Is it known for sure that bases feel slippery because of the production of soap/surfactant? Discussion around the question Why does bleach feel slippery? has started me thinking about the saponification explanation for the slippery feeling of basic solutions.
According to Wikipedia:

Alkaline drain openers can dissolve hair (containing proteins) and fats inside pipes via alkaline hydrolysis of amide and ester respectively:
$$\ce{RCONH2(amide or proteins)+ OH− → NH3 + RCOO−}$$
$$\ce{RCO2R’(ester or fats)+ OH− → R'OH + RCOO−}$$

and just above that:

Essentially, the hydroxide ions from the basic lye attack the carbonyl carbons of the fat, which eventually kicks off the hydrophobic tails of the triglyceride (tristearin/fat) to isolate glycerol.

So I am not sure if the "saponification" explanation for why a small amount of a fairly strong base feels slippery between thumb and forefinger really says that the reaction has produced a true soapy surfactant or that it should really say that it is lubrication by glycerol-like compounds that generate the intense slippery feeling.
Wikipedia also says:

The Stratum corneum (Latin for 'horny layer') is the outermost layer of the epidermis, consisting of dead cells (corneocytes). This layer is composed of 15–20 layers of flattened cells with no nuclei and cell organelles. Their cytoplasm shows filamentous keratin. These corneocytes are embedded in a lipid matrix composed of ceramides, cholesterol, and fatty acids. (emphasis added)

Corneocytes are keratinocytes and named as such because of the abundance of keratin protein filaments they contain.
I seem to remember my very smart high school chemistry teacher telling us that the breakdown of protein was also important, but I don't remember the production of soap as being so central. So I'd like to ask if there is any actual empirical evidence - ideally with a scientific reference, that demonstrates which molecules present between the fingers in this situation is primarily responsible for the slippery feeling. Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing inter-cellular lipid layers that would be exposed by the removal of successive layers of cells within the stratum corneum.

below: Epidermal layers, from here.

Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing [inter-cellular lipid layers][5] that would be exposed by the removal of successive layers of cells within the stratum corneum.

Very good guesses but none of the above actually (i think, what does "finger lipids" mean?). But the reason bases feel slippery to bare skin no matter how much you scrub first, is bases attack the phospholipid bilayer of your cells quite readily not only saponifying them, but breaking the inter-molecular bonds of molecules that already resemble a soap. Take for example phosphatidylcholine:

The phosphochloline already has all of the elements to make a soap and when base is added, the molecule does not need to interact with other molecules to achieve local charge balance, but now have the ions from the base to balance, which breaks intermolecular bonding creating a small layer of soap, making the base feel slippery.


Safe silver electroplating solutions Are there any practical, sane and safe solutions that can be used for silver elctroplating/electrolysis that are purchasable?

Objective is to create silver crystals, but without having evil cyanide floating around the house... copper I can handle with copper sulphate.  Silver has me stuck...

Any ideas?

I've never tried to grow big crystals or anything, but we commonly use 30 mM silver nitrate in 1 M $\ce{NH4OH}$ for electrodepositing silver onto metal (usually gold) electrodes.
If you don't care what kind of surface finish you end up with, it will work, but it makes a fairly rough film that can be somewhat easily removed with abrasion.
The tricky part is trying to get very smooth and adherent films.
It seems that there are still not any good replacements for electroplating from cyanide-containing solutions, for decorative or engineering purposes where the deposit will be subject to wear or has to look nice.


Is the Kekule structure of benzene completely wrong? I've found out about something called resonance but I'm getting the impression that this indicates the kekule structure is possibly correct as it uses double bonds (I thought that the delocalised model was the only correct one).

I'm a little confused on this so could someone clarify?

In benzene, all of the carbon-carbon bond lengths are equal.  Therefore, the Kekule structure shown below is an incorrect representation of benzene.

It is incorrect because it suggests that there are two different types of carbon-carbon bonds in benzene, a carbon-carbon double bond and a carbon-carbon single bond. Typically, the bond length in a carbon-carbon double bond is around 135 pm, a carbon-carbon single bond is around 147 pm long, while the carbon-carbon bond length in benzene is roughly in between at 140 pm.

Benzene is often drawn as follows to indicate that all of the carbon-carbon bonds have the same length and the pi electron system is continuous and uniformly distributed.

Sometimes, when one is discussing the resonance properties of benzene, two Kekule structures are drawn with the double headed resonance arrow between them as shown below.  This is meant to indicate that these two Kekule structures are resonance structures of benzene, and that taken together (e.g. in one Kekule structure a carbon-carbon bond is a single bond, in the other structure the same bond is a double bond; so "real" benzene is a simultaneous, superposition of these two structures where each carbon-carbon bond is in between a single and double bond), they represent the true electronic and physical structure of benzene.


How to determine if a given reaction requires heating to initiate? At room temperature, some reaction propagate merely upon intermixing the reagents (e.g. autoignition of triethylborane), while others require additional heat to initiate (e.g. crystallization of calcium chloride).

How can I determine if a given chemical reaction would require heating above room temperature to initiate? And to what temperature?

The ease of initiation of a reaction depends on the activation energy required. In order for reactions to proceed (at a reasonable rate), the forces binding the individual reactants must be overcome, which may require energy, solvent, catalyst or other input.

For example, $\ce{H2O2}$ is quite stable at room temperature, but exposure to horseradish peroxidase, silver oxide or other catalyst causes rapid decomposition. Iron filings do not oxidize quickly, but apply a bit of salt water and they rust so quickly they make an effective heater. $\ce{NI3}$ has an extremely low activation requirement!


Why is the reason for color different in these two cases? I have read in a book (NCERT India) about flame color for salts of alkali metal and color of transition metal complexes.

In the first case, it says that during excitation of electrons to higher energy it absorbs energy and then when it falls back it is emitted again giving color to the flame.

In the second case, it says that during d-d transition it absorbs a wavelength of photon and complementary color is shown. So, why can't there be emission in the second case like the first case.

These at two different scenarios and two different phenomena. Flame color is due to atomic emission and complexes are colored because of light absorption in the solid state and in solution.

In a high temperature flame, an atom is considered to be a free and independent entity. With heat, the electrons in the atoms are excited to higher energy levels. However, selection rules determine where an electron can land e.g. s electron goes to p, p electron can go to d, however s electron cannot go to d.

In case of transition metals, Hans Bethe (Nobel Prize) developed a theory called called crystal field splitting. In a free gaseous atom, all five d orbitals orbitals have the same energy (called degenerate orbitals), so electronic transitions do not make any sense. It is also forbidden. However in the presence of ligands, the d- orbitals are no longer of the same energy. Electronic transitions are possible within d orbitals. Search crystal field theory if you are interested...


Minimum heating temperature for dehydration of Cobalt Chloride I've been looking on line and have found several sources that state 100°C for a rapid dehydration of cobalt chloride in its various forms. I'm wondering if that's a minimum, or would it be possible to dehydrate it at a much lower temperature, provided ambient humidity is controlled.

The minimum I could find is 110℃(Chemical forums) although not verified

Next is 150℃ from this link. Although, two temperature are written(110℃ and 150℃).

  mp 86 °C; loses $\ce{6 H2O}$ at 110 °C

  Anhydrous cobalt chloride is blue in color. It can be prepared by
  dehydration of the hexahydrate either at 150–160 °C in vacuo.

Next is 160-170℃. (Chemiday)

  $$\ce{CoCl2•6H2O ->[\Delta] CoCl2 + 6H2O}$$
  The thermal decomposition of hexahydrate cobalt(II) chloride to
  produce cobalt(II) chloride and water. This reaction takes place at a
  temperature of 160-170°C in a flow dry $\ce{HCl}$.

Next is 200℃. From this paper:

  Thermal deaquation of cobalt chloride hexahydrate has been subjected
  to investigation by numerous authors because it is one of the salts
  that shows distinct steps in the loss of water of crystallization in
  the ambient temperature range  of  200℃.


Half reaction method justification I have to say I have never really understood the half reaction method. What bothers me is that it seems to be too artificial. Too many little rules to remember!
Why don't we use the mass conservation law to justify the process and use sistematically a set of linear equations to do the balances?

Thanks in advance!

The concept of the half reaction has nothing to do with balancing redox reactions; how one arrives at a balanced reaction is not really important. The interesting thing about redox reactions is that electrons do not necessarily have to be transferred directly between the oxidized and reduced species. In many types of electrochemical cells, the anode and cathode are isolated from one another by a salt bridge and the species in each half-cell never approach one another. In techniques such as potentiometry, often one cares only about one half of the reaction, e.g. in a combination pH electrode, the response from the half reaction that contains hydronium ions produces the analytical signal and the the other half is a reference electrode reaction whose job is to simply produce a constant potential. In things like voltammetry, one is often only concerned about the half reaction occurring at the working electrode and the half reaction that balances it at the counter electrode isn't analytically significant (and may in fact be more than one different reaction).

Because the two half-reactions need not necessarily occur at the same place and one half reaction can generally be substituted without affecting the other, it makes sense to think of them separately. This lets us do things like think about different lithium batteries. There are many types and they all use the $\ce{Li -> Li+ + e-}$ oxidation half-reaction, but it's coupled with different reduction half-reactions based on cobalt, iron, etc.


Determine the ΔH for the reaction in kJ/mol of ethanol? A $1.000~\mathrm{g}$ sample of ethanol, $\ce{C2H5OH}$, was burned in a bomb calorimeter whose heat capacity had been determined to be $2.71~\mathrm{kJ/^\circ C}$. The temperature of $3.000~\mathrm{kg}$ of water rose from $24.284~^\circ\mathrm{C}$ to $26.225~^\circ\mathrm{C}$. Determine the $\Delta H$ for the reaction in $\mathrm{kJ/mol}$ of
  ethanol?

I said :

\begin{aligned}
q_\mathrm{cal} &= C_\mathrm{cal}\cdot T\\
        &= (2.71\ \mathrm{kJ/^\circ C})(1.94\ \mathrm{^\circ C})\\
        &= 5.26~\mathrm{kJ}\\
\end{aligned}

Then the $q_\mathrm{r}=-5.26\ \mathrm{kJ}$

Then calculated the number of mol by dividing the mass g over molar mass and I got $0.02171~\mathrm{mol}$.

Finally I divided $-5.26\ \mathrm{kJ}$ by $0.02171\ \mathrm{mol}$ and I got $= -242.3~\mathrm{kJ/mol}$.

