Conductometric titration of Strong Acid And Strong base Consider conductometric titration of Strong Acid with strong base say HCl + NaOH.

My question is why initially the conductance decreases. If H+ of HCl is consumed by OH- of NaOH, forming water and hence conductance decreases.  I agree to these.
But On other hand NaCl is also forming in same amount which can give rise to Conductance. So in my point of view Conductance must remain same or differ slightly. Instead of decreases considerably.

You correctly identify that the number of cations do not change in solution, so from a charge perspective it appears that there should be no significant change in the conductivity.

However, consider two things:

Sodium ions are much heavier and larger than free protons and so should travel through water more slowly. Charge moving more slowly
means lower conductivity.
In addition to their smaller size and lightness, Protons can also "hop" from water molecule to water molecule and so can move through the solution much more quickly and easily. Other water molecules do not act as obstacles

Point 2 is called the Grotthuss Mechanism.

simple..proton is light than sodium so the mobility of proton is greater than sodium ion thats why conductance decrease


Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature? I'm reading this answer which refers to both red fuming nitric acid (RFNA) and white fuming nitric acid (WFNA).
Those two Wikipedia articles say:

Red fuming nitric acid (RFNA) is a storable oxidizer used as a rocket propellant. It consists of 84% nitric acid (HNO3), 13% dinitrogen tetroxide and 1–2% water. The color of red fuming nitric acid is due to the dinitrogen tetroxide, which breaks down partially to form nitrogen dioxide. The nitrogen dioxide dissolves until the liquid is saturated, and evaporates off into fumes with a suffocating odor. RFNA increases the flammability of combustible materials and is highly exothermic when reacting with water.
White fuming nitric acid (WFNA) is a storable liquid oxidizer used with kerosene and hydrazine rocket fuel. It consists of nearly pure nitric acid (HNO3). WFNA is commonly specified as containing no more than 2% water and less than 0.5% dissolved nitrogen dioxide or dinitrogen tetroxide.
WFNA as an oxidizer has somewhat less performance than red fuming nitric acid (RFNA) but is considerably safer (though extremely corrosive), as it has little to no dissolved nitrogen tetroxide, which is an extremely toxic and volatile chemical. If not inhibited, it will form nitrogen tetroxide on contact with most metals and some organic materials. WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours. (emphasis added)

I'm asking about that last sentence; why would red fuming nitric acid convert to white fuming nitric acid by leaving it out at low temperature?

You can convert RFNA to WFNA by removing dissolved nitrogen oxides ($\ce{N2O4}$ and $\ce{NO2}$) from the RFNA.

Given the volatility of $\ce{N2O4}$ and $\ce{NO2}$ it might be possible to just let the gases evaporate from the solution at low temperature. However, since the boiling point of $\ce{N2O4}$ if about $21.69~\pu{°C}$, I do not think that this is a very practical method.

Another path that is theoretically possible is the reaction of residual water with excess $\ce{N2O4}$ and $\ce{O2}$ from air to $\ce{HNO3}$.

$$\ce{2 H2O + 2 N2O4 + O2 -> 4 HNO3}$$

Again, to me this seems not to be a very plausible way. Since the Wikipedia article does not give a source for its claim, we cannot decide which of the two paths, if at all, is the correct one.

$\ce{NO2}$ and $\ce{N2O4}$ mixture is an equilibrium mixture which depends on temperature. At low temp $\ce{NO2}$ dimerises more, so equilibrium will be more shifted to $\ce{N2O4}$ (colourless).


why CO2 is not organic this might be very childish question but i am comfused because they say that organic molecules are those that contain H and C if that is so how can you call carbontetrachloride organic, even it hasn't any H atoms

Carbon dioxide
"is" neither organic nor inorganic, it's just carbon dioxide.
This distinction is nothing but a convention to separate things
between the textbooks and professors.
Carbon dioxide forms carbonic acid in water and the usual salts.
This was/is reason to think of it as inorganic.
On the other hand, urea, the diamide of carbonic acid was thought
to be organic when Wöhler synthesized urea for the first time
in lab, today it is categorized as inorganic.

The C-H bond has a lower bond energy than
the carbon-oxygen bond in carbon dioxide,
making carbon dioxide more stable/less
reactive than the typical organic compound.
So, when you're determining whether a carbon
compound is organic or not, look to see
whether it contains hydrogen in addition to
carbon and whether the carbon is bonded to
the hydrogen.


Citric acid and ammonium nitrate Soil based question. I have soil with to much Ammonium Nitrate in it. I am wondering if I use citric acid and water mixed will that flush out the extra Ammonium Nitrate that has built up in the soil.

Will citric acid bond to Ammonium Nitrate?

Any incite on these to compounds will be greatly appreciated.

Ammonium nitrate is very soluble in water, so watering the soil will remove it, though, depending on the amount and dilution, it could then pollute streams, lakes and other bodies of water, possibly causing an algae bloom.

Your best choice might be to experiment with nitrate-tolerant plants, such as lettuce, ornamental cabbage and rhubarb, or perhaps bell peppers.

Citric acid will not help in any way, and would make the soil less likely to support life.

In nature, denitrifying bacteria will slowly reduce the $\ce{NH4NO3}$. In an anaerobic environment, it might be converted to $\ce{NH4+}$.

Citric acid will not help flush Ammonium Nitrate from soil, as the other poster says, water will do that fine. For a fertiliser spill on soil, after removing whatever fertiliser you can, add sugar (sucrose) first then water. Microbial action will clean it up.


How many total absorptions would appear on proton -NMR spectrum for this molecule? I count 7 equivalent absorptions total, but answer says 8. I don't see where this eighth absorption comes from?

Heres how I'm counting starting on right side of molecule:

1 is Carbox acid H

2 is the 1 H just left of the double bonded Oxygen

3 is the 3 H's of the methyl group next to the lone H

4 and 5 are the 2 sets of equivalent H's in the ring structure

6 is is the 2 H's to left of ring

7 is the 2 methlys on the isoproplyl (these are equivalent so same signal)

How do they get 8?

To answer this question properly one must understand that the field of the instrument is important. At 400 MHz, ibuprofen displays 7 distinct carbon-bound proton signals. Below is the spectrum from the SDBS website. Although the two methyls of the isobutyl chain are recorded [G] as a "singlet" at δ 0.889, they are diastereotopic and inherently different no matter how far they are removed from the asymmetry on the other side of the ring. If this spectrum were recorded on a 1,020 MHz 1H NMR, perhaps separation of the two methyl signals would occur. That the two methyls are different, imagine that one of them is a perdeutero methyl group. This compound is a diastereomer of the compound having the other methyl group as the perdeutero group. So there are actually 9 total distinct types of hydrogens (including the carboxyl hydrogen) that, given the right spectrometer, could appear in a spectrum of ibuprofen. It is a matter of resolution. [As a case in point, see the 1H NMR of cholesterol at the SDBS website and note that the two methyl groups at the end of the side chain have different chemical shifts even though the are separated from the nearest center of asymmetry by three methylene groups.]
Acknowledgment: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

There are 8 non-equivalent protons as shown.


At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? This one is troubling me, and I even got it wrong in my exam:

  at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?

Please explain this one.

At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.

Let's consider the first order reaction $\ce{A <=>[k_f][k_b] B}$

So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.

Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:

Because at equilibrium state forward and reverse reactions occur at same place but they occur in opposite directions so reaction don't stops..


Intuition for why effective nuclear charge increases down a column My textbook (Chemistry the Central Science, edition 12, Brown et al.) says that effective nuclear charge increases down a column. Indeed, this is consistent with the values provided by Clementi in the 1960s.

Is there intuition or some qualitative explanation for why effective nuclear charge ($Z_\mathrm{eff}$) for the outermost electrons increases down a column (as opposed to being roughly equal)?

Initially, let's take the electrons as probability density fields instead of classical particles.

The increase on the number of protons in the nucleus increases the electric force on all electrons, but the core electrons suffer higher attraction due to higher proximity and lighter electronic blindage, so their probability density field is forced nearer the nucleus.

Reminding of Electric Force:
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$

Consequently, the outer electrons will suffer lower blindage force from these core electrons, parallelly to the nucleus attraction, allowing them to get closer to the nucleus as well.

As effect, of course, the outer electrons also force the inner electrons towards the nucleus, but their probability densities reach a minimum repulsion energy in such a distance that electrons on the same level are forced closer to the nucleus than on the atomic element above, going down a column in the periodic table.

It is due to the less efficient shielding from d and f orbitals.

