Term symbol for atomic nitrogen: why can't S be -1/2? The spectroscopic notation for an element is given by $^{2S+1}L_{J}$. I want to use this notation for nitrogen with configuration

$$\mathrm{[1s^22s^22p^3]}$$

This is what I did. $l=1$ for p orbital. There are three electrons in p orbital, two electrons: spin up and spin down, and the remaining electron in spin up position, so that my $$S = \frac 12 + \left(-\frac{1}{2}\right) + \frac12 = \frac12$$ which gives

$$J=l+s=1+1/2=3/2$$

My query is can't the remaining electron in p orbital be in spin down position so that my $S$ becomes $-1/2$ and so that $J=l+s=1/2$.

Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?

In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.

The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.

To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons

$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$

Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get

$$S_{12} = 1, 0$$

and now if we couple this to the spin of the third electron we get

$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$

Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:

$$\begin{align}
S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\
S_{12} = 0 &\Longrightarrow S = \frac 12
\end{align}$$

Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.

In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.

Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.

$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$

$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$

and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.

1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.

It can be spin down, but that would give you a doublet state, compared to a quartet if all electrons were spin up. The ground state is the one with the largest multiplicity.


Can solutions of polar covalent compounds conduct electricity? I learned in class that solutions of polar covalent compounds are weakly conductive, while ionic solutions are strongly conductive.

But I'm getting different answers online. According to this lecture, "They do not conduct electricity in the liquid state, or when soluble in water, do not conduct electricity in aqueous solution."

I saw in lab, however, that a dextrose (D-glucose) solution weakly conducted electricity, and so did that of levulose (D-fructose).

So can solutions of polar covalent compounds conduct electricity? If so, how? Since it's covalent, it's not like anything can dissociate to form ions, right?

Consider for example this lab where polar covalent bonds are said to be strong, weak or non-electrolytes : ELECTROLYTES AND NONELECTROLYTES CHEM 151

The part of the molecule which dissociates has strongly ionic character but the molecule itself is described as being polar covalent. Not not all bonds in the molecular have the same level of covalent character, but most are covalent and it is polar.

There are also other ways for molecules to allow conduction, conjugated systems can do it by a very different process see for example the wiki on Conductive Polymers


Reactivity of different alcohols with NaNH2 and CH3I Identify the final product (B) in the following reaction:

I'm not sure how $\ce{NaNH2}$ and $\ce{CH3I}$ will react with the different types of alcohol present here. According to my solution,ether was formed in place of $\ce{CH2OH}$ in the end, whereas the answer says that $\ce{CH2OH}$ group will remain unaffected. What am I doing wrong? Can anyone explain to me the full mechanism?

I suspect, possibly just to a minor extent, that you have also managed to produce an expected photolysis product here (see below).
Interestingly, the photo sensitive methyl iodide, in the presence of a suitable catalyst ($\ce{CO2, Cu,}$..) with lab light (rich in UV) may have commenced in a radical reaction chain in the presence of ethanol:
$$\ce{CH3I ->[$h \nu$] ^.CH3 + I^.}$$
$$\ce{C2H5OH + I^. -> HI + C2H5O^.}$$
$$\ce{C2H5O^. + C2H5O^. -> (C2H5)2O2}$$
$$\ce{(C2H5)2O2 + ^.CH3 -> (C2H5)2O + CH3O^.}$$
....(the reaction chain continues)
where, I have stopped at the formation of Ether, albeit possibly created in very small amounts but detectable by its smell, nevertheless.
Repeat the experiment controlling for light/catalyst and see if you can still smell the Ether.
My advice, any time working with a photoreactive iodides, be mindful of light exposure.
[EDIT] The implication of my analysis, check your product yields, reagents and procedure, as a minor photo-induced side product may not (but still could) be connected to your results. In particular, I would start with the $\ce{NaNH2}$ as per Wikipedia:

$\ce{NaNH2 + H2O → NH3 + NaOH}$
$\ce{4 NaNH2 + 7 O2 → 2 Na2O + 4 NO2 + 4 H2O}$

which suggests that either exposure to moisture/water/air can degrade the compound.


Electron flow: How does osmium tetroxide react with alkenes? This is a question that has bothered me for some time. I have always thought that the part of one molecule will only attack part of another molecule if it has a higher positive charge. So with this in mind, why does the alkene double bond attack the $\ce{O}$ atom of $\ce{OsO4}$. Isn’t the $\ce{O}$ atom negatively charged?

I have a suggestion that could explain the attacK of electron pair of alkene pi bond to the oxygen of osmium tetroxide : The electron pair of alkene pi bond repels the electron density of the oxygen when the osmium tetroxide approaches the alkene, creating a temporary δ+ on the oxygen that is closer to the pi bond of alkene. The electron-rich alkene then attacks the electron-poor oxygen.

Dimitrios Kariotoglou

Yes. But check the formal charge on the Osmium. It is so very positive that it actually steals electron density from the Oxygen.


What are the maximum number of electrons in each shell? In my textbook, it says that the maximum number of electrons that can fit in any given shell is given by 2n². This would mean 2 electrons could fit in the first shell, 8 could fit in the second shell, 18 in the third shell, and 32 in the fourth shell.

However, I was previously taught that the maximum number of electrons in the first orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the sixth orbital.  I am fairly sure that orbitals and shells are the same thing.

Which of these two methods is correct and should be used to find the number of electrons in an orbital?

I am in high school so please try to simplify your answer and use fairly basic terms.

Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n.

To answer your question...

In the first shell (n=1), we have:

The 1s orbital

In the second shell (n=2), we have:

The 2s orbital
The 2p orbitals

In the third shell (n=3), we have:

The 3s orbital
The 3p orbitals
The 3d orbitals

In the fourth shell (n=4), we have:

The 4s orbital
The 4p orbitals
The 4d orbitals
The 4f orbitals

So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher n. The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell.

Now for the different kinds of orbitals
Each kind of orbital has a different "shape", as you can see on the picture below. You can also see that:

The s-kind has only one orbital
The p-kind has three orbitals
The d-kind has five orbitals
The f-kind has seven orbitals

Each orbital can hold two electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold two electrons because they each have only one orbital.

The 2p, 3p, 4p, etc., can each hold six electrons because they each have three orbitals, that can hold two electrons each (3*2=6).

The 3d, 4d etc., can each hold ten electrons, because they each have five orbitals, and each orbital can hold two electrons (5*2=10).

Thus, to find the number of electrons possible per shell

First, we look at the n=1 shell (the first shell). It has:

The 1s orbital

An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons.

The n=2 (second) shell has:

The 2s orbital
The 2p orbitals

s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons.

The n=3 (third) shell has:

The 3s orbital
The 3p orbitals
The 3d orbitals

s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons.

Therefore, the formula $2n^2$ holds! What is the difference between your two methods?

There's an important distinction between "the number of electrons possible in a shell" and "the number of valence electrons possible for a period of elements".

There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell.

The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this:

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$$

An easy way to visualize this is like this:

The first shell can carry up to two electrons, the second shell can carry up to eight electrons.

The third shell can carry up 18 electrons, but it is more stable by carrying only eight electrons.  There is a formula for obtaining the maximum number of electrons for each shell which is given by $2n^2~\ldots$ where n is the position of a certain shell.


Justifying the tremendous −I effect of trimethylammonium group by molecular orbital theory The $\ce{-[NMe3]+}$ group is ranked almost at the first position in the −I effect series, which is often given at the very beginning of the teaching of general organic chemistry.

I know this can be justified by hyperconjugation, but I am interested in a rigorous justification by molecular orbital theory.

Powerful electron withdrawing substituents, like $\ce{-NO2}$ can be interpreted as having a large effect on the $\pu{\pi}$-system of an aromatic ring. Resonance structures can be drawn that put a positive charge at the ortho and para positions and put the electron out in the nitro group, on an oxygen.

This must be adjusted for $\ce{-N(CH3)3+}$, because it has no $\pu{\pi}$ electrons, and we can't draw structures with a double bond between the nitrogen and the ring carbon. It works thru a very strong $\ce{-I}$ effect. However, if we use the concept of hyperconjugation, and donate a $\pu{\pi}$ electron from the ring to the nitrogen and then a methyl group (making it $\ce{-CH3}$), unbounded to any other atom, we can imagine the oxygens in the picture replaced with methyls and will get the same distribution of charge in the ring: a positive charge at the ortho and para positions. This will result in meta substitution if any reaction does occur.