But the correct value is $-1365~\mathrm{kJ/mol}$. What's wrong?


Reaction mechanism for 7-hydroxy-4-methylcoumarin synthesis with an iodine catalyst 7-Hydroxy-4-methylcoumarin can be synthesized from ethyl acetoacetate and resorcinol with iodine as a catalyst. What is the mechanism for this? From what I understand it follows the Pechmann reaction but I'm having trouble to understand how it works in this particular case.

There is a paper about this exact reaction, but in addition to the use of $\ce{I2}$ it mentions the use of microwaves. In the reaction, $\ce{I2}$ acts as a Lewis acid, catalyzing both the transesterificaiton and electrophilic aromatic substitution reactions. It can be seen acting as such in the first step of the mechanism:

$\hspace{2cm}$

Next, a proton transfer occurs:

Next, the resorcinol derivative can act as a nucleophile, and ethanol is ejected, completing the transesterification reaction:

Next, $\ce{I2}$ coordinates to the keto carbonyl, activating the carbon towards electrophilic addition. Because this resorcinol derivative is so activated by the $\ce{-OH}$ and $\ce{-OOCR}$ groups, a weak Lewis acid like $\ce{I2}$ is sufficient to catalyze the reaction:

$\hspace{1.5cm}$

$\hspace{1.8cm}$

$\hspace{1.6cm}$


NaHCO3 required to raise pH of weak HCl solution from 3.5 to 7.2 (doubting my calculations) I'm a dentist, working on a medical device that will make shots at the dentist office hurt a lot less. The device mixes a drug (2% Lidocaine) that has been acidified with $\ce{HCl}$ to pH 3.5 for storage, with $\ce{NaHCO3}$ to bring it to physiological pH (about 7.2) just prior to injection.

The current method is by manually mixing an 8.4% $\ce{NaHCO3}$ solution with the drug in a 1:9 ratio V/V. In medicine, an 8.4% solution typically means 84mg of drug / ml of solution. So, the current protocol of mixing 100 ml of bicarb solution with 900 ml of drug solution, means the doctors are delivering 8400mg of $\ce{NaHCO3}$ /L , which is a LOT more than the 27mg that my current calculations (please see below, and please let me know if I'm off on the wrong track) say they need to do the job. You can see why I was (and am) skeptical of my calculations.

Assuming the calculations below are correct, then the next problem is:

If I want to deliver the $\ce{NaHCO3}$ in the form of a fluid (whether a solution, slurry, or any other way to make it flow), what is the best way to do that, and how much $\ce{NaHCO3}$ can I fit into a vessel with volume of 35 cubic millimeters (pretty small, and all the room I have available in my device)? In other words, what is the maximum concentration of $\ce{NaHCO3}$ that will stay in solution at a standard temperature and pressure with no head space in the vessel? Or, can I make a slurry that will hold an even higher concentration that remains fluid, and what would that concentration be? I only need to fit 27mg/L x 2ml = 0.054mg of $\ce{NaHCO3}$ into the space, and it would be a lot easier to manufacture if it were a liquid rather than a powder.

Also, I'm in need of some physical testing of the device to confirm that mixing has properly occurred at the tip of the needle when the prototype is built in a few weeks. There is some thought that we could use excess $\ce{NaHCO3}$ in a chamber, and have bypass mixing occur as the drug passes through the chamber and into the needle, instead of trying to get the whole volume (2ml) mixed prior to moving any drug out of the needle. If there is someone on here that might be interested in performing those validation experiments to help pass FDA gates, then please let me know how to find you. Thanks, Scott

The chemical reaction: $\ce{HCl + NaHCO3 → NaCl (aq) + CO2(g) + H2O(l)}$

The initial hydrogen ion concentration is: (pH = 3.5) [H+]i = 10-3.5 = $\pu{3.16228 x 10^-4 M}$

The required final hydrogen ion concentration is: (pH = 7.2) [H+]f = 10-7.2 = $\pu{6.309573 x 10^-8 M}$

moles $\ce{H+}$ reacted = moles of $\ce{HCl}$ reacted = $(\pu{3.16228 x 10^−4} − \pu{6.309573 x 10^−8}) = \pu{3.1616467 x 10^−4 mol}~\ce{HCl}$

moles of $\ce{NaHCO3}$ required = moles $\ce{HCl}$ reacted = $\pu{3.1616467 x 10^−4 moles}$

mass of $\ce{NaHCO3}$ required = $\pu{3.1616467 x 10^−4 moles}~\ce{NaHCO3} \times \pu{84g}~\ce{NaHCO3} / \pu{1 mole}~\ce{NaHCO3} = \pu{0.027 grams￼}$

The approach I would suggest is a bit different.

First I would calculate the nominal concentration of HCl (i.e. the number of moles of HCl added, divided by the total volume of the solution after the addition, regardless of what happens to the resulting chloride ions and protons) in your initial solution, using all the necessary equations (with the approximation that activity equals concentation, although the figures are a bit large here).

1) dissociation constant of lidocaine:

$K_{a,lido}=\frac{[L] \cdot [H^+]}{[LH^+]}$

2) the total nominal concentration of lidocaine must be the sum of all its forms:

$C_{lido}= [L]+[LH^+]$

3) charge balance:

$[Cl^-] + [OH^-] = [H^+]+[LH^+]$

4) the total concentration of chloride is the same as the nominal concentration of HCl, as the dissociation is complete and $Cl^-$ does not react with anything else:

$C_{HCl}= [Cl^-]$

5) the ion product of water:

$K_w=[H^+] \cdot [OH^-]$

The system of these 5 equations can be solved for $C_{HCl}$, after eliminating all chemical species (except $[H^+]$, so in total 4 unknowns):

$C_{HCl}= [H^+] - \frac {K_w}{[H^+]} + \frac {C_{lido}}{1+ \frac {K_{lido}}{[H^+]}}$

Plugging in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-3.5} M, K_w = 10^{-14} M^2$

results in:

$C_{HCl}= 0.08566 M$

i.e. to reach a pH of 3.5 starting from freebase, we used just slighly more than the stoichiometric amount of $HCl$. Here we are assuming as well that the added volume of HCl has not changed the nominal concentration of lidocaine very much. If it has, then you need to account for that.

Now, when you add sodium bicarbonate, all the equations above except 3 are still applicable; you need to add the dissociation constants for carbonic acid (I'll consider it like hydrated $CO_2$, and that none of it is escaping as gas, to keep things simple) and hydrogenocarbonate, and the mass balances for sodium and the various carbonate species.
Here they are:

$K_{a,carbonic acid}=\frac{[HCO_3^-] \cdot [H^+]}{[H_2CO_3]}$

$K_{a,hydrogcarbonate}=\frac{[CO_3^{2-}] \cdot [H^+]}{[HCO_3^-]}$

$C_{bicarb}= [Na^+]$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}]$

The charge balance now includes more ions:

$[Cl^-] + [OH^-] + [HCO_3^-] + 2 \cdot [CO_3^{2-}] = [Na^+]+[H^+]+[LH^+]$

We have now 9 equations (5-1+5). Like before, eliminating all chemical species except $[H^+]$  (i.e. 8 unknowns), gives a single expression. It is a bit complicated, but essentially it contains only $[H^+]$, $C_{bicarb}$ and $C_{HCl}$, and known constants.

In our case, as we know what the nominal concentration of $HCl$ is from the previous calculation, and the final pH we want to reach (7.2), we just need to solve for $C_{bicarb}$ and plug in the numbers:

$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-7.2} M, C_{HCl}=0.08566 M, K_{a,carbonic acid}=4.3 \cdot 10^{-7}, K_{a,hydrogcarbonate}=5.6 \cdot 10^{-11} $

resulting in:

$C_{bicarb}=0.1142 M$

So for each litre of your initial solution you should add 0.1142 mol of $NaHCO_3$, which is 9.59 g.

This is closer to the 'official' quantity you mentioned (even quite a bit more in fact), although I have trouble following all the figures. Maybe they did this experimentally and found that some $CO_2$ escapes, making the pH rise without the need for more base. Just a hypothesis.

In a way, even without all these involved calculations, one could say: if you have a 0.085 M solution of lidocaine hydrochloride and want to react all the $HCl$ with sodium bicarbonate, use $0.085 mol/L \cdot 84 g/mol = 7.14 g/L$ of $NaHCO_3$. The excess used in your official method compared to this figure is probably due to the slight excess of $HCl$ in the initial solution, and the need to take the pH closer to the desired 'physiological' value you mentioned.

EDIT: addendum to answer further questions by the OP

The OP asked what would happen by adding more sodium bicarbonate.

The answer is: as long as there is no solid precipitate of sodium bicarbonate (and always assuming that lidocaine stays in solution), the above equations remain valid; as soon as the nominal concentration of sodium bicarbonate is sufficiently high for it to precipitate, new equations apply, where all the above equations still apply, except for a correction of the mass balances related to $C_{bicarb}$, in particular by adding the contribution of the solid:

$C_{bicarb}= [Na^+] + [NaHCO_3]_s$

$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}] + [NaHCO_3]_s$

(where $[NaHCO_3]_s$ represents the solid salt expressed in the units of concentration, for homogeneity with the other terms).

And the equation for the solubility product of $NaHCO_3$ must be added:

$K_{s,bicarb}=[Na^+] \cdot [HCO_3^-]$

From the data table in this site it seems that $K_{s,bicarb}=10^{-0.55} M^2 = 0.2818 M^2$ (although from my own calculations based on a solubility of 96 g/L at 20°C it should be $1.277 M^2$, but OK, I will use the 'literature' value).

First, the limiting value of $C_{bicarb}$ after which residual solid bicarbonate is observed at equilibrium must be calculated.

This can be done by solving (numerically) the system of all 10 equations described above for all the chemical species plus $C_{bicarb}$, after setting $[NaHCO_3]_s=0$. The result is:

$[C_{bicarb}=0.5467 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

meaning that this system is only valid for $C_{bicarb} > 0.5467 M$.

Then the system can be solved using any value of $C_{bicarb}$ greater than the above threshold, and, interestingly, it gives always the same values, except for $[NaHCO_3]_s$, that increases accounting for the undissolved salt.

$C_{bicarb}=0.7 M$

$[[NaHCO_3]_s=0.1533 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$

$C_{bicarb}=7 M$

$[[NaHCO_3]_s=6.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

$C_{bicarb}=70 M$

$[[NaHCO_3]_s=69.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$

So it appears that, once solid sodium bicarbonate is present, the system is determined in a specific state, where all chemical species have a fixed concentration, regardless of how much solid there is.

