How concentrated does the analyte solution need to be in order to perform a (chelatometric) titration I’m currently investigating a method to chemically pretreat cellulosic waste as bisorbents for enhanced uptake of lead in water. I intend to measure the initial and final concentration of water to record the change in concentration of aqueous lead via a chelatometric titration.

In my experiment, I’ll be performing filtration in the batch mode at three different initial concentrations (5 mg/L, 20 mg/L, 50 mg/L). To visualize the magnitude of these concentrations, I’ve converted the units to the more conventional mol/L (M).

Initial concentrations of water samples I’ll be testing on in mol/L (M)

My question is, are these initial concentrations too low to be conducting a titration on. If so, what would be a good initial concentration of lead to be working with (considering after filtration, the concentration of lead will diminish significantly). If these initial (and very small post filtration) concentrations are measurable for a titration, what is a good  concentration for the EDTA titrant. Should the titrant concentration be significantly less than the analyte concentration?

All the best!

Sorry to tell you a bad news. Classical methods, like titrimetry or gravimetry do not work very well at low concentrations. Failure would be a strong word. The beauty of EDTA is that it forms 1:1 complex with most metals. Pb is not an exception. If your lead solution is 0.000001 M, your EDTA solution should be 0.000001 M. Such concentrations are never recommended in classical methods.

For ppm level, spectrophotometric methods can work quite well. Go to Google Scholar and search the methods in Google Scholar


How to identify lewis acids and bases What is the general methodology to identify Lewis acids and bases in an acid base reaction? Do we need to draw structures to identify which species is the electron donor and which one the acceptor, or is there any other way of doing this?

If you consider $\ce{HClO_4}$, oxygen $\ce{O}$ has lone pair(s), so there is a confusion whether it can donate the pair to other species or not. So in that case, although it is an acid, one may say it is a base. Similarly, in $\ce{H2S}$, sulfur $\ce{S}$ has got lone pairs and I am confused whether it will act as base or acid.

You need to draw the structure and identify the most acid and most basic site(s).
Notice how I didn't say "the most acid molecule".
Boron tribromide ($ \ce{BBr3} $) contains both a super-strong acidic site (the trivalent boron atom) and potentially basic sites (the lone pairs of the three bromine atoms).

Whether a molecule (e.g. $ \ce{H2S} $) will act as an acid or as a base depends on the reagents.


What would be the identification reactions of selenate(IV) ion? I have not been able to find any $\ce{SeO3^2-}$ identification reactions anywhere. Are there any?

You can use reducing agents to reduce the $\ce{SeO3^2-}$ salt to elemental selenium. This paper1 discusses the use of iron(II) salts in acidic medium (phosphoric - hydrochloric acid) as reducing agent. The reaction proceeds at r.t.
You can use other reducing agents like hydrochloric acid, sulfur dioxide, hydroxylamine hydrochloride, hydrazine hydrochloride. The reduced selenium has a red color which turns greyish-warm on warming, but be careful on overheating (boiling or evaporating turns to serious losses of selenium by forming volatile $\ce{SeCl4}$).
References

Rajkumar Kalaparthi , Srija Korapu , Padmarao Chekuri, 2020, New Spectrophotometric Determination of Sodium Selenate [Selenium (IV)] and Sodium Selenite[Selenium(VI)] with Iron(II), and Their Quantitative Analysis of Selenium (IV) – Selenium (VI) Present in A Binary Mixtures, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 09, Issue 05 (May 2020), DOI: 10.17577/IJERTV9IS050094
Vogel, A. I., & Svehla, G. (1996). Vogel's qualitative inorganic analysis


Salt concentration and electrical permittivity of water We know that the electric flux density $D$ can be related to electric field intensity $E$ as

$$D = \epsilon(\omega) E$$

where

$$\epsilon(\omega) = \epsilon_0 \epsilon_r(\omega)$$

and $\epsilon_0$ is the permittivity of vacuum and $\epsilon_r(\omega)$ is the relative permittivity of the material.

At the low-frequency limit, I would like to know how the value of $\epsilon_r (\omega)$ of water changes with temperature and salt ($\ce{NaCl}$) concentration. Please explain,

the rationale of the variation (from quantum mechanics or statistical mechanics etc)
observed trend holds good for any liquid or only for highly polar solvents
any references that you know of which talks in detail about these variations

The relative electric permittivity of water at $20^oC$ under the effect of a static or low frequency is $\epsilon_r\approx80$. Hasted (1948) conducted an experimental study on the dielectric properties of salt-water solutions and observed a dielectric decrement with salt concentration. That is, the addition of sodium chloride to water results in a drop in electrical permittivity.

Source: Hasted, Ritson and Collie (1948).

This can be understood by examining the following diagram.

The sodium and chloride ions dissociate in solution, resulting in an an electric field between them (shown in bold line). The water molecules, which are polarised, get oriented under the effect of the electric field so that their oxygen atoms (carrying a partial negative charge) face toward the sodium ion and their hydrogens (carrying partial positive charges) face toward the chloride ion. This orientation of the polar water molecules causes its own electric field (dashed lines), which cancels out most of the electric field that would exist if the ions were in a vacuum. The end result is that the sodium and chloride ions are effectively 'sheilded' by a 'hydration shell' of polarized water molecules which effectively lowers their to the applied external field and hence lowering the dielectric constant (electric permittivity) as the ionic concentration increases ().

In dilute solution (less than 2.0M) the dielectric decrement is linear. At higher salt concentrations, the dielectric decrement is observed to saturate at about $\epsilon_r=45$. The effect of sodium chloride concentration on electric permittivity is shown in the graph below at various temperatures.

Source: Gavish and Promislow (2012).

The above graph indicates that for a given salt concentration below about 3M, the electric permittivity decreases with temperature.

REFERENCES:

Hasted, J.B., Ritson, D.M., Collie, C.H., "Dielectric Properties of Aqueous Ionic Solutions. Parts I and II." Journal of Chemical Physics 16, 1 (1948). http://dx.doi.org/10.1063/1.1746645

Gavish, N., Promislow, K, "Dependence of the dielectric constant of electrolyte solutions on ionic concentration", http://arxiv.org/pdf/1208.5169.pdf


Why does a Fluoride ion only have a -1 charge and not a -2 charge or more? Transition elements can form ions with different charges. Why can't elements other than transition elements form ions with different charges? If it is a Fluoride ion, why does it have to be an anion of charge -1? Why can't it have a -2 charge, or more?

The octet rule is usually used to explain the charges that atoms in groups I, II, VI, and VII have in ionic compounds, by saying something like "noble gases are stable, therefore having 8 valence electrons must be stable for some reason." This is true, but understanding why noble gas configurations are stable is important, and without knowing that, it does seem like magic that some atoms don't adopt a noble gas configuration in ionic compounds.

The octet rule is a useful way to quickly predict ionic charges and write Lewis structures, but it's not a magical law that atoms follow. They don't "want" or "try" to obtain specific electronic structures, although we often talk like they do. What is really important is the total energy of the electron configuration. The electron configuration with the lowest energy is the most stable one.

  "Stability" really means "lowest energy"

The total energy of the electron configuration is governed by two things:

The energy of the electron orbitals
The effective nuclear charge that is "seen" by the electrons

The orbital energies determine the order in which atomic orbitals will be populated by electrons in the ground state - in other words, which orbitals the electrons will exist in for a given atom. We know from empirical evidence and from quantum mechanical calculations that the order goes like this for most elements (not for some transition metals and heavier elements):

Electrons will occupy these orbitals from the "ground up" (called the Aufbau principle). We know, therefore, that for 4th-period elements, the $3d$ sub-shell will fill before the $4p$ sub-shell will.

The atomic orbital diagram indirectly tells us something else. The principle quantum number, $n$ (which for the valence shell is equal to the row or period number), roughly corresponds to the distance from the nucleus of the orbital's highest electron density region. In other words, any sub-shell with $n=4$ will have most of its electron density further away from the nucleus than a sub-shell with $n=3$. This means that $3d$ electrons are closer to the nucleus than $4s$ electrons, even though $3d$ electrons initially have  higher energy.

  The distance of the electron from the nucleus is roughly proportional to $n$

Since the electrons have a negative charge, and the nucleus is positively charged, there is an electrostatic potential energy between the electrons and the nucleus. If the electrons were uniformly distributed throughout the atom, we would expect that the nuclear charge "felt" by each electron would be equal to the number of protons - the atomic number. However, because of the "rules" of quantum mechanics (the allowed solutions to Schrodinger's equation), electrons are not uniformly distributed - they occupy orbitals in a specific arrangement. Since some electrons are closer to the nucleus (lower $n$) than others, this means that the nuclear charge is partially screened by the inner electrons, and the outer electrons experience an effective nuclear charge that is somewhat less than the full nuclear charge. This diagram from wikipedia demonstrates the idea:

And this one shows the idea using the Bohr model (note that this is a schematic - not an accurate picture of what electrons "look" like in orbitals - only the relative distances from the nucleus is right)

When the valence electrons experience a large effective nuclear charge, they are harder to remove (the energy cost for removing them increases), and when the effective nuclear charge is lower, they are easier to remove. The actual effective nuclear charge is very difficult to accurately calculate, but it is very easy to estimate - all you need to do is take the number of protons, then subtract the number of "screening" electrons. Basically, we are assuming that each electron between the valence shell and the nucleus cancels out one proton:

$$
Z_{eff} = Z - S
$$

where $Z_{eff}$ is the effective nuclear charge, $Z$ is the actual nuclear charge, and $S$ is the screening constant, here assumed to equal the number of "core" electrons.

To find $S$, all you have to do is count the number of electrons with $n$ less than the valence shell (highest $n$).

Taking fluorine as an example, the electron configuration is:

$$
1s^22s^22p^5
$$

We see that $Z=9$ and $S=2$, giving an effective nuclear charge of +7. Compare that to lithium: $3-2=+1$. The valence electrons on fluorine are roughly 7 times harder to remove than the valence electron on lithium. In fact, any element to the left of fluorine has a lower effective nuclear charge for the valence shell, and will give up electrons more easily than fluorine will. In other words, it's not so much that fluorine won't give up electrons, it's that nothing can take them from it.