P orbitals electrons shield less the outermost electrons from the S orbitals electrons increasing the effective nuclear charge.!But p orbitals electrons shield the valence electrons pretty well and we dont take them in account.


Is activated carbon classified as organic or inorganic? Organic compounds are typically defined as “molecules containing carbon”. Wikipedia states that there for some historical (read: non-logical) reasons,

  a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon, and cyanides, as well as the allotropes of carbon such as diamond and graphite, are considered inorganic.

I thus wonder: is activated charcoal (also known as activated carbon) typically classified as an organic or inorganic material?

If I follow the list of exceptions given by Wikipedia, it should be organic (it's not an allotrope of carbon, in particular), but I get the feeling that most people in the field of porous materials would classify it as inorganic. So, I'm looking for an authoritative reference on this question.

While nomenclature is of particular interest to organic chemists to specify an exact compound, the classification of X into broad category Y or Z isn't a precise science, and not really of practical use.  The article cites a textbook by Seager to this effect, stating

  The distinction between "organic" and "inorganic" carbon compounds, while "useful in organizing the vast subject of chemistry... is somewhat arbitrary"

Even if you find a source that says "charcoal is (in)organic", you may just as well find one stating the opposite.  Just like the coal from which it may have been produced, it was once biomass and decidedly organic, but so was graphite and diamond, or CO2 and CO32−. I think it's overly pedantic and unproductive to try to come up with definitive judgements for these decidedly edge case scenarios.

After all, it's just a chemical on the shelf, what one does with it is far more relevant.  I don't use it on a daily basis, but it seems more like a tool than a reagent.  The fact it contains carbon seems beside the point; it's value isn't in the chemical composition but rather its extraordinary adsorptive properties.

Activated carbon treatment removes such impurities and restores plating performance to the desired level. it is a main purification technique for removing organic impurities from bright nickel plating solutions. so its an organic product.


Are `dTDP-4-dehydro-6-deoxy-alpha-D-galactose` and `dTDP-4-keto-6-deoxy-D-glucose` the same thing? Are dTDP-4-dehydro-6-deoxy-alpha-D-galactose and dTDP-4-keto-6-deoxy-D-glucose the same thing?
I am looking for the function of this enzyme in uniprot. It says

Catalyzes the synthesis of dTDP-4-amino-4,6-dideoxy-D-galactose (dTDP-Fuc4N) from dTDP-4-keto-6-deoxy-D-glucose (dTDP-D-Glc4O) and L-glutamate.
dTDP-4-amino-4,6-dideoxy-alpha-D-galactose + 2-oxoglutarate = dTDP-4-dehydro-6-deoxy-alpha-D-galactose + L-glutamate.

Note that glucose in the first line and galactose in the second line.
Or it is a typo there?

There are four stereocentres in hexoaldoses giving rise to a set of $2^4 = 16$ different molecules with the same connectivity; you can group these to eight pairs of enantiomers (perfect mirror images of each other). The enantiomers are labelled D and L depending on the orientation of the stereocentre which is farthest from the aldehyde group. In linear orientation, these are the following molecules:

(All images taken from Wikipedia; all D-isomers.)

In the order of reading these are allose, altrose, glucose, mannose, gulose, idose, galactose and talose. If you compare glucose (3rd in the first row) with galactose (3rd in the second row) you will realise that they differ only in the orientation of the hydroxy group in 4-position. The posh terminology is epimerism — D-glucose is the C4 epimer of D-galactose and vice-versa.

You asked about two labels where the sugar is preceded either by 4-dehydro-6-deoxy or 4-keto-6-deoxy. The agree in the 6-deoxy part which means that the oxygen on C6 is missing; you will find a methyl group $\ce{CH3}$ there instead. The labels 4-keto and 4-dehydro may also be confusing. The first says that a keto group is there instead of the hydroxy group, while the second says that two hydrogens have been removed from that position. The result is the same, instead of having $\ce{CHOH}$ as fourth carbon we have $\ce{C=O}$. This also deprives us of the stereocentre (a carbonyl group is never stereogenic).

Putting all of this together: Yes, the two are identical. If you start from the C4-epimers and remove the stereocentre at C4 it does not matter which you started from. Both galactose and glucose are common sugars so both choices are understandable (even though glucose is probably known even a tad more). I would, however, strongly object agains 4-keto-D-talose in the place of 4-keto-D-mannose (mannose and talose being C4-epimers) since mannose is much more known than talose.

Glucose and galactose are two different carbohydrates and these molecules are actually different. Both actually are isomers of each other, both with the formula of C6H12O6.

So the two biomolecules are different, not the same.


Can ozone have a triangular structure? The structure of ozone is traditionally depicted using two resonance forms:

However, what I am wondering is instead of using such an idea to explain the structure why can't we show the structure of ozone like this?

Some people have told me that such a structure is not possible but they fail to mention the reason. What I have in mind is maybe electron repulsion. But I would like to know exactly what mechanism would prevent this from happening?

Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\pi_\mathrm u$ orbitals are both singly occupied.

Note: 6a$_1$ and 2b$_1$ have the incorrect sign on the outer p-orbitals.

If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\mathrm{6a_1}$ and $\mathrm{2b_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.

So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\mathrm{1a_2}$ orbital and the $\mathrm{4b_2}$ orbital start rising in energy, eventually rising above the $\mathrm{6a_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\circ$.

This same diagram can be used for other molecules, such as $\ce{CO2}$. Try using it to figure out why $\ce{CO2}$ is linear.

O3, Ozone forms a hexagon like cyclohexane, both chair & boat conformations via 3 bent banana bonds or 3 single indirect covalent bonds like in the Pauling bent bonds. You need to see/use my revolutionary, new, unique Molecular Models set to show these banana bonds formed of 2 pairing free radicals, each bond.


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

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

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

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

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

I think it has to do with the equation heat dq=TdS so since gibbs dG=SdT+VdP+mudN we know S or entropy is partialG partialT and also gibbs= -nFE so S=partialE partialT and heat is TdS so TpartialE partialT* nF. In short TdS is the reversible heat


Difference between reversible and irreversible thermodynamic process I have been reading up (on this website) about the difference between reversible and irreversible thermodynamic process and it defines them saying that reversible processes are ideal processes while irreversible processes are the natural processes. Are these definitions correct?

If not how are they defined?

The simplest answer in thermodynamic terms is "a reversible process is one in which the entropy generation term is zero."

  Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.

For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.

These are not definitions of reversible and irreversible processes, but it is a description of one of the differences between the processes.

In real world situations, however, for some processes, the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle. (see here)

  A (...) definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings. In thermodynamic terms, a process "taking place" would refer to its transition from its initial state to its final state. again, wikipedia


Is Hydrogen Bonding a Type of Dipole Dipole Interaction? I understand that dipole dipole forces is due to the attraction of the different partials charges of atoms in different molecules due to their different electro-negativities.

For hydrogen bonding, my understanding of it is that they are just a special case of dipole dipole forces. In certain molecules in where the difference in electro-negativities between the different atoms is high enough, then even stronger dipole dipole forces will be form which are known as hydrogen bonding. For example, in $\ce{H_2O}$ where the difference in electro-negativity of H and O is 1.24, hydrogen bonding occur. This is why hydrogen bonding can only occur between hydrogen and nitrogen, fluorine or oxygen.

However, my school teacher says that hydrogen bonding in fact is not a type of dipole dipole force as she says that hydrogen bonding occurs differently to dipole dipole forces. She said that hydrogen bonding occurs to the attraction of the lone pairs in oxygen, fluorine and nitrogen atoms to atoms with a positive partial charge. Hence this is why hydrogen bonding is much stronger than conventional dipole dipole forces.

But this doesn't make sense because if hydrogen bonding is due to the attraction  of H to the lone pair of electrons, shouldn't hydrogen bonding also form between H and Cl or I or any other atom that has a lone pair and is more electronegative than H?

Edit 1

From the comments below, it appears that the actual answer is more complicated than the simple explanation of hydrogen bonding given above. After doing some research, I have became aware of the Advanced Theory of the Hydrogen Bond which states that the hydrogen bond actually contains 10% covalent nature. However none actually explain how they are formed. Could someone please explain how hydrogen bonding actually in reality form.

Edit 2

So in response to the excellent answer by jheindel, is the reason why hydrogen bonding is considered partly covalent in nature is because the let say in water, oxygen's MO partly overlaps with the MO of the hydrogen atom in the other molecule. Also this is why they are much stronger than conventional dipole dipole forces?

Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got.