In particular, the pH seems to be fixed at $7.60$. This is significantly lower than the pH of a saturated solution of $NaHCO_3$ ($8.31$) in water, probably because of the $H_2CO_3$ liberated in the reaction with $HCl$ (which, we are assuming, is remaining in solution).


How does Dalton's Law relate to scuba diving? My assignment is to create a report that emphasises on the important of chemistry more specifically the gas laws in scuba diving. I know that Dalton's Law is when the total pressure of a gas mixture equals the sum of the partial pressures that make up the mixture but i don't know how it relates to scuba diving. Can someone give me some hints?

In scuba diving it is essential to consider the partial pressures of different gases for the following reasons:

The partial pressure of oxygen should be greather than 0.16 bar to avoid hypoxia (at the surface this partial pressure is obviously 0.21 bar).
Oxygen is toxic at high partial pressure. For the dive one want to stay as close as possible to 1.3-1.4 bar (maximum), while decompression is made with maximum 1.6 bar.
Nitrogen is narcotic at high partial pressure and we usually try not to go higher to partial pressure similar to what we have at 30-35m.

As an example consider a deep technical dive to 80m (9 bar). If you have air in your tank (21% oxygen, 79% nutrogen), the partial pressure of oxygen will be
$$pp_{\ce{O2}} = p_\text{tot} \times x_{\ce{O2}} =  9\times0.21=1.89 > 1.6,$$
thus you are in serious danger because of oxygen toxicity (visual changes, nausea, anxiety, irritability, confusion, dizziness, tonic-clonic seizure, ...) which can lead to death. If you want to have an optimal oxygen partial pressure of 1.4 you will need to reduce your oxygen content to 15% (less than air). But since the partial pressure of this new mixture at the surface is 0.15 < 0.16 you should not start your dive with this tank.

In conclusion, what it is really important for your body is the partial pressure of the gas you are breathing, not the absolute pressure. Dalton's law is necessary to compute the different partial pressures.

This argument is quite advanced for scuba diving an you will have to search for technical diving in order to find the necessary informations. You may also want to look at other interesting physical phenomena that are important for scuba diving (Boyle's law, Charle's law, ...).


Can the idea of entropy be extended to orbitals? The forward reaction for: $\ce{HF<->H^+ +F^-}$ is entropically favourable; but energetically unfavorable: as there is too much electron density for $\ce{F^-}$ ion to cope up. That's a reason why $\ce{HF}$ is a weaker than $\ce{HI, HCl}$ etc.

But, can there be another explanation for weak acidity of $\ce{HF}$?

How about we introduce the idea of entropy to orbitals. Let me explain, in $\ce{HF}$ the elctron is in a molecular orbital of greater region than in $\ce{F^-}$. So, the valence electron pair is in an entropically more favorable state in $\ce{HF}$ than in $\ce{F^-}$: where it is contained in little space. That's why, $\ce{HF}$ is more favorable than   separated $\ce{H^+}$ and $\ce{F^+}$.

Maybe, for other hydracids of same kind eg. $\ce{HI, HCl}$ this effect isn't so important. That's why they are stronger acids.

Share your thoughts.

No.

The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally likely to be populated, then the number of possible microstates for the system, $W$, is large and hence $S = k \ln W$ is large.

An electron in an orbital isn't whizzing about, and an orbital most certainly isn't a container of variable size for the electron to move around in. An orbital is simply one (electronic) state of an electron.

Therefore, it doesn't make sense to talk about the translational states of an electron that is moving about in an orbital. It also doesn't make sense to talk about the translational entropy of an electron as being larger in a larger orbital.


Elimination reaction of 1-bromo-2-phenylcyclopentane In the following reaction, treating the trans-1-bromo-2-phenylcyclopentane with ethanolic KOH gives an alkene. But in this case, the double bond is on the left side (less substituted) instead of being the more substituted one.

But why does this happen? Zaitsev's rule favours the highly substituted alkene. Even if a carbocation is formed, a hydride shift would have landed it on the carbon with the phenyl group, as the carbocation would be stabilised by resonance. Does the trans configuration forbid such reaction?

There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.

$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.

$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.

Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.

As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.


Skin permeability, for all chemicals or just some? On skin permeability, according to Wikipedia:

"Human skin has a low permeability; that is, most foreign substances are
unable to penetrate and diffuse through the skin. Skin's outermost
layer, the stratum corneum, is an effective barrier to most inorganic
nanosized particles.[22][23]"

However there is little to be found on exactly what chemicals can permeate the unbroken skin.

This paper http://oem.bmj.com/content/61/4/376.full states the following:

"Many chemicals can cross the unbroken skin."

"However, the highly lipophilic nature of most solvents can also result
in dermal uptake when deposited on the skin."

Is Wikipedia's statement incorrect? What chemicals can permeate the skin, which ones cannot, and why?

"Skin's outermost layer, the stratum corneum, is an effective barrier to most inorganic nanosized particles."

But skin is quite permeable to many organic compounds. DMSO is one example, and it is warned that DMSO is a good solvent for other compounds that could be more toxic than DMSO.

Another example: we all (should) know that mercury is toxic. However, it's not going to hurt you much if you roll it around in your hand for a brief period (don't breathe in, though!). But organic mercury compounds are terribly toxic and skin permeable, including the infamous dimethyl mercury which killed a famous researcher of toxins.


If d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block? If I understand correctly, the shielding effect of d- (and f-) electrons seems to be much poorer than those of s- and p-electrons, due to the fact that they are less penetrating, have less electron density near the nucleus and so on (as can be seen from radial distribution functions).

However, trends for ionization energy, electron affinity, and electronegativity are MUCH more gradual than for s and p-blocks. This means that the effective nuclear charge ALSO increases much more gradually in the d-block, as confirmed by Slater and Clementi values. If d-electrons were poor shielders, surely this would mean that Z-eff would be much higher?

Doesn't this mean that, say for period 4, 3d-electrons are much more effective at shielding the 4s-electrons, than the 4s-electrons are at shielding 4p-electrons? Why does this happen?

Is there another factor at play? Is it because we are adding electrons to an "inner shell" that the shielding becomes more effective? Is Z-eff = Z - S an incorrect guideline to use?

Your question is similar to what I had in mind like a week back.

I know some textbooks do state a contradiction — that d-electrons have a poor screening effect, but then quickly use the explanation that due to increasing screening effect of nuclear charge from left to right in d-block elements (due to more electrons coming in the d-orbitals), trends are gradual.

Let me clear your mind of this false statement. Remember: it is not the screening effect that leads to a gradual change in trends across d-block elements, but  rather the repulsion between d-electrons and outermost s-electrons.

The follwing example from the 1st series of d-block elements should make this clear to you.
     As nuclear charge increases, electrons come in the 3d-orbitals, which is not the outermost. The outermost is the 4s orbital, already filled up with 2 electrons.

Now, from left to right in this period, the increased nuclear charge will tend to hold on to the outermost 4s electrons tighter, but the d-electrons have an opposite effect. They repel these 4s electrons, and hence net force felt by the 4s electrons decreases. So, atomic size is increases more than expected, and hence it is easier to remove these electrons too (less ionisation energy)

Hence, the increase in ionization energy, reduction is size, etc. is much more gradual while traversing from left to right in a period.


Measuring or calculating the pH of 4M NaOH? I prepared the solution in Milli Q water I prepared $4\ \mathrm{M}$ solution of $\ce{NaOH}$, and the glass electrode I use cannot measure it the $\mathrm{pH}$ correctly. What are the other options do I have to exactly know the solution's $\mathrm{pH}$?

For $\mathrm{pH}>12$ and $\ce{Li+}$ or $\ce{Na+}$ concentrations greater than $0.1\ \mathrm{M}$, glass electrodes experience alkali error. Because the concentration of $\ce{H3O+}$ ions in solution is so low, interference from alkali metal ions (like $\ce{Na+}$) becomes noticeable and causes the electrode to read lower than the true $\mathrm{pH}$ of the solution. $\ce{K+}$ typically cause less error than $\ce{Li+}$ or $\ce{Na+}$ due do its larger size.

Since sodium hydroxide is a strong base and dissociates completely into $\ce{Na+}$ and $\ce{OH-}$ ions, though, it's actually quite simple to just calculate the $\mathrm{pH}$.

$$K_\mathrm{w}=\ce{[H3O+][OH-]} \Rightarrow$$
$$ 1 \times 10^{-14} \ \mathrm{M^2}=\ce{[H3O+] \cdot 4\ \mathrm{M}} \Rightarrow $$
$$\ce{[H3O+]}=2.5 \times 10^{-15}\ \mathrm{M}$$
$$\mathrm{pH}=–\log(2.5 \times 10^{-15}\ \mathrm{M})=14.60$$

If you are worried about whether or not you truly have a $4\ \mathrm{M}$ solution of $\ce{NaOH}$, you should try titrating it with a known concentration of acid (in the past I always used potassium hydrogen phthalate). This process is called standardization in acid-base parlance.


Ion-selective electrodes Some ion-selective electrodes can be expensive (starting from $500). Is there any relatively inexpensive do-it-yourself alternatives that would still give very accurate results? How does one make an ion-selective electrode?

Ion selective electrodes are a group of electrodes that are very different from each other.

The common bits are, a selective membrane, a "sensing" electrode and a reference electrode to measure against.

The details of the membrane and sensing electrode depend highly on what selective electrode you are trying to build.

Here is an old patent on a Cl selective electrode for reference:
http://www.google.com/patents/US3740326

Regarding getting accurate results with a DIY: this really depends on which electrode you are trying to build. In the Cl example, I don't see any reason why a DIY would perform worse.


Can ideal gas law be applied to water vapour? How to accurately estimate the molar concentration of pure water vapour at 298 K? Can ideal gas law be applied to water vapour to calculate the molar concentration?

Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour.

However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $T=298\ \mathrm K$ alone is not enough. The ideal gas law does not give you a value for the second quantity.

For example, assuming equilibrium conditions for saturated steam at a temperature of $T=298.00\ \mathrm K$, the corresponding pressure is $p=3141.7\ \mathrm{Pa}$.

