Terminology of atomic spectroscopy: Difference Among Term, States and Level In A Primer on
Quantum Numbers and Spectroscopic Notation
Contents, the concept of a term and level is described as

  a) The level is the set of 2J+1 states with specific values of L, S,
  and J. The difference in the energy between two levels gives the
  wavelength or frequency of an atomic transition.

  b) The term is the set of levels characterized by a specific S and L.

1) Is there a better way to understand the subtle difference among the concept of a level, a state, and a term. A example would be helpful.

2) A side question is what are the German equivalents of term, state and level. What do they call it in atomic spectroscopy literature?
Thanks.

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of  Term, Level or State.

The Configuration such as  $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied.

These are split with electrostatic (Coulomb) coupling to form Terms.

The Term Symbol informs about the angular momentum. The number of levels in each Term is $(2J+1)$, the number of States is $(2L+1)(2S+1)$ .

The term symbol is

$$\large ^{2S+1}L_J$$

where $S$ is the total spin and $L$ the orbital angular momentum.  $J$ represents the angular momentum as quantum numbers $(J=L+S)$ and is determined by adding individual spin numbers according to Glebsch-Gordon series, i.e. if $q_1,\,q_2$ are angular momentum quantum numbers  then the series is all terms (separated by unit steps) with values $q_1+q_2 , q_1+q_2-1,\cdots \to |q_1-q_2|$. (There may be only be term).

The Terms are split with a magnetic interaction such as spin-orbit coupling, i.e coupling of spin angular momentum with that of the orbital. It is not present for S orbitals because here $L=0$.

Levels are distinguished by total angular momentum $J$, and are $2J+1$ degenerate.

An external magnetic field will remove all remaining degeneracy to form States. These are distinguished by $m_J$ and there are $2J+1$ of them.


Which equilibrium constant is appropriate to use? I have learnt that the standard free energy change is related to the equilibrium constant of a reaction by,
$$\Delta G^\circ = -RT \ln K$$

Here, does $K$ refer to $K_p$ or $K_c$?

Also, please give me the derivation of this formula.

On the net, I saw that we can use either $K_p$ or $K_c$. But doesn't that give two different free energy values?

Moreover, if the formula was derived for gaseous reactions (using $K_p$), how can we just extend this to other reactions saying that we can use $K_c$ as well?

As noted in this previous question, the correct definition of the equilibrium constant $K$ depends on activities. If you are interested in the derivation of the equation $\Delta G^\circ = -RT \ln K$ (which requires "proper" thermodynamics), read Philipp's answer to that question. For a reaction

$$0 \longrightarrow \sum_i \nu_i \ce{J}_i$$

(this is a fancy way of writing the reaction which makes sure that the stoichiometric coefficients $\nu_i$ of the reactants are negative), the equilibrium constant is defined as follows:

$$K = \prod_i a_i^{\nu_i}$$

where $a_i$ is the activity of the species $\ce{J}_i$ at equilibrium. (Note that since $\nu_i$ for reactants is negative, the terms for the reactants will appear in the denominator of a fraction.)

The question is, why are $K_c$ and $K_p$ taught at an introductory level despite them not actually being the equilibrium constant $K$? The answer is likely partly because it is difficult to introduce the idea of an activity, but also because $K_c$ and $K_p$ are good approximations to the actual equilibrium constant, $K$. To see why, we have to look at the mysterious quantity called the activity and see how it is related to the concentration of a solution, or the partial pressure of a gas.

If we assume certain ideality conditions, then we can write the activity of a gas as:

$$a_i = \frac{p_i}{p^\circ}$$

where $p_i$ is the partial pressure of the gas and $p^\circ$ is the standard pressure, defined by IUPAC to be equal to $1~\mathrm{bar}$. Therefore, $a_i$ is nothing but the partial pressure of the gas in bars, but without the units. If we therefore define $K_p$ as follows:

$$K_p = \prod_i p_i^{\nu_i}$$

then, as long as the pressures you use are in bars, the numerical value of $K_p$ will be the same as that of $K$, the true equilibrium constant. As such, if you ignore the fact that you are taking the natural logarithm of a quantity with units, the value of $\Delta G^\circ$ that you obtain will be "correct".

Similarly, for a solution, we can write:

$$a_i = \frac{c_i}{c^\circ}$$

where $c_i$ is the concentration of the species and $c^\circ$ is the standard concentration, again defined by IUPAC to be equal to $1~\mathrm{mol~dm^{-3}}$. So, the same idea applies. If you keep all concentrations in units of $\mathrm{mol~dm^{-3}}$, then the numerical value of $K_c$ will match that of $K$.