As you said, it has been found that the hydrogen bond (specifically in water) is about 90% electrostatic and 10% covalent. See this link for a sort-of-review/overview of hydrogen bonding and some relevant research.

Upon following the paper trail, I found the following research article: E.D. Isaacs, A. Shukla, P.M. Platzman, D.R. Hamann, B. Barbiellini, C.A. Tulk, J. Phys. Chem. Solids, 2000, 61, 403-406. (Mirror)

Essentially, they use some method which I don't really understand and find quite conclusive evidence of the (word of the day) anisotropy of hydrogen bonds. That means that the bond is indeed directionally dependent. Bond direction is one defining characteristic of covalent bonds.

That raises the question of what is going on with the orbitals that allow for hydrogen to participate in more than one covalent bond at a time.

A simple explanation of the orbital interactions for a water dimer is to view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer. This superposition which includes the hydrogen then explains the way that hydrogen is participating in one covalent bond and another partially covalent interaction.

Furthermore, this explanation of what the orbitals are doing fits well with what we would expect from a hydrogen participating in two different covalent interactions, which is that the energy curve has two minima where one is much deeper than the other.

Further evidence for the covalent nature of the bond is the fact that the O-H bond elongation and O---H hydrogen bond length always change together. That is, as the O-H bond lengthens, the O---H interaction length shortens. This implies a shift in electron density towards hydrogen which is most easily explained by a delocalization of electrons participating in a covalent interaction.

So, I hope that's a good start toward understanding the covalent nature of the hydrogen bond--it was for me!

But to answer the simple part of the question which you clearly already know is that yes dipole-dipole is a significant part of hydrogen bonding, and it certainly should be, but as is always true of chemistry, there's a lot more going on than what one might initially think.

EDIT (06/01/2016)

Since I posted this answer, I think I have developed a better understanding of the situation, and feel obliged to update my answer. First, let me split hydrogen-bonding into two obvious parts, electrostatic and covalent characteristics. This is an obvious split, but it is not completely clear yet what we mean by covalent, so let us try to define that by looking at some examples. Furthermore, we must be careful when talking in about hydrogen-bonding because in every hydrogen bond there is a donor and acceptor, and these two roles are often played best by different chemical species.

Hydrogen-Bonding in thiols

Here we go breaking yet another rule. They told us that sulfur couldn't hydrogen bond. Well, that's not exactly true. Okay, but that hydrogen-bonding has got to be basically negligible right? Not exactly.

The first thing we need to do here is get some information. First, sulfur has an electronegativity of 2.58 on the Pauling scale. This is roughly the same as the electronegativity of carbon which we do not consider to hydrogen-bond well. So, if there are hydrogen-bonds involving sulfur, we should expect that they will be primarily covalent as any dipole interactions will be pretty small.

As it turns out, sulfur is a much better hydrogen-bond acceptor than it is donor, and at times is a comparable acceptor to oxygen. Here is a table comparing certain dimers from a great source you can reference at the bottom of the page$^3$.

As you can see, when the sulfur in these dimers (which are all named in the caption of the table) acts a hydrogen bond acceptor, the shift in vibrational frequency of the $\ce{O-H}$ and the binding energy are comparable to when an oxygen is the acceptor atom.

Looking at other research$^4$ involving sulfur hydrogen bonds reveals that approximately 70% of the interaction energy between sulfur and oxygen when sulfur is the acceptor comes from dispersion. That is the key physical insight in all this. Despite the loss of electronegativity, sulfur gains something in its greater dispersion interactions which makes it a good hydrogen bond acceptor. It should be noted, however, that the electronegative component of these hydrogen-bonds maintains their directionality.

Also, to compare, according to that wikipedia article I linked, the interaction energy of a water dimer is only 24% dispersion. In water electrostatics dominate and induction and exchange play meaningful roles.

Conclusions

I have more I could say, but it's perhaps a longer explanation than is worth going into. To conclude, what is meant by covalent interactions in hydrogen-bonding varies based on the species you are considering. In something like water, we can think about this in the usual way by saying that there must be some orbital overlap and hence electronic exchange which will be stabilizing. With other species, we can see that what we mean by covalent interactions might be different. That is, there is still some orbital overlap, but we must also consider non-electrostatic interactions like dispersion.

3 Noncovalent Forces-Chapter 2

4
Wennmohs, F., Staemmler, V., & Schindler, M. (2003). Theoretical investigation of weak hydrogen bonds to sulfur. The Journal of chemical physics, 119(6), 3208-3218.

Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of other molecule. These are special type of dipole-dipole interactions. Here the hydrogen atom is linked with one atom through covalent bond and linked with other atom through hydrogen bond.however hydrogen bond is much weaker than the covalent bond. When hydrogen atom bonded to a strongly electronegative atom the bond pair of electrons displaced towards electronegative atom. Thus the only electron of hydrogen lies away from it, its nucleus get exposed and behaves as a bare proton and exert a strong electrostatic force of attraction on the strong electronegative atom of adjacent molecule this interaction is referred as hydrogen bond.


Are there bonds which share single electron? Are there bonds which share single electron?

During chemical bonding, I asked our teacher why there are only 2 electrons shared in a bond. He said that because it's that one orbital can accommodate only two electrons. Two electrons make it stable. He said that there are bonds which share a single electron which has some term I don't remember and is out of our understanding for now. So, I am curious whether there are such bonds, what are they called, some examples, and are they unstable due to the reason aforementioned?

There exists an interesting molecule dicarbon with bond order two and two pi bonds instead of the usual one sigma and one pi. This is the case when 2s-2p mixing is allowed in molecular orbital theory. However when the aforementioned condition is not allowed, dicarbon exists as a molecule with one sigma and two pi bonds of, interestingly, one electron each. So in a way, one electron bonds do exist.


Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane How can I determine the number of possible pairs of diastereomers here?
My first guess was two: RR with RS and SS with RS. But what about the potential chirality center middle carbon? There are two identical branches attached to it; however, those two branches can change R/S. So, is it a chirality center?

I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is non-stereogenic but it may be chirotopic depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a pseudo-asymmetric carbon atom. Carbon-2 is not a stereocenter, because it does not have four different "things" attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.

Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):

A: (1R,2s,3S)-1,3-dichloro-1,2,3-triphenylpropane

B: (1S,3S)-1,3-dichloro-1,2,3-triphenylpropane

C: (1R,2r,3S)-1,3-dichloro-1,2,3-triphenylpropane

F: (1R,3R)-1,3-dichloro-1,2,3-triphenylpropane

The small r and s are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the r/s designations, but I believe that the R configured substituent receives higher priority than the S configured substituent, and Cahn–Ingold–Prelog system is followed from there.

I don't think it is hindered enough to prevent rotation around the central carbon, so I think it only has two stereogenic centres, leading to four possible stereocentre combinations as you have identified: RR, SS, SR, RS

This results in more than two pairs of diastereomers though:

1) RR with SR
2) RR with RS
3) SS with SR
4) SS with RS


How to clean out a contaminated aluminum automotive radiator I have a very expensive custom made automotive aluminum radiator that has had a mouse or mice nesting inside the top tank.  I have removed most of the nesting material through the top inlet but I want to clear the radiator tubes that may be plugged up with the nesting material (looks like cotton of some sort) and be sure the radiator is flowing water through it as it should. As it is an aluminum radiator I am told they cannot remove the tank and "rod out" the radiator tubes.  The only solution offered by the company is replacement.  The radiator cost about $1200.
I have thought about an enzymatic drain cleaner but am not sure that it will not harm the aluminum radiator internally.  What ideas do you have?

Well I actually tried a small jar of the Enzymatic product with a sample of the nesting material I already removed as well as small piece of .023 6061 aluminum... Overnight the aluminum was completely decomposed.  The nesting material seemed unfazed.  So much for that idea!


What happens to gunpowder when it gets wet? What happens to the gunpowder and other explosives when they get wet? I know that they stop "working" (won't explode anymore) but for what reason? Is it anyhow similar to the fact, that wet wood is burning worse than the dry one?

In general, putting explosives in water will not prevent them from exploding, because most explosives already contain all of the necessary reactants to produce an explosive chemical reaction.  They do not need to draw oxygen from the air, and the reaction happens much too fast for the water to act as a heat sink.

Traditional "black powder" is an exception; it doesn't explode when wet.  I don't know exactly why this is, but two plausible reasons are:

Unlike most other explosives, black powder is shock-insensitive; it has to be initiated with a flame or spark.  Water can easily soak up all the heat energy from a small flame, like the one you get from a percussion cap.  This is the same as how it's harder to set fire to a living tree than a dry log, but if the fire gets hot enough it will still catch.
Black powder is a "low" explosive, meaning that the chemical reaction happens relatively slowly (for an explosion), which gives the water more time to absorb heat energy and quench the reaction.