What if we add an electron? For $\ce{F-}$, the electron configuration becomes:

$$
1s^22s^22p^6
$$

$Z_{eff}$ is still +7, since the number of core screening electrons hasn't changed. This means that it is energetically favorable for fluorine to pick up an electron, since the additional electron would still "see" a large positive nuclear charge.

What if we add another, to make $\ce{F^{2-}}$? Now the configuration looks like this:

$$
1s^22s^22p^63s^1
$$

$Z_{eff}$ is now $9 - 10 = -1$ - note that the number of screening electrons jumped to 10 because the "new" valence shell has $n=3$. In other words, because of the allowed orbital configurations the new electron would "see" a negative charge on the nucleus - it would require energy to keep it.

If you work through this process for all of the "fixed charge" elements, you find similar things - the effective nuclear charge is large for elements that gain electrons, until they gain too many (an octet), at which point it becomes negative. For fixed-charge metals, something else happens. Let's look at magnesium.

$\ce{Mg}$: $12-10=+2$

$\ce{Mg^1+}$: $12-10=+2$

$\ce{Mg^2+}$: $12-2=+10$

Once you remove two electrons, the effective nuclear charge jumps to +10 for the remaining valence electrons. The same is true for all the group I and II metals - once you get to the noble gas "core" - the octect - the effective nuclear charge seen by the valence electrons jumps way up.

This explains why "fixed charge" elements always have the same charge in ionic compounds, but what about transition metals?

Let's look at chromium. The electron configuration looks like this:

$$
1s^22s^22p^63s^23p^64s^23d^4
$$

Which electrons are the valence electrons? The ones with the highest $n$ - the 4s electrons. The effective nuclear charge they "see" is $24 - 22 = +2$ - note that the 4 electrons in the $3d$ sub-shell are counted as screening, since they have $n=3$ which is less than 4, putting them closer to the nucleus than the valence shell, despite their initially higher energy.

If we remove an electron, we get:

$$
1s^22s^22p^63s^23p^64s^13d^4
$$
(This isn't exactly correct, there would really be no $4s$ electron and 5 $3d$ electrons, but the reason is complicated and it doesn't matter to this explanation)

Note that we removed a valence electron - a $4s$ electron. Now $Z_{eff} = 24 - 22 = +2$ - it hasn't changed. We can remove one more to make $\ce{Cr^{2+}}$ and get:

$$
1s^22s^22p^63s^23p^63d^4
$$

Now the definition of "valence" becomes a little less clear - we know that the highest $n$ value corresponds to the valence shell, but what about $l$, the orbital shape? It turns out that d orbitals are larger than s and p orbitals, and so the s and p electrons end up screening the nucleus from the d electrons. This means that d electrons would be removed next, and the effective nuclear charge "seen" by them is $24 - 18 = +6$. This is a larger charge, but not so large that other elements (notably oxygen and the halogens) can't still pull the d electrons away. This gives us a range of chromium charges (oxidation states) from +1 to +6. For $Cr^{6+}$, the electron configuration would be:

$$
1s^22s^22p^63s^23p^6
$$

Now the effective nuclear charge is $24 - 10 =  +14$ - much too strong for anything else to "win" against.

As you can see, the rules of quantum mechanics determine the relative size of atomic orbitals and the order in which they are filled, which in turn determines the effective nuclear charge that valence electrons "see" for any given neutral atom or ion. The combination of the two leads to constraints on the charges that atomic ions can have, with the practical result being that group I, II, VI, and VII elements adopt a "filled octet" or "noble gas" configuration, while the other groups have more flexibility.


Origin of diastereoselectivity in Simmons–Smith cyclopropanation of allylic alcohol In Organic Chemistry by Clayden et al., the following reaction is shown:

The cyclopropane ring is formed on the same side as the alcohol group. Is this because of the interaction between Zn and the OH group, which directs the reagent to approach from the same side as the OH group?

Yes. Right after Clayden mentions this stereoselectivity, he proceeds to state that this is due to "coordination between the zinc atom and the hydroxyl group in the transition state" (pp. 1067-8, 1st ed):


Wade's rules for Zintl ions How do you apply Wade's rules to Zintl polyhedral cluster anions ?

For example, I believe $[\ce{Bi_3}]^{3-}$ is a triangle. My approach to justify would be:

5 electrons per each Bi atom
3 electrons for the charge
18 total electrons

$18 = 4\times3+6 \therefore$ arachno cluster: trigonal bipyramid ($n+2$ polyhedron) with two missing vertices $\therefore$ linear.

If it was a nido cluster $(4n+4\ \ce{e-})$, then it would be a tetrahedron with one missing vertex $\therefore$ a triangle. Am I mistaken about the observed geometry or is my method incorrect?

Let the number of atoms be n.

If the number of skeletal p electrons is 2n + 2, the structure is closo (a deltahedron with n vertices).

If the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).

If the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).

If the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).

see Excursions Beyond the Zintl Border and Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary

For example:

Neutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons.

$\ce{Sn_5^{2-}}$, $\ce{Pb_5^{2-}}$, $\ce{Ge_5^{2-}}$ and $\ce{Ge_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).

$\ce{Sn_9^{4-}}$ and $\ce{Ge_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. (gyroelongated square bipyramid with one square pyramid apex missing)

$\ce{Bi_4^{2-}}$ and $\ce{Sb_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).

Also, quoting Farley and Castleman's Observation of Gas-Phase Anionic Bismuth Zintl Ions J. Am. Chem. Soc. 1989, 111, 2734-2735

  Since bismuth and antimony are typically trivalent, anionic clusters
  of these elements very quickly exceed the maximum 2n + 8 electrons
  accountable under Wades rules for “hypo” compounds. In particular, the
  previously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and
  $\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are
  particularly intriguing due to the inability of current theories to
  account for them and lack of other corresponding metallic compounds
  with which they may be compared.

Now going back to the example $\ce{Bi_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:

The structure is isoelectronic with species known to be bent in the ground state including $\ce{O3}$, $\ce{S3}$, $\ce{S2O}$, $\ce{SO2}$ and allyl anion.
Experimental determination published in Stabilization of Ozone-like [Bi3]3- and in a more recent article that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).
Theoretical work in Zintl Anions Analogous to O3.

Though ozone and like species are found to be bent, it is also theorized that cyclic (equilateral triangular ring) ozone represents another higher local potential energy minimum, and this would be the same for $\ce{Bi_3^{3-}}$ too.

$\ce{Bi_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal pyramid.  A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.


Why are chelating ligands often used to treat poisoning by heavy metals (e.g. mercury, lead)? Why are chelating ligands such as EDTA often used to treat poisoning by heavy metals (e.g., mercury, lead)?

I have a vibe that it has something to do with the lone pairs of electrons that are able to bond to the heavy metals to form a stable coordination complex but I'm not sure about the in-depth process or explanation behind this.

Also, does it have to do with it being a polydentate ligand and not a monodentate ligand?

Lead, mercury and other heavy metals form complexes with both monodentate and polydentate ligands. But the complexes made with polydentate ligands like EDTA are much more stable than with monodentates. Usually metallic complexes become toxic if the central metallic ion (non-complexed) is released or not entirely surrounded by ligands. Monodentate complexes do not surround completely the central metallic ion, and they may be decomposed by the acidity of the stomach. Polydentate complexes resist to this acidity, and often they surround the metallic ions much better than monodentate. So if a person has ingested one of these heavy metal ions, its toxicity will be hidden by a complexing agent like EDTA.


Why is silicon oxide giant covalent structure? Why is silicon oxide giant covalent structure? Why it doesn't bond like carbon dioxide?

Carbon dioxide is a molecule with double bonds.
Elements of 3+ rows are not happy with double bonds, they prefer single bonds, even in compounds like $\ce{H2SO4}$ 'double bonds' are actually dative single bonds.
The reason for it is that pi-orbitals for 3+ rows are very diffuse, making overall overlapping with p-orbitals of 2-rows extremely ineffective
For this reason, $\ce{SiO2}$ forms structure with only singular Si-O bonds. It happens so that for it the only way to form one is to make 3-d infinite structure.
For similar reason $\ce{N2O5}$ is exactly like this in gas phase while $\ce{P2O5}$ forms tetrahedral $\ce{P4O10}$  and $\ce{SO3}$ loves to form trimers and chains.


Why HF is said to be formed from H-1s and F-2pz overlap? According to Wikipedia, HF is formed by the interaction between hydrogen's $1\mathrm{s}$ orbital and fluoride’s $2\mathrm{p}_z$ orbital:

  In hydrogen fluoride HF overlap between the H 1s and F 2s orbitals is
  allowed by symmetry but the difference in energy between the two
  atomic orbitals prevents them from interacting to create a molecular
  orbital. Overlap between the H 1s and F 2pz orbitals is also symmetry
  allowed, and these two atomic orbitals have a small energy separation.
  Thus, they interact, leading to creation of σ and σ* MOs and a
  molecule with a bond order of 1.

So:

If there is a too large energy separation between H-1s and F-2s, isn't the energy separation between H-1s and F-2p even larger?
Can it be experimentally verified that the bonding is indeed via the F-2p and not the F-2s?

You fell into a wrong analogy with the picture of energetically far apart. Let me break down your picture into something visible:

Imagine the orbitals of fluorine at different heights of a mountain (I’ll use the Zugspitze, because it feels like home). 1s is at the very peak of the mountain (yeah, I’m turning this around just for simplicity). The Eibsee (c. $1000~\mathrm{m}$ over sea level) would correspond to the 2s-orbital. Garmisch-Partenkirchen, the town in the valley before the mountain, ($700~\mathrm{m}$) would be the 2p orbital.

We need to place the 1s-orbital of hydrogen somewhere so that the height difference to the Eibsee is too large for an interaction. One possibility would be the Höllentalanger hut at around $1300~\mathrm{m}$. That would render the energy difference to both fluorine’s 2s and 2p too large. But in fact, hydrogen’s 1s is better placed at the height of Oberau, $650~\mathrm{m}$. That is way too far away from the 2s but rather close to the 2p.

This is confirmed by energy diagrams of $\ce{HF}$ like that presented in this lecture (in German). (Just search for HF on the page, it’s not far down. Ignore the text around it and look at the scheme.)