An example

Consider the Haber process (all species gaseous):

$$\ce{N2 + 3H2 <=> 2NH3}$$

We would define the "true" equilibrium constant as:

$$K = \frac{(a_{\ce{NH3}})^2}{(a_{\ce{N2}})(a_{\ce{H2}})^3}$$

and you would define $K_p$ as:

$$K = \frac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3}$$

Let's say (just as an example - the numbers are not correct!) that at a certain temperature $T$, we have the equilibrium partial pressures:

$$\begin{array}{c|c|c}
p_{\ce{NH3}} & p_{\ce{N2}} & p_{\ce{H2}} \\ \hline
20~\mathrm{bar} & 50~\mathrm{bar} & 50~\mathrm{bar}
\end{array}$$

Then you would have $K_p = 6.4 \times 10^{-5}~\mathrm{bar^{-2}}$. Now you can see why it is not correct to use $K_p$ in the equation above: you would have

$$\Delta G^\circ = -RT \ln(6.4 \times 10^{-5}~\mathrm{bar^{-2}})$$

which makes absolutely no sense since you can't take the logarithm of a unit. In a simplistic treatment you would just ignore the fact that a unit existed and just take the logarithm of the number.

However, you will not run into that problem if you use the "true" equilibrium constant, as you really should. If we assume ideality, i.e. $a_i = p_i/p^\circ$, then we find that $K = 6.4 \times 10^{-5}$. Since activities are dimensionless, you will find that $K$ is dimensionless and you can now take the logarithm. As I mentioned earlier, the numerical quantity of $K$ is exactly equivalent to $K_p$ as long as you keep your pressures in units of bars.

I have seen some questions where students are requested to determine $K_c$ for a gas-phase reaction using the ideal gas law and hence $\Delta G$:

$$\begin{align}
pV &= nRT \\
c = \frac{n}{V} &= \frac{p}{RT}
\end{align}$$

The value of $c_i$ and $p_i$ will therefore differ by a factor of $RT$. Therefore, the numerical values of $K_c$ and $K_p$ in gas-phase reactions will, in general, differ. (If you have the same number of moles of gas on both sides, then the factors of $RT$ will cancel out and numerically $K_c = K_p$, but that is not always the case.) As established earlier, because of the equation for the activity of an ideal gas, only the use of $K_p$ will lead to the correct numerical value for $\Delta G^\circ$.

If a question requires you to do so, then do it. Just bear in mind that it is not only wrong in terms of taking the natural logarithm of a quantity with units; it will also give you a wrong numerical result for $\Delta G^\circ$.


Reduction of acids with borane What is the mechanism of the reduction of organic acids with borane? Borane reductions don't work well with electron deficient carbonyl groups. Why is that so, and how isn't the carboxyl group an electron deficient group?

What is the mechanism of the reduction of organic acids with borane?

Source:
Henry Rzepa. Chemistry with a twist. Mechanism of the reduction of a carboxylic acid by borane: revisited and revised. October 16th, 2011. http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114
DOI: 10.15200/winn.142795.57029 provided by The Winnower, a DIY scholarly publishing platform.


Analytical solution to first-order rate laws Does anyone know the analytical solutions to the following rate laws:

1) $\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3)$

2) $\frac{\mathrm{d}\rho_\text{T}}{\mathrm{d}t} = \rho_\text{W} \cdot K_2 - (K_4+K_5) \cdot \rho_\text{T}$

3) $\frac{\mathrm{d}\rho_\text{C}}{\mathrm{d}t} = \rho_\text{W} \cdot K_3 + \rho_\text{T} \cdot K_5$

As an example, I have determined the following analytical solution for this rate law:

$\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3) \qquad \Rightarrow \qquad \rho_\text{W} = \rho_{\text{W}0} \cdot \operatorname{e}^{-(K_1+K_2+K_3) \cdot t}$

where $\rho_{\text{W}0} = $ initial density, $K = $ reaction rate constant, $t =$ time

The scheme for the reactions are:

$\ce{wood} \xrightarrow{K_{1}}{} \ce{gas}$
$\ce{wood} \xrightarrow{K_{2}}{} \ce{tar}$
$\ce{wood} \xrightarrow{K_{3}}{} \ce{char}$
$\ce{tar} \xrightarrow{K_{4}}{} \ce{gas}$
$\ce{tar} \xrightarrow{K_{5}}{} \ce{char}$

Any help with the three rate laws above would be greatly appreciated.