Worse, black powder can be permanently ruined by water.  This is because it's a mixture of saltpeter (potassium nitrate, $\ce{KNO3}$), sulfur, and charcoal, and saltpeter is water-soluble.  If you get it wet enough, the saltpeter can be completely washed out of the mixture.  This doesn't always happen; the 19th century industrial manufacturing process for black powder involved deliberately dampening it, both to make it less dangerous during manufacture, and to make saltpeter crystals form inside each particle of charcoal.  But they were careful to use only just enough water.

For combustion to take place, there needs to be an oxidizer, fuel, and heat. When a substance gets wet, it is surrounded with water and is not in contact with atmospheric oxygen, arguably the most common oxidizer. In the absence of any other oxidizers, combustion stops. Underwater fuses incorporate strong oxidizers like perchlorate salts so the reactions can continue in the absence of atmospheric oxygen. Water is also an incredible heat sink and will steal heat from the reaction and dampen it in that way.


Is sodium acetate strong or a weak electrolyte? Since it is formed of a strong base and a weak acid it should be a strong electrolyte but a chemistry teacher said that it is a weak electrolyte.....

Sodium acetate is a strong electrolyte in water.  The status of sodium hydroxide as a strong base triumphs; it makes even a weak acid such as acetic acid dissociate essesitally 100% into ions (by stealing the protons).  And when something is essentially 100% dissociated into ions, by definition it's a strong electrolyte.

Strong or weak electrolytes will depend on the hydration enthalpy of products in water. That's why any salt like $\ce{NaCl}$ will completely dissolve in the water, though $\ce{NaCl}$ has high bond energy in powder form. In the above case, the hydration of the $\ce{Na+}$ ion will be so dominating that the salt will dissociate (hydrated) completely despite less hydration enthalpy of the anion.
So, as a rule of thumb strong acid/base salts will have more hydration enthalpy but it is just a correlation and will depend on the solvent. The reason for partial or full dissociation is in hydration enthalpy and entropy change.


Why are some salt solutions coloured? I notice that salt solutions of $\ce{NaCl}$ and $\ce{KCl}$ are colourless while those of $\ce{CuSO4}$ and $\ce{FeSO4}$ are coloured.

I got as far as figuring that it has to do with the transition metal ions, but I can't explain why the salt solution of $\ce{ZnSO4}$ is colourless even though zinc itself is one of the transition metals.

You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to Copper Sulphate). Thus, metals with no empty $d$ orbitals of comparable energy become colourless.

OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this:

Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there.

All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about octahedral complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're "mixing" these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called "d-orbital splitting" {*}

Now, electrons may "jump" between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition.

Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is before the ligand bonds)

*For tetrahedral complexes, the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency)

$\ce{ZnSO4}$ is colorless because it has completely filled d orbitals while $\ce{FeSO4}$ is green because its d orbitals are incompletely filled.


The effect of pressure on equilibrium constant What will happen to the $K_\text{eq}$ in the following reaction if volume is increased?

$$\ce{2NO(g) + O2(g) <=> 2NO2(g) + energy}$$

From le Chatelier's principle I know that it must be constant unless temperature is changed, however it doesn't make sense because if

$$K_\text{eq}= \frac{[\ce{NO2}]^2}{[\ce{O2}][\ce{NO}]^2},$$

then by increasing the volume we know that equilibrium will shift left and favor reactants. Hence concentration of reactants would increase while concentration of products would decrease. If we plug that back into the equilibrium constant equation we see that if numerator decrease and denominator increase that must yield a smaller number than the initial number. Hence the equilibrium is not constant. Why this occurs?

First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.

Volume??? Wow. You may get better answers if you learn about a thermodynamic property called "pressure". After you learn about it, I suggest you use it when asking about the concentration of gas phase reactions. For gas phase reactions usually concentration is measured not in "moles per liter" but as pressure, invoking Dalton's Law of partial pressures. So, your Keq would be in units of (pressure units)-1, say Pa-1.
 Lets say Keq = a²/(bc²) and a+b+c= P. You seem to be saying that the only way you can change P is to change Keq. Well, no.   You have three equations and 3 unknowns, I'm not seeing the problem. (The unknowns are the new a, b, & c...you could call them a', b', c') (Assuming you knew the new P').


Epsom salts - heavy metal contamination myth? This recipe for home-made handwash (liquid soap) consists of bar soap, water, Epsom salts, and coconut oil.

The author specifically warns against using 'artificial' Epsom salts, preferring the naturally occurring type, because the 'artificial' type is supposedly contaminated by heavy metals incl. arsenic.

  Note – Be sure to get natural salts as some are man made and a full of
  unwanted chemicals. A lot of Epsom salts are made from synthetic
  materials, they are manufactured in factories from neutralized
  Sulfuric Acid, which is very high in heavy metals and arsenic. The
  REAL Epsom salt is naturally occurring mineral deposit in spring.

According to Wikipedia, Epsom salts are made by reacting sulfuric acid with magnesium oxide or magnesium carbonate.

I find it difficult to believe that 'artificial' Epsom salts would have appreciable amounts of heavy metal contamination, as neither sulfuric acid nor magnesium oxide are known to be frequently contaminated with heavy metals. A quick search on Google Scholar gives no results on the topic of heavy metal contamination in Epsom salts.

The only way I am aware that heavy metals might enter the process is if the sulfuric acid was produced from smelter stack gases, which do have appreciable heavy metal content. This is done at Mount Isa Mines, where $\ce{SO2}$ from the copper smelter stack gas is scrubbed out to make sulfuric acid, but the entire output of this plant goes straight to fertilizer production.

Is the above claim about heavy metal contamination of Epsom salts credible, or just an urban myth?

Here in the US, they sell different grades of Epsom salt, from USP down to plant food.  I'd presume that depending upon what the product will ultimately be used for, the manufacturer may use MgO with different purity levels - the lower the purity, the cheaper the product.  Even the "bath salts" grade contain trace amounts of arsenic and selenium (see the first link below).  Among the "medicinal" offerings, some are labeled "for external use" and others are labeled "for external and internal use."  I suspect that anything with the "for external and internal use" label would be more than safe enough to use in handwash.

Here are some links you might find helpful:

http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf

http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html

Well I'm no expert but I wanted to do a liver flush and bought some Epsom salts (magnesium sulfate heptahydrate), apparently safe for ingestion and after seeing this page further searching led to this:
http://www.fao.org/fileadmin/templates/agns/pdf/jecfa/cta/68/Magnesium_Sulfate.pdf

Section 4.2 is relevant and the key phrase is "limited to those that originate from the starting material".


Are all exothermic reactions spontaneous? $\ce{GaAs (s)}$ is used in solar cells deployed on satellites found in the orbit of Earth. The best way to make these materials is through a technique called "molecular beam epitaxy", in which the following spontaneous chemical reaction takes place:
$$\ce{Ga(g) + As(g) -> GaAs(s)}$$
Pick the statement which best describes this process:
(A) The process is endothermic.
(B) The process is exothermic.
(C) The process leads to a decrease in the entropy of the universe.
(D) The process leads to an increase in the entropy of the system.
(E) The process leads to a decrease in the temperature of the system.

The key says the correct answer is B. I am wondering if this is because all exothermic reactions are spontaneous?

No, it is not the case that all exothermic reactions are spontaneous.  More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also specifying the concentrations of reactants and products, since for a certain relative concentration of reactants and products, the reaction might be spontaneous in the forward direction, while for another set of relative concentrations it might be spontaneous in the reverse direction. [See illustration below.]

One could, however, make the general statement that chemicals will react spontaneously if, in so doing, the entropy of the universe increases.  This, however, is not one of the options you are given.

Why, then, is "B" the correct choice?  Answer: It's because this Ga and As are forming a chemical bond, and bond formation is exothermic. In addition, there is a phase change here from a gas to a solid, and that is also exothermic.  Furthermore, since Ga and As are both monatomic gases in this reaction, no chemical bonds need to be broken to allow the Ga and As to combine. Thus, since we have chemical bonds being formed but not broken, and a phase change from gas to solid, the reaction is exothermic.

Here's a graphical illustration of the general concept articulated in the first paragraph:

A exothermic reaction has the energy to proceed spontaneously. It may or may not depending on the kinetics. The reaction could be sterically hindered for example, and need a catalyst to happen at all.