Using the ideal gas law yields a concentration of

$$\frac nV=\frac p{RT}=\frac{3141.7\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298.00\ \mathrm K}=1.2680\ \mathrm{mol\ m^{-3}}$$

The actual value for water vapour at this temperature and pressure is

$$\frac nV=1.2701\ \mathrm{mol\ m^{-3}}$$

Thus, using the ideal gas law introduces an error of about $0.2\ \%$ in this case, which could be acceptable for many practical purposes.


How is phosphate stable disobeying octet rule? Doesn't the phosphate now have 10 electrons mutually shared between the oxygen atoms? I am unable to understand how the phosphate bond would be formed if phosphate does not obey the octet rule.

Thank you!

Reference: http://www.chemguide.co.uk/organicprops/aminoacids/makedna.gif

The octet rule is just a rule. Once you leave the 2nd row of the periodic table, the octet rule doesn't apply. Further reading: https://en.wikipedia.org/wiki/Hypervalent_molecule


Why must the total increase in oxidation numbers equal the total decrease of them in redox reactions? In my book, under the topic of redox reactions, it is given,

  In all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation number.

Is this a consequence of "Charges can never be created. The total charge of a system remains constant", or something else?

For the reaction,
$$\ce{
10 \color{red}{FeSO4} + 2 \color{red}{KMnO4} + 8 H2SO4 -> 5\color{red}{Fe2(SO4)3} + 2\color{red}{MnSO4} + K2SO4 + 8 H2O
}$$

(Reactants and products in red undergo oxidation or reduction)

For the above reaction, I tried whether the property stated above works or not.

The oxidation state of Fe and Mn in the reactants side are +2 and +7 respectively. The oxidation states of these elements in the product side are +3 and +2 respectively. The increase in OS is 1 and the decrease is 5.

Finally, the increase and decrease in oxidation states are not equal.

Kindly give an explanation for this fact.

The reaction is with an advantage written in the simplified ion form:

$$\ce{5 Fe^2+ + MnO4- + 8 H+ -> 5 Fe^3+ + Mn^2+ + 4 H2O}$$

Total oxidation numbers $+17=+17$

The oxidation number is formally the net charge an atom within a molecule would have held, if we had broken all bonds and bond electrons  had sticked to atoms according the atom electronegativity.

Therefore, the constant total sum of oxidation numbers across all involved atoms directly relates to the constant total sum of valence electrons and to the law of charge conservation.


Does expansion of an ideal gas not always represent an increase in entropy of the system? Expansion of an ideal gas not always represent an increase in entropy of the system. Statement is True or False?

What I thought is that whenever an ideal gas expands the randomness of gas molecules increases. So, entropy must increase.

Entropy does not depend on just the volume the gas has available.  It also depends on the energy the gas has to exploit that available volume.

Suppose you expand the gas by pushing against a piston.  The gas is doing work to "earn" its extra volume, so even as volume increases the gas is expending energy.  Thus unless heat flow into the gas is allowed the gas will cool.  This loss in energy tends to decrease random motion, and thus tends to decrease entropy, even as having more volume would have the opposite effect.  We have a "tug of war" here.

In an idealized version of such an expansion,  where we aallow no heat flow into or out of the gas (adiabatic) and we maintain equilibrium with no "losses" due to friction ("reversible") then the loss of energy to doing work must effectively balance the gain in volume, leaving zero entropy change.  With no heat flows through the insulation and no heat generation due to friction in this idealized scenario, there must be zero entropy change in both the gas and the surroundings.

The lesson:  Any process (such as expansion  of an ideal gas) that is  reversible (passing through equilibrium states) and adiabatic (no heat flow) has zero entropy change.


Deriving rate law from a reaction mechanism for a chain reaction using steady-state approximation I am trying to derive the overall rate equation for the photocatalysed halogenation (chlorination) of an alkane that proceed via this 4 step chain reaction:

$\ce{Cl2->2Cl^. \quad(1) initiation}$
$\ce{Cl^. +RH->R^. + HCl \quad(2) propagation}$
$\ce{R^. + Cl2-> RCl + Cl^. \quad(3) propagation}$
$\ce{R^. + Cl^. ->RCl \quad(4) termination}$

I have calculated the differential rate equations for $\ce{R^.}$ and $\ce{Cl^.}$ and set them to zero at steady state.
[The free radical dot symbols are omitted from this point onward for brevity]
This gives me $$\ce{For Cl ,2k1[Cl2]-k2[Cl][RH] +k3[R][Cl2] -k4[R][Cl]=0 \quad(5)}$$
$$\ce{For R, k2[Cl][RH] -k3[R][Cl2] -k4[R][Cl]=0 \quad(6)}$$
Adding these two equations,
$$\ce{[Cl]=\frac{k1}{k4}\frac{[Cl2]}{[R]}}$$
Then I tried substituting this into equation (6), which gives me a quadratic equation. The solution of the quadratic equation is of the form $$\ce{[R]=constant +\sqrt{\mathrm{constant [RH]}}}$$
The overall rate law from this expression is a very complicated equation.
This looks wrong because the rate law for radical halogenation is usually of the form $\ce{k[RH]\sqrt{\mathrm{[Cl_2]}}}$ (Wikipedia)
Where am I going wrong?

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms
By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the termination reaction).

Where am I going wrong?

I'm not sure why your steady state approximation is less appropriate than that of the textbook. I'm sure that there is some experimental data (and some fitted rate constants) that would speak to that. Also, the appropriate approximation depends on the initial conditions.

[Comment from Poutnik] The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order.

The derivation in the book is for the thermal reaction, but the OP asked about the photocatalyzed reaction, so there is a discrepancy there.


Why reduced atoms are more energetic? For example, in glycolysis, $\ce{NAD+} $ is reduced to $\ce{NADH}$ and then it's more energetic. Why adding electrons to an atoms gives it more energy?

All electrochemistry is relative.  And any time you have a reduction or oxidation reaction, you are dealing with an electrochemistry problem.

NADH is "more energetic" than NAD+ because the hidden assumption is that energy levels are relative to reaction with the atmosphere.  NADH can react with oxygen in the atmosphere to release energy.  But on a planet with no oxygen and lots of hydrogen in the atmosphere, NAD+ would be the "more energetic" molecule.

Another less exotic example is anaerobic fermentation in yeast.  NADH builds up in glycolysis, and must be converted back to NAD+ through formation of ethanol + CO2.  That process consumes energy; thus, in anaerobic environments, often times NAD+ can be thought of as a higher-energy molecule than NADH.


How to remove sulfuric acid from sample I have problem with my experiment. My experiment is acid hydrolysis of ethyl benzoate. The acid that I used is sulfuric acid. My problem is when I get the precipitate  (impure benzoic acid) I have to filter it from the solution (of course) and I need to remove the sulfuric acid from my sample or else my filter paper will not dry. But the major problem is sulfuric acid takes longer time to dry. It could takes even month. I've tried leave it for a month and it still wet. My lecturer asks me to rinse my sample + filter paper with cold water (distilled water). And when I did my sample which is the precipitate is LESS. Maybe because I used large amount of cold water. Is there are suggestions or methods to remove the sulfuric acid?

Sulfuric acid is an excellent dessicant and is also very non-volatile itself (it boils at 337 °C).  In other words it would likely still have been there in yet another month.

Your instructor was correct to suggest that you rinse the sulfuric acid with some cold water.  Benzoic acid does have some non-negligible solubility in water (1.7 g/L at 0 °C).  Unfortunately, it is even more soluble in most organic solvents so cold water may well be your best bet.  The solubility in 25 °C water is 3.4 g/L, twice the solubility that it has at 0 °C, so definitely keep it cold.

Additionally, use as little water as necessary to get the sulfuric acid out.  If you are working with a gram or so of benzoic acid, I would start with a dropwise addition of about 10 mL. If it’s still gooey, or still appears wet the next day, you need to repeat the rinse. It’s best if you can use a vaccum filter system of course, and spread the material thin and even before rinsing.


Will the volume occupied by a mixture of nitrogen and water be same as the volume occupied by them individually? I am currently studying gaseous state and I encountered the following problem.

  A container with a volume 3 liter holds $\ce{N2(g)}$ and $\ce{H2O(l)}$ at $\pu{29 ^\circ C}$. The pressure is found to be $\pu{1 atm}$. The water is then split into hydrogen and oxygen by electrolysis according to the reaction
  $$\ce{H2O (l) -> H2 (g) + 1/2 O2 (g)}$$ After the reaction is complete, the pressure is $\pu{1.86 atm}$. What mass of water was present in the container? The aqueous tension of water at $\pu{29 ^\circ C}$ is $\pu{0.04 atm}$.

This is one of the steps of the solution given along with the question (edited to show the units):

  Amount of substance of nitrogen:
  $$n(\ce{N2}) = \frac{\pu{0.96 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.116 mol}$$
  Amount of substance of water:
  $$n(\ce{H2O}) = \frac{\pu{0.04 atm} \cdot \pu{3 L}}{\pu{0.0821 atm L mol-1 K-1}\cdot \pu{302 K}} = \pu{0.00484 mol}$$

The thing I don't understand here is that how can the volume of nitrogen and water be taken as $\pu{3 l}$? Both of them would occupy some volume not the whole volume will not be occupied by one of them.

The idea behind the question is that you know the partial pressure of N$_2$ is 0.96 atm (= 1 atm - 0.04 atm). And you know that the partial pressure of all the H$_2$ that can be produced from the water is thus (1.86 atm - 0.96 atm)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the number of moles of H$_2$, which is equal to the number of moles of water = 0.6 atm *3 L/(302 K * 0.0821 atm L / mol K).

Note the number of moles of water given in your answer isn't the total no. of moles of water, it's the number of moles of water vapor prior to the electrolysis.


Why can you not make a compound with a quaternary alpha carbon using malonic ester? So my textbook tells me that using a malonic ester to make a quaternary alpha carbon is impossible (I get that this might be hyperbole). However, it doesn't explain why and I don't understand why.

My thought was that after the first addition the tertiary compound would be too bulky to act as a nucleophile. However in the next section, an ethyl acetate is treated with LDA and acts as a nucleophlile 3 times to create a quaternary alpha carbon.

I also considered that you'd be making a tertiary carbon anion to make a second addition to a malonic ester's alpha carbon. However you also make a tertiary carbon anion using LDA on ethyl acetate and then alkyl halides.

I've added a screenshot of my textbook. Skip to the part in parenthesis if you want to read it.

Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!

Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23

EDIT 1

The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide in situ:

\[\ce{2Na + 2ROH -> 2NaOR + H2}\]

I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.