For $\ce{-N(CH3)3}$, it's like hyperconjugation in reverse. Interestingly, while $\ce{-N(CH3)3}$ makes it very difficult to substitute, and always meta, $\ce{-NH3+}$ is only a weak meta director.

Reference

Advanced Organic Chemistry, L.F. Fieser, 1961, p628

$\ce{Me}$ is a pushing group which pushes electron towards the nitrogen atom. As a result electron concentration is towards the nitrogen atom.

The anti bonding electrons of nitrogen are shifted to the bonding orbital of nitrogen. This is because the bonding orbitals are closer to the more electronegative atom which in this case is nitrogen.

So nitrogen becomes more "electron hungry" and in order to stabilize itself tries to pull electron towards itself and thereby renders a -I effect.


Why are halogens ortho para directing even though deactivating? Halogens bonded to benzene ring has three lone pairs. These three electron pairs can cause resonance in benzene ring. But, halogens are also highly electronegative and thus they have strong -I effect. So, they are deactivating groups. But, why are they ortho para directing?

Let's take as an example a typical halogen, chlorine. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through the inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution

Image courtesy: NCERT Chemistry Class 11th

Through resonance, a halogen tends to stabilise the carbocation, and the effect is more pronounced at ortho- and para- positions. However, the inductive effect is stronger than resonance here, and causes net electron withdrawal and thus net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack.

Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

We know that halogen have - I effect which deactivate the benzene ring but halogen are also carried +M effect which is produminished -i effect because +M effect  is greater than -i effect and as we know that all +M-effect operating group are ortho and para directing because in case of +M effect operating group the more electron density are found in ortho and para possition hence further the electrophilic substitution the elecrophile attack in ortho and para position.


What is the electronic configuration of La⁺? Second IP of La is most likely to second IP of which element:

  (1) Be
  (2) Ba
  (3) Ca
  (4) Zn

The given answer is calcium. I don't understand if the 6s electron or the 5d electron leaves on ionization. Any help is appreciated.

During ionization only the electrons in last shell leaves.So its 6s


Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene? Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene?  Shouldn't the former have a higher boiling point because it possesses a dipole moment? I think the van der Waals forces should be equal in both cases.

The difference between the melting points is small: ortho 180 °C, meta 172 °C, para 174 °C, so they are basically the same. It is just not possible to argue bigger or smaller with hand-waving arguments about dipoles or polarity etc. with something so complicated. However, when the thermal energy is increased, whatever intermolecular interactions exist become less important relative to the average thermal energy, and which now has a dominating and disruptive effect. Thus I would expect all isomers to have a similar boiling point.

P isomers has symmetrical structure .  Reasulting in high attracting forces b/w molecules and it give raise to high melting point


How to determine the chemical state of a reactant or product in a chemical equation? I am asked this question:

Write a balanced equation, including state symbols, for this reaction:
$\ce{calcium + hydrochloric acid -> calcium chloride + hydrogen}$

I answered this:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(s) + H2}$

This was the correct answer:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(aq) + H2}$

I thought that the $\ce{CaCl2}$ was a solid precipitate (ppt.) instead of aqueous(aq).
Here are the known chemical states:

solid(s)
liquid(l)
gas(g)
aqueous solution(aq)

Is there any method to work out the chemical state of a reactant or product in a chemical equation, if I know the chemical states of all of the other reactants?

According to these basic solubility rules:

Calcium chloride (CaCl2) is soluble. Hence, you write CaCl2(aq) in the equation.

You must first learn two things before hoping to solve any chemical problem:

You must learn how to write a balanced equation, with the correct numbers of atoms, before guessing about their physical state. For example, in your first equation, there is one $\ce{Ca}$ on the left-hand side and $\ce{2 Ca}$ at the right-hand side. This is not balanced.
You should realize that $\ce{CaCl}$ and $\ce{CaCl2}$ are not the same substance. When you say that you "typed $\ce{CaCl(s)}$ instead of $\ce{CaCl(aq)}$" you still are wrong. You should have said, "instead of $\ce{CaCl2(aq)}$".

Now to answer your question, there is no way of obtaining the physical state from the periodic table.

Simply speaking, you should here know that pure $\ce{HCl}$ is a gas at room temperature, and it is very soluble in water. In the pure state, $\ce{HCl}$ is called hydrogen chloride. Dissolved in water, $\ce{HCl}$ is called hydrochloric acid.


How is the change in Gibb's free energy a condition for spontaneity for reactions not happening under constant pressure conditions? According to what I have studied, the change in Gibb's free energy is only a condition for spontaneity for processes happening at constant pressure and temperature conditions.

Consider a reversible reaction happening in a closed vessel at a constant temperature.
We check the feasibility of the reaction at any instant using the following equation:

$\Delta G = \Delta G^{.} + RT\ln(Q)  $

As the reaction happens, the partial pressures of each gaseous component will keep on changing as well as the total pressure. Then how can we use Gibb's free energy for checking spontaneity?

Gibbs energy have an another equation which is:-

So according to this equation spontaneity of a reaction depends both on the enthalpy change as well as entropy change.
We all know that a system attains stability as the system's energy is lowered that is it has to release energy. If in a reaction energy is released enthalpy change is negative.
We also know that a reaction is favoured in the direction of increased chaos or a positive entropy change.

So from this we can see that Enthalpy change is negative and entropy change is positive for a spontaneous reaction to take place and the overall equation of

becomes negative that is Gibbs free energy must always be negative for a spontaneous reaction.....Can be applied to any reactions.
Hope this helps.


Why are there 3 columns in Group VIII of the Periodic Table? I'm currently reading in-depth about the layout of the Periodic Table, and I wondered why the table has 3 columns in its Group VIII:

As I understand, this is an old notation of the table, now deprecated. But why was Group VIII a triple-sized group?

The answer might be trivial, but it's interesting.

In the beginning of the $20$th century, the periodic table had $8$ columns, and not $18$ as today. Electrons and of course $spdf$ electrons were unknown. The numerous transition elements were included in this table by doubling the line between Argon and Krypton, then between Krypton and Xenon. It was going this way.
The $4$th period was made of two lines. So each square (each box) contained two elements written on two lines. The first seven atoms $\ce{_{19}K - _{25}Mn}$ occupied the columns $\ce{I - VII}$, and their oxidation state may always have been equal to the number of the column. Then the three atoms $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ did not belong to any column. We will come back to them later on. Then the seven next atoms $\ce{_{29}Cu - _{35}Br}$ had to be put in columns $1$ to $\ce{VII}$, under the corresponding elements : $\ce{_{29}Cu}$ in the same box and under $\ce{_{19}K}$, $\ce{_{30}Zn}$ in the same box and under $\ce{_{20}Ca}$, etc. up to $\ce{_{25}Mn}$ above $\ce{_{35}Br}$. With this system all atoms had at least one oxidation state equal to the number of their column. The same procedure was adopted for $5$th period, with $\ce{_{37}Rb}$ and $\ce{_{47}Ag}$ in the first column,$\ce{_{38}Sr}$ above $\ce{_{48}Cd}$, etc.
This table was useful, except nobody knew what to do with the three atoms  $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ having all valence $2$, and of course of the six corresponding atoms of the platinum group, immediately under them : $\ce{_{44}Ru}$ to $\ce{_{46}Pd}$ and $\ce{_{76}Os}$ to $\ce{_{78}Pt}$. There was plenty of discussion at that time about what to do with them. Usually these elements were "rejected" in the 8th column, with the noble gases. or they were put in a separate column made specially for them. This problem was finally solved with the discovery of the $spdf$ electrons.

Dimitri Mendeleev designed his periodic table so that elements with similar properties were grouped together and group VIII demonstrated the most similar characteristics within each element, more so than other transitional metals.

Things such as the oxidation state of +2 and +3 being common among those specific metals-this is the best reason I could find and they are more consistent with the 18 electron rule than other transitional metals, particularly Fe, Co, and Ni.

The old system was removed because we thought there was sound reasoning to Mendeleev's decision (maybe he saw things we didn't), but there wasn't, and neither did we find enough cause to keep it that way. It has only led to people being misled or distracted because their immediate thought is: "why are they together?". When really, there is no solid justification.

From what I could find, the rationalisation for it being removed wasn't discussed at length or maybe even at all, so the above paragraph is just my assumption.

References:

https://answers.yahoo.com/question/index?qid=20060609045851AAOi9a7

p.s. I'll add to this as/if I discover more.