This is the solution I get:

$$\begin{align}\rho _T(t) &= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$

Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$.
Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$


How to determine gerade & ungerade symmetry of a MO orbital? J.D.Lee writes in his book Concise Inorganic Chemistry:

  [...] An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes remains the same,  the orbital is gerade, and if the sign changes, the orbital is ungerade.

Now, it can be detected easily which MO, formed due to the overlapping of s-s or p-p orbitals, is gerade or ungerade as shown in these pics:

But what about the molecular orbitals formed due to the overlapping of s-p orbitals or p-d orbitals?
mo due to to overlapping of s-p orbitals:

mo due to overlapping of p-d orbitals:

If I rotate about the first orbital about an axis perpendicular to the inter-nuclear axis, the smaller lobe goes to the right only; the sign always remains same; just the position of the lobe has been changed. But in the first picture, only the bonding mo is gerade while the anti-bonding mo is ungerade; I'm not getting how by rotating along the axis perpendicular to the inter-nuclear axis, the sign of the antibonding mo changes making it ungerade.

Can anyone please help me how to apply the procedure for the molecular orbitals formed due to the overlapping of s-p & p-d orbitals?

The better way to do it is to check what happens under inversion ($i$ or $\bar 1$). If the orbital stays the same, it is g, otherwise u.

However, as Orthocresol mentioned in the comments, checking that is only possible if the entire molecule contains inversion symmetry. Not all point groups do, and the vast majority of molecules does not (partly because $C_1$ is likely the most prevalent point group out there).

For example, consider transition metals’ d orbitals. In octahedral complexes, they are labelled $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In tetrahedral complexes, which do not have inversion symmetry, they are $\mathrm{t_2}$ and $\mathrm{e}$.


Metal Compounds that bond covalently I would like to ask if anyone has a list or knows which covalent compounds have metals in them. For example, beryllium and aluminium are both metals but they bond covalently with chlorine to form beryllium chloride and aluminium chloride respectively. Other than these 2 compounds, I do not know any other and thus would like to ask this question.

There is no sharp line between ionic, metallic or covalent bonds.

Most of transition metal oxides and sulfates have strong covalent characteristics. For example, the whole reason of using ligand field theory instead of crystal field theory is that the ionic description (CFT) breaks down, and the covalent effects are important in all but the simplest cases.

If you want to take even more covalent molecules, look for organometallic compounds, metal carbonyls, metallocenes, etc.


How to calculate the enthalpy change of water? Calculate the $\Delta H$ of the reaction where $2.6\:\mathrm{g}$ of water, $C_\mathrm{s}= 4.184 \mathrm{\frac{J}{g\:K}}$ is heated, raising the temperature increases from $298\:\mathrm{K}$ to $303\:\mathrm{K}$.

I used the following equation: $\Delta H=mC_\mathrm{s}\Delta T$

$m = 2.6~\mathrm{g}$, $C_\mathrm{s} = 4.184~\mathrm{J/g}$, $\Delta T= 5$

What I got: $(2.6)(4.184)(5)= 54.392~\mathrm{J/g}$

Did I solve this correctly?

Yes, but your answer should be just in joules not $\mathrm{J/g}$, and your units are off.

$$(2.6~\mathrm{\color{\red}{g}})\left(4.184~\mathrm{\frac{J}{\color{\red}{g}\ K}}\right)(5~\mathrm{K})= 50~\mathrm{J}$$


Doubt about Van Der Waals equation I got how the real pressure would be less than ideal pressure due to net force of attraction between gas molecules acting towards the center.
But, why is the real volume greater than ideal volume? I thought the real volume would be less than ideal one since the ideal volume is the volume of container (volume of gas particles negligible) while the real volume is the available volume for gases to move excluding their individual volumes i.e. Vi = Vr + volume of gas particles

In the ideal gas equation $$p_iV_i = nRT$$ $p_i$ is the pressure if there was no intermolecular forces between the molecules and $V_i$ is the free volume available for the molecules to do motion in, which is the volume of the container.
In real gases however, the pressure that you observe (real pressure) is less than the expected pressure. This is because of the intermolecular forces. So you convert the $p_r$ to a pressure which the gas would exert if there were no intermolecular forces.  $$p + a\frac{n^2}{V^2}$$
The free volume available to real gas is less than the volume considered in ideal gas equation. That is because Kinetic theory of gases made an assumption that the volume occupied by molecules of an ideal gas is negligible. But in real gases, the molecules are of considerable size. Thus, the full container will not be free for motion, some volume of the container will be occupied by the molecules. Let the volume occupied by one mole gas molecules be $b$, for n mole gas this is $nb$, so free volume available to gas is less than the volume of container $$V_i =  V_{\text container} - nb$$
Edit: In your question, the $V_r$ they're talking about is volume of the container. And $V_i$ is actual free volume available to the real gas.