With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go.

EDIT 2

For the example given above, no further direct alkylation at that position is possible, and this is exactly, what ron pointed out in his
comment to the question:

  Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.

He also suggested the viable solution, which I had outlined in another question on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great synthons to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.

Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.

In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using two bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a strong base like $\ce{LDA}$ is used. (Exactly this is outlined in the textbook for the alkylation of an ester).


Is an ionic bond more like a covalent bond or an intermolecular force? I have asked a question loosely asking this, where I confused terms and did not specify what I wanted to know here, so I formed a new question.

What are the differences and similarities between ionic bonds and covalent bonds and ionic bonds and dipole-dipole interactions?

It seems to be that an ionic bond is nothing but a stronger dipole-dipole interaction. The only difference is where the electrons are.

For example, let's take a sodium cation and chloride anion bond. The sodium had its electron taken away from it. It is now positive. Chloride had an electron added to it and it's feeling negative. Now the two will attract because of these charges.

Now let's take two water molecules. The hydrogen of one will be positive because it is less electronegative and thus the electron density will be greater around the oxygen, which is more electronegative, making it negative. Now the hydrogen of one and the oxygen of another will attract due to the charges.

It seems like the only difference is that if you split the sodium chloride and the water's hydrogens and oxygen, the sodium and chloride will still have a charge, but the hydrogens and oxygen will not.

By textbook, a dipole-dipole interaction is an intermolecular force while a covalent bond, as well as an ionic bond, is a chemical bond. It seems to me that, yes, the forming of ions is a chemical reaction, but the ionic bond is an intermolecular force.

An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time.

At the lowest-detail end of the spectrum, you have ionic bonds - we usually model these as simple Coulombic (a.k.a. charge-charge or electrostatic) interactions. This is good enough to describe bonding in most cases where the bond is between a metal and a non-metal. We model the bond as a complete electron transfer from one atom to another, which results in an electrostatic charge. The Coulomb force between the charges holds the ion together, and so we call it a molecule. What we really mean is "these atoms are now stuck together."

As we move along the periodic table and the metal atoms become less metallic and the non-metals become more metallic, at some rather arbitrary point we say "the ionic approximation is no longer good enough" and start seeing the bond as covalent. What we really mean is that the electrons are not fully transferred between atoms, and so charge-charge interactions don't explain bonding well enough. At that point we can no longer ignore quantum mechanics, and have to acknowledge that the electrons exist in molecular orbitals, not atomic orbitals.

That is the main difference between ionic and covalent bonds. As for dipole-dipole interactions, it again goes back to what the electrons are doing. In a molecule with a dipole moment, what this means is that within the molecule there are polar covalent bonds, and the geometry is such that the electron density on the surface of the molecule is lopsided - one side has more electrons than the other. The dipoles can align and will be attracted to each other via the Coulomb force, but this is not the same as how we model an ionic bond - in that case, one atom gives up one or more electrons completely. In a molecule with a dipole moment, one (or more) atoms just gets "more" of the electron density. Then the two separate molecules, each with their own dipole, are attracted to each other.

We call ionic bonds "intramolecular" forces because they are what hold molecules together - without ionic bonds, you couldn't have a salt crystal, for example. When you break apart sodium chloride, you get sodium ions and chloride ions.

We call dipole interactions "intermolecular" because they are what make separate molecules stick together. When you pull apart two water molecules, you still have two water molecules.

Thinking about it this way, an easy way to distinguish between intermolecular and intramolecular forces is to ask yourself whether you still have the same substance after pulling things apart. If you do, it must have been an intermolecular force. If you don't, it must have been a bond or an intramolecular force.

note - For very large molecules like proteins it gets a little tricky - you can have forces between sites on the same molecule that act like what we normally describe as intermolecular forces.


Ring expansion of two fused rings to a larger ring I saw the following reaction mechanism in paper Tetrahedron Lett. 1976, 17 (33), 2869–2872.:

I'm not able to understand the following parts:

How did the conversion of 27 to 29 take place? I've never seen that kind of a ring expansion, and a detailed explanation would help - maybe including the necessary transition state(s) and why it happens.
27 to 28 and 30 to 31 aren't clear as well; why should such rearrangements happen and how do they happen? What is the mechanism?

I think the possible mechanism for the intermidiate conversions can be as following.Here thr $\ce{R}$ is actually $\ce{-OCHO}$.


Comparing strength of the C-H, N-H, and O-H bonds $$
\begin{align}
\ce{CH4(g) + Cl(g) &-> CH3(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-14 kJ mol-1}_\mathrm{rxn}\\
\ce{NH4(g) + Cl(g) &-> NH2(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-36 kJ mol-1}_\mathrm{rxn}\\
\ce{H2O(g) + Cl(g) &-> OH(g) + HCl(g)} &\quad ΔH^\circ &= \pu{+40 kJ mol-1}_\mathrm{rxn}
\end{align}
$$

  41. Based on the data above, what can be concluded regarding the strength of the $\ce{C-H}$, $\ce{N-H}$, and $\ce{O-H}$ bonds in the molecules shown?

  (A) The $\ce{C-H}$ bond is the strongest.
  (B) The $\ce{N-H}$ bond is the strongest.
  (C) The $\ce{O-H}$ bond is the strongest.
  (D) Nothing can be concluded without knowing the strength of the $\ce{H-Cl}$ bond.

I am not sure how to approach the problem 41. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. So would the answer be B because the $ΔH$ is exothermic and is the greater than $ΔH$ of -14 for A?

The answer should be C.
One way would be to write the thermochemical equations, however since we just want to compare, we can do it without going into that.
The formation of the H-Cl bond, (like all bonds) is exothermic. And the breaking of the C-H, N-H and O-H bonds is endothermic (since their formation is exothermic too).
Since the contribution of the H-Cl bond to the enthalpy is the same for all three, we can focus on the C-H, N-H and O-H bonds alone.

Now, the stronger the bond, the more energy it would need, and hence make a positive contribution to the enthalpy change. Hence we can see that since the total enthalpy of the reaction is highest(most positive) for the reaction involving O-H bond, the O-H bond is the strongest. (Infact, it is stronger than the H-Cl bond too as the enthalpy change is positive).

Note that this question also demonstrates the order of electronegativity, which is
O>Cl >~ N> C>H. Since the bond with most electronegativity difference would be the strongest, it would be the O-H bond.


GAMESS and Gaussian : what is the maximum efficient number of supported nodes? I have looked at GAMESS and Gaussian manuals but can not find the maximum number of efficient nodes. As I tried GAMESS on our supercomputer, I can add as many CPUs as I want but it really does not Improve the calculation time and efficacy . On the other hand just adding maximum number of CPUs is just waste of CPU-Hour time and resources.

Is there any criteria for selection of best number of CPUs for optimum resource usage ?

Is there any criteria for selection of best number of CPUs for optimum
  resource usage?

The criteria is basically the same as for choosing the level of theory: benchmarking. Parallel scaling depends on many factors: the level of theory, the size of the molecule, the machine architecture, etc., so there is no golden rule.

Though from my experience I can say that Gaussian 09 single point energy, geometry optimisation and frequencies calculations:

scale well for HF & DFT up to 16 physical cores;
scale well for MP2 up to 8 physical cores;
almost do not scale for CI & CC.

Also note that Gaussian 09 is about 2 times faster for almost any kind of calculations than the previous version (Gaussian 03) and doesn't suffer from some awkwardly strange memory effects (see e.g. here).


Except pure alloys, are there any compounds with more metal elements proportion of atoms than nonmetal elements in proportion of atoms? Except pure alloys, are there any compounds with more metal elements in proportion of atoms than nonmetal elements in proportion of atoms?

For example, Aluminium oxide has 2 metal elements & 3 nonmetal elements. Do we have a compound with X metal elements & Y nonmetal elements, where Y is more then X.

There are some rare but very interesting cases in which two metals react when mixed instead of forming an alloy, creating a non-metallic compound made only with metal atoms. Examples are caesium auride ($\ce{CsAu}$), caesium platinide ($\ce{Cs2Pt}$) and several barium platinides ($\ce{BaPt}$, $\ce{Ba2Pt}$, $\ce{Ba3Pt2}$, and possibly others), all of which best described as ionic salts rather than alloys. These are a remarkable consequence of relativistic effects in heavy atoms.


Oxidation of acetophenone to benzoic acid According to the solutions manual, the oxidation of acetophenone to benzoic acid can be achieved through addition of chromic acid and application of heat.

The solutions manual refers me to a section in the book which explains that when a benzylic hydrogen is present, a strong oxidizing agent combined with heat can perform oxidative cleavage.

Problem is that I don't see any benzylic hydrogens on acetophenone. Neither does its enol tautomer have any benzylic hydrogens ...

So, how is the oxidation of acetphenone to benzoic acid possible given what the book and solutions manual has told me? Self-consistency aside is this even a possible reaction?

From a SciFinder search, there are ~10 examples of oxidations of aryl methyl ketones resulting in a carboxylic acid using a chromium based reagent (usually dichromate but also chromium trioxide). This suggests the reaction is at least possible, although I think including it in a textbook is unwarranted.

One reaction slipped through my search that is informative (below). Notice that in the presence of a benzylic position that does contain a hydrogen, the methyl ketone is unchanged, while the other group is oxidized.

Taken together, it seems that chromium oxidation of an acetophenone to a carboxylic acid is slower than alkyl benzylic oxidation.


What is the chemistry behind the reaction of KNO₃, K₂CO₃ and Sulfur? I'm preparing a chemistry show where I end the show with yellow powder bang.

The powder is a mixture of:

$$\ce{K2CO3 + KNO3 + S} $$
My question is, what's the reaction between these chemicals?

Here's just a guess, but I do have some confidence that I'm right:
$\ce{KNO3 + S -> K2SO4 + K2SO3 + K2S + K2O + SO2 + N2}$
$\ce{K2CO3 ->[\Delta] K2O + CO2}$
The first, unbalanced equation is the reaction between potassium nitrate and sulfur, producing quite some gas as product, and a variety of potassium compounds left behind, with the composition depending on your formulation of the powder. Of course there might be some sulfur trioxide produced, but the major oxide of sulfur should be the dioxide. In the second equation is the thermal decomposition of potassium carbonate releasing carbon dioxide. This is the part that I'm not very sure of, but it may be to release more gas to give your reaction more puff. However, IIRC the decomposition requires a very high temperature. In contrast, bicarbonate has a much lower decomposition temperature and also release more gas(carbon dioxide and water), but the products would likely be absorbed by potassium oxide to form hydroxide and carbonate, so  bicarbonate might not help much.