Does potential difference or current drive electrolysis? Would a high voltage ( of the order of a few score kV ) drive electrolysis?

Or, does it require a large current and low voltage? Alternatively, does electrolysis require both large voltage & large current?

For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.

However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.

For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.

Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:

Work produced by electrical energy, $W = V\cdot I\cdot t$

For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}$

Where $E^\circ_\text{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\Delta G^\circ$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).

For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.

$\ce{H2O(aq) ->[\text{elect}] H+(aq) + OH- (aq)}$

At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.

$\ce{4OH- (aq) -> 2H2O + O2 + 4e- }$ (anode half-cell reaction)

Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.

$\ce{2H+(aq) + 2e- -> H2(g)}$ (cathode half-cell reaction)

The overall cell reaction is therefore:

$\ce{2H2O(aq) -> 2H2(g) + O2(g)}$ ; $\Delta G^\circ=237.2\ \mathrm{kJ/mol}$

The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.

The minimum (theoretical) voltage required is:

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23\ \mathrm{V}$

However, in practice, a slightly larger voltage of 1.48 V is required since the  enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25 V.

Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.

High voltage and low current is more efficient but must be oscillated at specific frequencies to be effective low cost electrolysis. it is the oscillation that breaks the bond of hydrogen and oxygen not the current or voltage. Current can be below .5 amp at millions of volts and still be non life threatening like a stun gun. But you need to experiment with the frequency of oscillation to find the most efficient electrolysis.


Accuracy of FP-LAPW vs Plane Wave Pseudopotential: examples I am trying to learn about FP-LAPW software. I would like to know how well FP-LAPW outperforms Pseudopotential PW calculations. What are some properties/calculations/quantities in which FP-LAPW gives consistently better results than Pseudopotential PW theory? I know that FP-LAPW should give better results on almost everything (after all, it is an all-electron calculation). I am specifically looking for calculations where FP-LAPW gives much better results than Pseudopotential PW.

In all cases FP-LAPW will be better than Pseudopotential + Plane Waves calculations. However, FP-LAPW calculations are better at calculating properties which involve core electrons (e.g. core level shifts, hyperfine structures). Such calculations are often not possible with Pseudopotentials + Plane Waves, or are more complicated.This is because PP+ PW calculations do not treat core electrons explicitly.  Furthermore,  FP-LAPW calculations do not depend on pseupotentials. Although one can test pseudopotentials to check for transferability, there is no guarantee that the pseudopotential works for the system that you are simulating. As a result, "artifacts" due to pseudopotentials can be avoided with FP LAPW.


What is the difference between ΔG and ΔrG? Consider the reaction

$$\ce{A -> B}$$

The reaction Gibbs free energy, $\Delta_\mathrm{r} G$ is given by the following equation
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q$$

Now what is the difference between $\Delta G$ and $\Delta_\mathrm{r} G$ ?

Which one of the above represents $G_{\text{products}} - G_{\text{reactants}}$ ?

I read in my book $\Delta_\mathrm{r} G$ is the slope of $G$ plotted against the extent of reaction at different instants. And at equilibrium, it is zero. So can we say that it has value at a particular instant while $\Delta G$ is for a process ($\ce{A}$ to $\ce{B}$ in my question) i.e. is for a time interval?

Does $\Delta G$ for the above reaction represent change in Gibbs Energy when 1 mole of $\ce{A}$ reacts completely to form $1$ mole of $\ce{B}$ ?

Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.

The Gibbs free energy change, $\Delta G$

You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:

$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$

If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.

But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.

The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.

So, a more proper question would read like this:

  $\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})]
\end{align}$$

We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.

$$\begin{array}{c|c}
\text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline
\ce{NH4NO3(s)} & -183.87 \\
\ce{N2O(g)} & +104.20 \\
\ce{H2O(g)} & -228.57
\end{array}$$
$$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$

(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:

$$\begin{align}
\Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\
&= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\
&= -1073.62~\mathrm{kJ}
\end{align}$$

Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.

Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

  In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. (These numbers are made up.)

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.

In general, one could write, for a chemical reaction,

$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$

where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.

The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$

As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.

However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.

First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.

Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.

Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have

$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$

and our expression for $\Delta_\mathrm{r} G$ is

$$\begin{align}
\Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\
&= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\
&= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}}
\end{align}$$

Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.

Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.

Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.

Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.

A diagram to sum up

Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.
Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!
Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.
If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.
The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat

Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.

In this context $\Delta G$ and $\Delta_\mathrm{r} G$ are different notations for two similar things, both are changes in Gibbs Free Energy. The "r" means, that you can only interpret it for a very specific equation.


Why the formation of PH4+ is difficult compared to NH4+? My textbook asks this as an objective question.
What I had in mind was that the lone pair on phosporous was more distributed as compared to nitrogen(since nitrogen is more electronegative and smaller in size ) and hence the statement. But the option given in the textbook says the answer is lone pair of phosphorous resides in almost pure s orbital.
How do you know if the lone pair is placed in pure s orbital or pure p orbital

The bond angle in $\ce{PH3}$ is about $92-93^\circ$, and by the percentage $s$ character formula i.e $\cos(\theta) = s/(s-1)$ you can calculate the $s$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $p_x$ ,$p_y$, $p_z$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.

We think that both P and N contain lone pair of electrons ,but this is not true if it comes to you donating that lone pair to hydrogen ion AKA H+
Here comes back bonding into action it simply means that if an atom has got lone pair ,it would distribute it to a group to which it is bonded if that group contains (or is) an atom  which has its orbital unfilled or empty or has an incomplete octet
First two reasons have got almost same meaning.
P's lone pair get distributed over its own 3D orbital hence availability of lone pair decreases in its pure orbital (lone pair can only exist in pure orbitals)so bond to H + is well ..... prohibited.

This explanation has been interpreted from Concise Inorganic Chemistry by J.D. Lee


Why is the molecular geometry of a molecule with 5 bonds not uniform? I'll note that I'm a student in a rather introductory course, and today we learned about molecular geometries.

I'm curious as to why certain geometries (say, $\ce{CH4}$, tetrahedral) are uniform (evenly distributed in a sphere) while other geometries, say "$\ce{SbCl5}$," (5 bonds to the central atom) are less-uniformly distributed - two of the atoms are linear and 3 of them of trigonal planar.

Is this just a simplification for teaching introductory students, or is this how the molecule actually behaves?  I would think that to achieve "equilibrium" from the $\ce{Cl}$ atoms repelling each other, the distribution of the $\ce{Cl}$ atoms would be perfectly normal around the sphere.

You may want to look into VSEPR theory. To a good first order approximation you can think of electron pairs and external atoms on a central atom as taking up about the same space.
This explains why water's bond angle (104 degrees) is very close to the 109.5 degrees exact equidistant angle associated with methane, or $\ce{AlCl4+}$.

Your example with $\ce{SbCl5}$ is a good example where the theory gets a bit more complicated than simply maximizing distance between steric centers. The bonds to the chlorines are not equal. And yes this is how the chemical would look: see the Wikipedia article on orbital hybridisation.

Now, as your example has 5 identical groups, inter-conversion can occur, where the axial group "switches" with an equatorial group. But at any moment there will be two distinct chemical environments around the compound.


Are the Phosphate and Sulfate ion diagrams with double bonds an invalid picture? Whenever you see an image of phosphate, it's always shown having 5 bonds. 3 single bonds to anionic oxygen atoms and a double bond to an extra neutral oxygen.

Not only that, but they're shown being arranged in a tetrahedral manner. All of this additionally applies to sulfate, only that it has 2 double bonds instead of  just one.

Now, from what I've been able to gather from the net, including this website is that this can be explained by invoking sp3 hybridization and a d orbital. The 3 anionic oxygens who already have a charge of -1 pair up with the 3 partially filled hybrid orbitals in phosphorous, while the remaining filled sp3 orbital on phosphorous donates it's electron pair to the empty orbital on an extra oxygen.

Once again, this is exactly the same for sulfate, except that it can only bond to 2 anionic oxygens while also being able to donate 2 electron pairs to other oxygen atoms.

This leaves us with 4 single bonds to oxygen atoms. This doesn't seem unusual at all. Nitrogen can easily take on 4 bonds just like phosphorous, and seeing how row 3 elements have a looser hold on their outer electrons, it makes sense that they can donate their complete electron pairs much easier than row two elements, seeing how their electron pairs would "stick out" more, making them much more available for bonding.