Now you may be asking yourself why the hydrogen’s 1s energy is so high compared to fluorine’s orbitals. Compare ionisation energies, which are typically understood as the energy difference of the highest orbital to ‘an electron in vacuum’. For hydrogen, this energy is $1312~\mathrm{kJ/mol}$ while for fluorine this is $1681~\mathrm{kJ/mol}$. The reasoning behind it is fluorine’s very high electronegativity, attracting the electrons towards its nucleus. Fluorine’s nucleus has a $+9$ charge which is much better at drawing electrons in than hydrogen’s $+1$ (even though it needs to draw nine electrons in rather than one). This stabilises all of fluorine’s orbitals; and those which are ‘closer to the nucleus’ (s-orbitals) most.

I am not well-educated in modern physical chemistry methods, but I heard that orbitals were visualised. If that is the case, it won’t be long until we can visualise the $\ce{HF}$ orbitals, and indeed show that the bonding orbital (highest energy σ type) will have a nodal plane through fluorine.


Comparing SN2 reaction rates I've read in a book that the main factor for determining SN2 reaction rate is steric hindrance. The lesser it is, the faster the reaction.

So consider this question:

  $\ce{KI}$ in acetone undergoes SN2 reaction with each of $\ce{P,Q,R}$ and $\ce{S}$. The rates of reaction vary as:

I do not understand how $\ce{S}$ is the fastest. In my opinion it should be the slowest as it is very sterically hindered. What is going on here?

Source: Joint Entrance Exam (JEE) 2013 India

There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction.  Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.

In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors.  Considering only steric factors for a moment we would order the compounds as P>S~R>Q.  But now let's consider electronic factors.

Considering the allyl system in compound R, the transition state looks something like this

image source

The p orbitals in the adjacent double bond overlap with the p orbital at the $\mathrm{S_{N}2}$ reaction center (remember that $\mathrm{S_{N}2}$ transition states are roughly $\ce{sp^2}$ hybridized) and stabilizes the transition state through resonance delocalization.  Note that the transition state is electron rich, there is a full unit of negative charge in the transition state.

When a carbonyl is placed next to the $\mathrm{S_{N}2}$ reaction center, we replace the $\ce{C=C}$ double bond in the above drawing with a $\ce{C=O}$ double bond.  Again there is overlap and resonance stabilization of the transition state.  But in the carbonyl case the stabilization is even greater than with the $\ce{C=C}$ double bond because the carbonyl carbon is positively polarized and inductively stabilizes the electron rich (remember that negative charge) transition state even more.

image source

While we would expect the allyl case (R) to be faster than the ethyl case, in your series of compounds apparently it is not faster than the methyl case.

(Note: often the allyl system will react a bit faster than the methyl analogue, but the two rates are usually close)

However, the additional inductive stabilization from the carbonyl allows it to react faster than the methyl compound so our final reactivity order becomes S>P>R>Q.


Attempting to balance a redox reaction using half reactions leaves an unbalanced component I have this redox reaction that I can't balance using half reaction method.

$$\ce{Mg + HNO3 -> Mg(NO3)2 + N2O + H2O}$$

What I get at the end is

$$\ce{2Mg + 6HNO3 -> 2Mg(NO3)2 + 3N2O +3H2O}$$

I can't get the $\ce{N}$ balanced.

It seems likely that you are starting with the wrong half reactions.

Since this reaction happens in water, separate the ionic species into their ions:

$$\ce{Mg(NO3)2 -> Mg^{2+} + 2NO3-}$$

$$\ce{HNO3 -> H+ + NO3-}$$

Now, one half reaction should involve magnesium (and have no nitrogen anywhere). The other half reaction should involve nitrogen only.

$$\ce{Mg -> Mg^{2+}}$$
$$\ce{NO3- -> N2O}$$


Calibration Ammonia Sensor and Gas Concentration My name is Felipe and this is my first post here.
I'm working on a project to calibrate an electrochemical ammonia sensor. But I have a lot of doubts because I am a layman in chemistry. I'll list them and I'd be grateful if you could respond them.

The system I have here is basically a cylinder with ammonia gas of known concentration connected by a 0.5 lpm regulator and a tygon tube to a chamber where the sensor is.
Here are my doubts:

When I open the regulator the gas starts to flow into the chamber, the concentration of oxygen(or other gases) in the chamber can affect the concentration of ammonia gas? Or even if the gases mix, there is a fixed concentration of ammonia?
I've seen a lot of calibration videos. Why does nobody care about the tube length? I think as long as the tube length more is the waste of gas, right?
When I stop with the ammonia gas flow and disconnect the tube from the chamber. What are the precautions I should take with the contaminating ammonia in the air? Let's say I'm using a calibration gas of 50ppm

The calibration gas will purge the air out of your chamber, so that it contains 50 ppm NH3.  This assumes that there are no reactions that change the mixing ratio of the ammonia (e.g. reaction with water in your chamber).  The length of the tube doesn't matter unless the gas reacts with your tubing, though length will decrease the pressure of the gas coming out of the tube.


Specific Heat affected by Pressure I know the specific heat depends (slightly) on the conditions present. For example, methane's specific heat is 2.5 at 400 K, but 4.5 at higher temperatures.
How might specific heat depend on pressure, though. -- if we are under super low pressure (say 6 mbar rather than earth's typical 1014 mbar), how would specific heat change. I am expecting it to differ along with $PV = nRT$, but not sure.

Also, $\Delta H = mc\Delta T$ -- would I need to plug $\Delta PV/nR$ in for $\Delta T$?

At low pressures like 6 mb and even 1014 mg, the heat capacity is very insensitive to pressure (and the ideal gas value can be used).  At higher pressures, where non-ideal gas effects set in, the enthalpy and the heat capacity are, in general, both functions of pressure as well as temperature.  The partial derivative of enthalpy with respect to pressure at constant temperature for an arbitrary substance is given by:

$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
This is equal to zero for an ideal gas.


Why are metals conductive in terms of orbitals? Back in high school I was taught a definition of metals:

  Metals are a lattice of positive nuclei embedded within a sea of delocalised electrons

Metals are conductive because the electrons aren't bound to any particular atom, and hence can flow freely throughout the entire metal.

However, I was recently taught about electron orbitals, and I'm not sure how to reconcile the two pictures. If the metal atoms were sharing pi or sigma bonds, I'm not sure how any of the electrons could be truly delocalised, and how conductivity arises. In terms of the atomic orbital theory, how does conductivity arise in metals?

It's not that the electrons are completely unbound from the atoms - they still attract each other and form the bonds with other atoms. But the attraction between the atom and its electron is pretty weak so it's easy to detach the electron from the atom.

The properties of the metals are better to be described via ionization energy - the energy needed to remove an electron from an atom. Elements to the right of the periodic table have high ionization energy - they are not willing to lose electrons. Therefore these elements are bad conductors (and are not metals).

Metals have lower ionization energy - their valence shell just started to be filling and the strength of the proton is not enough to keep them tight. Therefore these electrons are free to flow with minimum efforts applied - which would form electrical current.

The intention of the "sea" model is to describe (in simplified manner) the fact that electrons are floating around from one atom to another. And while they participate in bonding, this is a dynamic structure that breaks and forms all the time.

Here is a nice video from Khan Academy on the metallic nature.


How do electronegativity and lone pairs affect bond angles? How do bond angles vary in molecules with a lone pair and central atom of different electronegativity, but in the same period so that electronegativity matters more than orbital size? Let's assume that the molecules I am comparing have the same substituents and one lone pair.

Will the molecule with the more electronegative atom in the middle have the lone pair electrons closer to the nucleus, so they take up more bonding space and the angles for the other substituents will be smaller?

The electronegativity of the central atom does not really influence bond angles in any meaningful way. The size of atoms and the orbitals used and useable are much more important.

Sometimes, students may be taught concepts such as electronegativity or ‘lone pairs require space’ to explain why $\ce{H2O}$ has a smaller bond angle ($105^\circ$) than $\ce{NH3}\ (107^\circ)$ or similar. However, careful analysis of the molecular orbitals reveals that sterics are a much more important reason than ‘lone pairs requiring more space and thus making tetrahedral $\mathrm{sp^3}$ contract’. Indeed, as soon as a lone pair exists on a central atom, there is a tendency to have the lone pair occupy an orbital with an s-contribution as high as possible and to have bonding orbitals with a maximised p-contribution. p-orbitals, having a direction, can form stronger σ bonds due to better overlap.

This is why the bond angle drops quickly when comparing second-period to third-period central elements to close to $90^\circ$ but then hardly changes down the group. On the other hand, the reduced bond angles observed from ammonia to water are mainly due to more steric stress on nitrogen — which needs to accomodate an additional hydrogen and thus needs to space them out more.

Indeed, if you compare the bond angles of different $\ce{OX2}$ compounds, the only general trend is ‘the larger the second element’s atoms, the greater the bond angle.’ See the table below (all values taken from Wikipedia).

\begin{array}{ccc}\hline
\ce{X} & \angle(\ce{X-O-X}) & \angle(\ce{X-N-X})\\ \hline
\ce{H} & 104.5^\circ & 107.8^\circ \\
\ce{F} & 103^\circ & 102.5^\circ \\
\ce{Cl} & 110.9^\circ & -/- \\
\ce{Br} & 112^\circ & -/- \\ \hline \end{array}

The drop in bond angles from hydrogen to fluorine is sufficiently explained by the much longer bond lengths of the $\ce{OF2}/\ce{NF3}$ species when compared to the corresponding hydrogen species due to fluorine using p-orbitals for bonding where hydrogen uses its 1s orbital.

While the bond lengths also increase when going from fluorine to chlorine and bromine, the angle also does. Note especially $\ce{OBr2}$ whose bond angle is larger than the tetrahedral bond angle. Any argument involving electronegativity should assume a smaller angle here since the electronegativity of the partners drops from fluorine to bromine.


Balancing and combining the half equations in redox I can understand redox equations of the following form, I break these down into half equations and combine them.

$$\ce{MnO4^- + C2O4^{2-} -> Mn^{2+} + CO2}$$

Where I have issue is with the following (unbalanced) equation:

$$\ce{H2 + NO -> NH_3 + H2O}$$

I am asked to show balanced half equations and the final combined equation.

