury conduct electricity? It is stated that mercury is a liquid at STP because the high relativistic mass of its valence electrons stabilizes their orbits and thus prevents the electrons from being shared. If so, liquid mercury would seem to have no "free" electrons; how, then, does it conduct electricity?

Keep in mind that "stabilizing" orbitals does not "prevent" electrons from being shared, only that they would be shared to a smaller extent.  The very fact that mercury is a liquid and not a gas shows that there is some bonding in mercury, even if it is weaker than the bonding in most metals.  So, it is conductive the same way other metals are, just less conductive.  You can look up the conductivity of mercury, and you will find it to be lower than most metals.


What are contrast agents in MRI and how do they work? By reading some Wikipedia pages about MRI and MRI Contrast agent, I understand that contrast agents are mediums that reduce the relaxation times of the body tissues thereby increasing the contrast of the image obtained by MRI. My questions:

Are contrast agents just used for targeting a special tissue, such as a tumor, or they can be used the increase the contrast of the image of the whole body?
I know they decrease the relaxation time of the hydrogen atoms in body, but how do they do that?

Thanks

1) Contrast agent is used to create a sharp contrast on an otherwise fuzzy image. Consequently, its distribution must have sharp gradient, i.e it must have strong affinity to some type of tissue or physiological liquid. The target may be widely represented in the body, but it still should be a clear target. Or, alternatively, it should be injected into a physiological liquid and have no time to distribute evenly.

2) Wikipedia provides simplistic description. The idea is that to relax efficiently, a magnetic nucleus (including a proton from a hydrogen atom) needs to interact efficiently with chaotic microscopically inhomogeneous magnetic field. Such field may be produces by another microscopic magnetic objects. $\ce{Gd^{+3}}$ has seven unpaired electrons, giving the atom the strongest magnetic field of all single atoms available in reasonable chemical conditions.


Why is the bond order of CO+ 3.5? According to my books the bond order of $\ce{CO+}$ is  $3.5$. But shouldn't it be $2.5$? On googling this, I found the following answer that is on Stack Exchange but its only talks about the bond length.

I am unable to understand why it is $3.5$ as I am in class 11.

For a long time it was taught in school and universities that the HOMO of carbon monoxide is anti-bonding. Without more context it was also often taught that the bond order in CO is three, since there are eight electrons in bonding orbitals and two in anti-bonding orbitals. $$\text{Bond order} = \frac12(\text{bonding} - \text{anti-bonding})$$
By assuming that the HOMO is anti-bonding (it is not!) and taking away one electron, the  bond order has to increase to 3.5. This is wrong.

When we have a look at the MO diagram, a calculated version can be found here, we know that the HOMO, i.e. 3σ, is a bonding orbital, while the anti-bonding orbital is the 2σ. Upon ionisation, we would indeed remove one bonding electron and therefore the bond order has to decrease to 2.5 as you suggested.
However, it is not that easy. Strictly speaking the below MO scheme is, as well as MO theory itself, an approximation, and only one possible configuration. While we do not have to use resonance structures with MO theory, we have to consider other configurations (analogous to excited states). So naturally the bond order of CO is not strictly 3. And removing an electron does not mean we are removing it from only one orbital, rather than decreasing the electron density. Therefore we cannot accurately predict the bond order with these simple considerations.
Experimental observations and theoretical calculations suggest that the bond indeed becomes stronger when removing an electron. See the linked question and Philipp's answer within for more detail. (Don't look at the other answers, they are as wrong as they could be.)

In short: The bond order of $\ce{CO}$ is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$.

An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there are no non-bonding orbitals.


How can ionic substances have an infinitely-repeating lattice structure? The title's pretty much it. At school, we're taught that ionic substances are composed of an infinitely-repeating lattice, with atoms at fixed angles from other atoms. How is this possible when I can get an individual grain of salt? Clearly if there's an individual grain, it can't be infinitely repeating.

To be technically accurate, it's "practically" infinite.

It's not infinite since as you mention a grain of salt (or even a large crystal) has some finite size.

You have to remember the actual atomic dimensions $10^{-10} \:\mathrm{m}$. So while it's not truly infinite, there are an incredibly large number of atoms in a crystal.


Do molecules containing only carbon exist? Why doesn't carbon form bonds with itself to form a molecule? Carbon shows the property of catenation, so why doesn't it form a cyclic molecule such as:
 C = C
 ‖   ‖
 C = C

They do form.  Besides fullerenes, there are carbon molecules with varying numbers of atoms around some stars. Such molecular forms are, of course, less stable than the macromolecular structures we more commonly see under ambient conditions.


Can we add an electron to an atom? Say that we have an atom which is inert, or stable. How would one go about adding one electron. Is it possible?

F-center defects in crystals are due to unpaired electrons and can be caused by ionizing radiation; these defects can accumulate over thousands of years. This can be used to find the age of minerals with retroactive inclusions, or to intentionally color gems.

An electron gun, as in a cathode ray tube (CRT), will also do the job. Older televisions had CRT's, and to prevent electrons building up on the front of the tube, they were siphoned off to ground. "Static" electricity on the faceplate of a CRT can be felt; it is due to the build-up of electrons.

Of course, one could question if electrons in the above examples are bound to a particular atom, are interstitial, or are free to move, but as mentioned above by @MaxW, an ionized gas (plasma) does have particles with excess electrons, as well as those with a deficiency. In interstellar space, an ion could last for years.