However, in Phosphate when you get to the 5th bond you have to explain it by being a little more creative. You could say that a p orbital on one of the oxygen atoms donates it's electron pair to an empty d orbital on phosphorous, creating and extra bond between phosphorous and one of the oxygens. The same thing happens in sulfate only that for some reason it can form two double bonds. (which I really wonder if that would be pushing the limits of stability at all. I mean wouldn't the two d orbitals kind of overlap at an awkward angle causing some kind of repulsion?)

Now, apparently from what I've heard, experimental evidence has shown that there is very little d orbital involvement in molecules like phosphate or sulfate, and they're practically all single bonds. So my question is, is this true? And if so, then why is the double bond picture still in practically every textbook, diagram, and piece of scientific literature that you come across? How did this picture of these ions emerge? I know it had something to do with linus pauling, but was this just accepted based on theory alone and no experimental evidence at all? Was there some experimental evidence for it in the beginning?

Sorry if it sounds like I'm asking too much, but I just find this whole situation so confusing. I'd appreciate your answers.

Recent experimental evidence shows that d orbitals are involved in the extra double bond of sulfate or phosphate ions. Therefore a stable sulfate or phosphate ion has 2 double bonds instead of only one to make the formal charge on sulfur or phosphorus zero. Any Lewis structure with expanded octet (periods 3 and higher) should have zero formal charge on the central atom.


Is it possible to measure light absorbance for analysis of lead concentration without photospectrophotmeter I’m designing an experiment which requires I measure final and initial lead concentrations for the determination of aqueous lead uptake by different mediums. Because lead concentrations would be measured on the ppm scale, I’d need to use a spectrophotometer to measure light absorbance by the water samplings. I could then use Beer Lambert’s Law to determine concentration.

With recent protocols surrounding COVID-19, it’s been difficult to contact an institution. I was wondering if there was any way I could measure light absorbance of my samples at home. Is there some other apparatus I could look into using (or perhaps even purchase)?

All the best!

Another possibility is the grease spot photometer (aka Bunsen photometer) German Wiki page. This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light. Also the darker the room, the better.

The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.

If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).

Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48

How to measure photometrically

so schematically we have:

o--U----|----o

+---d---+

with light sources o, cuvette U and the spotted paper |. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.

We have $I_{transmitted} \sim \frac{1}{d^2}$

Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:

$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$

$a$ and $b$ are determined by calibration - I use here $a$ instead of $2 \lg d_0$ because several choices are possible and in general the fitted value should be used rather than the theoretical one since it includes some effects in which practice differs from theory (e.g. differences between % transmission in the grease spot from % reflection outside).

Standard addition instead of calibration would be possible as well.

Perhaps you can add H2S to create PbS (as a starting step). Then, based on an extract from Wikipedia on Lead Sulfide:

  Although of little commercial value, PbS is one of the oldest and most common detection element materials in various infrared detectors.[12] As an infrared detector, PbS functions as a photon detector, responding directly to the photons of radiation, as opposed to thermal detectors, which respond to a change in detector element temperature caused by the radiation. A PbS element can be used to measure radiation in either of two ways: by measuring the tiny photocurrent the photons cause when they hit the PbS material, or by measuring the change in the material's electrical resistance that the photons cause. Measuring the resistance change is the more commonly used method. At room temperature, PbS is sensitive to radiation at wavelengths between approximately 1 and 2.5 μm. This range corresponds to the shorter wavelengths in the infra-red portion of the spectrum, the so-called short-wavelength infrared (SWIR). Only very hot objects emit radiation in these wavelengths.

some ideas. Perhaps examining a comparative study (based on known concentrations of Pb) of the changed in measured electrical resistance of a sample of PbS suspension taken from the solution after treating with the same volume of H2S followed by irradiation may provide a path. This not meant to be a complete answer, but a suggestion on a possible path to be explored.

[EDIT] To be clear, I am assuming that one cannot measure light absorbance, due to lack of equipment, for the purpose of establishing Pb concentration. But, one can perhaps measure electrical resistance. The question is can one construct a path (actually, more likely a graph to which a curve is fitted) to determine Pb concentration from the change in electrical resistance recorded upon applying a fixed amount of infrared radiation to a solution where varying amounts of PbS is introduced. If experimenting with known starting Pb concentrations, converted to PbS, suggests a detectable and measurably path with some degree of goodness-of-fit, then yes, else no (which is the only answer provided so far).

Here is some background on PbS based photodetectors where irradiation exposure varies and the amount of PbS concentration remains fixed.

  PbS is a standard SWIR semiconductor detector (1 - 3.3 µm) whereas PbSe is used in the MWIR range (1 - 4.7 µm when uncooled; up to 5.2 µm when cooled). Our lead salt detectors are photoconductive; the detector resistance is reduced during illumination. The crystal structure is polycrystalline and is produced via chemical deposition.

Please note the actual size of the PbS based photodetector.

Note, my proposed study of feasibility avoids the use of a nephelometer (which can measure the concentration of some inert suspended particulates in say a liquid colloid by employing a light 'source' beam and a second light detector at 90° to the source beam) by taking advantage of the properties of irradiated PbS particulates. Also, it should be noted that a spectrophotometer, based on a quantitative measurement of the reflection or transmission properties of a material as a function of wavelength, based again on light, is also presumably not an available option.

[EDIT EDIT] Just found a reference, THE RELATIONSHIP BETWEEN ELECTRICAL RESISTANCE AND DISPERSED PHASE CONCENTRATION IN OIL IN WATER EMULSIONS to quote:

  The electrical resistance of a series of oil‐in‐water emulsions has been measured, and used to test the validity of four equations relating resistance with the concentration of dispersed phase. A modification of one of the equations was found to give the best relationship.

In the current context, we have a suspension of irradiated PbS in water (releasing electrons, per this 2013 article 'Photoinduced electron transfer from PbS quantum dots to cobalt(III) Schiff base complexes: light activation of a protein inhibitor'), which I would have guessed is more promising in determining the PbS concentration than say of oil in a water emulsion.

Also found a new method to quote a source:

  In summary, we have demonstrated a practical way to measure nanoparticle concentration in a colloidal solution using quartz crystal microgravimetry. Application of a small drop of the nanoparticle colloid in a volatile organic solvent to the crystal surface leaves a dry nanoparticle film after solvent evaporation. Crystal resonant frequency shifts obeyed Sauerbrey’s equation for the dry nanoparticle concentrations up to 1300 µg/mL, as calibrated using a set of serial dilutions of Si and Ag nanopowders in methanol, rhodamine B in methanol, and ferrocene in cyclohexane.


How to calculate the pH of a buffer after dilution You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?

The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:

$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Adding water means diluting the buffer.

Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.

Calculate the number of moles in the original buffer solution - you should get around 0.26 (that is using ur n=cv equation), now add up the new volume to the old one which would give u 0.884L then use your other mole number you got earlier to use in the n=cv equation to get new concentration, it should give around 0.294. No use pH=pKa+log([A-]/[HA]) equation bearing in mind that you just calculated the new concentration of ([A-]/[HA]) so just substitute it with the 0.294 and u should get the new pH to be 4.22 or the 4.2.


Is there a simple way to separate deuterium oxide from tap water? I understand that tap water contains minute amounts of deuterium oxide (heavy water). Is there a simple way to separate, concentrate the $\ce{D2O}$?

It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)

Electrolysis of water strongly favors "H" being converted to hydrogen gas rather than "D", by a factor of about 8 to 1 (depending upon the electrodes).

Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$ Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method.  The article says the process is suitable for "those who have occasion to prepare or concentrate heavy water for research, demonstration or
advanced student laboratory use".

In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains.  This increased the original concentration of deuterium by a factor of 12.

By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage.

Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level.  They used three stages.  Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.

Because, Deuterium is twice as Heavy as Protium (the more abundant form of Hydrogen) it gives the molecule of water a greater moment of inertia and slows down its average rate of rotation.   Consequently it does not bond to an growing ice crystal quite as rapidly as the lighter molecules of water.  So if you bring water to a boil (to break up polymerization and increase the number of single water molecules) and then cool it rapidly to the point of fusion so there is less time for them to participate in polymer groups the ice will have a greater percentage of light molecules than normal water and the water that is not frozen will have a slightly higher percentage of deuterium.  Instead of 110 ppm you might get it to 112 parts per million.  So the water that didn't freeze is brought to a boil again and then rapidly cooled to the point of fusion removing even more of the lighter molecules.  A very slow process with the only difficulty being the rapid cooling needed to avoid excessive polymerization. It requires far simpler apparatus than distillation or electrolysis as given in the first answer, but is much less efficient and the boiling phase would eventually use more energy.  However, the question gave simmplicity as a criteria and this boil and freeze process is simple.