I can see the following changes in oxidation states
reduction: $$\ce{ H -> H^{1+} + e^{1-} }$$ (for both the $\ce{H}$ in water and ammonia)
oxidation: $$\ce{ N^{2+} + 5e- -> N^{3-} }$$

My issue is then with balancing these and combining them, I am not sure if the half equation involving $\ce{H}$ should have $\ce{H2O}$ as the product or the $\ce{H3}$ from $\ce{NH3}$.

How should I go about breaking this into half equations?
Is it ok for both the half-equations to have molecules of the same compound in the product?
Should I ever have water as the product when trying to show a half equation (ie $\ce{H -> H_2O}$)?

Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.

$$\ce{H2 + NO -> NH3 + H2O}$$

Step 1: Ionise the required compounds and remove spectator ions

None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.*
$$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$
*-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.

Step 2: Split into Oxidation and reduction halves

This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$  (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1).
So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3}
\end{array}
$$

Step 3: Balance only those atoms undergoing redox

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\
\text{(because Hydrogen is } & \text{(because only Nitrogen } \\
\text{undergoing redox but} & \text{is undergoing redox,} \\
\text{Oxygen is not.)} & \text{it's already balanced)} \\
\end{array}
$$

Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O}
\end{array}
$$

Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> [H+ + OH- ] + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> \qquad H2O \qquad + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\hline
\ce{H2 + \qquad-> \qquad \qquad \qquad \quad 2H+} & \ce{NO + 5H+ -> NH3 + H2O}\\
\hline
\end{array}
$$

Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$

This step must follow Step 5a and means nothing on it's own.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\
\ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\
\hline
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad  }\\
\hline
\end{array}
$$

The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.

Till the above steps mass has been balanced, now charge balancing.

Step 6: Balance charge in each side of each half by adding electrons

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half}  \\
\hline
(acidic)\quad \ce{H2  -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\
\hline
(basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\
\hline
\end{array}
$$

→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.

→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half.
eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons.
Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')

Step 7: Make 'net electrons produced' equal to 'net electrons consumed'

Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)

$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
(acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\
\hline
(basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\
\hline
\end{array}
$$

Finally add the two halves and cancel out the common cmopounds and you get the balanced equation
Acidic Medium
$$\begin{matrix}
\ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\
\ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\
\end{matrix}$$

Basic Medium
$$\begin{matrix}
\ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\
\ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\
\end{matrix}$$

In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$

You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.


Stability of α-chlorocarbocations I'm trying to compare the relative stability of two carbocations:
$$\ce{CH3-\overset{+}{C}H-Cl}\quad\text{vs}\quad\ce{CH3-\overset{+}{C}H2}$$
Does the $\ce{Cl}$ exert a +M effect to stabilise the carbocation, or does it exert a −I effect to destabilise it?

As Oscar Lanzi suggested, both $+M$ and $-I$ applies here, but $\ce{Cl}$ stabilizes carbocation, meaning $+M$ is more effective than $-I$. This fact was confirmed by this peer-reviewed paper (Ref.1):

  The lowering of $\ce{C_\beta–H}$ stretching frequencies in carbocations 1a–d and 2a–c induced by hyperconjugation was tested as a possible probe for estimating the electron donating ability of $\alpha$-substituents. Conclusions are based on the results of high level quantum chemical calculations confirmed with experimental FT-IR spectra. Because the decrease in the $\ce{C_\beta–H}$ stretching frequency is comparable in 1b and in 1c, and in 2b and 2c respectively, it follows that $\alpha$-substitution $\color{red}{\text{by a methyl group}}$ or $\color{red}{\text{by chlorine}}$ stabilizes a carbocation with $\color{red}{\text{almost the same effectiveness}}$.

The effect of the chlorine $n$-electron back donation is evident by a partially double bond character of the $\ce{C+–Cl}$ bond in $\alpha$-chlorocarbocations observed experimentally in IR spectra (Ref.2). In this experiment, FT-IR spectrum of $\ce{Cl3C+}$ cation has displayed the $\ce{C–Cl}$ stretching frequency at $\pu{1040 cm–1}$, which is $\pu{250 cm–1}$ higher than in neutral alkyl chlorides (e.g., characteristic $\ce{C–Cl}$ stretching frequency of $\ce{CCl4}$ is $\pu{785 cm-1}$). This is indicative of partial double bond character of $\ce{C+–Cl}$ bond suggested by the resonance structures:

$$\ce{Cl2C+-Cl <-> CH2C=Cl+}$$

The pioneering work of Olah and coworkers has also predicted this phenomena by $\ce{^{13}C}$-NMR studies of trihalocarbocations (Ref.3):

$$
\begin{array}{c|c|c|c}
\ce{HCX3} & \delta\ce{^{13}C}\text{ of }\ce{CHX3}  & \ce{^+CX3}  & \delta\ce{^{13}C}\text{ of }\ce{^+CX3} & \Delta \delta\ce{^{13}C} \\
\hline
\ce{HCCl3} & 77.7 & \ce{^+CCl3} & 236.3 & 158.6 \\
\ce{HCBr3} & 12.3 & \ce{^+CBr3} & 207 & 194.7 \\
\ce{HCI3} & -139.7 & \ce{^+CI3} & 95 & 234.7\\
\hline
\end{array}
$$

Olah has suggested that the decreasing trend of $\Delta \delta\ce{^{13}C}$ by going $\ce{I}$ to $\ce{Br}$ to $\ce{Cl}$ is in agreement with the positive charge stabilization by back-bonding in order of $\ce{Cl > Br > I}$.

References:

Milan Mesić, Igor Novak, Dionis E. Sunko, Hrvoj Vančik, "$\ce{C–H}$ Hyperconjugation in $\alpha$-chlorocarbocations," J. Chem. Soc., Perkin Trans. 2 1998, (11), 2371-2374 (https://doi.org/10.1039/A805772I).
Hrvoj Vančik, Ksenija Percač, Dionis E. Sunko, "Chloromethyl cations in cryogenic antimony pentafluoride matrixes and the generation of carbocations from hydrocarbons," J. Am. Chem. Soc. 1990, 112(20), 7418–7419 (https://doi.org/10.1021/ja00176a065).
George A. Olah, Ludger Heiliger, G. K. Surya Prakash, "Stable carbocations. Part 276. Trihalomethyl cations," J. Am. Chem. Soc. 1989, 111(20), 8020–8021 (https://doi.org/10.1021/ja00202a056).


Why use glacial acetic acid in bromination of anisole? In my organic chemistry lab (next week, the experiment has not been conducted yet) we are being asked to react anisole with bromine to get 2-bromoanisole (as well as 2,4-dibromoanisole and all the others).  In our lab instructions we are told to add anisole, bromine and glacial acetic acid to our mixture.  No mention of other compounds (such as $\ce{FeBr3}$) was made.  I was under the impression that bromine is not reactive enough to brominate an aromatic without $\ce{FeBr3}$.  What purpose does the acid serve, and how does it make this reaction work without the iron compound?

[...] bromine is not reactive enough to brominate an aromatic without FeBr3

That's true for benzene, but the methoxy-group in anisole isn't just o/p-directing.
It is activating as well. Actually, it's that much activating that anisole will react with bromine even without the presence of a Lewis acid.

So, why glacial acetic acid? It's a polar solvent and thus nice to stabilize the Wheland intermediates. Moreover, it's easily removed in aqueous workup.

Edit 1
Does the acetic acid react with the (mostly undissociated) acetic acid? At least a polarization of bromine ($\ce{Br^{\delta+}-Br^{\delta-}...HOAc}$) is conceivable and I remember that $\ce{BrOAc}$ has been postulated as the electrophilic species.

Edit 2
I remembered some brominations on dimethoxy-substituted hydroxymethylnaphthalines I did back in the days. The reaction was performed at 60 °C in $\ce{CHCl3}$ in the presence of about 3 equivalents of anhydrous sodium acetate. Excess of bromine was reduced by addition of aqueous sodium sulfite. The yield was in the 85% range and I don't find any notes on dibrominated products.

Edit3

  And with excess Bromine you would (theoretically) react all of the anisole?

Pretty much, with 4-bromoanisole as the main product.


What do "USP 26" and "USP 42" mean? I am being asked by a colleague if the Polyvinylpyrrolidone (PVP) we buy in our lab is "USP 26" or "USP 42". I've never heard those terms before. I searched up online but found nothing.
I'm sure its something concerning the US Pharmacopeia, but I've found their site to be extremely walled up and unhelpful.
Can anyone tell me what do those terms mean, and how to find out if my PVP is either? The manufacturer's Safety data sheet didn't seem to mention it.
Note: I don't work at a food grade or drug grade lab. I'm also not in the USA.

I was able to reach the manufacturer and ask her directly. USPXX is the "edition" of the US Pharmacopeia's "exams" that a product must pass in order to be considered pharmaceutical grade. As time goes on, these exams may get updated and improved. So a higher number USPXX means that a product passed a more recent (and likely, more demanding) USP set of exams.
Therefore, USP 26 is a cheaper PVP, that may not be able to pass the USP 42 standards. Or at least, wasn't tested on them.
On the other hand, USP 42 is a more expensive PVP,  that can and did pass the standard.
As for the actual differences between both standards, I don't know them. So I can't say if the difference between then it's significant or not. It seems USP guidelines are not available freely.
Edit: Todd Minehardt's comment had the right answer it seems.


Does diamagnetic anisotropy only apply to a proton bonded to triple bonded carbon? Lecture tells me that double bond, with the sp^2 bonding will have more s-character, which will hold the electons closer to nucleus, which means a more deshielded proton which means downfield chem shift.
But it doesnt follow the same idea with triple bond, with sp bonded carbon, and this is where diamagnetic anisotropy becomes a factor.... But is D.A. a factor with double bond also and why? Thats what I dont understand, because both double and triple will have pi bonds.. thanks for any help

In principle, diamagnetic anisotropy can be invoked for any bond or any functional group; however, there's only a few situations where it's commonly used to understand patterns of shielding/deshielding.  A decent discussion of this can be found in Silverstein's textbook (Spectrometric Identification of Organic Compounds), but I seem to recall that a couple of his figures are drawn in a confusing manner...