Real Volume: Volume of the container
Ideal volume: Free space available
Real pressure: Observed pressure
Ideal pressure: Pressure if the gas didn't have any intermolecular forces.


Does changing EDTA concentrations affect the amount of calcium ions in milk? For my chemistry experiment I am titrating EDTA against milk to find out the concentration of calcium ions in milk. I will be using EriochromeT as the indicator. The lab tech said that if I use temperature as my independent variable the end-point of titration will fluctuate. She suggested that I should change the concentration of EDTA instead. Is it a good suggestion? What is a better independent variable to use?

Since this post was a while ago, I'm assuming an answer has already been decided on, but I'll take a shot at answering anyway.

If I were you I would rely on concentration changes of EDTA as the independent variable since it allows for more precision in your titration. If you were to vary temperature, it would have to be the temperature of the titrant such that volume changes could be accounted for (since analyte is usually in an approximate volume of water). While volume fluctuations of titrant are theoretically possible, they are rather difficult in a laboratory setting and not exactly worth the effort unless you have lots of time and expensive equipment to control the temperatures exactly. Not only that, but volume changes of water are only approximately 5% from 0 Celcius to 100 Celcius.1 Thus, as I said, changes in [EDTA] would be more efficient and easier to manage.

Thermal Expansion of Water


Why is it that increasing the molar mass generally increases the Entropy? I've just found read that entropy increases through these reasons:

Changing the phase of the elements to a gas.
Increasing the molar mass of that specific element increases the entropy.

I understand the first one, but I can't grasp the second one.
Because naturally you would think that $\ce{H2}$ would have a higher entropy than $\ce{Ne}$ because one mole of it is generally lighter.

But however it is said that Ne has a higher entropy because of its lower molar mass.

Why?

The allowed energy levels for a particle of mass $m$ in a box of length $X$ are given by
$$E_i=\frac{h^2}{8mX^2}i^2\quad i=1,2,\ldots$$

(Already this equation gives a hint that an ideal gas of high molar mass has more thermally accessible translational states than a gas of low molar mass. Therefore, the molar entropy of a gas of high molar mass is greater.)

In this case, the partition function for translational motion in the $x$-direction is approximately
$$z=\left(\frac{2\pi mkT}{h^2}\right)^{1/2}X$$

The partition function for translational motion in three dimensions is
$$\begin{align}
z&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\
&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V
\end{align}$$

For a system of $N$ particles

$$\begin{align}Z&=\frac{z^N}{N!}\\&=\frac{1}{N!}\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\right]^N\end{align}$$

The statistical entropy is
$$\begin{align}
S &=\frac{U-U_0}{T}+k\ln Z\\
&=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\
&=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\
&\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\
&=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\
&=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\
&=\frac{U-U_0}{T}+n R \ln \left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}V}{nN_\text{A}}\right] \\
\end{align}$$

The internal energy of a monoatomic ideal gas is

$$U=U_0+\tfrac{3}{2}nRT$$

which leads to the Sackur–Tetrode equation:

$$\begin{align}
S &=\frac{U-U_0}{T}+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=\tfrac{3}{2}nR+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=nR\left\{\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]+\tfrac{3}{2}\right\}\\
&=nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}^{5/2}V}{nN_\text{A}}\right]
\end{align}$$

This equation implies that the entropy $S$ of a monoatomic ideal gas depends on the mass $m$.