Does anyone know what program was used to make this molecular structure picture? I am a quantum chemistry beginer, and this picture color scheme is classic. I wonder how to make such pictures by myself.

The first type of image is very easy to make. I would suggest Avogadro with POV-ray. Avogadro can be used to draw the structures (or open a file containing the information, like .mol2, .sdf etc.) and then high-quality images can be generated using POV-ray.
To do this, install Avogadro, then draw a structure, and then File> Export > POV-ray. A new subwindow will open, where you can specify the location of the POV-ray executable.

In my case, the executable is pvengine64.exe, it might be different in different platforms, or in different versions.
You can also change the resolution here. Finally, click "Render", and a new windows with POV-ray should open, and it should display the image, after a few seconds.

This is the image of the molecule that you showed in your question. The background looks like that because the "Alpha transparency" option is turned on in the Avogadro subwindow, which makes POV-ray render only the molecule, and there is no background.
As for the second picture, it looks like something that would be used in crystallography to represent probability ellipsoids of atoms. I don't know much about this, so I can't help you there.


Why are imines more susceptible to reduction than carbonyl compounds? Why are imines more susceptible to reduction than carbonyl compounds? Imines react with $\ce{NaCNBH3}$ while the carbonyls don't.

Generally speaking, I think it is the reverse - carbonyls are reduced faster than imines with various sources of $\ce{H^-}$.  The reason usually cited (see pages 2-4 of this reference) for this behavior is that the imine double bond is less polarized than the carbonyl double bond.  In the figure below resonance structure II contributes less when X=N as compared to the case X=O, due to the decreased electronegativity of nitrogen compared to oxygen.

However, if the reaction is run under mildly acidic conditions, since the imine nitrogen is more basic than the carbonyl oxygen, more of the imine nitrogen will be protonated than the carbonyl oxygen.  This will make the protonated imine carbon much more electrophilic (positively polarized).  Consequently, under acidic conditions, an imine may be more reactive than a carbonyl to nucleophilic reduction (in addition to the above reference, also see this reference).


Naming alkyl groups of an ester $\hspace{6.4cm}$
I would like to know how to name the alkyl groups connected to the above molecule. For example, would the far left alkyl group be considered a methyl ester group? Would the alkyl group bound to the oxygen be a methylene ester group?

You are going about naming them incorrectly, try breaking it into two parts: the conjugate base of a carboxylic acid and an alkyl group.

$\hspace{5.9cm}$

Combining the two, we get:

$\hspace{6.9cm}$

Both the carboxyl and alkyl carbon chains can be numbered, with the carboxyl chain starting at the carboxyl carbon, and the alkyl chain starting at the carbon that places its bond to the ester oxygen at the lowest position. See below:

$\hspace{5.5cm}$

To refer to a specific carbon, simply say "at position $x$ on the carboxyl/alkyl chain."


Second order reaction This is an example in my textbook that I am trying to figure out

$1/[\ce{A}] = 1/[\ce{A}]_0 + kt$

In this case:
$[\ce{A}] = 4.62\times10^{-2}\ \mathrm M$
$[\ce{A}]_0 = 0.217\ \mathrm M$
$k = 0.540\ \mathrm{M^{-1}\ s^{-1}}$

so I get this
$1/(4.62\times10^{-2}\ \mathrm M) = 1/(0.217\ \mathrm M) + (0.540\ \mathrm{M^{-1}\ s^{-1}})(t)$

the answer should be 31.5 seconds

I can’t get this answer I must be missing something
in the order that it should be.

When you manipulate the equation to solve for t you'll want to subtract 1/[A]0 from 1/[A] and then divide by k.


Does Uranium ore count as a mixture? I'm doing an assignment where we must research the separation of a mixture to obtain a useful final product. I thought about the separation of uranium from its ore after it has been mined, but I did not know if the ore counted as a mixture, or a compound.

Yes, uranium ore is a mixture.

Uranium compounds such as pitchblende (uraninite) are only present as a small part (usually $<1$%) of the ore.

Extensive purification is needed. Uranium ore purification would be a good choice for your project.

See for example:

http://www.areva.com/EN/operations-677/uranium-processing-from-ore-to-yellow-cake.html

http://www.iaea.org/sites/default/files/publications/magazines/bulletin/bull23-2/23204882428.pdf

image source: http://chemwiki.ucdavis.edu/Development_Details/Approaches/ChemCases/Nuclear_Chemistry_and_the_Community/08._Uranium_Production


About alternative magnetic effect on porcelain (for induction furnace purposes) I am trying to use the Induction furnace mechanism (like described via this video), for heating part of the Muffle furnace, as one dental porcelain furnace prototype.
So, my plan is to put copper coil with frequency around 10 ~ 4000 HZ, and placing some iron bar inside the ceramic container (one sample shown below!):

.
I have googled the induction muffle furnace, and find only one patent by this topic (you can see below):

So, I am in doubt that some part of this project must be wrong. So, I asked here about the effect of changing the magnetic field on the characteristics of dental porcelain, if that effect were too much and create some deformation via dental porcelain coating, this prototype would not be proper for this purpose.
As I am not familiar with chemical terms and ..., If possible, suggest some proper tag to faster finding the effect of the alternative magnetic on the dental porcelain coating.
Update:
For clarifying question it must be said:

induction furnace mechanism are working fine, and for creating high
temperature the plan have had some additional metal bar to become hot
(based of induction mechanism), and create the desired heating for
melting the porcelain coating. But my question is about the effects of
alternative magnetic fields on desired characteristics of porcelain
(target).

Thanks for your consideration.

On reason you don't see much information is that your plan will not work. The copper wire will get hot by resistance and conduction and at some temperature it will react with the refractory = fail. Copper has essentially no use as a heating element because the oxide spalls off after each heat cycle. Use Nichrome wire ( this alloy family has a few different names ). However , as I understand induction heating it needs an conductive target. I am guessing your dental porcelain has poor electric conductivity , so will not heat by induction.


Decarboxylation of Sodium Pyruvate Can Sodium Pyruvate decarboxylate to acetic acid + CO2 in the presence of just moisture and heat?  If so, what kind of temperature is required?

As pointed out in the comments above, the expected product would be acetaldehyde, not acetic acid.

If you stop and think about the mechanism of any thermal decarboxylation, you immediately stumble upon the issue. The decarboxylation would lead to a d1 type synthon, which from organic chemistry you should realise is the wrong polarity for a carbonyl group (the natural polarity of the C=O bond is to have a partial positive charge on carbon and partial negative charge on oxygen).


Dehydration of aldoximes What is the mechanism for dehydration of aldoximes by acetic anhydride?

My attempt:
I doubt if it's correct.

My proposed mechanism is a nucleophilic addition/elimination mechanism without a tautomerism step.

It is all concerted.

The first step, addition, features the oxygen in the (E)-propionaldoxime acting as a nucleophile to attack one of the carbonyl carbons in the acetic anhydride, which pushes electrons the carbonyl oxygen, making it negatively charged.

Then, the negatively charged oxygen strikes back, initiating a chain of concerted cyclic electron rearrangement which cleaves quite a few bonds, resulting in three molecules, completing the elimination step.


What is the hybridization of chromium in chromate and dichromate ions? I was recently reading about the transition metals and some important compounds of them. I came across a rather interesting diagram in my book... a tetrahedral structure for the chromate ion. The only tetrahedral structure I know arises from an sp3 hybridization of the central atom, but as far as chromium is concerned, I think that a d3s would be more practical(?).. But then again, which specific orbitals of 3d would hybridize to give a tetrahedral structure?(this is beyond me, i seriously do not understand the shapes of the d orbitals)

The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:

$\mathrm{t_{2(g)}}: \mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$

In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).

In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:

Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.

The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).

Don’t worry, these schemes will come to haunt you at one point when studying chemistry.


What is non toxic to fish, has a low freezing point, and floats on water? I am looking for a liquid that floats on water that will prevent the water from freezing (or at least slow it down).  The application would be a small area (less than a square foot).  It has to be non-toxic as I do not want to harm any critters.  Thoughts???

There are oils sold for swimming pools to slow or stop evaporative cooling . That would be a problem for air breathers like mosquito larva. I don't know anything about it because I went with 1" thick Styrofoam panels. That is what I currently have on my pond ( 10' X 5 '), works very well .I cheat a little and have 150 watt aquarium heater in the 700 gallons. My pond is currently 59 F while nearby open water is 49 F. For your tiny area the Styrofoam is fine leave about 1/2" on all sides for air . On the other hand why not do the obvious and put in a 25 watt ( if they make them that small) aquarium thermostat/heater ?


Effective Nuclear Charge of Noble Gases As stated in my textbook, the effective nuclear charge can be calculated by taking the non-valence electrons away from the atomic number, as also stated in the first calculation on Wikipedia: https://en.wikipedia.org/wiki/Effective_nuclear_charge

The textbook also states:

If we assume that the noble gas core is completely shielding, then the 10 inner electrons
of the neon core make the effective nuclear charge 10 less than the nuclear charge.

My understanding is it implies noble gases have 0 effective nuclear charge.

However, as many sources state noble gases have 8 valence electrons, shouldn't their effective charges be +8?

I think that this is basically an accounting problem. The full electron configuration of $\ce{Ne}$ is: $1s^22s^22p_x^22p_y^22p_z^2$. So, what does one now define as core electrons? It could be: "all electrons not in the outermost shell", yielding the $1s$ electrons. An alternative is "all electrons in completely filled shells", yielding all 10 electrons as core electrons. (Note that these two definitions yield identical results for almost all non-noble gas elements, with some difficulties surrounding the $f$-block.) Both definitions have some merit, I would prefer the first one, giving an effective charge of $+8$, although I do not know what one does with that number next.

A word on effective nuclear charge
As noted on the wikipedia page linked in the question, to arrive at more meaningful effective nuclear charges (as might be helpful for the systematic creation of basis sets for computational/quantum chemistry), one has to take the effective shielding strength of the different orbital shapes into account. The $d$-block-contraction, as can be seen in the somewhat different chemistry of $\ce{As}$ compared to $\ce{P}$ and $\ce{Sb}$, is due to this. Thus, in the end, a pen-and-paper calculation as alluded to above can be useful to think about some concepts in a qualitative fashion, but more extensive calculation is needed for quantitative reasoning.