Measurement of cell constant Why is cell constant value measured through indirect methods (like using a solution whose conductivity is already known)? Isn't it easier to  measure the length and area of cross section of a conductivity cell directly?

The mathematical description of the cell constant ($l/A$) does not include factors like fringing fields (the ionic current does not travel precisely in a well-defined path between the electrodes) and the actual surface area for rough surfaces.
The cross section of a round electrode may have different fringing fields depending on its separation from its companion. And if the two electrodes are of different geometry, you have to decide how to average the areas out. I think there was more standardization in the choice of electrode materials, size and spacing than has been carried over into the beautifully simple formula $l/A$, so when the formula is used outside of certain narrow bounds (and I don't know what they are), it's only an approximation, and resistance or conductance can be measured more precisely than the cell constant from geometry. So you might as well calibrate against a known cell or solution.

The conductivity value of any cell changes with the area of cross-section of the electrode. When we use an electrolyte of concentration within the desired area, the conductivity value varies. When a standard value of specific conductance is given, then from the observed conductance value we can calculate the cell constant.


Can any other common reagent other than than H3PO4 cause this conversion? Can any other common reagent other than than $\ce{H3PO4}$ cause this conversion? What about (conc) $\ce{H2SO4}$ or (conc) $\ce{HNO3}$ or (conc) $\ce{HCl}$?

Eaton's reagent - $\ce{P2O5}$ in Methanesulfonic Acid - is a good alternative.
JOC 62(1997) 3552

Conc. Nitric acid risks nitrating the aromatic ring. I have never seen conc. Nitric or conc. $\ce{HCl}$ used for this

I think any acid mentioned by you should work. It's just a electrophilic aromatic substitution.


Why did a former emulsion of oil and water precipitate after heating and cooling when heat made it originally? I added 100 g coconut oil, 100 g distilled water and 500 mg lecithin to an 1800 watt blender.  The motor from the blender has has a passive heating action.  I blended it at full for four minutes.  The result seemed to be a hot, uniform substance with a thin layer of foam on top.

I poured this mix into a smaller container and placed it in the fridge.  After a while it set into a solid, uniform substance with no apparent precipitation.

I then heated the substance in the microwave  for thirty seconds.  When I removed it, the components had separated and precipitated.  Cooling it in the fridge again resulted in a delicate solid sheet at the top which broke through to liquid underneath.

Why did the mix precipitate (presumably de-emulsify) after being microwaved but not occur when blended?

The simple explanation is that the speed of de-emulsification depends strongly on physical state and temperature.

If you have created a solid emulsion by cooling the initial blend, the de-emulsification will be very slow as there is little scope for the components to move around at all never mind separate. warming this solid up so it liquefies creates two factors that make separation easier: the mixture liquefies and the temperature rises, creating many easier ways for separation to occur. Once separation has occurred only re-blending will restore the emulsion.

Emulsions often separate slowly (milk, for example, slowly forms a layer of creamier fat on top but it doesn't happen quickly after thorough mixing). In your case the initial cooling "froze" the emulsion but the microwave heating enabled de-emulsification to occur by heating the mix. Cooling the result without blending is not going to restore the emulsion without further blending.

How long did yoi leave it to settle? Maybe it takes yet longer time to settle down than what you gave it?

Another idea is that either microwaves or heat influenced the emulsion structure and broke the oil-water complex?


Why do things want to lower their energy? Like if we consider a container of gas molecules as a system, Why does it want to lower its energy?

I have read that always more spontaneous events are favored. So Gibb's Free energy should have a negative value i.e ΔG=ΔH-TΔS<0. So Entropy Contributes towards more spontaneity. But Randomness contributes towards more entropy and having more Internal energy causes more random motion.
So overall more energy should contribute towards more Spontaneity which is favored in nature. But we observe that things tend to lower their energy. Please Explain.

Edit: Well I guess I might have found the answer -
When a system lowers its energy, ΔH becomes negative ( due to it being exothermic) so at moderate temperatures, the entropy contributes to lower spontaneity but ΔH is more negative so ΔG becomes more negative(more spontaneous). While at high temperatures, endothermic processes are favored (ΔH positive)
Thermodynamics is so beautiful!

Maybe my shortest answer ever:

Energy is loosely a measure of how disturbed a system it is from its natural, grounded, non-excited state, and it is from here that the "less energy, more stable" thing comes from. Your question reverse-engineers that.

It is an observation coming from the study of the gravitation. If you allow an object situated on a table to move somewhat, it may happen that the object is at the edge of the table, and fall to the ground. The contrary never happens spontaneously.


Constitutional isomers for four carbon thioesters Question:

  How many isomers are there for thioesters with the formula $\ce{C4H8OS}$?

Answer given is four.

I can see that the following two are possible constitutional isomers:

But what are the other two?

One way to break things down in by the number of carbon atoms on either "side" of the thioester.

There are 4 carbon atoms to work with, of which one will be taken up by the thioester moiety.  Thus there are three left to work with.

On the carbonyl "side" of the thioester, between zero and three carbons are possible.  However, on the sulfur "side" of the thioester, at least one carbon atom is required; otherwise the compound is not a thioester but a thioacid.  Thus the possibilities for carbon atom distribution are:

one carbon on S side, two on CO side.
two carbons on S side, one on CO side.
three carbons on S side, zero on CO side.

You have drawn the only structures for the first two possibilities.

For the last possibility, there are two possible ways to attach a three-carbon alkyl chain to the thioester sulfur:

as a $n$-propyl group
as an isopropyl group

Thus, the four possible thioesters are:

methyl thiopropanoate
ethyl thioacetate
n-propyl thioformate
isopropyl thioformate

if you get stuck drawing more than 2 isomers what i do is take a carbon out and put it somewhere else if you need any more help lmk and on 3 i messed up on the chemical equation but the structure is correct


Why is the cyanide ion toxic? As the title implies, what is the molecular basis of cyanide toxicity? I did some searching around at the CDC and it only states that it prevents cells from using oxygen. I also read how it could take as little as $131~\mathrm{ppm}$ to kill. How can a simple $\ce{CN-}$ ion do so much damage?

Cyanide is a pretty good ligand for coordination compounds. The electron pair on carbon (which, incidentally, also carries the Lewis structure’s formal charge) is located in the HOMO — much like as in $\ce{CO}$, whose molecular orbitals can be found in this answer by Martin (replace oxygen with nitrogen to arrive at $\ce{CN-}$) — making it a good Lewis base and a good σ donor.

It is remarkable for being a strong-field ligand, i.e. one that often creates low-spin complexes even for 3d-metals (which generally prefer high-spin if there are no competing reasons for low-spin). While one could draw analogies to $\ce{CO}$ or even $\ce{NO+}$ whose π-antibonding orbitals are strongly Lewis acidic and participate in metal to ligand backbonding, it turns out that for cyanide the main reason is a high covalency of the metal–ligand-σ-bond due to the relevant orbitals being very similar in energy. Thus, cyanide is unique in that it both stabilises higher oxidation states by being an anion and low-spin complexes due to its binding modalities.

One might think, in analogy to $\ce{K4[Fe(CN)6]}$, that cyanide preferentially binds to haemoglobin’s iron(II) centre and thereby inhibits oxygen transport, but that is in fact not the case as this centre requires a less-oxidised iron(II). Oxygen (due to the resulting redox-mechanism) and carbon monoxide are much better ligands for haemoglobin.

What cyanide does strongly inhibit is cytochrome c oxidase, the last enzyme in the respiratory cycle which collects electrons from four cytochrome c’s and transfers them to an oxygen molecule to create two water molecules:

$$\ce{4 Fe^2+-cytochrome~$c$ + 4 H+ + O2 -> 4 Fe^3+-cytochrome~$c$ + 2 H2O}$$

This is done by a plethora of metal centres consisting of two haems, to cytochromes and two copper centres. I can’t tell you exactly which one, but there will be at least one which needs accessable iron(III) at some point in the catalytic cycle to which $\ce{CN-}$ can bind well and strongly inhibiting the catalysis of the enzyme.