Effect of substituents on arene oxide-phenol rearrangement When I was learning about the reactions of epoxides from my textbook, I came across the following mechanism for arene oxide-phenol rearrangement in the presence of acid:

NIH Shift - National Institute of Health Shift (c.f. Wikipedia)

In the above example, there is no substituents on the ring and hence the final product remains irrespective of which $\ce{C-O}$ bond in the arene oxide cleaves to open the ring. Later, the author presented the following reaction where the ring has an electron donating group - methyl group:

Here the final product depends on which of the two $\ce{C-O}$ bonds in the arene oxide is broken. Based on the general reaction mechanism provided by the textbook, I derived the following mechanism to form the product prescribed by the author:

Before writing the above mechanism, I first arrived at a different product by following the mechanism given below:

The reason I initially chose to cleave the left $\ce{C-O}$ bond in "Reaction Mechanism - 2" was, it gives rise to the carbocation 2(a) where the methyl group is at the para position. I've learnt that methyl group stabilizes a positive charge by hyperconjugation. Since, resonance effect is felt only at the ortho and para positions and not at the meta position, the carbocation 2(a) is more stable than 1(a) from "Reaction Mechanism - 1". Moreover, in the first mechanism, NIH shift involves a 1,2 Deuterium shift whereas in the second it's a 1,2 Hydrogen shift. I find a deuterium shift to be slightly energy expensive process due to its higher mass and stronger bond with carbon compared to a hydrogen shift.

The only reason why I think the first mechanism is better than the second is the product 1(b) is more stable than 2(b). But I don't understand why is the second reaction not preferred over the first even though it has two factors supporting (stability of 2(a) over 1(a) and hydrogen shift over deuterium shift) it.

I faced similar contradictory results while writing the mechanisms for the following reactions (from the same textbook):

My questions are:

Are the reactions provided by my textbook correct?
If yes, why is the cleavage of one $\ce{C-O}$ bond of the epoxide preferred over the other even though it produces a comparatively less stable carbocation? Are there any other factors which determine which bond cleaves?
Is arene-oxide phenol rearrangement "specific" or highly "selective"? Or in simpler terms, does the rearrangement exclusively form only one product or does it form some small amount of the other product too?

Image source: My own work :) | Made using Marvin Sketch.
Textbook: Organic Chemistry by Solomons, Fryhle, and Snyder | Chapter 10: Alcohols and Ethers | Section 10.14: Reactions of Epoxides

What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.

First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.

Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:

Hence, mechanism 1 should be the preferred pathway

I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:

However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.

Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.

o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well

Clarification

I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:

First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1

Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1

Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all


Which side of an aluminum foil should I put in contact with food? Aluminum foil has two sides: a matte one, and a shiny one.
My question is: which one should I put in contact with food?

Criteria are lowest risk of toxicity and best qualities for food preservation (e.g. do not favor oxidizing, or react with acid nor alkaline ingredients)

For information, I've come across these several explanations; they are however contradictory (and might not be accurate):

Both side have same properties. The difference in terms of appearance is only due to the manufacturing process. (Two foils are laminated at the same time, so each foil is in contact with the rolls on one side only — hence the asymmetrical look). In this case, we can use both side of the foil.
Aluminum foil is actually made of two layers: one of aluminum, another that is coating. The surface of the aluminum layer that is in contact with atmosphere oxidizes and turns into aluminum oxide — hence the matte side.In that case, we should use the shiny side, since aluminum oxide is not that healthy.

After some thought, I guess I'll go with the first argument, that seems plausible (though I don't think it's accidental). As for coating one side with a layer, that seems like a bit too much.

Assuming the second argument is correct, then it's most probable that it doesn't matter which side you use. Because if one side is a 'coated', that 'coat' has got to be safe. Since well, aluminium foil is used pretty much only for packaging food, I don't think people could go about selling Al foil coated with some toxic film for the past couple of decades or so without someone pointing out something very wrong.

Now if it's the first argument, then it still doesn't matter which side you use. Aluminium oxide isn't really as dangerous as out make it out to be, especially if you're dealing with an oxide film that's still on the foil (Hogging a tablespoon of powdered aluminium oxide is another story altogether...). We sometimes call that process (aluminum developing an oxide coat) Passivation, because the foil is rendered 'passive' or 'inert'. Why else do you think it takes so long to get aluminium to react with acids or alkalies?

In a nutshell, regardless of which argument is correct, it'd be perfectly safe to use either side of the foil.

The reflective surface will reflect heat and the matte side will absorb heat. If you're baking or defrosting, the matte side will absorb more radiant heat and reflect less infrared heat while the shiny side will reflect more of both, so it makes more sense to bake and defrost with the matte side facing up. Conversely, during freezing, it makes more sense to reverse the sides as the shiny side face up would reflect heat away from the food.

I've always felt that one of the more practical reasons for creating two visually dissimilar sides on the foil, is to make it easier for you to identify which side you had initially used to wrap some food in. Say, you had just hurriedly unwrapped a cheeseburger and you don't feel like finishing it off. If you re-wrapped using the wrong side, you just get greasy fingers, which is quite annoying.