Can liquids be made a lot purer than by fractional distilation? Suppose you have a fused quartz tube with one closed end coiled into a spiral with a diameter of more than 10 m and you submerge it into a mixture of water and ethanol in a way that gets rid of all air then reposition it in such a way that it's partly submerged with the axis of the coil going nearly horizontally and slightly up in the direction from open end to closed end. Furthermore, suppose that the coil then starts moving and spinning along its length in the direction from open end to closed end and once the closed end passes the height where cavitation occurs, the closed end is cooled to recondense liquid there and after cavitation occurrs again on the next loop, liquid is condensed again in the same way? Is it true that if the concentration of water is above a critical concentration, it will become more concentrated after it cavitates and is recondensed because a smaller fraction of ethanol vapour will condense than water and if the concentration of water is below that amount, the water will become less concentrated after it cavitates and then is recondensed because a smaller fraction of water will evaporate than ethanol? Could really pure water be gotten in that way because fused quartz is extremely insoluble in water?

Since ethanol and the water-ethanol azetrope boil at a lower temperature than water, if you start with a lot of water with a smaller amount of ethanol  then you can "distill" (however you do it...) all the ethanol out and be left with "pure" water.

But water and ethanol form an azeotrope which is 95.5% ethanol by weight and boils at 78.1 °C. For the water-ethanol azeotrope the wt. % ethanol in the liquid is the same as the wt % of ethanol in the vapor. Thus there is no way to exploit liquid to gas phase transition to enrich the ethanol more than that.


Why do s-block elements impart color to a flame? Flame tests are a common way of identifying trace amounts of alkali metals (and to a lesser extent alkaline earth metals) in mixtures. Certain salts of $\ce{Ba, Sr}$ are also used in fireworks for their ability to impart a colour to the flame.

Although predominant among s block elements this behaviour is also seen in other elements like $\ce{Cu}$ and $\ce{Pb}$. My question is, what is so special about these elements that lends them this characteristic?

All elements colour the flame somehow. The salt that is held into the flame dissociates homogenously (i.e. you’re not creating unpaired ions but rather single metal atoms and their counter-radicals), the heat energy excites the electrons and when they fall back down they emit photons at certain wavelengths, corresponding to the energy differences between the orbital an electron was excited to and the one it falls down to.

The further away from the nucleus the electrons are, the easier they are to excite. Thus valence electrons are most likely to be excited. The more valence electrons you have (i.e. the further right you go in a given period) the harder it gets to excite one of them because the whole atom contracts and the energy levels of the outer shells are lowered. Therefore, it is a lot easier for electrons of elements on the left half of the periodic table to be excited.

Furthermore, the more electrons you can excite, the more complicated the resulting spectrum will be. For only one or two electrons, you usually expect (and observe) distinct lines of definite wavelengths; the more you have, the larger the possibility of broader bands.

Therefore, the easiest element to recognise would be hydrogen, which only creates very few single, distinct lines. Alkaline metals, having few valence electrons, will still create easy spectra with distinct lines (albeit more). It is similar for alkaline earth metals, although the spectra tend to get more complicated (see barium for an example of something with broad bands). For many, if not all other metals, the spectra just get too difficult to interpret and consistently assign very quickly. They will all seem white-ish to some degree with hardly any possibility of distinguishing characteristic lines easily. Thus, in the typical inorganic chemistry ion lottery lab course, typically only alkalines and alkaline earths are detected via flame colouring.

I also remember from that course that copper halides seem to provide a more characteristic green colour (that I didn’t even try recognising — I relied on other methods to detect copper). I think that there is a mulitude of combined effects in action to give this combination its characteristic colour. In any case, I don’t know what exactly is going on there.


Ideal gas derivations For an ideal gas, how can someone obtain this:

$$ \bar{V}dP = RdT- P \bar{dV}$$

from which it can de deducted, that for a constant temperature

$$ \frac{dP}{\bar{dV}} = -RT $$

This looks like the differential of the ideal gas equation. I'm going to use $V_m$ for the molar volume.
$$PV_m=RT$$
Now we convert each side to the differential. We keep $P$ and $V_m$ together since they are both variable. We can move $R$ out of the differential because it is a constant.
$$d\left(PV_m\right)=RdT$$
We can evaluate $d\left(PV_m\right)$ using the product rule:
$$d\left(PV_m\right) = PdV_m + V_mdP$$
Now:
$$PdV_m + V_mdP = RdT$$
Which is equivalent to your first equation.
**However, at constant temperature $dT=0$, so at constant temperature, the equation simplifies to:
$$PdV_m + V_mdP=0$$
$$V_mdP = - PdV_m$$
$$\frac{dP}{dV_m}=-\frac{P}{V_m}$$
Which is not equal to your second equation. The closest I can get is the following:
$$\frac{dP}{dV_m}=-\frac{P}{V_m}\cdot\frac{V_m}{V_m}=-\frac{PV_m}{V^2_m}=-\frac{RT}{V^2_m}$$
Which is what we would get by evaulating:
$$\left(\frac{\partial}{\partial V_m}\right)_T \left(P=\frac{RT}{V_m}\right)$$


IUPAC nomenclature of protonated aminoacids What is the IUPAC systematic name for the above compound.

So I know that carboxylic acids have highest priority. Now the length of carbon chain is 7. So I think the name should be 7-ammonioheptanoic acid. Am I correct?

Interestingly, Ammonium salts come above acids in the seniority table for functional groups (Reference: Wikipedia). Also, the ion in question has an alkyl group attached to it, which makes the suffix aminium rather than the commonly used ammonium. (Reference: Gold Book, Wikipedia)  This means that the correct name of the compound is 6-carboxyhexan-1-aminium. This was also checked with ChemDraw online, as mentioned in the comments.


How do i create a significantly correct world with large crystal formations in my novel? I have uploaded an shorter and less broad Question which is largely the same but requires a different kind of answer. What i want to say is that i re-asked my question in a different way because i do not want to delete the answers i got  here so far and because the answered also would be incorrect in the context of my re-asked and re-written question
link: Which crystals can exist permanently (and do not break down) under normal earth like outside conditions?

Hi I am not really all that well versed in chemistry or molecules for that matter but I am trying to find a way to "create" a scientific and realistic (trough probably unlikely) world with large crystal formations on the surface. The reason I want this is because I want to create a species for my novel that has its culture based around the large crystal formations living on the planet.

Now to limit my question a bit I would like the answer to these two points:

What kind of crystal(s).
How do you think the process should work.
What limitations it creates.

The criteria are that:

The crystals must be permanent without breaking down in normal earth conditions/atmosphere.
Preferably no interference required by intelligent lifeforms.
There will be oxygen and carbon dioxide in the atmosphere.
They must be able to grow quite large. I’m thinking about large hexagonal crystal columns.

What are the tools?:

Well you can more or less use anything in nature, you can make suns explode if need be. But a more probable start would be selecting the chemical composition and what would be needed to form crystals from that point.

For instance if you need a long period of very high heat and then cooling down to “normal” temperatures you can make the system sun go red giant and heating the planet before becoming a white dwarf.

Do you mind a scenario somewhat less dramatic than exploding suns? ;)

Just let selenite ($\ce{CaSO4*2H2O}$) cristallize very slowly und undisturbed in a system of caves for a long time.
Let the inhabitants of your world discover the cave system or let parts of the cave system collape to make it more accessible.

A model for this already exists: the Cave of the Crystals, connected to the Naica mine in Mexico.

Update

Note that I'm not a geoscientist, but to me, large hexagonal columns that can exist under earth's atmosphere cry basalt, which in turn indicates vulcanic activities.

The English wikipedia has a list columnar structures resulting from vulcanism. Some examples are

Giant's Causeway in Northern Ireland
Fingal's Cave in Scotland
Devil's Postpile in California


Tartaric acid chirality In the image above, four variations of tartaric acid are shown:

The variant in the top left is the "basic" skeletal formula of tartaric acid.
The one on the top left is D-(-)-tartaric acid.
The one on the bottom right is L-(+)-tartaric acid.

How would you name the one on the bottom right using the dextro- and levo- method instead of the S and R method?

You can’t really name it according to D/L.

This nomenclature requires you to draw a Fischer structure wherein the skeletal atom labelled ‘1’ by nomenclature rules is at the top; then, the orientation of the bottom-most functional group determines whether the compound is dextro, D (pointing to the right) or levo, L (pointing to the left. Below you can see Fischer depictions of all three tartaric acids.

Figure 1: D-tartaric acid, L-tartaric acid and meso-tartaric acid from left to right.

Now do one of the following: either rotate your PC screen/mobile device by $180^\circ$ or rotate your head by $180^\circ$.* You should notice that both D- and L-tartaric acid still have the bottommost hydroxy group pointing to the right or left, respectively. Both carboxylic acids are identical, so we would have to mark one carbon atom by using $\ce{^13C}$ to distinguish the top and bottom. The orientations lead to the same deduced stereochemistry.

This is not the case in the third molecule. Before rotation, both hydroxy groups are pointing to the right-hand side, afterwards they are pointing to the left-hand side. But, as I already stated, the two depictions are identical in the absence of isotope marking. Therefore, this compound eludes the D/L nomenclature and cannot be named unambiguously in accordance with it.

Thankfully, though, we have another descriptor that does not directly stem from D/L: it is the meso descriptor. It is used in case a molecule with asymmetric centres also features a plane of symmetry causing it to be identical with its mirror image and therefore achiral. While it is not part of the D/L terminology, it is the commonly accepted name for meso-tartaric acid.

*: Please only perform gymnastics if you can handle them. I decline all responsibility for injuries that may occur. Stack Exchange cannot be held responsible.


Why is the addition of a singlet carbene to an alkene stereospecific? Why is the addition of a singlet carbene to an alkene stereospecific, but not when a triplet carbene is added?

Here is a diagram that might help explain things.

Carbenes have two electrons that, for the moment, are not being used for bonding.  These two electrons can exist in the same orbital with their spins paired (singlet carbene) or in separate orbitals with their spins unpaired (triplet carbene).  Normal hydrocarbons exist as singlets, spins are paired in all bonds, as they must be according to the Pauli exclusion principle.