Since the inhibited process is directly responsible for cellular respiration by transferring electrons to molecular oxygen (and being the only enzyme dealing with molecular oxygen in the respiratory chain), inhibition of this protein will cause the breakdown of the entire respiratory chain and lead to cellular suffocation.

Considering that an amount of $200~\mathrm{mg}$ can be toxic according to Wikipedia, and also considering that a human being contains some 10 trillion cells also according to Wikipedia, that’s still some 400 million cyanide ions per cell demonstrating the efficiency of binding and inhibition.

Note that popular antidotes also directly address the fact that cyanide is a strong ligand:

One antidote, hydroxycobalamin, presents the cyanide ion with a $\ce{Co^{III}}$ centre and a weakly bound, easily displaced hydroxido ligand. Hydroxide is displaced by cyanide creating cyanidocobalamin, a variant of vitamin B12 that the body has no problems dealing with. It is ejected via the kidneys. Injected into the circulatory system, it scavenges cyanides before they can enter cells.
Another antidote, also injected into the circulatory system, is 4-dimethylaminophenol (not to be abbreviated as DMAP to prevent confusion with 4-dimethylaminopyridine). It is an oxidising agent designed to oxidise haemoglobin’s iron(II) to methaemoglobin (containing iron(III)) thereby creating a much better acceptor for cyanide. Since there is a lot of iron(II) (a quick estimate with $8~\mathrm{mmol/l}$ gives $40~\mathrm{mmol}$ haemoglobin in human blood) in the blood for oxygen transport, small amounts of this can be sacrificed to capture cyanide before that reaches the respiratory cycle.

Cyanide is made up of a nitrogen triple bonded to a carbon with a lone pair, [C≡N]−. (Carbon has 4 valence electrons and Nitrogen has 5; pairing will result in a triple bond, a pair of electrons around the nitrogen, and an extra electron pairs with the 4th electron around the carbon). Carbon does not like to have a 5th electron, it is unstable, even the electron charge is shared between the Carbon and Nitrogen. The resulting negative charge (and triple bond) means that cyanide will be very reactive and want to form another bond.

As a result, cyanide is particularly deadly because it will bond with iron, (usually Fe3+ or Fe2+). In our body, iron is found in protein in the electron transport chain. When cyanide inhibits the electron transport chain by binding to iron, the cell can no longer produce energy.

Cyanide is also deadly when it bonds with hydrogen to create hydrogen cyanide, or HCN. HCN can be inhaled, where it will bond with iron, and result in cell death.


Why are there free electrons in a metal? Recently, we covered metallic bonding in chemistry, and frankly, I understood little.

I understand that:

Metals bond to each other via metallic bonding
Electricity can flow via free or delocalized electrons

But, I do not understand why the metal atoms turn into ions and delocalize the electrons, why don't the metal atoms stay as atoms?

When electricity flows, the electrons are considered "free" only because there are more electrons than there should be, and because the transition metals, such as iron, copper, lead, zinc, aluminum, gold etc. are willing to transiently accept and give up electrons from the d-orbitals of their valence shell.

Transition metals are defined in part by their stability in a wide range of "oxidation states"; that is, in several combinations of having too many or too few electrons compared to protons. This is thought to be because of the d orbital in their valence shells. Compared to the s and p orbitals at a particular energy level, electrons in the d shell are in a relatively high energy state, and by that token they have a relatively "loose" connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material, usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely). This impetus can be caused by many things, from mechanical impact to chemical reactions to electromagnetic radiation (aka light, though not all of it visible); antennas work to capture radio frequencies, because the light at those frequencies induces an electric current in the wire of the antenna. Now, in the absence of a continuous force keeping the electron in this higher energy state, the electron (and the metal atoms) will naturally settle into a state of equilibrium. Electricity is generated when just such a force is acting on the metal, giving energy to the electrons in the d orbital and forcing them to move in a certain direction.

This impetus can come from many sources, as discussed, be it the movement of a magnet within a coil of wire, or a chemical redox reaction in a battery creating a relative imbalance of electrons at each of two electrodes. The end result is that the electrons, given additional energy from this voltage source, are ejected from their "parent" atom and are captured by another. The "holes" left behind by these electrons are filled by other electrons coming in behind them from further back in the circuit. Thus, the energy provided by the voltage source is carried along the wire by the transfer of electrons.

The analogy typically made is to the flow of water, and it generally holds in many circumstances; the "voltage source" can be thought of as being like a pump or a reservoir, from which water flows through pipes, and the amount of water and the pressure it's placed under (by the pump or by gravity) can be harnessed to do work, before draining back to a lower reservoir. The pipes are similar to wires in many ways; the larger the diameter, and the smoother the inside of the pipe, the more and the faster water can flow through it (equivalent in many ways to the thickness and conductivity of the metal wire), and when under enough pressure (high enough voltage), the pipes will actually expand slightly and hold more water than they would at low pressure (this is a property of wires and other electrical conductors called "capacitance"; the ability to store a charge while under voltage and to discharge it after the voltage is released).

In the view of physics,there are many atoms in the given element  for instance copper ,in which has no energy gap between valence band and conduction band .when external energy applied to the given element.the free electrons will more energy and gets  free from valence band and jump in to conduction band and moves.


What is hyperconjugation? What is hyperconjugation and how do you know if there is hyperconjugation in a compound? Which orbitals are involved in hyperconjugation?

In the ethyl cation, one can write the following resonance structures:

The structure on the right can be actually written 3 times since there are 3 hydrogens available. Note also that the "ethylenic-like" structure on the right involves a $\mathrm{sp^3-p}$ bond (the $\ce{C-H}$ bond is $\mathrm{sp^3}$ hybridized and the cationic carbon has a $\mathrm{p}$ orbital), so it is not a true pi bond ($\mathrm{p-p}$) like that found in ethylene.  The resonance structure on the right is meant to show the "hyperconjugation" effect of the $\ce{C-H}$ bond ($\mathrm{sp^3-s}$) with the $\mathrm{p}$ orbital located on the adjacent cationic carbon.  The drawing is a shorthand way of saying that when the $\ce{C-H}$ bond is rotated into the same plane as the $\mathrm{p}$ orbital on the cationic carbon, there can be a stabilizing overlap - and it is called "hyperconjugation." In the 2-propyl cation, there are 6 of these hyperconjugative resonance structures (there are 6 $\ce{C-H}$ bonds on the 2 methyl groups adjacent to the cationic carbon and in the t-butyl cation there are 9 such resonance structures. This can be used to explain why tertiary carbonium ions are more stable than secondary carbonium ions, which in turn are more stable than primary carbonium ions.

Hyper-conjugation is in general a stabilizing interaction which involves the delocalization of σ-electrons belonging to the $\ce{C-H}$ bond of an alkyl group attached to an atom with an unshared p-orbital. The σ-electrons of a $\ce{C-H}$ bond of an alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p-orbital.


Solubility of methanol and ethanol in water Which substance is the most soluble in water?

  $\ce{CH3OH}$
  $\ce{CH4}$
  $\ce{C2H6}$
  $\ce{C2H5OH}$

I wasn't able to decide between 1 and 4, as both have hydroxides, but the answer is 1. How?

For primary alcohols, the trend is the longer the chain, the less soluble.

However, at room temperature, methanol, ethanol and propanol are all completely miscible with water.

At lower temperature, below the freezing point of water, there is finite solubility.

See the phase diagram that is Fig. 1 in CRYSTAL GROWTH OF WATER ETHANOL MIXTURE and that in Fig. 1 of Solid-Liquid Phase Diagram of the System Methanol-Water

Which alcohol would you choose for windshield deicing solution?

How does a alcohol get soluble in water? The $\ce{H2O}$ molecule acts as bipolar and the positive pole forms some weak hydrogen attraction (name this 'bond' if you like)  with the oxygen of alcohol. Now it's obvious, the more bulkier the group around oxygen, the less space around oxygen to form this bond. Hence $\ce{CH3OH}$ is more soluble in water than $\ce{C2H5OH}$. After all, $\ce{-C2H5}$ group is more soluble than $\ce{-CH3}$ group.


If capsaicin is insoluble in water, what makes chilli water taste spicy? I have always been told that the 'heat' from chillies came from capsaicin which is oil based. Recently I began wondering how environmentally friendly chilli based insecticide/repellent sprays could work if they are water based. A chemist friend confirmed for me that capsaicin is not soluble in water.