Also, radical anions are easily produced, for example the reduction of naphthalene by an alkali metal.


Which is more electronegative: hydrogen or astatine? Which is more electronegative - hydrogen or astatine? I'm a bit confused because my teacher taught me that for an atom to be electronegative it has to be on the top of the periodic table and/or on the right. But hydrogen is on the top-left and astatine is on the bottom-right of the periodic table.

How does one determine which is more electronegative?

An interesting way to determine which of $\ce{A}$ or $\ce{B}$ is more electronegative is to take their binary compound $\ce{A_xB_y}$ and observe its dissociation. Note that it can be $$\ce{A_xB_y -> xA^y+ + yB^x-} \tag{1}\label{a}$$ or $$\ce{A_xB_y -> yB^x+ + xA^y-} \tag{2}\label{b}$$

Reaction $\ref{a}$ indicates that $\ce{A}$ is less electronegative than $\ce{B}$, because in the dissociation of $\ce{A_xB_y}$, $\ce{A}$ carries the positive charge, while $\ce{B}$ carries the negative charge. Reaction $\ref{b}$ instead indicates that, for similar reasons, $\ce{A}$ is more electronegative than $\ce{B}$ 1

Going by the same logic, we may tend to observe a hydrohalo acid in this case, specifically, the hydroastatic acid/hydrogen astatide. Quoting from its Wikipedia page,

  This chemical compound can dissolve in water to form hydroastatic acid, which exhibits properties very similar to the other four binary acids, and is in fact the strongest among them. However, it is limited in use due to its ready decomposition into elemental hydrogen and astatine (my emphasis) Because the atoms have a nearly equal electronegativity, and as the $\ce{At+}$ ion has been observed, dissociation could easily result in the hydrogen carrying the negative charge. (my emphasis) Thus, a hydrogen astatide sample can undergo the following reaction:
  $$\ce{2HAt -> H+ + At- + H- + At+ -> H2 + At2}$$

which also explains the spontaneous decomposition into elemental hydrogen gas and the astatine precipitate.

If you also have a look at their electronegativities, both have a Pauling electronegativity $2.2$, further complicating the comparison.

Therefore, based on these evidences, I can say, without the least hesitation, that theoretical comparison of the electronegativities of hydrogen and astatine is not possible.

[1]: a shrewd observer would notice that the apt way to write the second reaction's reactant would be $\ce{B_yA_x}$ instead, since $\ce{A}$ is more electronegative


In which of the following chemical compounds are the building blocks molecules? In which of the following chemical compounds are the building blocks molecules?
(A) $\ce{AgNO3}$
(B) $\ce{HNO3}$
(C) $\ce{KNO3}$
(D) $\ce{NaNO3}$

(B) is the correct answer, but I have no idea why. Could you briefly explain why this is the correct answer?

Answer B is correct because nitric acid is the only compound whose building blocks satisfy the IUPAC Gold Book definition of a molecule, a neutral (uncharged) entity consisting of at least 2 or more atoms. The building blocks of all other given compounds are atomic ($\ce{K+}$, $\ce{Na+}$ and $\ce{Ag+}$) or molecular ions ($\ce{NO3-}$), which by their nature do not fulfill the requirement of being electrically neutral. Per definition, they are molecular entities, i.e. distinct particles which can be identified as separate entities. This term is more general and also includes charged species like ions.


How does cullet reduce the production temperature in glass production? In my book, I read that broken glass pieces are reused in the production of glass, known to be cullet. This on one hand saves the production cost and also reduces the production temperature.

My question is: how does adding cullet reduce the production temperature? I see no reason why the materials would react any differently with or without cullet.

No expert here, but cullet would react faster than sand with the alkali, developing into a more uniform fluid that can be mixed at a (slightly) lower temperature so as to eventually dissolve the sand.

It is like beginning the reaction to produce glass at 25% (or however much cullet is used), rather than 0%, so that raising the temperature is not needed to to speed up the reaction. Lower temperatures can be used without needing to increase the time, and thus wear and tear on the equipment can be reduced.

So it's not that the cullet causes a reduction in temperature, but rather allows it.


Naming of substituted amines and amides While naming substituted amines and amides, should we consider the "N" while deciding the order of substituents or simply place the substituents attached to nitrogen in the beginning?

E.g. Should the compound be named:

3-chloro-N-ethyl-4-methyl-N-propylhexanamide

or

N-chloro-N-propyl-3-chloro-4-methylhexanamide.

Substituents are listed alphabetically by the name of the substituent. The N in front of the substituent is a locant, and thus is not used when alphabetizing:

  3-chloro-N-ethyl-4-methyl-N-propylhexanamide is correct because the substituents are in correct alphabetic order:

chloro, ethyl, methyl, propyl


Relative stability of carboxylic acid derivatives This picture depicts the relative stability associated with resonance structure.

My questions are:

1.1. Teacher says anhydride is less stable because of two reasons above.
but it seems there's no reason the resonance of anhydride would be less likely compared to acetate.
1.2. and about reason #2, the actual structure is the hybrid of these resonances so it's hard to agree with this reason.

Why won't Cl give its e- pair to adjacent carbon for resonance structure.