How does Le Chatelier's principle work in a gaseous system? My textbook said Le Chatelier's principle normally applies to gases but there is no explanation.

I did some Googling and it said if you increase the pressure, it will shift to the side with less moles of gas.

What does this really mean? Could someone clarify Le Chatelier's principle for me?

Le Chatalier's principle states that if a stress is applied to a system at equilibrium, then  the system, will shift its equilibrium position in order to bring the system back to its equilibrium.

So the stress applied to the system could be temperature, pressure and the concentration of one of the components in the system.

I will choose to explain about pressure because apparently you have the problem understanding that.

Pressure.
If you add pressure to a system then the stress will be to decrease the pressure by shifting the equilibrium position towards the side with less number of particles.
If you  remove pressure from a system, then the stress will be to increase the pressure of the system by shifting the equilibrium position towards the side with the higher number of particles.

So it is true that when you increase the increase the pressure in a system, then the system is stressed to reduce the stress. In order to do that, the system's equilibrium will shift to the side with less number of particles.

For example:

2Fe$_2$O$_3$ + 3C -----> 4Fe + 3CO$_2$

So if you add pressure to this above system, then the stress is to reduce the pressure. On your reactants side the sum of the coefficients is 5 and on your products side, the sum is 7. Since you need to bring it down then the shift will be to the side with the least sum which is to the left-hand side. Therefore, the shift will be to the left-hand side.

Hope this helped you with the problem.

Just as if it was a chemical reaction of aqueous solutions, a gaseous chemical reaction has its reagents and products and these exist in an equilibrium kind of state that can be shifted either way depending on the conditons the reaction is in and the amount of reagents and products you have (in moles). Adding pressure to the reaction is like compressing the gasses into a smaller volume and this causes more molecule collisions. Since the space between the molecules involved in the chemical reaction is decreased, you have more molecule collitions among them, this makes the molecules to react with each other more rapidly and speeds up the reaction. The reaction will go the way the equilibrium constant tells it to go, but with this addition of pressure it will do it faster. Molecule collitions also creates and destroys molecule's bonds, thus this can create or absorb heat depending on the reaction that is taking place.


Why does 2+ oxidation state become more stable relative to 3+ oxidation state for first row of transition metals? I viewed an image showing all the possible oxidation states of each element in the first row of transition metals, and the main oxidation states highlighted in a different colour.

I noticed all elements Sc to Cr had in common a   3+ main oxidation state,
and from Mn to  Zn 2+ was the common main oxidation state.

The reason stated for this in my good textbook was 2+ oxidation state become more stable relative to 3+ oxidation state, I can't quite understand why.

So  Q : why does 2+ oxidation state become more stable relative to 3+ oxidation state  for first row of transition metals?

I had drawn the electron - in -boxes configuartion for all elements in the first row, and looked at Hund's rule and partially and fully filled orbtals, but I have not really seen a clear, fundamental explanation of why, for this question.

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.


Role of methanol in NaBH4 reduction In reduction of carbonyls (aldehyde and ketones) to alcohols using NaBH4/Methanol, alkoxides are formed with the Boron which usually gets protonated on addition of water/acid.

My question is what is the function of methanol in this reaction as methanol reacts with NaBH4 to form sodiumtetramethoxyborate and hydrogen gas.

Some sources say methanol protonates the alkoxide ion and after doing series of reduction experiments I find that the alkoxide formed only gets protonated to an alcohol on addition of water.

So what does methanol do in this reaction/what is the effect of methanol?

I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.

You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)


Why is tetraamminecopper(II) a square planar and not a tetrahedral species? In $\ce{[Cu(NH3)4]^2+}$, the $\ce{Cu^2+}$ ion has 9 electrons in the $\mathrm{3d}$ orbital with only one unpaired electron. How is it a square planar geometry? Where is that unpaired electron going?

It is very convenient to use crystal field theory to discuss this.

It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.

The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc.

Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate.

However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound.

This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.

All Cupric complexes irrespective of the kind of ligands is always dsp2 square planar hybridization . Now the question is that why this is so?
In tetrahedral complexes when Cu is in +2 state the one electron from 4s2 and one from 3d orbital gets removed and so it a 3d9 complex .Now from 3d9 complex one electron goes to 4p orbital and in this way it now becomes 3d8 and dsp2 complex.


Has the molecular diatomic oxygen trianion been observed or predicted? Has diatomic oxygen ever been observed with a 3− charge, i.e. $\ce{O_2^{3-}}$? Have any ab intio calculations been done on this ion?

Update:

Just to summarize some of the answers below and add my two cents, it seems that no studies have reported observing this ion experimentally, either in isolation or as part of a condensed phase. Ab initio calculations suggest that $\ce{O_2^{3-}}$ is unstable in isolation, meaning that the loss of an electron is exothermic. However, first-principles calculations have predicted that a molecular oxygen trianion can exist as part of a solid; see Phys. Rev. B 85, 035210 (2012).

According to "Small Multiply Charged Anions as Building Blocks in
Chemistry" Acc. Chem. Res. 1996, 29, 497-502

  it is concluded that all isolated diatomic dianions or
  other multiply charged diatomic anions are not stable
  to loss of an extra electron or fragmentation in their
  ground electronic states

So even $\ce{O2^{2-}}$, let alone $\ce{O2^{3-}}$ cannot exist in isolation.

Yes, it is, in $\ce{H3O2}$. This is from an a CCB bond. However, this is highly unlikely.


Why is this PH3 considered to be an unhybridized molecule? Also, don't hybrid orbitals appear in all molecules?

The answer by "ron" says "the molecule can be viewed as being unhybridized":
How does lone pair of a central atom affect the dipole moment?

I'll try to explain this by comparing this with the case of NH3.
N is a better electron withdrawing group than P. So in the case of NH3 the electron cloud is more closer to N. And therefore the repulsion in the bonds causes the 3 H atoms to be farther away from each other resulting in a bond angle of about 109°. But in the case of PH3 the electron cloud is farther away from the central atom(P) than in the case of NH3. So the bond pair - bond pair repulsion is comparatively lesser, causing the 3 H atoms to move closer together to an angle of almost 90°, resembling the px, py, and pz orbitals, as a consequence it's lone pairs are present in almost pure s - orbital. So it is is said that PH3 can be viewed as unhybridized.

It is due to greater difference in energy of atomic orbitals of phosphorous such as 3s and 3p.


Why is Charcoal such an excellent adsorbent? Charcoal (also activated charcoal) is known to adsorb a huge variety of substances including a variety of paints, dyes and many different kinds of ions. Moreover, the amount of adsorption at normal pressure and temperature is high. I know physisorption requires high capacity of Van-der-waal bonding with the adsorbate which is favoured by surface atoms with large, easily-dispersed electron clouds and chemisorption requires actual chemical bond formation which requires highly reactive surface atoms. Charged species get adsorbed to form a double layer when the surface is charged, either due to ionization or due to polarity.

  But charcoal possesses none of the structural specialities required to
  be a great adsorbent. Neither it has charged centres (unlike colloidal
  proteins, or metal sols) nor it has a very high reactivity (Charcoal
  is almost the final product of combustion of wood, bone, sugar etc and
  is supposed to be immune to further oxidation even at considerable
  temperature). Why is it then an excellent adsorbent?

Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface area to mass ratio.

Coal is more effective and efficient in adsorbing fumes due to several aspects:

We can increase the surface area so it can adsorb higher volume of fumes according to the low of mass and volume; we can use that to develop the efficiency of coal to adsorb a large amount of adsorbent matter
Low cost
Low transportation
Doesn't react with fumes


Order of Reactivity of Halogens in Electrophilic Addition What is the order of reactivity of halogens in electrophilic addition reaction via formation of cyclohalonium intermediate?
(A) $\ce{F2}$> $\ce{Cl2}$> $\ce{Br2}$> $\ce{I2}$
(B) $\ce{I2}$> $\ce{Br2}$> $\ce{Cl2}$> $\ce{F2}$
(C) $\ce{I2}$> $\ce{Br2}$> $\ce{F2}$> $\ce{Cl2}$
(D) $\ce{Br2}$> $\ce{Cl2}$> $\ce{I2}$> $\ce{F2}$

I tried to predict it using the mechanism, but couldn't reach a conclusion. Is this based on thermodynamic and kinetic data?
Here's the mechanism involving bromine, for reference:

Could someone provide a detailed explanation for the above problem, with data if possible?