I minced a chilli and added this to a coffee mug of boiling water, left overnight then strained through a coffee filter. The resulting chilli water is definitely spicy, but why? What has dissolved into the water (or is the mixture some sort of emulsion of tiny oil droplets?).

You got me curious, so I poked around a bit on this.

First there is insoluble and then there is insoluble. The Wikipedia article on capsaicin lists its solubility as $\pu{0.0013 g}/\pu{100 mL}$ which is $13$ parts per million. So capsaicin is "relatively insoluble", but not wholly so. Second the Wikipedia article also points out that capsaicin itself is just one of the various capsaicinoid chemicals which give peppers their "hot" taste.

At the end of the (old) Wikipedia article on the Scoville scale, a taste test for hotness, it states

  [...] A measurement of one part capsaicin per million corresponds to about 16 Scoville units. [...]

So, 13 pmm multiplied by 16 Scoville units per ppm is about 200 Scoville units.

A table on hot sauces and peppers lists the Original TABASCO® brand Pepper Sauce from McIlhenny Company as being 2,500 to 5,000 Scoville units. So 200 Scoville units would be about one tenth strength sauce. That should still be enough to tingle your tongue.

Lets make it simple with simple quantum mechanics. water is consists of Hydrogen and Oxygen. Oxygen have bigger nucleus so it pull electron in hydrogen but hydrogen won't let it go so Hydrogen trying to pull it in but unfortunately, Oxygen is heavier and have stronger nuclear force than hydrogen. Oxygen even able to pull 2 Hydrogen atom.

Okay, we know Oxygen is such a muscular macho savage guy with a mustache and every atoms are such a thug compton gangster but what exactly happen ?. Those make water polar. because the hydrogen electron is pulled towards oxygen. The unequal distribution of force inside hydrogen atom make it slightly positive due to the induced dipole during the bodning process. That's why we classify water as polar covalent compound. And of course oxygen have slightly negative polarity because it must maintain the stability of its own atom and bonding.

What is capsaicin ?. Capsaicin is huge hydrocarbon. based on the nucleus force analysis it is nonpolar. They have polarity but not much. they have negative polarity in C=O double bond. THey also have O-H in one of its end. And a little bit positive but in really-really tiny amount of polarity of N-H binding.

So, what will happen if I put capsaicin in the water. well, water is friendly Rey Mysterio and Capsaicin is unfriendly Big Show. Water attach to some of capsaicin polar ends and try to pull it. Unfortunately to pull one capsaicin molecule, Water need more energy because it is heavy and repulsive in much ends. But, C=O ends is seems attracted to water so, it would basically able break apart and bonding with water. C=O Double bond have negative polarity, Hydrogen in water have positive polarity so, it able to pull out Oxygen from capsaicin. Yes, without any energy given to the reaction it's can be (roughly) 95-99 % unsoluble in water.

But we know, of course, reaction sometimes requires energy. Say we need to shake it, Blend it, Boil it, Electrolysis, or even burn it. Just give it energy. The solubility able to increase. But, How far it would likely to increase ?. As you notice, in repulsion of homopolarity there's energy needed to destroy this capsaicin. It's the weakness, just put up an effort to have certain amount energy which is exert the repulsion energy between capsaicin and water.

So, what happen to you is natural. When you heating up water. The orbitals of water is changing due to the electron vibrations. There's some amount of kinetic energy absorbed and force electron to move. So, the shielding between electron in orbitals can be forced and the orbital shape is slightly changing.

That's why some of capsaicins are able to dissolved in water.

But, in your case you have coffee. Coffee is complex mixture of chemical compounds. Coffee is not only consists of caffeine it's also have a bit protein and a bit minerals. Such as potassium. So, the chemical reaction between capsaicin in coffee mixture is really complex.

But let's see why it's more dissolveable. Let's check caffeine. Caffeine is have double negative polarity C=O bond. It's enough to attract water and makes caffeine soluble in water. So, you try to dissolve capsaicin in strengthen water molecule. It's a handicap match between coffee + water againts the big show. So, who wins ? I predict the tag team win. The solubility will increase but still  I predict, it will not be completely dissolve in this mixture unless you put on some milk.


Does chemistry need machine learning? In many fields of science (e.g. biology, medicine, psychology, statistics, physics), machine learning and artificial intelligence techniques are becoming more and more popular to analyze data. Is it also so in chemistry?

The short answer is yes. Machine learning, data mining, AI and other techniques are highly useful in chemistry.

I completely agree with Fred's answer that lots of machine learning, expert systems and statistical analysis in chemistry goes back a long time. This is particularly true in analytical chemistry - match a mass spec or NMR or IR against a library of known compounds.

Now I saw your tag "computational chemistry" and there are some newer applications here. Basically the question is whether you can predict properties (e.g., the heat of formation, dipole moment, etc.) that would normally require quantum mechanics.

There are a few groups attempting this, but one of the more successful attempts so far has come from Anatole von Lilienfeld.

  The researchers found that with a landscape of more than 5000 molecules, the error for predicting atomisation energies of new molecules drops below 10kcal/mol, approaching the 5kcal/mol accuracy of hybrid DFT. 'Calculating a molecule's atomisation energy using hybrid DFT would take on average one hour on a single CPU,' says von Lilienfeld. 'With machine learning, it's milliseconds.'

There are more conventional approaches to use machine learning, evolutionary algorithms and the like to perform parameterization of force fields, semiempirical quantum mechanics, etc.

Basically, if you have a lot of data, machine learning techniques can be effective ways to analyze the data and use it for other purposes.

If we don't use artificial intelligence in chemistry , then chemistry will be the only field amongst all , not to have used that .
But the truth is that lot of artificial intelligence has already been depolyed in chemistry , without us being known that it is being used.

Artificial inteligence can solve almost all the problems in the chemistry.

An interesting article for this can be found in wikipedia
https://en.wikipedia.org/wiki/Artificial_intelligence_in_chemistry

Many chemist think that artificial intelligence is tough , but in fact it is very easy. A chemistry person with basic computer knowledge can learn it in 6-7 days , and can deploy it in millions of ways in Chemistry.

An application of artificial intelligence has been developed by Wiley .It is a software called chemplanner .It uses is that it can synthesize any organic molecule in the world that a human can think of , in matter of seconds , using literature of already existing  reactions in the world.
So , with the help of this artificial intelligence software by wiley , you can synthesize any organic molecule in the world .


Does ethanol "decompose" or "remove" phosgene from chloroform? A question I was asked was: "Why is ethanol added to chloroform bottles?". I know that chloroform can be slowly converted to poisonous phosgene gas, and that ethanol is added to prevent phosgene levels from rising. However, the options I was given were:

It decomposes phosgene by converting it into ethyl carbonate.
It removes phosgene  by converting it into ethyl carbonate

Is there a difference between the two wordings, and is one more correct than the other? The answer is supposedly "decomposes".

It's a weird way of phrasing it, but I think they are trying to lead you to this

Basically, chloroform is stabilized with ethanol or amylene, but if a sample is dried, it will no longer contain stabilizer. To actually remove phosgene, you can wash it with NaHCO3. This would essentially leave you with entirely harmless products.

This is my best guess as to what they are trying to get you to understand. Please let me know if this helps.


Decrease of volume by half in equilibrium - with specific numbers- how to solve? $$\ce{CO (g) + Cl2(g)<=>COCl2}(g)$$
  We have
  2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
  contained in 1 L.
  The system is currently in equilibrium.

  We then decrease the volume by half to 0.5 L

  What will be the new concentration in equilibrium of CO?

I tried calculating at first the constant of concentration;

$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$

I found it to be $4.8825$

then I calculated the concentration with the new Volume:

$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$

$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$

$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$

obviously this wasn't in equilibrium so I went to do this:

$\ce{CO + Cl2<=>COCl2}$

$4M-x$, $2M-x$,  $19.53+x$

(as the number of mols in the products is smaller and the system wants to fix the increase in pressure)

When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M ,
and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible?

I would be very glad for help with calculations and technique of solving as well!

I solved the quadratic equation by Wolfram Alpha

The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have
1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.