Someone says it's about electronegativity but it contradicts with resonance associated with oxygen. And someone says it's about the high energy orbital of Cl (n=3) but I can imagine a acetyl fluoride and this logic doesn't solve this problem anymore.

Firstly "stabilised" is very ambiguous. "Stable" is always defined with respect to something. The proper phrasing here should probably be stable towards nucleophilic attack.

For the anhydride - yes the actual structure is a hybrid. That just means that both C=O have to "share" the central oxygen's lone pair - in a sense each of them get half resonance each - which means that both C=O's are not very stabilised.

I don't buy the "less likely" explanation. Your teacher is trying to say that there's a "50% chance" that at any one point in time a particular C=O is stabilised, but resonance isn't something that you flip a coin and decide which way it goes.

For the acyl chloride, it is really to do with the leaving group ability of Cl-. See this question for a discussion: Do acyl fluorides exhibit resonance donation or inductive withdrawal?

Essentially the idea is that since $\ce{Cl-}$ is a better leaving group than $\ce{OMe-}$, for the acyl chloride (X = Cl), resonance form 2 contributes more to the actual structure than it does in the ester (X = OR):

As you can see, resonance form 2 makes the C=O more susceptible to nucleophilic attack because of the positive charge associated with the carbonyl group. Therefore, any substituent X that favours resonance form 2 will make the carbonyl group less stable towards nucleophiles.


Orbital conservation in Molecular orbital theory MOT is really confusing me.

I was taught that the no. of orbitals remains conserved.That means, the number of atomic orbitals used is equal in number to the molecular orbitals formed.

But how is it possible in the following case?

For example in Case 1, I take a positive lobe of a p orbital and combine it head on with the positive lobe of another p orbital I get a bonding MO.

In a different scenario i.e Case 2, I take a positive lobe of a p orbital and combine it head on with the negative lobe of another p orbital to get an antibonding MO.

So overall I have used 4 atomic orbitals (2 in each case) to get 2 MO( one bonding and one antibonding).
So what is happening here?
Am I missing something?

In the linear combination of atomic orbitals (LCAO) there would be a bonding orbital and an antibonding orbital when two atomic orbitals combine to form a bond. The bonding orbital would be lower in energy that either atom contributing an orbital, and the antibonding higher than either.

One twist comes from the number of electrons. If there are just two electrons, then they both go into a bonding orbital. If there are three electrons then two go into a bonding orbital and one into the antibonding orbital.

There are three p orbitals so two atoms could be bonding with 1-3 p orbitals. If two p orbital bonds were being used then both p molecule bonding orbitals should be lower in energy than either antibonding orbital.

The whole concept of having a bonding orbital and an antibonding orbital is necessary to explain for instance electronic transitions. Imagine an oxygen atom in free space. If an electron is added, the oxygen molecule will accept the electron to form an ion $\ce{O^{-}_2}$. The electron doesn't go to one oxygen or the other, it is shared between both in a molecular orbital. Look at the diagram below for $\ce{O2}$. We'd predict that the electron will fall into one of the antibonding pi orbitals ($\pi^*$). From this we'd predict that a free oxygen atom could actually take three extra electrons and be stable. However the fourth electron would fill all the orbitals and $\ce{O^{-4}_2}$ would have no more electron binding energy than two free $\ce{O^{-2}}$ atoms.


Spectroscopic Determination of an Equilibrium Constant I got values of $\pu{5.0E-4}$, $\pu{2.5E-4}$, and $183$ for the following problems. However, these seem to be inaccurate and I'm not sure what to do.

Here is the procedure:

  Obtain 6 disposable test tubes. Label the test tubes 1 through 6. To each of the $\pu{6 mm}$ test tubes, add $\pu{5.0 mL}$ of $\pu{2.0E–4 M}$ $\ce{KSCN}$. If you make a mistake, you must rinse out and thoroughly dry the test tube before starting over.

  To test tube 1, add $\pu{5.0 mL}$ of $\pu{0.20 M}\; \ce{Fe(NO3)3}$ and stir. The resulting solution is your standard solution of $\ce{FeSCN^2+}$.

  For the other test tubes, do the following: Measure $\pu{10.0 mL}$ of $\pu{0.20 M}$ $\ce{Fe(NO3)3}$ in a $\pu{25 mL}$ graduated cylinder, fill to $\pu{25.0 mL}$ with distilled water, and stir thoroughly to mix. Measure $\pu{5.0 mL}$ of the resulting $\pu{0.080 M}$ $\ce{Fe^3+}$ into test tube 2.

  Discard all but $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$, refill the graduated cylinder with $\pu{10.0 mL}$ of the $\pu{0.080 M}$ $\ce{Fe^3+}$ solution to $\pu{25.0 mL}$ with distilled water, and stir thoroughly. Add $\pu{5.0 mL}$ of the resulting $\pu{0.032 M}$ $\ce{Fe^3+}$ to test tube 3.

  Again discard all but $\pu{10.0 mL}$ of the contents of the graduated cylinder, refill to $\pu{25.0 mL}$ with distilled water, stir, and add $\pu{5.0 mL}$ to test tube 4.

  Repeat above procedure with test tubes 5 and 6.