EDIT: I've read @Vedant's answer and he seems to have provided a good explanation as to why the two surfaces are dissimilar. But I wouldn't entirely dismiss the reason I've mentioned (identification), since, from my personal experiences...I'm pretty sure my keyboard would be covered in grease by now.

But all the same, I've included @Vedant's explanation as well, to give you the complete package.


Does H2O2 break down into H2O + O or H2O + O2? In hydroponics, Hydrogen Peroxide is used to sanitise nutrient solution from pathogens. Allegedly it does not take sides, oxidising aerobic as well as anaerobic, including damaging the roots, so I was told. I believe this argument, since O as a free radical is highly reactive. I would appreciate your input on this. However, this isn't the main question.

The main question is that manufactures also claim that it will provide DO, dissolved oxygen, to the solution. I find this hard to believe. Reason being, DO refers to dissolved $\ce{O2}$ (Dioxide) not Monoxide. When $\ce{H2O2}$ breaks down it breaks down to $\ce{H2O + O}$ (free radical), right?

Can you please put some clarity into this?

$\ce{H2O2}$ -> $\ce{H2O +1/2O2}$

This might make it easier to understand.

$\ce{H2O2}$ will homolytically cleave for form two $\ce{.OH}$ radicals. Radicals are very reactive and will start a chain reaction, but ultimately you will end up with water and oxygen products from $\ce{H2O2}$.

(Why does it form two hydroxy radicals? because the O-O single bond is weak and unstable - see MO theory)

How do we know this? Look at a bottle of $\ce{H2O2}$ that has been left out of the fridge for too long. There will be a slight pressure build up of $\ce{O2}$ gas and the % concentration $\ce{H2O2}$ will be down due to formation of $\ce{H2O}$and loss of $\ce{H2O2}$.

For sake of completion,
You can alternatively balance the equation thus:
$\ce{2H2O2 -> 2H2O + O2}$


Why can sulfur have an expanded octet but not oxygen? In my search to understand the bonding in structures like $\ce{SF6}$ I found many sources that said it was because sulfur has d orbitals to accommodate an expanded octet, which made sense to me. But I also found sources like the paper by Reed and Weinhold (1986) that say d orbitals contribute very little to the bonding in $\ce{SF6}$. An alternative explanation is that $\ce{SF6}$ has three-center/two-electron bonds. What I don't understand is why compounds like $\ce{SF6}$, $\ce{ClF5}$, $\ce{PCl5}$ exist for third row element centers, but not $\ce{OF6}$ if it has nothing to do with d orbitals? References would be great.

sulfur has 2 electrons in its 3s subshell , and 4 in 3p subshell. On absorbing energy the paired electrons get excited to to the next higher energy level which is the 3d subshell. So after exciting it's electrons six unpaired electrons are present which can take part in bonding via overlapping of orbitals. Other compounds can also be explained similarly. Also the d- orbitals contribute to only two bonds out of six bonds( which would have been meant in the paper by Reed and Weinhold (1986) ). Oxygen has no empty orbitals with it . It only possess orbitals upto 2p which contains 4 electrons. So it cannot expand its octet by exiting it's electrons.


Molecular orbitals used with CCSD(T) geometries Suppose I am given some geometry data (say, of a water dimer) from CCSD(T).  If I were to do a single-point energy calculation to generate orbitals for the system, wouldn't CCSD give something more representative of the correlation in the system than DFT?

I see "CCSD(T) orbitals don't exist - used DFT orbitals instead" kind of statements in papers (quotations below), and wasn't sure why they don't even mention CCSD.  Perhaps it's just due to speed/availability, but that would lead to another question "why is it not worth it to use CCSD orbitals with CCSD(T) geometry".

The same question can be asked for DFT orbitals being used with MP2 geometries, but MP2 perturbs about HF ... so clearly DFT is needed to capture correlation.

[edit: more context]
What I'm trying to do is "export" these orbitals to a diffusion quantum Monte Carlo (DMC) calculation which doesn't do geometry optimization by itself but augments the imported orbitals with explicit correlations (eg. something that depends directly on electron pair distances).  Usually the final DMC energy is not very sensitive to whether the orbitals are from CC or DFT (I have outputs from both), but I was checking here to make sure I wasn't missing any "deeper" reason.