If we look first at the pi electrons in ethylene, their spins are paired since ethylene is a singlet in the ground state.  When we add a singlet carbene to ethylene, a bond is formed by using one electron from the carbene and one from ethylene.  Those two electrons are selected such that their spins will be paired in the newly-formed bond (Pauli exclusion principle).  Either 1) a second bond is formed at the same time using the two remaining spin-paired electrons (one on ethylene and one on the carbene) to form cyclopropane in a spin-conserving, concerted addition or 2) a transient (short-lived), intermediate 1,3-biradical is formed that still contains two spin-paired electrons is formed, since the spins are paired on these two remaining electrons in the biradical and since they are held in relatively close proximity they can close quickly to form a three-membered ring without violating the Pauli exclusion principle.

Let's now contrast this to the triplet carbene case.  The spins are unpaired in the triplet carbene.  When it forms a bond with one of the ethylene pi electrons, in order to obey the Pauli exclusion principle we must select our electrons so that they are paired in the newly formed bond.  This requires that the two remaining electrons in our 1,3-biradical have their spins unpaired.  Before the biradical can form the final bond and close to cyclopropane, we must wait for one of the electron spins to flip so that we have a singlet biradical.  The triplet cannot close directly since this would violate the Pauli exclusion principle.

This "waiting" for the spins to pair in the triplet biradical means that the triplet biradical has an appreciable lifetime.  Had we run our reaction using cis-2-butene instead of ethylene.  We could see the effects of biradical lifetime by observing the stereochemistry in the resultant products.  The singlet biradical, which either reacts in one concerted step or passes through a short-lived, singlet biradical would produce only cis-dimethylcyclopropane.  In the case of the triplet carbene, the triplet biradical exists long enough (waiting for one electron spin to flip) that rotation about the various carbon-carbon bonds is now possible.  As a result, we obtain both cis- and trans-dimethylcyclopropane.


Estimation of nickel ions I was doing experiment of estimation of copper and nickel ions by means of volumetric and gravimetric analysis respectively.

Suppose we analysed $\ce{Cu^2+}$ ions by precipitating with $\ce{H2S}$
But still there are some copper ions remaining in solution.

Is there possibility that they ($\ce{Cu^2+}$) can form complex with DMG
and interfere with Ni-DMG complex.

Copper can form a stable DMG complex (if it is there in the mixture in high concentration), it may also effect the color of the resultant Ni-DMG complex. Just go through Vogel's quantitative inorganic analysis book once.


Difference between ideal & ideal-dilute solution These two terms has been frequently used in Physical Chemistry by Atkins. He describes Raoult's Law using ideal solution while Henry's law using ideal-dilute solution.

I thought, at first they were same but Atkins in order to contrast the two laws presented graphs in  which one curve was for ideal solution & another was for ideal-dilute solution. Though he wrote two different sections for these solutions, I really couldn't contrast what the difference actually is between ideal solution & ideal-dilute solution. I googled it but to vain! Can anyone please help what he is trying to say by these two terms?

Raoult's Law is for ideal solutions (all of it: solvent and solute):

  Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components." (all quoted from above link)

As stated here concerning ideal-dilute solutions:

  "Many liquid mixtures deviate from the ideal behaviour predicted by Raoult's law. Ideal-dilute solutions are those for which the solvent obeys Raoult's law and the solute Henry's law. The difference in behaviour is because the solvent molecules tend to be surrounded by other solvent molecules so that they behave in an almost ideal way. However, the solute molecules tend also to be surrounded by solvent molecules so that their environment and therefore their thermodynamic behaviour is very unlike that of the pure substance." (all quoted from above link)

In summary: Ideal solutions are those which obey Raoult's Law; ideal-dilute solutions are ones where the solvent obey's Raoult's Law and the solute obeys Henry's Law.


Connection of term symbols with specific microstates for atomic carbon I'm currently studying atomic term symbols. I wanted to try it on a simple atomic carbon with the electron configuration $1s^22s^22p^2$.

I know, that only open-shell electrons are involved in the term symbol classification, so that leaves us working with solely 2 electrons in $2p$ sub-shell.

First of all, we have ${6}\choose{2}$$= \frac{6!}{2!(6-2)!}=15$ possible microstates for the $1s^22s^22p^2$ configuration.

The microstates are listed in the following table:

And, if I understand it correctly, we can categorize them into several "subsets" labeled with the corresponding term symbols.

The atomic term symbol is defined as ${}^{2S+1}L_J$.

In our case, the possible values of S, L and J are following:
$$\max(M_L) = \max(L) = +2 \Rightarrow L = 0,1,2$$
$$\max(M_S) = \max(S) = +1 \Rightarrow S = 0, 1$$
$$J = L+S, L+S-1, \ldots, |L-S| = 3,2,1,0$$

Considering possible values of $S$, multiplicity can be $1$ or $3$, i.e. singlet or triplet. Values of $L$ enable symbols $S,P,D$.

So, our presumed set of states is ${}^3D, {}^1D, {}^3P, {}^1P, {}^3S, {}^1S$.

${}^3D$ "contains" $S=1$ and $L=2$. In that situation, both electrons would have to be spin-up and positioned together in the orbital with $m_l = +1$. That is impossible, as it contradicts Pauli exlusion principle. I.e. the state ${}^3D$ does not exist.

${}^1D$ is possible ($L=2,1,0, S=0$), so we can compute the number of corresponding microstates $N$  with the formula $N = (2L+1)(2S+1) = 5\cdot 1 = 5$.

Using the same formula, ${}^3P$ is going to contain 9 microstates.

With the "substraction method" described in this video, we can arrive to the conclusion, that both ${}^1P$ and ${}^3S$ won't exist.

And finally, ${}^1S$ will contain the last one state.

Question

Now I know the possible term symbols, but I'm not sure, which microstates belong to them specifically.

It's clear, that microstates 13 and 15 will belong to ${}^1D$, but how can I determine it for the other states? ${}^1D$ should contain 3 more microstates with $S=0$ and $m_L$ being equal to -1, 0 and +1, but there are multiple candidates for every configuration.

So how could I distinguish between, e.g. microstates 7 and 9 or 8, 11 and 14?

I don't understand this point, as even $J=2$ for ${}^1D$, so it can't be used to distinguish among the "similar" microstates and to assign them properly.

To add on to what Feodoran has mentioned, you might be interested in looking at the Clebsch–Gordan coefficients, which are essentially the coefficients of the microstates in the term symbols. You can look these up in e.g. the appendices of Atkins Molecular Quantum Mechanics, 5th ed. Note that each term symbol is a collection of states, so a $D$ term (for example) which has $L  = 2$ would comprise five states: one each with $M_L = -2, -1, 0, +1, +2$. These five states can in general be denoted as $|L, M_L\rangle$.

$L$, which is a total angular momentum, is obtained by a coupling of two individual angular momenta. For two electrons in p-orbitals, these individual angular momenta are $l_1 = 1$ (with $m_{l1} = -1, 0, +1$) and $l_2 = 1$ (with $m_{l2} = -1, 0, +1$). The microstates in the table specify both values of $m_{l1}$ and $m_{l2}$, so they are simultaneous eigenstates of $m_{l1}$ and $m_{l2}$, i.e. they can be denoted by $|m_{l1},m_{l2}\rangle$.

Now, in general each coupled representation $|L, M_L\rangle$ will be a linear combination of the microstates, or uncoupled representations $|m_{l1},m_{l2}\rangle$. It is in general not possible to construct a state which is simultaneously an eigenstate of $L, M_L, m_{l1}$, and $m_{l2}$, because the relevant operators do not commute (refer to Atkins or any other QM text on coupling of angular momentum). There are examples of one-to-one correspondence between the coupled and uncoupled representations – for example, microstate 13 perfectly corresponds to $|L = 2, M_L = +2\rangle$, but it is not general, so:

$$ |L, M_L\rangle = \sum_{m_{l1},m_{l2}} c^{L, M_L}_{m_{l1},m_{l2}} |m_{l1},m_{l2}\rangle$$

The Clebsch–Gordan coefficients are simply the $c^{L, M_L}_{m_{l1},m_{l2}}$ in the expansion above. Usually the notation is slightly different – so instead of $(L, M_L, m_{l1}, m_{l2})$, it is usually $(j, m_j, m_{j1}, m_{j2})$ – but the concept is entirely the same.

Using the coefficients for $j_1 = j_2 = 1$ in Atkins (which is relevant to the $\mathrm{p^2}$ configuration of atomic carbon), we find that the $^1S$ term is given by

$$|L = 0, M_L = 0\rangle = \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |m_{l1}= -1, m_{l2} = 1\rangle}_{\text{a combination of microstates 8 and 11}} - \underbrace{\frac{1}{\sqrt 3} |m_{l1}= 0, m_{l2} = 0\rangle}_{\text{microstate 14}}$$

It's slightly more confusing because there is a spin component as well, which the coefficients do not cover, and this is why I've given a handwavy "combination of microstates 8 and 11". This "combination" is needed to generate the singlet spin wavefunction $2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle)$.

As an aside, the familiar singlet and triplet spin wavefunctions

$$\underbrace{|\alpha\alpha\rangle, |\beta\beta\rangle, \frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{triplet spin wavefunctions}}, \underbrace{\frac{1}{\sqrt 2}(|\alpha\beta\rangle + |\beta\alpha\rangle)}_{\text{singlet spin wavefunction}}$$

are also coupled representations. These four wavefunctions are, respectively,

$$|S=1, M_S=+1\rangle, |S=1, M_S=-1\rangle, |S=1, M_S=0\rangle, |S=0, M_S=0\rangle$$

(Different notation again, but it is the same concept.) The uncoupled representations are the states in which each electron has been assigned a spin, i.e. $|\alpha\alpha\rangle, |\alpha\beta\rangle, |\beta\alpha\rangle, |\beta\beta\rangle$. For example:

$$|m_{s1} = +1/2, m_{s2} = +1/2\rangle = |\alpha\alpha\rangle$$

and so we can see a simpler example of how the coefficients link the two representations:

$$\begin{align}
|S=1, M_S=+1\rangle &= |m_{s1} = +1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=-1\rangle &= |m_{s1} = -1/2, m_{s2} = -1/2 \rangle \\
|S=1, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle + \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle  \\
|S=0, M_S=0\rangle &= \frac{1}{\sqrt 2}|m_{s1} = +1/2, m_{s2} = -1/2 \rangle - \frac{1}{\sqrt 2}|m_{s1} = -1/2, m_{s2} = +1/2 \rangle \\
\end{align}$$