  Calculate the initial concentrations of $\ce{Fe^3+}$ and $\ce{SCN-}$ for the six solutions to be prepared in this experiment (i.e., the 6 solutions prepared in test tubes 1–6). These are the concentrations just after the two solutions are mixed in the large test tube (thereby diluting each other) but before any $\ce{FeSCN^2+}$ has formed. Assume that the volumes are additive and that the $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated. What is the concentration of $\ce{Fe^3+}$ in test tube #5?

  $$
\begin{array}{c|cc}
\hline
\text{Test tube} & \text{Initial}~[\ce{Fe^3+}] & \text{Initial}~[\ce{SCN-}] \\
\hline
1 & & \\
2 & & \\
3 & & \\
4 & & \\
5 & & \\
6 & & \\
\hline
\end{array}
$$

  For the chart in question 1, what is the molar concentration of $\ce{SCN-}$ in tube #4?
  If the equilibrium $\ce{[FeSCN^2+]}$ in test tube 2 was determined to be $\pu{8.3E-5 M}$, calculate the equilibrium $\ce{[Fe^3+]}$.

You must first calculate dilution factors. Assuming the volumes are additive, we can calculate dilution using $M_1V_1=M_2V_2$ equation. Also, assuming $\ce{Fe(NO3)3}$ and $\ce{KSCN}$ are each completely dissociated, we can say:

$$\ce{[Fe(NO3)3] = [Fe^3+]} \text{ and } \ce{[KSCN] = [SCN-]}$$

Thus, initial $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ are $0.20$ and $\pu{2.0 \times 10^{-4} M}$, respectively.

In test tube one: $\pu{5.0 mL}$ of $\pu{0.20 M} \; \ce{Fe(NO3)3}$ and $\pu{5.0 mL}$ of $\pu{2.0 × 10^{−4} M} \; \ce{KSCN}$ are added together, thus dilution factor is 1:2. Therefore, using $M_1V_1=M_2V_2$ equation you can calculate initial concentrations of $\ce{[Fe^3+]}$ and $\ce{[SCN-]}$ after the addition (but before they react with each other) as follows:

$$\ce{[Fe^3+]} = \pu{\frac{0.2 \times 5.0}{10.0} M} = \pu{0.10 M} \text{ and } \ce{[SCN-]} = \pu{\frac{2.0 × 10^{−4}  \times 5.0}{10.0} M} = \pu{1.0 × 10^{−4} M}$$

As a matter of fact, since the original solution of $\ce{KSCN}$ did not undergo any dilution maneuver before final 1:1 dilution, $\ce{[SCN-]}$ of all 6 test tubes are same, and its value is $\pu{1.0 × 10^{−4} M}$.

On the other hand, the original solution of $\ce{Fe(NO3)3}$ has undergone serial dilution of $\pu{10 mL}$ to $\pu{25 mL}$ (hence dilution factor is $10/25$ or $2/5$). Therefore, if original concentration of each solution is $x$ before add together in test tubes, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$ is:

$$\ce{[Fe^3+]}_a = \frac{x \times 2.0}{5.0} \: \pu{M}$$

Thus, the final $\ce{[Fe^3+]}_a$ before add with $\ce{[SCN-]}$:

$$\ce{[Fe^3+]}_2 = \pu{\frac{0.20 \times 2.0}{5.0} M}= \pu{0.08 M}, \, \ce{[Fe^3+]}_3 = \pu{\frac{0.08 \times 2.0}{5.0} M}= \pu{0.032 M}, \,  \ce{[Fe^3+]}_4 = \pu{\frac{0.032 \times 2.0}{5.0} M}= \pu{0.0128 M}, \,  \ce{[Fe^3+]}_5 = \pu{\frac{0.0128 \times 2.0}{5.0} M}= \pu{0.00512 M}, \,  \ce{[Fe^3+]}_6 = \pu{\frac{0.00512 \times 2.0}{5.0} M}= \pu{0.002048 M}$$

Finally, those concentrations were further diluted in 1:1 ratio in test tubes, and therefore, initial $\ce{[Fe^3+]}$ are half of those $\ce{[Fe^3+]}_a$ values. I tabulated all of them accordingly in following table, as answer to your question 1:

$$
\begin{array}{c|cc}
\text{Test Tube #} &\text{Initial $\ce{[Fe^3+]}$} &\text{Initial $\ce{[SCN-]}$} \\\hline
1 & \pu{0.10 M} & \pu{1.0 × 10^{−4} M}\\
2 & \pu{0.04 M} & \pu{1.0 × 10^{−4} M}\\
3 & \pu{0.016 M} & \pu{1.0 × 10^{−4} M}\\
4 & \pu{0.0064 M} & \pu{1.0 × 10^{−4} M}\\
5 & \pu{0.00256 M} & \pu{1.0 × 10^{−4} M}\\
6 & \pu{0.001024 M} & \pu{1.0 × 10^{−4} M}\\\hline
\end{array}
$$