From ChemPhysChem 2018, 19 (15), 1886–1894:

  Since MP2 orbitals are nonexistent, the charge-transfer interactions were computed using the B3LYP functional with the aug'-cc-pVTZ basis set at the MP2/aug'-cc-pVTZ geometries, so that at least some electron correlation effects could be included.

and J. Phys. Chem. A 2017, 121 (49), 9544–9556:

  Due to the nonexistence of CCSD(T) orbitals, ΔE(del) was calculated with ωB97XD/aug-cc-pVTZ.

Pulling together some of my comments into an answer:

The principal question at issue is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" You can obtain an corresponding 1-particle density matrix from all of these methods, which can be diagonalized to yield natural orbitals (which will have non-integer occupation numbers. There are also certainly reference orbitals used to compute these methods (usually Hartree-Fock, but there are methods like orbital optimized-MP2 (OO-MP2) and Brueckner doubles (an orbital optimized CC variant)). Those cited statements are, to some extent, meaningless without the answers to those questions.

If you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy.

For your purpose, since you are just exporting to DMC, yeah wouldn't expect the results to vary too much by orbital, the DMC procedure just needs a mostly reasonable starting point. CC natural orbitals would probably produce a more compact wavefunction and therefore maybe run faster than DFT orbitals, but they are enough more expensive to compute that it's probably not worth it. Probably the orbitals you got from the CC calculation you did are just the HF reference orbitals. It's also worth pointing out that natural orbitals are probably not so different from the DFT orbitals unless there is a lot of correlation (in which case, DFT method will probably produce poor orbitals). For systems where the reference state is qualitatively wrong (e.g. diradical system that might suffer from a lot of spin-contamination in HF), OO-MP2 would improve that and those orbitals would probably be substantially better and that would be a case where the cost would probably be justifiable.


Do electrons only fill 'spin up' first? Or could it start filling 'down spins' first? Due to Hund's rule, electrons start filling up the orbitals without pairing up. When this is happening, do the electrons all fill up the 'up' spin? Could they fill in the 'down' spin? Why do they prioritize the 'up' spin if they do?

No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this for all the orbitals you want to fill.

But: The convention is that you start with 'spin up' as this usually is taken to represent the $\frac{1}{2} \! \hbar$ eigenvalue of the $z$ component of the spin angular momentum operator and you should adapt to that in order to avoid confusion.


How to figure out if the reaction proceeds via SN1 or SN2 mechanism? I have been assigned with the task of figuring out whether a reaction is $\ce{S_{N}1}$ or $\ce{S_{N}2}$. The information that I have is, that the yield is dependent on solvent polarity. More specifically, the higher the solvent polarity, the higher the yield.
The problem is that both $\ce{S_{N}1}$ or $\ce{S_{N}2}$ products are the same. Both result in charge separation (therefore both are favored by polar solvents).

I proposed to repeat the experiment. I proposed to make the site of substitution chiral. That way it is possible to check for $\ce{S_{N}2}$ inversion.
Another proposal was to use the fact that the rates of $\ce{S_{N}2}$ and $\ce{S_{N}1}$ were different. However, I am not sure how to use this fact and perform an experiment.

What should I do in this situation? The people that I am working for do not want to repeat the experiment at all. Using the rates approach seems to be the best option since there won't be a need to obtain new materials. But I have no idea on how to do this. What are your suggestions? How can I figure out whether the reaction proceeded through $\ce{S_{N}1}$ or $\ce{S_{N}2}$ by using rates approach?

As you have the information that the yield depends on solvent polarity. It is the key point to be considered in analysis of type of reaction. In case of a $\ce{S_{N}1}$ reaction, the polar solvent favors the reaction. $\ce{S_{N}1}$ is a two stepped reaction, in which the 1st step is carbocation formation, such that there appears the dipole dipole interaction between the carbocation and polar solvent which favors the reaction.

Whereas in the case of $\ce{S_{N}2}$ reaction, the dipole dipole interaction develops between polar solvent and the nucleophile rather than with intermediate product. It is because the charge is dispersed in two centres in intermediate while charge is centered in single center in nucleophile. This deactivates the $\ce{S_{N}2}$ reaction and the reaction cannot proceed forward smoothly.

Since you have another specific information that the higher the solvent polarity, the higher the yield. Since higher solvent polarity favors the $\ce{S_{N}1}$ reaction, the higher will be the yield through a $\ce{S_{N}1}$ mechanism. So the reaction with tendency with higher solvent polarity and higher yield is a $\ce{S_{N}1}$ reaction.