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate.
The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity.
The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]:

It is not normally possible to add fluorine directly to alkenes as the reaction is so exothermic that bond fission occurs. Many alkenes will not add iodine directly either and when the reaction does occur it is usually readily reversible.

As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:

Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003 ISBN-10:8177584332

ans is d               Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes place
due to more E.N of Cl2 compare to Br2. It does not give e pair to carbocation


Supporting electrolytes and increased conductivity? Hi a question about supporting electrolytes (SE) I hope you can help with.  I found some related questions here through searching but looking for a simple answer (if there is one!):
If SE are electroinactive (in the potential range of interest) and so do not pass or receive electrons at the electrodes to complete the circuit then how to they "increase conductivity"?  I am referring to typical DC experiments, e.g. chronoamperometry.  It's not like electrons hop through the solution from one electrode to the other using SE ions as a shuttle.
I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring (e.g oxidizing an enzyme product), thereby facilitating these redox reactions that do occur to continue. Is this the answer?  So in just water + SE, for example, it is more conductive because they facilitate the electrolysis of water (H2 and O2 generation (e.g for SE without Cl-) by this method?
I also understand that movement of ions is considered current, but with DC wouldn't this migration slow down and exhaust as time went on? Is it this transient conductivity due to ion movement that they refer to?
Thanks!

Thanks for your answer Maurice.  What you describe sounds like this (maybe not clearly written) part of my own possible answer to my question: "I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring,  thereby facilitating these redox reactions that do occur to continue".
I think my confusion arises because the IUPAC definition is:  "A supporting electrolyte, in electrochemistry is an electrolyte containing chemical species that are not electroactive (within the range of potentials used) and which has an ionic strength and conductivity much larger than those due to the electroactive species added to the electrolyte", which makes it sound like the the SE ions themselves are conductive.  Plus the most common explanation offered is along the lines of "because they are ions and ions conduct".
If I understand correctly now as discussed, they actually facilitate the reaction of the true electroactive species (which produce faradaic currents through redox - the current being a measure of conductivity), by maintaining neutrality around the electrode, rather than being "conductive" themselves (along with other roles such as making results more purely diffusion controlled, etc).
Put simply, the SE ions do not increase conductivity directly by contributing to the current (like an electroactive ion would), but increase conductivity by increasing the rate of faradaic processes of the actual electroactive species that are present (by using the SE ions to maintain electrical neutrality around the electrode).
Is this it?
Again, thanks for your time!

Remember that in electrochemistry whenever a direct current is passing through the solution there must be electrolysis occurring at the electrode. Now, all you want to see in electrochemical experiment, is current due to the oxidation or the reduction of the desired solute which is diffusing to the electrode in an unstirred cell.
The gist of your excellent and thought provoking question is:

My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

After giving it a little thought and some literature search, it is clear that supporting electrolytes increase conductivity is an erroneous conclusion which is rampant everywhere. In the electrochemical experiment, the flux of the supporting electrolyte must be zero, it should not be carrying any current, because by definition, it is non-reactive in that potential window. The mass transport of the electro-inactive species must be zero.
*Yes, separately, if we were to measure the conductivity of a solution to which an supporting electrolyte, the conductivity would be higher than the pure solvent..but that is a separate measurement. Remember that conductivity is measured by very high frequency AC current.
The book, Electrochemical Engineering by Thomas F. Fuller, John N. Harb, explains it better, pg 69:

It is not easy to understand what you are looking for. Maybe an example could help. Suppose you are using a solution of sodium sulfate $\ce{Na2SO4}$. The corresponding ions $\ce{Na+}$ and $\ce{SO4^{2-}}$are not discharged at the electrodes. On the contrary $\ce{H2O}$ is reacting at both electrodes. Let's start with the cathode.
At the cathode, $\ce{H2O}$ reacts with electrons coming from the outer cell, and produces $\ce{H2}$ and $\ce{OH-}$ ions. But this reaction can only happen if there is enough positive ions around the cathode to realize the electric neutrality of the solution. That is where $\ce{Na+}$ intervenes. Forget about $\ce{H2}$. If there is enough $\ce{Na+}$ at the cathode, $\ce{H2O}$ can react and produce $\ce{OH-}$ ions in solution. The rate of this reaction is proportional to the concentration of $\ce{Na+}$ ions in solution.  This is why the presence of the electrolyte increases the reaction rate. Simultaneously, the negative ions $\ce{SO4^{2-}}$ are repelled by the cathode and attracted by the anode.
At the anode, $\ce{H2O}$ reacts in a similar manner to produce electrons. To do it, $\ce{H2O}$ is decomposed and produces $\ce{H+}$ ions in solution (or $\ce{H3O+}$)  and $\ce{O2}$ gas. But this production of $\ce{H+}$ ions can only happen if there are enough negative ions in solution near the anode. If there are no negative ions, the first $\ce{H+}$ ions repell the next ones, and the reaction stops. This is where the sulfate ions intervene. The $\ce{H+}$ ions attract the $\ce{SO4^{2-}}$, and so the neutrality of the solution is maintained. The rate of the reaction is proportional to the concentration of the sulfate ions.
As a consequence, there is a general movement of positive ions towards the anode, and of negative ions towards the cathode. The amount of gases produced at the electrode is proportional to the concentration of electrolyte in solution.


Reverse electron demand Diels-Alder Why is the reverse electron demand Diels-Alder reaction much rarer than the normal one? By reverse electron demand, I mean one with an electron-rich dienophile and an electron deficient diene.

In the book by Clayden they have drawn a diagram with the electron deficient dienophile lower in energy and saying that it would prefer to give its LUMO to the reaction.

But, I don't get what's wrong with an electron-rich dienophile lending its HOMO.

Inverse electron demand reactions are inefficient and slow. Even the prototypical Diels-Alder reaction of 1,3-butadiene and ethylene is quite sluggish. From here you can see the reaction only proceeds to 93.5% yield when ran with ethylene in excess by 50:1 at 200ºC and 350 atm for 4 hours. Yikes!
Now consider that some normal electron demand D-A reactions can occur at room temperature. In fact, 1,3-butadiene spontaneously dimerizes to 4-vinylcyclohexene while in storage, meaning the dimer must be cracked by heating immediately before use. This should give you an idea of the sensitivity of D-A reactions to substituent groups. By lowering the HOMO-LUMO gap, orbital overlap is improved, lowering the activation energy and resulting in a faster reaction.
To supplement and support my answer, here is a paraphrased excerpt from Modern Physical Organic Chemistry by Anslyn and Dougherty (2006):

Electron-withdrawing groups on the dienophile lower its LUMO. This occurs because electronegative substituents have correspondingly lower energy MOs, and lower the energies of all MOs in which they participate. Inversely, electron-donating groups raise the diene MOs.

In support of this analysis, the log($k$) for Diels-Alder reactions correlate quite well with the inverse of the difference in energy between the ionization potential of the diene (related to the energy of the HOMO) and the electron affinity of the dienophile (related to the energy of the LUMO). To further enhance the electrophilicity of the dienophile, it is common to add Lewis acids that can complex the electron withdrawing groups on the dienophile, further lowering the LUMO.

In order to understand why the HOMO of the dienophile is less easily donated, let us again understand the original Diels-Alder properly.

In the original Diels-Alder reaction, there is a close similarity in the energies of the HOMO of the diene and the LUMO of the dienophile. As a result, they form the strongest orbital interactions. The presence of electron releasing groups on the diene and electron withdrawing groups on the dienophile essentially reduce this energy difference even further.

Moving on to the inverse electron-demand Diels-Alder: here, the LUMO of the diene and the HOMO of the dienophile are closest in terms of their energies. Because of this, they form the strongest orbital interactions now, and hence this reaction is able to take place.

As you might have already guessed, the chances of there being a close similarity in the energies of the LUMO of the diene and the HOMO of the dienophile is much less compared to the normal Diels-Alder case. Because of this, we have a greater number of reagents that can undergo the normal Diels-Alder than the inverse one. It's just a question of its rarity than its feasibility, because now you're looking for an electron deficient diene and an electron rich dienophile. In general, due to the resonance stabilisation in dienes, there's a good deal more electron rich dienes than deficient ones, and vice versa for the dienophiles...

I hope that this answers your question.


Effect of reduction potential on particle size of chemically precipitated silver particles Is there no mathematical relationship between redox potential and particle size saying that an $X$ potential should yield a $Y$ particle size? Right now its general that large redox potential results in smaller particles, but I want to know if, for example, for silver particles I use $\ce{AgNO3}$ with glucose reducing agent and say the redox potential is $\pu{0.33V}$, what the ballpark for the particle size should be based on that potential (assuming $\mathrm{pH}$ does not change)?