There are 7 different values in the table. But you have given one
relevant answer, and it is also incorrect.
Find the answer for $\ce{[SCN-]}$ in test tube 4 from the table, which is
identical in all 6 test tubes.
The concern equilibrium is: $$\ce{Fe^3+ + SCN- <=> FeSCN^2+}$$ Reaction mole ratio of $\ce{[Fe^3+]}$ : $\ce{[SCN-]}$ : $\ce{[FeSCN^2+]}$ is 1 : 1 : 1. If equilibrium concentration of $\ce{[FeSCN^2+]}$ in test tube 2 is $\ce{ 8.3×10^{−5} M}$, the equilibrium concentration of $\ce{[Fe^3+]}$ in test tube 2 is $\ce{ (0.04 - 8.3×10^{−5}) M}$ and that of $\ce{[SCN-]}$ in test tube 2 is $\ce{ (1.0×10^{−4} - 8.3×10^{−5}) M}$. Now, you may able to calculate equilibrium constent for the reaction using following equation:

$$K_\mathrm{eq} =  \frac{\ce{[FeSCN^2+]}}{\ce{[Fe^3+]}\ce{[SCN-]}}$$


Is there a very stinky chemical solution that isn't a health hazard? An effective way to disperse rioters & looters would be to send an armored vehicle in to spray them with a very stinky liquid. Is there a chemical that stinks but is not  a health risk and that breaks down after a few hours, or could be neutralized with another solution after a few hours?

My vote would be for butyric acid and related short-chain fatty acids with between 3 and ~8 carbons (isobutyric acid, valeric acid, isovaleric acid, caproic acid, etc.)

In pure form these chemicals would be dangerous acids, but in dilute aqueous solution buffered at pH 5 or so, they are not very corrosive or toxic but are still pretty stinky.

Skatole could be another example.  See the Wikipedia article for malodorants for some more ideas.


Combining acid dissociation constants to determine pH of diprotic acid If I have a diprotic acid with $K_{\mathrm{a1}}$ and $K_{\mathrm{a2}}$ as the acid dissociation constants, why can't I calculate the final $\mathrm{pH}$ using $K = K_{\mathrm{a1}}\cdot K_{\mathrm{a2}}$ and the following net reaction?

$$
\ce{H2A + 2H2O ->  A^2- + 2\,H3O+} \\
K = \frac{4x^3}{I-x} \\
\mathrm{pH} = -\log_{10}[2x]$$

where $I$ is the initial concentration of $\ce{H2A}$. For instance $I$ could be $1\ \mathrm{M}$, and we could use $K_{\mathrm{a1}} = 5.9 \times 10^{-2}$ and $K_{\mathrm{a2}} = 6.4 \times 10^{-5}$ (constants for oxalic acid).

Solving it the standard way, one acid dissociation at a time, gives $\mathrm{pH} = 0.67$ whereas the combined way (above) gives $\mathrm{pH} = 3.9$. In what way is using this combined equilibrium constant neglecting the total hydronium concentration?

This should be broken into:
\begin{align}
\ce{H2A + H2O &-> HA- + H3O+}\\
\ce{HA- + H2O &-> A^{2-} + H3O+}
\end{align}

With equilibrium equations:
\begin{align}
K_\mathrm{a1}&=\frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}\tag1\label{Ka1}\\
K_\mathrm{a2}&=\frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]}{[\ce{HA-}]}\tag2\label{Ka2}\\
\end{align}

Let us solve for a general case. From logic and general knowledge we know the water dissociation constant, that the masses must balance, and that the charges must balance. These are shown here:
\begin{align}
K_\mathrm{w}&=[\ce{H3O+}]\cdot[\ce{OH-}]\tag3\label{Kw}\\
I&=[\ce{H2A}]+[\ce{HA-}]+[\ce{A^{2-}}]\tag4\label{Mass Balance}\\
[\ce{H3O+}]&=[\ce{OH-}]+[\ce{HA-}]+2[\ce{A^{2-}}]\\
&=\frac{K_\mathrm{w}}{[\ce{H3O+}]}+[\ce{HA-}]+2[\ce{A^{2-}}]\tag5\label{Charge Balance}\\
\end{align}

Using equations \eqref{Ka1} and \eqref{Ka2}, we can solve for $[\ce{HA-}]$ and $[\ce{A^{2-}}]$:
\begin{align}
K_\mathrm{a1} &= \frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}\tag6\label{HA-}\\
K_\mathrm{a1} \cdot K_\mathrm{a2} &= \frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]^2}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\tag7\label{A2-}\\
\end{align}

Substituting into \eqref{Mass Balance}:
\begin{align}
I &= [\ce{H2A}]+\frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}
     +\frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\\
  &= [\ce{H2A}] \cdot \frac{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H3O+}]^2}\tag9\label{Sub Mass Balance}
\end{align}

Solving \eqref{Sub Mass Balance} for $[\ce{H2A}]$:

$$[\ce{H2A}]=\frac{I \cdot [\ce{H3O+}]^2}{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{10}\label{H2A}$$

Substituting \eqref{H2A} into \eqref{HA-} and \eqref{A2-}:
\begin{align}
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
\tag{11}\label{Sub HA-}\\
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{12}\label{Sub A2-}
\end{align}

Substituting equations \eqref{Sub HA-} and \eqref{Sub A2-} into equation \eqref{Charge Balance}:
\begin{align}
&&
[\ce{H3O+}] &= \frac{K_\mathrm{w}}{[\ce{H3O+}]}+\frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]+2K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\\
\Rightarrow &&
[\ce{H3O+}]^2 &= K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot ([\ce{H3O+}]^2+2K_\mathrm{a2}\cdot[\ce{H3O+}])}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{13}\label{Final Eq}
\end{align}

The above equation exactly calculates the $[\ce{H3O+}]$ for a solution of any diprotic acid. The same method used to solve for this equation can be applied to polyprotic acids as well. To make this into a more manageable equation, however, you may choose to assume $K_\mathrm{a1} \cdot K_\mathrm{a2}=0$ (but only in the case of weak acids).