How does the equilibrium shift when concentration of reactant and product are increased simulatenously Before answering my question, take a look at this example
$$\ce{2NOCl <=> 2NO + Cl2}$$
Let's suppose I have $\pu{2 mol/L}~\ce{NOCl},~\pu{2 mol/L}~\ce{NO}$, and $\pu{1 mol/L}~\ce{Cl2}$ in a closed system. If we were to calculate the equilibrium constant, it would be
$$K_c = \frac{(2)^2(1)}{(2)^2}$$
Now suppose I increase the concentration of both $\ce{NOCl}$ and $\ce{NO}$ by $\pu{1 mol/L}$. The equilibrium will stay the same,
$$K_c = \frac{(2+1)^2(1)}{(2+1)^2};$$
however, if I were to instead increase the concentration of both $\ce{NOCl}$ and $\ce{Cl2}$ by $\pu{1 mol/L}$, the equilibrium will shift to the product side
$$K_c = \frac{(2)^2(1+1)}{(2+1)^2}$$
Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

In terms of rate of reaction, at equilibrium the forward rate is proportional to the concentrations of the reactant (as if this were an elementary reaction). Likewise, at equilibrium the reverse rate is proportional to the concentrations of the products (as this were an elementary reaction). If you change the concentration of a single reactant and a single product by the same factor (i.e. multiply by the same number, not add the same number) and if the stoichiometric factors are the same (i.e. the order of reaction in this special case), and the reactants are sufficiently dilute, the system will still be at equilibrium. In the second change described by the OP, neither is the case (one concentration increases by 50%, the other doubles, and the exponents are different).

In terms of thermodynamics, if the Gibbs free energy of the reactants is changed by the same amount (i.e. add or subtract a number) the system will remain at equilibrium. This will happen if one species on either side is changed by the same factor because the Gibbs free energy changes with the logarithm of the activity of a species. Again, this only works if the stoichiometric coefficient of the two species are the same.

Nomenclature K vs Q

In the equations shown by the OP, different concentrations are substituted into the equilibrium constant expression even when the reaction is not at equilibrium. The correct name for these expresssions is reaction quotient Q. If Q matches K, the system is at equilibrium. In the examples provided, Q stays the same in one case (still in equilibrium) and changes in the other case (no longer in equilibrium). What the equilibrium would be needs not to be calculated. Knowing that Q is unequal to K is sufficient for showing that the system is in fact not in equilibrium.

When a reaction shifts to the right or left direction, it means that the reaction is trying to remove the equilibrium stress. So removing or adding reagent or product in a reaction at equilibrium will not change the value of the constant.

You said you would increase the amount of substance of $\ce{NOCl}$ and $\ce{NO}$, but you did not add it into $\ce{NO}$ concentration.

Adding products and reagents concentration would change the constant's value, but not shift the equilibrium towards any side.


What is the minimum pressure for existence of a liquid solution? The minimum pressure for the existence of liquid solution is?

  $0.6\;\mathrm{atm}$
  $0.8\;\mathrm{atm}$
  $0.77\;\mathrm{atm}$
  $0.9\;\mathrm{atm}$

Although this seems to be an informative question, I want to know the process by which we get this answer.

I know that the critical pressure for a liquid solution is $p={ a\over {27b^2}}$ but what is the value of a and b in that case?

I expect an easy to understand answer for high-school level.

The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the van der Waals equation for nonideal gases:

$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$

These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$.

$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$

However, these constants are empirically derived, and thus different for every substance, and so the critical pressure would be different for every substance.

For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is

$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$

Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).

The minimum pressure for the existence of a liquid solution is just below 1 atm. Because at 1 atm pressure(vapour pressure) every liquid gets boiled off corresponding to the temperature at which it attains the atmospheric pressure as it's vapour pressure.


Cyanide hydrolysis; could it yield an oxime? The mechanisms for both acid and base catalyzed hydrolysis of nitriles are clear;

(With the subsequent deprotonation from the hydroxyl) and

(Source)
In both cases, the hydroxyl adds to the carbon while the proton adds to the nitrogen.
I thought up an alternate mechanism, with the proton attaching to the carbon, and the hydroxyl going to the thereby positively-charged nitrogen, yielding an oxime.
Why doesn't this occur? I thought of the fact that the nitrogen atom, with a lone pair, would cause any nearby protons to come straight to it, but I'd like clarification on that.

Yes, since nitrogen is more electronegative than carbon, and also since it has the lone pair available for taking, the proton is likelier to go to it.


Hydrogen displaces lead? there was a reaction in my book. It is as follows:

$\ce{PbO + H2-> Pb + H2O}$

My question is Hydrogen is below lead in the reactivity series. Then how is this reaction possible?

The activity series you're referring to is constructed for ambient conditions where the temperature is $25°\text{C}$ and water is a liquid.  But, of course, thermodynamic properties can depend on temperature and especially on phase.  At higher temperatures where water is ordinarily a gas, it has a lot more entropy than the liquid.  That makes oxidation of hydrogen to water more thermodynamically favorable than what appears in your ambient temperature series.

For high temperature reactions where water mixes with hydrogen in the gas phase, the entropy effect pushes hydrogen up the series so it roughly matches iron.

Look at it from a redox reaction perspective. I'm not sure what the parameters the reaction is taking place under but the Pb is going from Pb2+ to just Pb. This does make sense because lead oxide is commonly used as an oxidizing agent and hydrogen gas is considered to be a powerful reducing agent.
Hope this helps

Just verified this with my advisor. It's an odd reaction but it does work.
All the best


How can I lower the flash point of stearic acid (Palm Based Fatty Acids) I am trying to build a environmentally neutral firelighter that does not use petroleum based accelerant. I have found that the flash point is way to high to be viable in Stearic Acid (Hexadecanoic acid / Octadecanoic acid). Is it possible to have a "safe", "environmentally friendly" and biodegradable additive to lower the flash point? I am open to suggestions and alternatives. Any assistance in this is appreciated.

could you have a small amount of a more volatile liquid with the hexadodecanoic acid - such as white spirit or kerosene? This would support a flame and raise the temperature to the point at which the hexadodecanoic acid acid will burn.

Do you mean it is "too high"? ESL student? Setting aside your use of the word "safe", sure: use ethanol. It's conceivable that you do not know that "flash point" is a measure of a materials' safety with respect to fire. By definition, a material with a lower FP will be less safe than one with a higher FP (ceteris paribus). I'd suggest you study the history of the match, if you haven't already. You need to explain why a butane lighter isn't sufficiently "safe", "environmentally neutral" and "organic" (and dare we say "natural") for you. Same with safety and kitchen matches.
In broad strokes, I can think of nothing which will be both exothermically reactive enough to work and be safe enough; adding a volatile to a non-volatile to make it more volatile seems to (obviously) fail the safety test. To initiate a flame, you need an (intense) energy source and a fuel. This reaction must (with 99.999% certainty) occur in the gas phase. Vapors are one possibility (the most efficient) the other is very, very finely divided powders. I can imagine a "firestarter" which uses a flint and a puff of powder to initiate the flame.
More specifically, it is only the rare azeotrope which will lower the minimum boiling point of the mixture below that of the lower boiling component. I doubt the difference will be enough to help you. Based on the chemical functional groups present, the slope of the vapor pressure vs temperature curves are indeed different. You want a steep curve, but you'd have to look into that correlation yourself. (You don't explain why you can't just enclose your "firestarter" to prevent evaporative losses...) One of the oldest "rules" in chemistry is "like dissolves like". In other words acetic or propionic acid might have significant solubility in stearic acid (but would be acid (corrosive to metals) and stinky, although mostly - imho - "non-toxic"). This type of mixture might be a way to lower FP. (or how about salts like ammonium acetate? (or other nitrogen bases) IDK...) Actually, flash point is a lousy way to quantify a compounds' fire properties. It is used because it is simple. Fire safety experts (which I am not) find measures of energy required to create a flame (not to sustain it) at a given temperature and (gas phase) composition much better. Because fuels provide the source of energy to sustain the flame reaction, it turns out that flame initiation it the process which requires the most energy input per molecule. Chemicals differ in the ease at which they start burning, at a given VP (of course, VP and Temperature are highly correlated in or near equilibrium, but that's not necessarily what you have when starting a fire). I'd look at it like this: 1. Find the most acceptable chemical with the lowest energy requirement at the temperatures the device "sparks" at. 2. find the most acceptable fuel to sustain that fire. (Note that here I mean sustaining for a second or two, long enough to start the "real" fire (or candle, etc.)). This implies that the first thing you need to do is locate a "environmentally neutral" source for your spark. (Do batteries meet your criteria? They wouldn't mine, how about ferrocerium or flint and steel? Or, what seems to be the "most" environmentally neutral energy source, a magnifying lens (only works when sun is shining, however.))