In short, this picture relies on the fact that pairs of electrons are diamagnetic, i.e., when immersed in a magnetic field, they get magnetized against the direction of the magnetic field:

$M=\chi B_0$

(M is the induced magnetization, B0 is the applied magnetic field, and $\chi$ is the magnetic susc., which is negative for diamagnetic objects).  Thus, any electron density in a stable molecule (ignoring radicals) will become magnetized against the magnetic field (this is why we speak in terms of shielding).

We can picture molecular orbitals or valence bond orbitals of interest to us simply as a cloud of electron density that becomes magnetized in a magnetic field.  Anisotropy means that effect depends on an angle: we're familiar with the picture of a magnetic dipole having magnetic field lines that curve around from the N end back to the S end – so something below or above the electron cloud will be in the (so called) shielding cone whereas things off to the side will be in the deshielding plane.  Note, this is true for anything that is diamagnetic - there's no need for mystical ring currents or anything like that; any electron cloud placed in a magnetic field will be magnetized and, because it's diamagnetic, will reduce the local magnetic field above and below the electron cloud, but increase it off to the sides.

However, the molecules are tumbling rapidly in solution, so most of this effect simply averages away to zero...  We need special situations where a proton of interest is held in a fixed position relative to the electron cloud of interest to get an noticable effect.  Common example:

Flat conjugated systems like benzene, but also larger annulenes and related molecules.  Most chemists are familiar with the argument that hydrogens outside the ring are deshielded by the "ring current" – while some may debate the existence of the ring current, the pi system is, by its nature, diamagnetic, so it's not surprising that any protons held in the outer edge are deshielded while methylenes held in a position above or below the ring are significantly shielded.  There are beautiful examples of annulenes with protons inside the large aromatic ring that have negative chemical shifts due to the large shielding anisotropy of the pi system.
Triple bonds have two pi bonds that are 90º to each other.  It's a little harder to "see", but you can look at various possible orientations in a magnetic field and see that you can rationalize some shielding of the carbons in the triple bond.
Cyclohexane derivatives.  While not as dramatic as the benzene example (and confusingly illustrated in the Silverstein book), you can picture the sigma skeleton of the cyclohexane as a large cloud of electron density, so it should have a shielding cone above/below the ring and a deshielding plane off to the sides of the ring.  This turns out to be true!  For a system that has a relatively fixed chair conformation and a diastereotopic methylene, you'll always find the equatorial hydrogen deshielded relative to the axial hydrogen.

I'm sure there are more examples, but those are the only examples that I have seen work consistently enough to be useful.


Does there exist any relation between metallic property and oxidation potential? I was having a revision on periodic table I was thinking about different properties I could connect most of I could not connect metallic property and oxidation potential .

The short answer is yes. The reason for this is that metals exhibit metallic properties because their electrons are relatively loosely bound to the atoms. A metallic bond is one in which the valence electrons are free to move throughout the lattice structure of the metal. Here's a schematic diagram from wiki commons:

The idea is that the positively charged metallic "cores" (which includes the nucleus and the more tightly bound inner core electrons) are sort of suspended in a crystalline structure within a "sea" of mobile electrons.

This model explains most of the  metallic properties we observe, from heat and electrical conductance, to malleability and ductility.

Now think about oxidation - in an oxidation-reduction (redox) reaction, one element gains electrons (it is reduced) and the other loses electrons (it is oxidized). Here's an animation, also from wikipedia:

So, for an element to be metallic, its valence electrons need to be relatively "loose" - it can't take too much energy to detach them. For an element to be oxidized, something else has to "want" the electrons more - the energy cost of removing the electron has to be lower then the gain from adding them to the reducing agent.

As a result there is a very strong correlation between the metallic nature of elements and their relative ease of oxidation. There is also a very strong correlation between both of these properties and the trends in ionization energy, electron affinity, and  electronegativity.

Of course, in chemistry, things are never that simple. There lots of details and it is tough to make general statements that are always true. For example, you can say that often times the more "metallic" an element is, the more easily it will be oxidized. However, when you compare elements using a standard electrode potential table (which gives us a way of measuring relative ease of oxidation) you find that it is easier to oxidize calcium ($-3.8 \space \rm{V}$) than it is to oxidize potassium ($-2.9 \space \rm{V}$), even though potassium is closer to the "metallic side" of the periodic table, it has a lower first ionization energy, and it is less electronegative.

This illustrates a limitation of thinking of elements in terms of metals and non-metals: it is really a shorthand way of describing all of the things that affect how tightly those valence electrons are bound to the atom, in comparison to how badly other elements "want" to remove them. With that amount of generalization, we lose too much detail to be able to make accurate predictions in every situation.


Can copper sulfate react with terpenes (such as limonene and terpinenes) at room temperature? I am just wondering, can I make a metalorganic or organometallic complex of copper and limonene/terpinene/phellandrene in the form of lime oil? If so, what is the formula of the complex formed? I tried the reaction with lime oil in iPrOH, and a copper sulfate solution. The result was a blue crystalline solid. What are the crystals? I'd like to know.

For the record, after M. Farook's agreement I am making my comment an answer.
Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate. Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.


What topics to review before starting organic chemistry? It's been over a semester since my most recent chemistry class, (general chemistry 2) because I could not take organic at my community college. It is only considered a sophomore level class unless taken in-residence at the university I am transferring to (which is understandable.)

I was curious if I could narrow down what topics I could begin reviewing before the start of the semester that would help me in organic chemistry.

As @inɒzɘmɒЯ.A.M pointed out in his comment, the usual prerequisite for an organic chemistry course is that a general chemistry course is already taken and successfully passed. The general chemistry is, of course, a rather large topic, so I'll just name a few subtopics which I feel are absolutely essential to start a venture into the world of organic chemistry.

Structure of matter.

Modern atomic theory (Schrödinger equation, atomic orbitals, spin,  Pauli exclusion principle, electron configurations).
The periodic table (periodicity, group trends).
Chemical bonding (covalent and ionic bonding, Lewis structures, resonance, VSEPR theory , introductory Valence Bond Theory and Molecular Orbital Theory).

Chemical reactions.

Nomenclature & stoichiometry (empirical and molecular formulas, chemical equations, mole, limiting reagents, yield).
Solutions (solubility, concentration, properties of solutions).
Acid-base reactions (definitions of acids and bases, pH, strong/weak acids and bases, trends).
Oxidation-reduction reactions (oxidation numbers, balancing oxidation-reduction reactions).

Thermodynamics.

Basic concepts (energy, work, heat, enthalpy, entropy, Gibbs free energy ).
The first and the second laws of thermodynamics.
Hess's law.

Kinetics.

Reaction rates (temperature dependence, Arrhenius equation).
Reaction orders (first and second order reactions, molecularity, rate laws).
Catalysis.

Chemical equilibrium.

Basic principles (the notion of equilibrium, the law of mass action, equilibrium constants, the direction of a reaction).
Shifting the equilibrium (changing amounts, pressure, volume, temperature).


Finding pH after dilution of acetic acid We have $\pu{10 L}$ of solution of $\ce{CH3COONa}$ with concentration $C = \pu{1 M}$ and $\mathrm{pH} = 10$. Then we add $\pu{90 L}$ $\ce{H2O}$. What will be the pH of the new solution?

My work:
\begin{align}
  \ce{CH3COONa &<=> CH3COO- + Na+}\\
  \ce{CH3COO- + H2O &-> CH3COOH + OH-}
\end{align}
We have $\pu{1 M } \ce{CH3COO-}$ and let $x$ denote the concentration of $\ce{OH-}$. If the $\mathrm{pH}$ of the solution is $10$, it's implied that
$$\mathrm{pOH}=4\iff-\log_{10} x=4 \iff x=\pu{10^{-4} M}.$$
I know I have to use that the amounts of substances didn't change, so $$C_1V_1 = C_2V_2,$$
but I don't know what values I have to put in there. Is the $x$ I found my $C_1$?

$$\ce{CH_3COO- + H_2O<=>CH3COOH + OH-}$$
Approximating that all the $\ce{OH-}$ is from this equilibrium, $[\ce{CH3COOH}] = [\ce{OH-}]$.
\begin{align}
  [\ce{OH-}]_\text{initial} &= [\ce{CH3COOH}]_\text{initial} \\
    &= \pu{10^{-4}M}\\
  \frac{[\ce{OH-}]_\text{initial}[\ce{CH3COOH}]_\text{initial}}{[\ce{CH3COO-}]_\text{initial}[\ce{H2O}]_\text{initial}}
  &= \frac{[\ce{OH-}]_\text{final}[\ce{CH3COOH}]_\text{final}}{[\ce{CH3COO-}]_\text{final}[\ce{H2O}]_\text{final}}
\end{align}
Then you can choose to approximate $[\ce{H2O}]$ as a constant value and omit it from both sides (or not if you need great accuracy).
\begin{align}
  [\ce{CH3COO-}]_\text{initial} &= \pu{1 M}\\
  [\ce{CH3COO-}]_\text{final} &= \pu{\frac{10}{100} M}\\
  x &= [\ce{OH-}]_\text{final} = [\ce{CH3COOH}]_\text{final}
\end{align}
Solve for $x$.


Benchtop NMR spectroscopy parameter explanation; tau and number of points I have just carried out an exploration of the effect of different solvents and conditions on the 1H caffeine relaxation by benchtop NMR spectroscopy.
When collecting the 1D 1H NMR spectra of the caffeine in CDCl3 solution, the parameter for the number of points was set up, but when the relaxation run for T1 and T2 were done, the parameter was replaced by the value of time delay, tau.
I think I kind of get a clear view from your discussion, as the number of points is needed in determining the right peaks for 1D 1H spectra. But, as for the relaxation time, the intensities behaviour during the time delay is now being the subject of interest, hence the np parameter is neglected.
This is what I understand by far.
#updated
They are the number of points (np) and tau.
I think I kind of understand how np works;

it's the discrete points of analogue FID that are being converted to
digital points along with the FID signal envelope. Therefore it's
associated with the resolution of the spectra which the lesser np used
to define the FID, the lower the resolution.
*Let me know if I interpreted this wrongly.

This question raised when I'm analysing my data though, would really appreciate it if someone can enlighten me.

AS Karl said read the instrument's manual and add the meaning of the symbols. However there is one important point in discrete Fourier transform (DFT).