As for the case you proposed, using $K_\mathrm{w}=1 \times 10^{-14}$, $I=1\ \mathrm{M}$, $K_\mathrm{a1}=5.9 \times 10^{−2}$ and $K_\mathrm{a2}=6.4 \times 10^{−5}$ in the equation $y=K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot (x^2+2K_\mathrm{a2})}{x^2 + x \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}-x^2$, and using a graphic calculator to find where the equation is equal to 0, we find $[\ce{H3O+}]=0.215508\ \mathrm{M}$ and $\mathrm{pH}=0.666537$.


Chemical reasons for choosing Salt Water or Chlorinated pool? I am currently researching as too which is better for maintaining the chemical balance in a swimming pool. I have obviously tried the standard Google search for Salt Water or Chlorinated Pools and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:

What chemical reasons are there for choosing a salt water pool over a chlorinated pool?

Note: Salt water meaning the chlorine is generated and chlorinated meaning that you add the tablets/pucks etc.

A salt water pool still has hypochlorous acid $(\ce{HOCl})$ as the active component so essentially it CAN be the same thing as a chlorinated pool. For a chlorinated pool the $\ce{HOCl}$ is directly added, whereas for a salt water pool the salt water is electrolysed to create $\ce{Cl2}$ which then reacts with water to form $\ce{HOCl}$.

However, there is indeed often a difference and the explanation is partly chemistry and partly chemical engineering. In terms of chemistry  the overabundance of chloramines $(\ce{NH2Cl})$ in chlorinated pools over salt water pools makes the difference. When hypochlorous acid, the active component in both, comes into contact with ammonia (from sweat, saliva etc) you get the formation of these chloramines:
$$\ce{NH3 + HOCl → NH2Cl + H2O} $$
Chloramines are the major reason for irritated eyes and skin.

Now the difference between salt water and chlorinated pools has a chemical engineering reason. Salt water pools have a constant generation of chlorine as opposed to a 'batch' delivery in the case of chlorinated pools. Therefore the control over the concentration of hypochlorous acid in the water is better, which also allows better control over the concentration of chloramines.

So to sum up: salt water pools allow higher control over the concentration of chloramines and therefore might be better, however, if this control is not properly applied, it could also even be worse!


Reaction of silver nitrate and hydrochloric acid Consider the reaction of $\ce{AgNO3}$ and $\ce{HCl}$. I read that silver chloride would be formed. But $\ce{H} > \ce{Ag}$ in reactivity, then how could $\ce{Ag}$ displace $\ce{H}$ from $\ce{HCl}$ ?

If there was a reaction, $\ce{AgCl}$ and $\ce{HNO3}$ were to form. But then these would react again to give us the original compounds back. The actual question was to write down the reaction between $\ce{AgNO3}$ and $\ce{HCl}$. I (wrongly) realized that the compounds wouldn't react because of the reasons stated.

One of the most difficult parts of chemistry is learning to recognize a type of reaction based solely on its reactants. This has to be done before you can apply a reaction pattern to the problem, and so it is critical to get this step right in the beginning.

In this case, you are looking at the reaction:

$$
\ce{AgNO3 + HCl ->  AgCl + HNO3}
$$

You already know the products, but are questioning how these products can be formed since the activity of H is greater than Ag, implying that Ag is easier to oxidize than H. This is true - but take a look at the oxidation numbers of each species in this reaction:

$\ce{Ag: +1 -> +1}$

$\ce{NO3: -1 -> -1}$

$\ce{H: +1 -> +1}$

$\ce{Cl: -1 -> -1}$

Since there is no overall transfer of electrons, this can't be a redox reaction, and that means activities won't matter here.

That means there must be another driving force for this reaction - another reaction pattern that fits better.

I don't want to give you the answer directly, but I will give you some advice that should help:

Write the full equation - including the phases.

$$
\ce{AgNO3(aq) + HCl(aq) ->  AgCl(s) + HNO3(aq)}
$$

See if you can find another reaction pattern that fits this equation better. Later, I'll update this answer to show how you can identify the correct pattern for aqueous reactions using only the reactants.

Update

As I mentioned, one of the hardest parts of chemistry is learning to identify reaction patterns based only on the reactants. This is a double-displacement (or metathesis, or precipitation) reaction. It is easy to see now, given that the reactants are aqueous and at least one of the products is solid (the precipitate). Formation of the solid is the driving force for this reaction - the quick explanation is that the forces attracting silver and chloride ions together are stronger than the solvation forces between those ions and water, as well as the forces holding them to sodium and nitrate ions.

This is good, but how do we predict that this is a precipitation reaction ahead of time?

The key is to have a good understanding of the three common types of salt/acid/base reactions that occur in aqueous solution, and to learn to recognize clues in the reactants.

Briefly, the three types that are most commonly seen are:

Precipitation
Acid/base (Arrhenius definition)
Redox (single displacement)

You can recognize each of these by the reactants if you know what to look for.

Precipitation - Two soluble salts (ionic compounds), or a salt with an acid or base.