Here is an example of a study of fish oil nanoemulsions which examines, to quote:

  The influence of surfactant-to-oil-ratio on particle size and physical stability was evaluated.

Here is yet another study to quote:

  Although the role of surfactant in the synthesis of particles in the nanometer range has frequently been documented...The surface coverage per surfactant molecule could be related to the surface tension of the latex, thus providing a relationship between particle size and stability.

And, another more expansive work to quote:

  It was reported that the formation process of these NPs could be divided into three stages: growth, incubation and Oatwald ripening stages. The higher boiling point of 300 °C of paraffin affords a broader range of reaction temperature and makes it possible to effectively control the size of silver NPs by varying the heating temperature alone without changing the solvent. Moreover, the size of the colloidal silver NPs could be regulated not only by changing the heating temperature, or the ripening time, but also by adjusting the ratio of oleylamine to the silver precursor.

While chemical reduction methods are connected to reduction potential, the multi-stage aspect of the creation process for NPs suggests only a partial correlation, so not a 'mathematical relationship', but perhaps a statistical one.


kJ per mole or kJ per 2 moles? From Chemguide:

Is the measurement unit $\ce{kJ*mol^{-1}}$ correct here? Shouldn't it be measured simply in kJ, because the $-572$ figure is for the whole reaction, i.e. for 2 moles of the product?

the whole reaction, i.e. for 2 moles of the product?

This is your mistake.

The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed.

The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.

That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.

What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that

If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.

That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.

  So what is the per mol for

As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.

However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.

That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.

What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.

Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be

$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$

You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:

$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$

then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$

Note how the answer comes out to be the same.

But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use

$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$

From Wikipedia of Standard Enthalpy of Reaction:

  The standard enthalpy of reaction (denoted $\Delta H_\mathrm{r}^\Theta$) is the enthalpy
  change that occurs in a system when one mole of matter is transformed
  by a chemical reaction under standard conditions.

For the above reaction, the stoichiometry of one mole is taken into consideration when one mole of hydrogen and half a mole of oxygen react to form one mole of Water.

  For a generic chemical reaction

  $\ce{$−v$_{A} A + $−v$_{B} B +\ \dots -> $v$_{P} P + $v$_{Q} Q\ \dots}$

  the standard enthalpy of reaction $\Delta H_\mathrm{r}^\Theta$ is related to the standard
  enthalpy of formation $\Delta H_\mathrm{f}^0$ of the reactants and products by the
  following equation:

$$\Delta H_\mathrm{r}^\Theta = \sum_B v_B \Delta H_\mathrm{f}^\Theta \left( B \right)$$

You can read more from here — Standard Enthalpy of Reaction


The meaning of permanent dipole moment I was asked in one question that which of all given molecules have a permanent dipole moment?

And $\ce{XeF6}$ was given as a correct answer quoting that it is a distorted octahedral, and hence the lone pair results in the molecule having a dipole moment. But since we know that the lone pair goes at each face of the octahedral one by one, the net effect should cancel out. Hence, is it correct to say that the molecule has a PERMANENT dipole?

I am having similar doubt in the case of the very fundamental ammonia example. Since it undergoes inversion do we say that it has a permanent dipole?

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1):

The abstract of Ref.1 tells them all:

  According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential scanning calorimetry, $\ce{XeF6}$ exists in at least six different modifications. Three of them are formed at temperatures above room temperature, one exists at room temperature, while two have been found at low temperatures. In the high temperature modifications $\ce{XeF6}$ forms a non-symmetric tetramer, better described as a cyclic trimer with a weakly associated monomer. The normal temperature modification is the previously described cubic phase IV, having disordered tetrameric and hexameric units. The low temperature modifications are regular tetramers. Only in presence of $\ce{HF}$ symmetric dimers are formed.

The tetramer $\bf{C}$ (ball and stick version of $\bf{A}$) represent the general version of all modifications, a cyclic trimer with a weakly associated monomer. The symmetric dimer $\bf{D}$ is the structure of $\ce{XeF6}$ in the presence of $\ce{HF}$ (highlighted by $\color{green}{\text{green}}$ ovals). Nonetheless, all of these modifications display distorted octahedral structure for individual $\ce{XeF6}$ molecule as represented by ball & stick 3D structure in the central box. None of $\ce{F-Xe-F}$ angles are equal to $90^\circ$, which is the normal angle of central atom with two adjacent ligands in regular octahedron. As a consequence, $\ce{XeF6}$ has net permanent dipole moment since all six $\ce{Xe^\delta+-F^\delta-}$ bonds are polar.

For your question about ammonia, look at Ivan Neretin's comment and here where there is an accepted answer.

References:

Sevim Hoyer, Thomas Emmler, Konrad Seppelt, “The structure of xenon hexafluoride in the solid state,” Journal of Fluorine Chemistry 2006, 127(10), 1415-1422 (https://doi.org/10.1016/j.jfluchem.2006.04.014).

The distorted octahedral structure of $\ce{XeF6}$ produces a permanent dipole.
The key word is distorted structure, which results from the $\ce{Xe}$ lone pair).

If a structure is distorted, this results in a (lowering) breaking of symmetry.
If a molecule is unsymmetrical with strong polarising groups, there will be a local permanent dipole moment associated with it. In a general sense, you can think of bonds as vectors.

Now in your $\ce{XeF6}$ example:

Consider the direction of the bonds as vectors.
Then calculated (or visually consider) the sum of those vectors.
If the sum of all vectors is not equal to 0, then you have a dipole moment.

Now consider an perfectly octahedral molecule. You will find that the vector sum is now equal to zero, and therefore the molecule does not contain a permanent dipole.

Now for your $\ce{NH3}$ example. $\ce{NH3}$ has a distorted tetrahedral shape, where the bonds are forced away from the lone pair. The mistake is to assume bonds and lone-pairs are the same. The electron negative charge within the lone pair is held closer to the centre of the atom within the lone-pair orbital, and so repels the electrons in the bonds to a greater extent.

To contrast $\ce{:NH3}$ (here, $\ce{:}$ represent lone pair) with $\ce{CH4}$ (which doesn't have a permanent dipole). All bonded vectors in $\ce{CH4}$ are equal (same atom-bond type) and are evenly distributed.


Why hydrogen bonds are stronger than Van der Waals forces? Why the hydrogen bonds are more intense than Van der Waals forces in the case where the molecules with the Van der Waals forces have a stronger dipolar moment than the molecules with the hydrogen bond?
For example, ethanol (hydrogen bond) has a boiling point of 78 ºC and ethanal 20 ºC (Van der Waals forces) but their dipole moments are respectively 1,69 D and 2,69 D.
I thought the intensity of the hydrogen bond was due to the high polarization of the bond between a small and very electronegative atom like N, O, F, but this is not the case in the previous example, as I understand.

Distance
In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ... O=C interaction. For one, the N-H group has a larger dipole moment than the C-H group. For another, the distance between the donor and acceptor are larger for the weak C-H hydrogen bond.
When the dipoles of two ethanol molecules approach, they can come fairly close because the hydrogen on the hydroxyl group lacks inner electrons. For ethanal, in contrast, the carbonyl carbon of one molecule has to come close to the oxygen atom of the other. Both atoms have filled inner shells (and more electrons in the outer shell), so that approach is limited by electronic repulsion.
Covalent character
The hydrogen bond has covalent character, explaining why it is a directional interaction. This covalent character (like all covalent bonds) can not be fully explained using classical (Coulomb interaction) arguments. So in a hydrogen bond, we have an additional attractive term that is lacking in pure dipole-dipole interactions. In extreme cases, you might think of the hydrogen bond as a 4 electron-3 center bond (the 4 electrons are the lone pair of the acceptor and the bonding electrons of the covalent bond of hydrogen with the rest of the molecule; the three atoms are the hydrogen, the electronegative atom it is covalently bound to, and the acceptor atom).
Nomenclature
Some use Van der Waals forces as an umbrella term for (permanent) dipole-dipole, dipole-induced dipole, and temporary dipole-induced dipole interactions. Others use Van der Waals forces (or dispersion forces or London interactions) as a separate and weaker kind of interaction and contrast it with the stronger dipole-dipole interaction.