I think I kind of understand how np works; it's the discrete points of analogue FID that are being converted to digital points along with the FID signal envelope. Therefore it's associated with the resolution of the spectra. *Let me know if I interpreted this wrongly.

I have been recently discussing this exact point with signal processing engineers and spectroscopist friend with a lot of good constructive discussions. No, the number of points cannot enhance the intrinsic resolution of the signals whether it is NMR or FTIR or any other FT spectroscopic technique. What you increase, when you collect more points, is the frequency resolution on the x-axis, and consequently more points per peak. That is it.
For an N-point DFT, the frequency step is
$$\frac{1}{N\Delta t}$$
For example if your sampling frequency is fixed to 1 s (an arbitrary number), in one experiment you collect 1000 points, the frequency step on the axis is
$$\frac{1}{1000 . 1 s}=0.001 Hz $$
In another experiment, you only collect 250 points with a time step of 1 s
$$\frac{1}{250 . 1 s}= 0.004Hz$$
Note the time to sample the signal in each case is different, because it is given by $N\Delta t$. So by chance, if your original experimental spectrum was "limited" by the number of points per peak by increasing the number of points, you might see an improvement in the peak shape and apparent resolution when you increase the number of points in an FID.


Reaction of alkyl halides with strong unhindered bases Please look at the image on this website: http://www.physicsforums.com/showthread.php?t=765853

For (b) and (c), what the solution manual says makes no sense to me.
You're reacting a primary alkyl halide with an unhindered base $\ce{EtONa}$. They say that the only product is the alkene. Why do they not consider the $\mathrm{S_N2}$ reaction that WILL take place in larger quantity?
I'm uploading the question too on the other website

I agree with you, I'd expect some $\mathrm{S_N2}$ product too. I'm not sure about the relative quantities of the $\mathrm{E2}$ and $\mathrm{S_N2}$ products though.

I wouldn't flat out say that you'd expect more $\mathrm{S_N2}$ product than $\mathrm{E2}$ product, because although the substrate is unhindered, the solvent is ethanol, which is protic. The solvent interacts with the nucleophile through hydrogen bonding.

This hinders the nucleophile. This slows down the rate of the $\mathrm{S_N2}$ reaction because now the nucleophile is dragging along lots of solvent molecules, and it also increases the bulk of the nucleophile.


Equivalent hydrogens in 2,3-dimethylbutane So while calculating for the relative yield of monochlorination for 2,3-dimethylbutane, I was counting the primary hydrogens, I counted 6, but in fact it was 12. Why is it 12? How do you find equivalent hydrogens? Since the compound is symmetrical doesn't the hydrogens on both sides be equivalent to each other?

The hydrogen atoms in a methyl group are equivalent to each other. There are 4 methyl groups and therefore there are 12 equivalent hydrogens. 2,3-dimethylbutane is symmetric as you stated, and so on one side you have 6 equivalent hydrogens, this implies that there are 6 more equivalent hydrogens on the opposite side.


Is bicyclo[1.1.0]butane chiral? Is bicyclo[1.1.0]butane chiral? I think it isn't, but if one of the hydrogen atoms bound to the tertiary or secondary carbon atoms is replaced with e.g. chlorine, the formed compound will be chiral, right?

Bicyclo[1.1.0}butane is an achiral molecule.

Any molecule containing a plane of symmetry must be achiral and bicylco[1.1.0]butane has 2 planes of symmetry.  One plane bisects the molecule and passes through the 2 tertiary carbons, the other plane bisects the molecule and contains the 2 secondary carbons.

Even replacing one of the secondary or tertiary hydrogens with a substituent does not make the molecule chiral.  Such a substitution will remove one plane of symmetry, but not the other one.

In the above bicylco[1.1.0]butane where we have substituted a chlorine for one of the tertiary hydrogens, the plane containing the 2 bridgehead carbons and the chlorine still remains.  However, if you replace one of the tertiary hydrogens and one of the secondary hydrogens, the resultant disubstituted-bicylco[1.1.0]butane will be chiral since both symmetry planes have been destroyed and no other plane or axis of symmetry exists.


Intermolecular forces in secondary amines compared to primary amines Secondary amines form hydrogen bonds, but having nitrogen in the middle of the chain rather than at the end makes the permanent dipole on the molecule slightly less. The lower boiling point is due to the lower dipole-dipole attractions in the dimethylamine compared with ethylamine.

Could someone please tell me why nitrogen, being in the middle of the molecule (as in secondary amines) causes the permanent dipole to be slightly less.

A primary amine has two $\ce{N-H}$ bonds and one $\ce{N-C}$ bond. A secondary amine has two $\ce{N-C}$ bonds and one $\ce{N-H}$ bond. Since $\ce{N-H}$ is more polarised than $\ce{N-C}$ we can expect ethyl amine to be more polar than dimethyl amine.


Synthesis of hydrazine with bleach and ammonia I am trying to make rocket fuel using hydrazine, hydrogen peroxide and home products. I want to put an action figure in orbit.

Is there any way to synthesize hydrazine with bleach and ammonia, using $\ce{H2SO4}$ as a catalyst?

I sympathize with your desire to launch an action figure. However, the chemistry you propose is really dangerous. You would be better off with black-powder based rockets.

  Is there any way to synthesize hydrazine with bleach and ammonia?

Do not do this.

My answer is not "yes" or "no". My answer is Don't. Do not try this without proper safety equipment and training, and then think twice or three times before doing so, and then have an ambulance standing by. Certainly don't do it at home, or by yourself.

This particular reaction is capable of producing hydrazine. However, it also producing a whole host of additional stuff, like chloramine, hydroxylamine, nitrogen trichloride, hydrogen chloride, and chlorine gas, all of which are either gases or volatile liquids, and all of which are poisonous. Hydrazine is poisonous. Nitrogen trichloride is a shock-sensitive explosive on the order of a sneeze can set it off. The reactions are as follows.

Chloramine:
$$\ce{NH3 + HOCl -> Cl-NH2 + H2O}$$

Dichloramine:
$$\ce{Cl-NH2 + HOCl -> NHCl2 + H2O}$$

Nitrogen trichloride:
$$\ce{NHCl2 + HOCl -> NCl3 + H2O}$$

Hydrazine and HCl:
$$\ce{NH3 + NH2Cl -> H2N-NH2 + HCl}$$

Hydroxylamine and HCl
$$\ce{NH2Cl + H2O -> NH2OH + HCl}$$

Chlorine gas:
$$\ce{HCl + HOCl -> H2O + Cl2}$$

Any organic impurities in the mix can lead to other deadly stuff being produced.

This reaction is not controllable to produce a single specific product. Everything it produces is poisonous and/or explosive. I will recommend against, and I will not help you find a recipe.


Proton affinities in machine readable format? Short: How to get proton affinities for a large list of chemicals? A database? A programming API?

Longer:

I have ca. 2000 chemicals (downloaded from MCM, mostly organics that occur in the atmosphere and their reaction products). I would like to try to get proton affinities for as many of the 2000 chemicals as possible.

I know how to look up the proton affinity for a chemical in the NIST Chemistry WebBook (for example isoprene). But I would prefer to get a machine readable database, or a Python API (or any programming language), to get these nicely in a program code.

What options do I have?

(I am not going to get 2000 values by hand, so my last option would be to write a small web scraper to make these 2000 web queries)

To my knowledge, a (Python) API for the NIST chemistry WebBook does not exist.

Rather than filling out the Search for Species Data by CAS Registry Number by hand, you might want to use mechanize.

But since you're only interested in Gas phase ion energetics data anyway, you might want to construct the relevant URLs directly.

Isoprene:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&Units=SI&Mask=20#Ion-Energetics

Styrene:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C100425&Units=SI&Mask=20#Ion-Energetics

The parameter pattern is obvious: ID is the CAS number of the compound minus the dashes, prefixed by C.


Depicting molecules by Natta projection Does anyone know the meaning of these two bonds in stereochemistry? These bonds are provided by the $\mathrm{\LaTeX}$ chemfig package. The creator of the chemfig package describes them as "hollow cram bonds".

Cram bonds are synonymous to the "wedged" bonds or, in this case, to "hollow wedged" bonds, which IUPAC [1, p. 1910] describes as follows (emphasis mine):

  There are a variety of obscure conventions for depicting arbitrary mixtures of diastereoisomers. The most common (but still obscure) of these conventions says that bold and hashed wedges should be used for absolute configuration, unwedged bold and unwedged hashed lines represent relative configuration and racemic character, while hollow wedges and dashed lines represent relative configuration only.

A bit more context is given in [2] (bond type 4):

  Central to the correct perception of a stereounit’s topography is the stylized generation of a vanishing point, realized by the graphic stereodescriptor, or
  stereobond, commonly called a “wedge” as illustrated in Figure 1 as 1a and 1b. Three such wedge symbols as in 3, 4, and 6 are in use, each of them capable of portraying the ligand Z in 1 and 2 as being located in the front (3 and 4) or behind (6) the picture plane. In an apparent search for further simplification, line symbols 5, 7, and 8 were invented as wedge substitutes.
  Lacking differentiation of their termini, line-type symbols appear, a priori, to be inherently incapable of serving such a purpose; nevertheless, they provide a useful function in the context of a convention. Accordingly, the stereogenic unit to which the line-type bond symbol is attached is considered to lie in the picture plane so that the symbol printed in bold (5) reveals its ligand Z in front of that plane, whereas the hashed symbols (7 and 8) are intended to show that the ligand is behind it. In other words, symbol 5 portrays the stereocenter at the vanishing point and hence functions such as 3 and 4.

  Figure 1. Perspective drawings of stereocenters with commonly used stereobonds.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897.
Maehr, H. Graphic Representation of Configuration in Two-Dimensional Space. Current Conventions, Clarifications, and Proposed Extensions. J. Chem. Inf. Comput. Sci. 2002, 42 (4), 894–902. https://doi.org/10.1021/ci025518w.


Is chloronitrobenzene and nitrochlorobenzene same? Is it 2-chloronitrobenzene or 2-nitrochlorobenzene?

If we take nitro group as the substituent, it will be 2-chloronitrobenzene. But if we take chloro group as the substituent, it is 2-nitrochlorobenzene.

What if we start numbering the carbon atom from the carbon to which the nitro group is attached, it should be o-chloronitrobenzene. But no, carbon atom numbering starts from the group to which chloro group is attached.