Acid/base - An Arrhenius acid and base (compound containing hydroxide)

Redox - a salt or acid and an elemental metal

There are many other types of reactions that occur in aqueous solution, and many variations of the acid/base and redox category, but these three cover the cases most commonly seen in a classroom.

Once you have identified the likely pattern that the reaction will follow, the next step is to predict the products using that pattern to see if they make sense. If they do, then you have likely chosen correctly.

You can use this algorithm for more advanced chemistry as well - in organic chemistry, for example, one of the major goals is to learn to predict reactions based on functional groups. Once you can identify functional groups and have memorized reaction patterns for them, it becomes possible to predict a huge range of reactions.


Any method for testing Nitrate and Nitrite in biological fluids like aqueous humour? If anyone knows the methods of detection of nitrate and nitrite in biological fluids such as aqueous humor, saliva of animals, please let me know. Any relevant literature for the same is available, kindly share.

There is 1982 publication describe nitrate, nitrite, and $\ce{[^{15}N]}$-nitrate in biological fluids. It's abstract describe the procedure briefly:

  Abstract: A new automated system for the analysis of nitrate via reduction with a high-pressure cadmium column is described. Samples of urine, saliva, deproteinized plasma, gastric juice, and milk can be analyzed for nitrate, nitrite, or both with a lower limit of detection of $\pu{1.0 nmol} \; \ce{NO3−}$ or $\ce{NO2−/ml}$. The system allows quantitative reduction of nitrate and automatically eliminates interference from other compounds normally present in urine and other biological fluids. Analysis rate is 30 samples per hour, with preparation for most samples limited to simple dilution with distilled water. The application of gas chromatography/mass spectrometry for the analysis of $\ce{^{15}NO3−}$ in urine after derivatization to $\ce{^{15}NO2}$-benzene is also described.

Thousands of samples of human and rat urine, human saliva, and plasma have been analyzed for nitrate and nitrite using this system. Typical ranges of nitrate and nitrite levels in biological samples are shown in the Table.

$$
\begin{array}{ccc}
\hline
\text{Biological fluid} & \text{$\ce{NO3-}$ Concentration, $\pu{\mu M}$} & \text{$\ce{NO2-}$ Concentration, $\pu{\mu M}$}  \\ \hline
\text{Urine} & 250-2000 & \text{ND}^a \\
\text{Saliva} & 200-600 & 30-210 \\
\text{Plasma} & 15-60 & \text{ND}^a \\
\text{Gastric juice} & 50-85 & 0.4-60^b \\
\text{Milk} & 20-30 & \text{ND}^a \\\hline
\end{array}
$$
$^a$Not determined. $\; \; ^b$Nitrite concentrations at the high end of this range are from people with gastric pathologies.

In general, for methods of the detection and determination of nitrite and nitrate, read Ref.2  (a review) and 3. However, Ref.1 claimed that the method in Ref.2 did not work for biological fluids.

References:

Laura C. Green, David A. Wagner, Joseph Glogowski, Paul L. Skipper, John S. Wishnok, Steven R. Tannenbaum, “Analysis of nitrate, nitrite, and $\ce{[^{15}N]}$nitrate in biological fluids,” Analytical Biochemistry 1982, 126(1), 131–138 (https://doi.org/10.1016/0003-2697(82)90118-X).
Qiu-Hua Wang, Li-Ju Yu, Yang Liu, Lan Lin, Ri-gang Lu, Jian-ping Zhu, Lan He, Zhong-Lin Lu, “Methods for the detection and determination of nitrite and nitrate: A review,” Talanta 2017, 165, 709–720 (https://doi.org/10.1016/j.talanta.2016.12.044).
M. P. Stainton, “Simple, efficient reduction column for use in the automated determination of nitrate in water,” Analytical Chemistry 1974, 46(11), 1616–1616 (https://doi.org/10.1021/ac60347a051).


Why don't metals form covalent bonds in bulk metal? I really don't understand why metals form metallic bonds. I mean, it makes no sense. It would make much more sense for them to form covalent bonds with themselves and have a 'pseudo-full' outer shell. How does freeing off electrons make them any more stable?

This is due to the low ionization energies of the metals. It's easier for them to release few electrons from the outer shell to obtain a noble gas configuration rather than consuming several ones. However, the difference between an ionic and a polar covalent bond is always fringe.

Again, it's not true that metals don't form covalent bonds at all. I guess you never heard of quadruple bond or δ-bond. There are several examples such as $\ce{K2[Re2Cl8]·2H2O}$ and Chromium(II) acetate hydrate.


Copper acetate synthesis without hydrogen peroxide Usually, when people synthesize copper acetate, they use a mixture of acetic acid and hydrogen peroxide. I assume this is because the acetic acid actually reacts with the various oxides on the copper, rather than with the copper itself. However, assuming I don't have access to hydrogen peroxide, can I do a similar thing (but slower), just by waiting until fresh copper surface is exposed and then taking it out. Then, I will heat it up (or just wait) until oxide layer forms again, and keep repeating until copper is corroded away.

I placed 1tbs salt, 1L vinegar, and 200g solid copper in a jar. After 3 months in a dark place the solution was extremely dark blue - so blue I couldn't see through it. This is what happened (bottom answer).
How may copper acetate ligands be manipulated to change colors?

Heat may be able to substitute for time and quicken the formation of copper acetate...I really haven't tried. Peroxide is dirt cheap. I would have used it if I had known.