Dipole moments are considered only in the absence of intermolecular hydrogen bonding.
Hydrogen bonding is much stronger than other dipole-dipole interactions (Van der Waals forces of attraction). Then energy liberated by dipole-dipole interactions pales in comparison to that liberated by hydrogen bonding. The stronger dipole moment argument is used where neither of the molecules that we are comparing has any form of hydrogen bonding (inter or intra).
Hope it helps!


Why does Co2+ have 7 electrons in the 3d orbital, and not 5 like Mn? I'm taking general chemistry 2 this half of the summer. We are currently going over coordination complexes, ligands and transition metal ions. In particular, this question is in regards to the ligand field stabilization energy.

Why does $\mathrm{Co}^{2+}$ have 7 valence electrons in the $3d$ orbital? Doesn't the superscript $^{2+}$ denote that it is missing 2 electrons, and therefore isoelectric with $\mathrm{Mn}$?

Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ here and there, but at the end of the day, you just need to * memorize it*.

  When d-block elements form ions, the $4s$ electrons are lost first. Source

The $n+l$ rule tells you the order in which atomic orbitals are filled, and according to the rule the $4s$ orbital is occupied before the $3d$ orbital because it has lower energy.
Thus, the electron configuration of $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$.
But, the $n+l$ rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration.
For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.

As it is sometimes explained, the statement that $4s$ orbital is lower in energy than $3d$ orbital is true only when the orbitals are unoccupied. But while you fill $3d$ orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the $4s$ orbital. Thus, when electrons are lost from $\ce{Co}$ atom, they are lost from the 4$s$ orbital first because it is actually higher in energy when both $3d$ and $4s$ are filled with electrons.

P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.

Electronic configuration of $\ce{Co}$ is as follows: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^7$.

When Cobalt loses 2 electrons to become $\ce{Co^{2+}}$ it loses the electrons which are in $4s^2$, not the ones in $3d^7$ because the electrons in $4s^2$ have a high reactivity.

So $\ce{Co^{2+}}$ still contains 7 electrons in it's valence shell. The new configuration of $\ce{Co^{2+}}$ is
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^7$.


How can semi-permeable membrane be so smart? According to the definition, a semi-permeable membrane is a membrane that allows only solvent molecules to pass through it; solute molecules are blocked.  Solvent molecules flow from the membrane side that has low solute concentration (i.e. high solvent concentration) to high solute concentration (i.e. low solvent concentration).

Lets take a case:

I take oil as a solvent in one side and water as a solvent in other side. I take the same solute in both sides.
Initially, the water side is filled to height $h$ and the oil side to height $2h$. The amount of solute is same in both the sides. So the solute concentration is more in water than in the oil.  Accordingly, oil will flow from the oil side to water side. Now, I add more water on the water side to reduce the solute concentration on the oil side.  Accordingly, water should now flow towards the oil side.

So what will happen? Will this happen? What makes the semi-permeable membrane so smart or intelligent?

Most of the time what you say about semi-permeable membranes is correct.

  a semi-permeable membrane is a membrane that allows only solvent molecules to pass between the membrane; solute molecules are blocked. Solvent molecules flow from the membrane side that has from low solute concentration (i.e. high solvent concentration) to high solute concentration (i.e. low solvent concentration).

However, you have to keep in mind several assumptions that are inherent in the definition as you have given it.  The most important of these is that thermodynamic activity differences in the membrane-permeable solvents determine the membrane behavior.  When phases are immiscible, such as the water/oil system in your example, it is essential to think in terms of activities, not in terms of concentration.

  I take oil as a solvent in one side and water as a solvent in other side. I take the same solute in both sides.

So we are imagining a solute that can be dissolved by both water and by oil.  What do we know about the solubility of the solute in these two solvents?  What is the partition coefficient?  And what can we say about the thermodynamic activity difference across the membrane for the three species?  Let's go step by step:

Water.  Water is not soluble in oil to an appreciable degree.  In equilibrated phases, the thermodynamic activity of each chemical species is the same in each phase.  Therefore in an oil-water mixture, the thermodynamic activity of water in the aqueous phase is the same as the thermodynamic activity of water in the oil phase. The extremely low solubility of water in oil means that if there were to be a difference in thermodynamic activities across the membrane, only a few molecules would need to cross over for the condition of equal water activities to be satisfied.
Oil.  Oil is not soluble in water to any appreciable agree.  Therefore the same argument applies.
Solute.  The solute could have a different concentration on each side of the membrane, but because we don't know the partition coefficient, we don't know what the equilibrium state is.  Suppose the solute were membrane permeable.  Even if it was twice as concentrated in water than in oil, if it's partition coefficient favored an equilibrium where the oil-to-water concentration ratio was 10, then it would flow from the water to the oil.  But, in your example the solute is not permeable to the membrane.

  Initially, the water side is filled to height $h$ and the oil side to height $2h$. The amount of solute is same in both the sides. So the solute concentration is more in water than in the oil.

Now hopefully the discussion above makes you realize that there aren't many chemiosmotic forces in your example.  The only driving force left is hydrostatic pressure.  So the oil will cross the membrane, not the water, just because of hydrostatic pressure differences.  This is regardless of the solute concentration.

  Now, I add more water on the water side to reduce the solute concentration on the oil side. Accordingly, water should now flow towards the oil side.

Well adding more water raises the height $h$ to some higher height and thus increases the hydrostatic pressure.  So the increased pressure forces water back through the membrane.

Semi-permeable membranes such as cell membranes use active transport to move things AGAINST a concentration gradient (Or through a membrane that would otherwise be impermeable), using energy to do so. You are right, for many materials, things simply balance out like your example. However, in active transport, only some materials are taken in.
Here's an article about active transport: https://en.wikipedia.org/wiki/Active_transport


Why does the solubility of Group II hydroxides increase and the solubility of sulphates decrease down the group? I know that, solubility of alkaline earth metal hydroxides increases down the group and solubility of alkaline earth metal sulfates decreases down the group.

I wish to understand the reason for these trends.

Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy.
Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy.

Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy.

Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group.

Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.

The solubility of Group 2 hydroxides increases down the group because as you go down the group the lattice energy decreases down the group and the same goes to enthalpy change of hydration causing the increase in solubility.


Identifying the number of bonding and lone pairs without a dot and cross diagram I've just learned how to predict the shapes of molecules in class today using VSEPR theory.
I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?

There is a simple, four-step calculation that you can perform:

count the valence electrons the atoms you are bonding have.
count the number of valence electrons the atoms would like to have for a noble gas’ valence shell. (i.e. eight for everything main group, two for hydrogen.)
substract $2.-1.$, i.e. the first (existing) from the second (desired). This is your number of bonding electrons. Divide by two for bonding electron pairs.
substract $1.-3.$, i.e. the third (bonding) from the first (existing). This is your number of free electrons. Divide by two for lone pairs.

Using a simple example such as sulphur dioxide:

$6\ (\ce{S}) + 2 \cdot 6\ (\ce{2 O}) = 18$
$8\ (\ce{S}) + 2 \cdot 8\ (\ce{2 O}) = 24$
$24 - 18 = 6$, i.e. three bonding electron pairs.
$18 - 6 = 12$ i.e. six lone pairs.

Unfortunately, that by itself does not allow you to write the structure; you need to know which element is in the centre and how they are connected. Here, we have a central sulphur bonded to two oxygens. The final result is:

$$\ce{O=S+-O-}$$

With two lone pairs on the left-hand oxygen, one on the sulphur and three on the right hand oxygen. The formal charges are important.

There is a very simple way to do it, but it is not recommended at all.
Also there are a lot of assumptions in this method.

Detect the central atom along with the peripheral atoms.
For Example:- In Methane, the Carbon atom is the central atom.

Count the valance electrons of the central atom and the peripheral atoms.
Exception: For Hydrogen , using "1" won't yield results. So specifically for Hydrogen use "7". This is one of the assumptions we have to make.

Divide the above value obtained by 8. The quotient gives the value of the number of sigma bond pairs and the remainder divided by 2 gives the number of lone pairs.

Explanation:
The number of valance electrons counted divided by 8 will give the number of sigma bonds formed. This is just like counting the number of atoms which are getting complete octets, i.e. forming sigma bonds to get octet complete and becoming stable. The remainder is the no. of non bonded electrons, which when divided by 2 gives the number of lone pairs.
Example:
In Sulfur tetrafluoride:
Number of Valence electrons$$1\times(6)+4\times(7)=34$$
No. Of Bond Pairs$$\frac{34}{8}=4$$
No. Of Lone Pairs$$\frac{34\%8}{2}=1$$
To be very clear, this method is not fool proof. It is just a workaround to save time if you do not want to make the structures as it works on majority of compounds.