Is it the fact that chlorobenzene forms ortho-para substituted product and it is named accordingly or according to some IUPAC rules?

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of citation of substituent prefixes and the numbering of the corresponding locants are laid down in two different rules.

Simple substituent groups that are named by means of prefixes (such as ‘nitro’ and ‘chloro’) are arranged alphabetically.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the compound given in the question is x-chloro-y-nitrobenzene (not x-nitro-y-chlorobenzene).

The locants x and y are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings (e.g. chlorobenzene or nitrobenzene); but in substitutive names of disubstituted benzene, numerical locants have to be used to distinguish the 1,2-, 1,3-, and 1,4-isomers.

Note that the letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4-isomers of disubstituted benzene (e.g. o-chloronitrobenzene). According to current IUPAC recommendations, this usage is strongly discouraged and it is not used in preferred IUPAC names.

The relevant rules concerning the numbering of locants for substituent prefixes are:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), the compound given in the question could be named as 1-chloro-2-nitrobenzene as well as 2-chloro-1-nitrobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-chloro-2-nitrobenzene rather than 2-chloro-1-nitrobenzene since chloro is cited first as a prefix in the name.


What is natural bond orbital theory used for? My understanding is that the NBO methodology transforms the molecular orbital picture of a molecule into a collection of pairwise interactions (bond orbitals). This is obviously handy for didactic and comprehension purposes as it distils what might be a very complex and non-local MO picture into manageable lobes conforming to the introductory theories of chemical bonding (VB, orbital hybridisation) taught in high-school and early university courses.

Does natural bond orbital theory have utility outside of this role?

Yes, along with a family of related methods. I honestly don't fully understand the mechanics behind it, but the orbitals obtained with the NBO method can be used for a treatment called Natural Resonance Theory, which gives you an idea of what sort of Lewis-structure resonance components there are and in what weightings, as well as bond indices, and a measure of covalency.

There are three papers from the developers of the technique, covering theory, usage, and some more examples. (I'd probably start with the usage paper.)

Also, along with NBO, there are the Natural Atomic Orbitals which are often lumped in because they're also computed by the NBO program, and Natural Population Analysis which you can use the Natural Atomic Orbitals to do. These give you an idea of what sort of atomic orbital character is contributing to your molecular orbitals, and also atomic charges and net spins.


Increasing partition coefficient in an extraction My friend had this homework question about supposing you had a slightly polar organic compound in an ether/water extraction with a partition coefficient in favor of ether. The question was:

  What simple method can be used to increase the partition coefficient?

My thought was to add salt to the extraction mixture. I was thinking that if you saturate the water with salt, there essentially isn't any room for the polar organic compound to stay in the water so it gets extracted into the organic layer. However, I'm wondering if there's a better explanation for why adding salt would increase the partition coefficient. Thank you for the help!

Generally, extraction of substances with organic solvents is influenced by various factors, among which there are the following:

nature of the extracted substance and the extractant (not a variable in your example);
temperature;
presence of electrolytes in aqueous solutions;
$\mathrm{pH}$ of the medium;
rate of agitation etc.

In your particular case probably only temperature and electrolytic presence are relevant.
Change in temperature often affects the distribution constant rather seriously, as not only the solubility of the extracted substances in each phase changes (unequally), but also the mutual solubility of the organic and aqueous phases changes too. When the temperature changes, the dissociation/association of the substance in the corresponding phase can change, influencing hydration (solvation) and therefore extractability. Ideally you might want to check solubility-temperature plots for your system, taking into account that this data often cannot be obtained by extrapolation.
Addition of highly soluble salts (electrolytes) to an aqueous solution of another substance can lower (salting-out effect) or increase (salting-in effect) its solubility in water. Increased high ionic strength also lowers activity of water in terms of depressing its competitive effects (first of all, solvation).
The salting-out effect of electrolytes depends on the nature and properties of the extracting substance and the concentration, charge and ionic radii of the salting-out ions. Smaller ions with larger charge have higher charge density than ions with a larger radius, therefore they are hydrated better.  $\ce{NaCl}$ and $\ce{Na2SO4}$ are commonly used for salting-out organic substances. However, this rule has a number of exceptions. Some non-electrolytes that are well soluble in water also have salting out-effect. For example, ethyl alcohol salts-out acetic acid from its aqueous solutions upon extraction with ethyl acetate.
The salting-in/out phenomenon is often explained based on the electrostatic model proposed by Debye and McAulay, reviewed in [1], and can be approximately described via equation:
$$c/c^\circ = 1 - K_D \cdot c_\mathrm{s} + A \cdot c_\mathrm{s}^{0.75},$$
where $c^\circ$ and $c$ are solubilities [$\pu{mol L-1}$] of non-electrolyte in pure solvent and solvent with salt, respectively; $K_D$ and $A$ - empirical coefficients; $c_\mathrm{s}$ - salt concentration.
Pretty much your explanation is on the same page with the original Debye theory:

This theory pictures the "salting-out" effect as a consequence of the aggregation of the water molecules of greater polarity around the ions owing to the field of the latter. The less polar non-electrolyte molecules tend to congregate in the portions of the solution most remote from the field of the ions, with the net result that an enhancement of the water molecules around the ions and of the non-electrolyte in the solution regions away from the ions occurs. A reduction of the solubility of the non-electrolyte, referred to the total water present, thus results, due to the increase of non-electrolyte-water ratio in those regions of the solution containing the non-electrolyte.
The criterion of polarity which Debye employs is the dielectric constant of the saturated aqueous solution of the non-electrolyte relative to water at the same temperature. On this basis if the saturated solution has a dielectric constant less than water, "salting out" occurs in the presence of the added salt. If the dielectric constant of the saturated non-electrolyte solution is above that of water, the non-electrolyte becomes more soluble in the salt solution or is "salted in". This means that the non-electrolyte tends to aggregate around the ions at the expense of the water, and so the nonelectrolyte-water ratio in regions removed from the ions is lowered and the total water present can hold more non-electrolyte in solution.

Also, in biochemistry so-called Hofmeister series are used to favor protein extraction by salting-out for given conditions.
References

Gross, P. M. The “Salting out” of Non-Electrolytes from Aqueous Solutions. Chemical Reviews 1933, 13 (1), 91–101. https://doi.org/10.1021/cr60044a007.


What is the most acidic neutral hydrocarbon? This is a question I've been thinking about quite frequently. I had seen that this hydrocarbon has a low $\mathrm pK_\mathrm a$ with reference to the highlighted H-atom

Loss of the central gives aromatic character to the three 5-membered rings in the compound, contributing to its acidity.

Are there any other neutral hydrocarbons which are stronger acids? How do their $\mathrm pK_\mathrm a$'s stack up against, say another organic acid such as acetic acid?

EDIT: As pointed out in the comments, $\ce{CH_5^+}$ (methanium) is a superacid, but is not a neutral hydrocarbon.

In this answer a hydrocarbon anion called Kuhn's anion has the formula $\ce{C_{67}H_{39}^-}$ has reported $pK_b=8.1$ which would correspond to the neutral hydrocarbon having $pK_a=5.9$.  The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem. 1990, 55 (3), 996–1002):


Chromene and cumarin nomenclature We have these nomenclatures of the same family :

4H-chromene
2H-chromene
Why do we use 2H and 4H to name them? And how ?

There are two isomers of chromene (1-benzopyran): 2H-chromene and 4H-chromene.

According to Subsection P-14.7 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name can be made specific by indicating the position of the extra hydrogen atom in the structure, i.e. the position where no multiple bond is attached. This is accomplished by adding to the front of the name an italic capital ‘H’ preceded by the appropriate numerical locant for this atom.

In 2H-chromene, the ‘indicated hydrogen’ locates one extra hydrogen atom in position ‘2’ of the chromene structure; and in 4H-chromene, the ‘indicated hydrogen’ indicates an extra hydrogen atom at position ‘4’, i.e. one hydrogen atom more than the number present if there were a double bond in the ring at that position.


Unit derivation concerning Nernst equation We know that Nernst equation is given by:

$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.

Now, A question asks me to derive the unit of  $\frac{RT}{nF}$.

Which I think can easily be derived this way:

$\frac{RT}{nF}=\frac{E-E^{\circ}}{\ln[\ce{M^{n+}}]}$,

Now, the unit of $E-E^{\circ}$ is volts (since it is emf), and $\ln[\ce{M^{n+}}]$ is just a constant numerical value.

Hence, I think the unit of $\frac{RT}{nF}$ is Volts.

This was my approach. Can anyone tell me if it is a correct approach?

M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables:

$R$ is a constant: $\pu{8.314 J mol-1 K-1}$
$F$ is also a constant: $\approx \pu{96485  C mol-1}$
$T$ is a variable: $\pu{K}$
$n$ is number of total electrons in the redox reaction involved, and hence a variable but unit less.

Now lets see the unit conversion:
$$\frac{RT}{nF} = \frac{\pu{J mol-1 K-1}\times \pu{K}}{\pu{C mol-1}} = \frac{\pu{J}}{\pu{C}} $$
By definition (Ohm's law):
$$ V = \frac{\text{potential energy}}{\text{Charge}} = \frac{\pu{J}}{\pu{C}} $$
$$\therefore \:\; \frac{RT}{nF} = \frac{\pu{J}}{\pu{C}}= \pu{V} $$


How to (numerically) model a phosphoric acid titration curve I've been learning how to simulate a 1M phosphoric acid titration curve using numerical methods in R.
So far has this has been the best curve:

Notice how it flattens out (artificially) towards the left.
I am not certain if this is a programming and / or a "chemistry" issue.