Can dust conduct electricity? Just out of curiosity, can dust conduct electricity? The source of dust varies, is there any form of dust that can conduct electricity?

The answer to this question hinges on two key questions:

What kind of dust are we dealing with?
How is the dust arranged (i.e. what is your experimental setup)?

Which Dust?

Firstly, I'm going to assume you meant domestic dust (since there are more kinds of dust than you'd imagine).

I quote Wikipedia:

  Dust in homes, offices, and other human environments contains small amounts of plant pollen, human and animal hairs, textile fibers, paper fibers, minerals from outdoor soil, human skin cells, burnt meteorite particles, and many other materials which may be found in the local environment.

Arrangement

The arrangement of the dust particles is key. Are we considering an aerosol, or are we carefully laying a line of dust on the table and connecting it with two electrodes?

Aerosol - We can safely assume that the conductance of air has a far higher impact on the measurement than the conductance of the particles in the aerosol (the dust). So here, dust does not conduct electricity. Note: I ignore the "conducting" via static electricity processes here, but they may play a role. I do not consider that "classical" conductance.

Line of Dust - Organic material is usually insulating, as is our skin (if it wasn't for the moisture that is needed for, you know, living), paper, textiles, hair and so on. So the main constituents of domestic dust does not conduct electricity. So also here, we will not observe conductance.

There is one exception to the above, where conductance is a possibility: Metal dust. This is kind of cheating, but consider the waste of some metalworking process, in which metal dust (or powder) is generated. This dust, being composed of metals and metal oxides, would likely conduct electricity (when laying it out in a line, still not as an aerosol).


Which reaction is more violent? As far I know fluorine is the most electronegative element in the periodic table and caesium is the most electropositve element. I had a conception that the reaction between the both would be the most violent as they have extreme opposite natures and perhaps they would do it faster than other elements.

But recently, I came across a book where I read that the electron gain enthalpy of chlorine is a bit higher than fluorine. This means when a reaction would occur with chlorine it would liberate more energy than that of the reaction with fluorine and therefore more violent. But, chlorine is less electronegative than fluorine.

So which reaction would be more sudden or fast and will liberate huge amounts of energy?

Although fluorine is more electronegative than chlorine it has less electron gain enthalpy because of its small size.

The electrons in the outer shell of fluorine repel the incoming electron more effectively than the electrons in chlorine. More closer the electron comes to the nucleus more energy is released. In case of chlorine, the electron comes more close to the nucleus.

Thus, chlorine has more electron gain enthalpy.

Size matters in chemistry. It is more dominating factor than electronegativity in most cases.

You would not expect enormous difference in enthalpy of formation of $\ce{CsF}$ and enthalpy of formation of other alkali halides. In case of $\ce{CsF}$, The bond length of the resultant salt will be larger than other alkali halides due to large size of cesium atom and small size of fluorine atom. The bond between them would not be very strong. Consequently, the reaction will not be strongly exothermic.

Putting the theory part aside, here is the data :

$\Delta{H^{\circ}_f}$  -  $\ce{CsF}$ $\rightarrow$ $-550\frac{KJ}{mole}$,

$\Delta{H^{\circ}_f}$ - $\ce{NaF}$ $\rightarrow$ $-575\frac{KJ}{mole}$,

$Electron\space gain\space enthalpy\space - Chlorine$ $\rightarrow$ $-349 \frac{KJ}{mole}$

The comparisons in your question are very different and they can only be compared by looking at the data. The prediction will be very inaccurate. If you want to compare two properties, you will at least need some similarity between the two properties. Like, you can compare enthalpy of formation of two different alkali halides or electron gain enthalpies of Halogens or chalcogens. Electron gain enthalpy and enthalpy of formation of alkali halides are very different things.


Is it possible to efficiently convert hydrogen to methane? According to this wikipedia page, hydrogen can combine with the carbon found in carbon steel to form some methane inside the steel, weakening it. Is is possible to convert hydrogen to methane by putting it through
graphite powder or activated carbon?

Note that I mainly want the reaction to occur at STP. The flow of the
gas through the graphite will be very slow (at the speed it is generated by electrolysis), so
maybe catalysts not needed?

The purpose of this is to form a more energy-dense fuel than pure hydrogen. Also, it will be much safer since methane is less flammable.

Is it possible to efficiently convert hydrogen to methane?

No. You'll end up putting more energy into the process than you'll get out of burning the methane less the energy of burning the hydrogen that you started from. So although you can make methane from hydrogen, it isn't "efficient."


Effect of water viscosity (increased due to NaCl concentration) on dissolution rate of CO2 We did an experiment involving different $\ce{NaCl}$ concentration(ranging, 1-4%) on 100 mL of water. Our aim of the experiment was to find the dissolution rate of $\ce{CO2}$in the brine solvent. We did some research and found that $\ce{NaCl}$ increases the viscosity of the solvent, which in-turn decreases the dissolution rate. We want to explain how the entire process works, in a microscopic level.

Answers with good source would be good.

By dissolving NaCl you increase the ionic strength of the solution, i.e. there is more electrostatic attraction between the layers of water increasing the viscosity. Carbon dioxide is non-polar, so by increasing the polarity of the solvent, the carbon dioxide tend to form less interactions with the solvent.