The system is defined by charge balance, mass balance, and equilibrium equations:
$$[\ce{H3A}] + [\ce{H2A-}] + [\ce{HA^{2-}}] + [\ce{A^{3-}}] = P_{CA}$$
$$[\ce{H+}] + [\ce{Na+}] = [\ce{H2A-}] + 2[\ce{HA^{2-}}] + 3[\ce{A^{3-}}] + Kw/[\ce{H+}]$$
$$K_{a, 1} = [\ce{H2A-}] [\ce{H+}] / [\ce{H3A}]$$
$$K_{a, 2} = [\ce{HA^{2-}}] [\ce{H+}] / [\ce{H2A^-}]$$
$$K_{a, 3} = [\ce{A^{3-}}] [\ce{H+}] / [\ce{HA^{2-}}]$$
I've posted a very similar question at SO and shared code for the simulation over there, using R's nleqslv::nleqslv (go have a look if you'd like to).
I thought this is also a good place for this issue. For instance, this answer (and the ones it links to) do seem to be useful, but I'm not sure how to solve that equation for $[\ce{H3O+}]$ (called $x$ over there).
I would like to know which is the best way to simulate a titration curve.
And also: are the equations I used correct? do you know what am I missing?
Thanks!

Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$.
Calculate fractions of respective phosphate forms as the function of $\ce{[H+]}$ and acidity constants.
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^3$$
$$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = [\ce{H+}] \cdot K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$a_4 =  K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot K_\mathrm{a3} = a_3 \cdot \frac {K_\mathrm{a3}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 + a_4$$
.. and then reuse it and its additive terms(*):
$$[\ce{H3A}]=c_0 \cdot \frac {a_1}{CD}$$
$$[\ce{H2A-}]=c_0 \cdot \frac {a_2}{CD}$$
$$[\ce{HA^2-}]=c_0 \cdot \frac {a_3}{CD}$$
$$[\ce{A^3-}]=c_0 \cdot \frac {a_4}{CD}$$
$c_0$ is the total phosphate concentration. From the fractions, you can get $\ce{[Na+]}$:
$$\ce{[Na+]} = \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]}$$

There is one important thing to consider: Titration dilutes the acid.
If  there is the volume $V_0$ of $\ce{H3PO4}$ with molar concentration $c_\mathrm{0,init}$, and if we spend the volume $V_1$ of $\ce{NaOH}$ solution of concentration $c_1$, then:
$$[\ce{Na+}]=c_1 \cdot \frac {V_1}{V_0 + V_1 }$$
$$c_0=c_\mathrm{0,init} \cdot \frac {V_0}{V_0 + V_1}$$
So perhaps the best approach can be:

There are given concentrations of respective solutions
For given volumes of respective solutions calculate $c_0$ and $\ce{[Na+]}$
For given $c_0$ and $\ce{[Na+]}$, calculate $\ce{[H+]}$ by finding the solution of the equation below ( $f([\ce{H+}]) =  0$ ) by numerical methods, like Regular falsi, or Newton one.

$$ \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]} - \ce{[Na+]} = 0$$
For $y_1=f(x_1)$, $y_2=f(x_2)$, $y_1 \cdot y_2 \lt 0$, there is the Regula falsi iteration:
$$x_3 =  \frac{x_2 \cdot y_1 - x_1 \cdot y_2  }{  y_1 - y_2}$$

(*) Note that the formulas are quite common, I have just rearrange them. They should be able to be found on internet. I have once derived them myself, but not this time, because I remember them.


Why isn't water an ionic compound? If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic?

This is what I'm thinking:

$$
\ce{(H^+)_2O^{2-}}
$$

Thank you

First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.


Chemical diagram for cummingtonite? Any suggestions on diagramming the chemical structure for cummingtonite?
$$\ce{(Mg,Fe)_7 Si_8 O_{22} (OH)_2}$$
As I understand it, the $\ce{(Mg,Fe)_7}$ says the molecule has seven atoms that could be either Magnesium or Iron, with probably more Magnesium than Iron, meaning the structure isn't precisely defined, i.e. a mineral not a specific molecule.  Is there a suitable way to diagram this that looks nice but not excessively complex?

In fact, I'm rather ignorant about chemistry, but started screen printing subtlety goofy neck ties.  I'm envisioning doing essentially this but vertically and adding a chemical structure diagram.  I'll do the final pattern in latex if possible.

p.s.  If I were making ties professionally, maybe I'd diagram an assortment of different cummingtonite molecules as a background for the formula.  I'm not doing that, way beyond my screen printing ability.  Just curious though, how does one know the probability distribution for the Magnesium and Iron?  Just varies by sample perhaps?

Crystalline structures of inorganic materials aren't usually represented by diagrams, but rather by 3D renderings. First, you need to find a crystallographic structure file, which are usually in CIF format. A Google search leads to this one for cummingtonite.

Then, you need some molecular visualization software. Here, I use CrystalMaker, which can create nice photorealistic 3D images, and has a simple tool to record animations. The result is here:

($\ce{Fe}$ is pink, $\ce{Mg}$ is yellow, $\ce{Si}$ is blue; both $\ce{Mg}$ and $\ce{Si}$ are at the center of the tetrahedra, and oxygen atoms correspond to all the tetrahedra edges)


Solubility of Noble Gases I read this statement that:

  Xenon is the most soluble noble gas in water.

My first doubt is:
Why does the solubility of noble gases in water increase down the group?

Even if this is the trend, why is Radon not considered most soluble?

Wikipedia gives the following solubilities for the noble gases in water at $25^\circ \text C$ in $\mathrm{cm^3~kg^{-1}}$:

$\mathrm{He: 8.61~~~Ne: 10.5~~~ Ar: 33.6~~~ Kr: 59.4 ~~~Xe: 108.1 ~~~Rn: 230}$

Why does solubility of noble gases increase down the group?

The intermolecular interactions of noble gases are in the form of London dispersion forces, which are strongly influenced by the polarizability of the atom which increases as the atomic radius increases. This can be thought of as resulting from the electrons being 'less tightly bound' to the nucleus and the atomic radius increases. Atomic radius increases down the group therefore solubility also increases. For more on polarizability see here.

Why is radon not the most soluble noble gas?

Radon (and other heavy radioactive elements) are often ignored when talking about trends because they are uncommon. Radon is in fact the most soluble noble gas but the half lives of its isotopes are on the order of hours or days so it is not really very useful.


Why do rare earth metal oxides vary in color so much? I only know a little about what causes different color, but I know it has to do with the electron orbitals and how they reflect or absorb/emit incoming light. The rare earth metals are in the lanthanide series so they all have an identical valence shell.

I've worked with lots of chemicals, but there's not much that is quite the same as the bright and exotic colors produced from the rare earth metal oxides. I worked with the oxides in powdered form in ceramic glazes and they each vary greatly in color, with colors including pink, light blue, black, creamy yellow, and white. (After firing, lime green and grey can also be introduced). Why do they each produce such different and vibrant visual colors?

TL;DR f-f transitions in lanthanide ions give rise to the colours of the ions. The electronic states can be derived from quantum mechanics, and their energies vary from ion to ion, which give rise to different colours.. The colours are pale because these transitions are Laporte-forbidden and therefore weak.

These absorption bands do indeed arise from electronic transitions, but the theory required to describe them is more sophisticated than simple crystal field theory where one speaks of d-d* transitions. Instead one would have to construct the symmetries of the electronic states via direct products. In the case of the transition metals, there is some description here. For example, a $\mathrm{d^2}$ octahedral transition metal complex would have a $\mathrm{^3T_{1g}}$ ground state and there would be three spin-allowed transitions to excited states ($\mathrm{^3T_{2g}}$, $\mathrm{^3T_{1g}}$, $\mathrm{^3A_{2g}}$). Not all of them may fall within the visible region, but often, more than one of them do.

As a concrete example, here's the spectrum of $\ce{[V(H2O)6]^2+}$.1 In wavenumber units the visible spectrum is roughly $14\,000\mathrm{-}25\,000~\mathrm{cm^{-1}}$:

You can hopefully already appreciate that it is not as simple as "d-d* transitions" which would imply that the complex only absorbs at one wavelength.

Now, for the lanthanides, exactly the same principles apply. The lanthanides characteristically do not interact strongly with any bound ligands, and therefore the effect of the ligand field is almost negligible. The relevant electronic terms are therefore the atomic term symbols, which are obtained by Russell-Saunders coupling, which is more fully described in numerous physical chemistry textbooks. For example, consider $\ce{Pr^3+}$ ($\mathrm{4f^2}$). The allowed term symbols are

$$\mathrm{{}^3H, {}^3F, {}^3P, {}^1I, {}^1G, {}^1D, {}^1S}$$

and following spin-orbit coupling (which is quite a significant effect in the lanthanides) one gets the electronic terms

$$\mathrm{{}^3H_4, {}^3H_5, {}^3H_6, {}^3F_2, {}^3F_3, {}^3F_4, {}^3P_0, {}^3P_1, {}^3P_2, {}^1I_6, {}^1G_4, {}^1D_2, {}^1S_0}$$

The ground state as predicted by Hund's rules is $\mathrm{^3H_4}$. There are therefore up to 12 different transitions that must be considered. In some other lanthanide ions, there are even more possible states.

The bottom line is that each lanthanide ion, with its unique electronic configuration, will have different excited states and different excitation energies. Therefore their absorption spectra differ and they display different colours. $\ce{Pr^3+}$ itself has the following spectrum.2 The lower excited states don't fall within the visible region.

As for why they are "bright", that is simply related to the selection rules. For transitions between atomic states, the requirements are

$$\Delta S = 0; \Delta l = \pm 1; \Delta L = 0, \pm 1; \Delta J = 0, \pm 1 \text{ except }J = 0 \not\leftrightarrow 0$$

and for centrosymmetric systems like atoms there is an additional constraint that

$$\mathrm{g} \not\leftrightarrow \mathrm{g}; \mathrm{u} \not\leftrightarrow \mathrm{u}; \mathrm{g} \leftrightarrow \mathrm{u}$$

which is the Laporte selection rule. The bottom line is that these "f-f transitions" (I am using this term to loosely refer to the collective set of electronic transitions) are parity-conserving and therefore Laporte-forbidden. Vibronic coupling, which is the pathway which allows this selection rule can be relaxed, is not of consequence in the lanthanide ions because of the lack of covalency; the colours are not intense, and therefore typically a pale colour is observed. (Compare $\ce{KMnO4}$ which has a fully parity-allowed LMCT transition; its dark purple colour is exactly the opposite of pale.)

Much greater discussion of the electronic spectra of lanthanides can be found in ref 3.

References

Oxford lecture notes.
Aspinall, H. C. Chemistry of the f-Block Elements; p 35.
Cotton, S. Lanthanide and Actinide Chemistry; pp 61 onwards.