Can bromide ion reduce ferric ion? Can bromide ion reduce ferric ion ?

Just a query.
In my opinion it can since bromide ion can undergo oxidation.
Is my answer correct?

You can predict this with the help of electrochemical series.

On going up the series reducing power increases, for you case it can be seen that Fe is a better reducing agent than bromide ion as Fe is above Br2 so bromide can't reduce ferric ion.


Why can't this molecule avoid being antiaromatic by becoming nonplanar? The peripheral system is clearly antiaromatic if planar, but we know that antiaromatic compounds are more unstable than non aromatic compounds.

So, I think that the molecule will lose its planarity and become non aromatic, but this molecule is antiaromatic according to my professor.

Isn't my reasoning correct?

Also, what would be the IUPAC name of this molecule?

The central atom is approximated to be tetrahedral and thus approximately $\mathrm{sp^3}$ hybridised. The three atoms of the outer ring it is bonded to are approximately $\mathrm{sp^2}$ hybridised as they are all part of double bonds.

If you build this, you will notice that while these three atoms are necessarily in a plane they have different orientations. It is simply not possible to make a planar ring without some significant distortions of the geometry. I thus challenge your professor’s assumption.

I would expect this system to have strongly localised double bonds, i.e. significant bond length alteration, in such a way that the double bond are twisted with respect to each other to minimise strain.


Why is quinoline less basic than pyridine? Quinoline is less basic than pyridine, even though it has more resonating structures. Why? Does the negative inductive effect reduce basicity of quinoline?

This is not due to the inductive effect, but rather, due to solvation. The positive charge of quinoline on the nitrogen is harder to solvate due to steric repulsion of the nearby ring, whereas in pyridine, the small hydrogen atoms in place do not pose a strong steric hindrance for the solvation of the ion.


Fitting multiple pKa's I have three histidine residues that are next to each other in a protein. When I compute the fraction of unprotonated states for each individual histidine as a function of $\mathrm{pH}$ my titration curve has a non-Henderson–Hasselbalch (HH) shape. Hence I can't fit it using the HH equation for 1 site:

$$\frac{1}{1 + 10^{n(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}$$

in order to solve for the Hill coefficient $n$ and $\mathrm{p}K_\mathrm{a}.$

Obviously the three histidine $\mathrm{p}K_\mathrm{a}$'s are dependent on each other since they are very close to one another and are probably sharing the proton. So how do I go about fitting the curves? Should I sum the total unprotonated states in all three histidines and fit the curve with the three site equation

$$\frac{1}{1 + 10^{n_1(\mathrm{p}K_\mathrm{a1} - \mathrm{pH})}} + \frac{1}{1 + 10^{n_2(\mathrm{p}K_\mathrm{a2} - \mathrm{pH})}} + \frac{1}{1 + 10^{n_3(\mathrm{p}K_\mathrm{a3} - \mathrm{pH})}}?$$

When I use this equation I still don't get a good fit and I get unreasonable $\mathrm{p}K_\mathrm{a}$'s $(-1.4, 7.7, -1.3).$ Any suggestions?

As requested below, the matlab code I used for the fit is the following:

ft=fittype('(1/(1+10.^(a1*(p1-x))))+(1/(1+10.^(a2*(p2-x))))+(1/(1+10.^(a3*(p3-x))))', 'independent','x','dependent','y');
opts = fitoptions('Method','NonlinearLeastSquares');
[fitresult, gof] = fit( xData, yData, ft, opts );
h = plot(fitresult,xData,yData)

I posted values for p1, p2, p3 which should report $\mathrm{p}K_\mathrm{a1},$ $\mathrm{p}K_\mathrm{a2}$ and $\mathrm{p}K_\mathrm{a3}.$

Given your hypothesis that the three $\mathrm{p}K_\mathrm{a}$'s are dependent on one another and that experimentally you do not seem to be able to identify which histidines are protonated and which are not at any given time, it would seem that the simplest model to use is that of a triprotic acid such as citric acid. The triple HH equation model you used is only appropriate for three completely independent $\mathrm{p}K_\mathrm{a}$'s (such that the $\mathrm{p}K_\mathrm{a}$ at site 1 is unchanged by the protonation state of site 2).

In the dependent $\mathrm{p}K_\mathrm{a}$ model, we have four species (vs the eight in your model): $\ce{H3P},$ $\ce{H2P},$ $\ce{HP}$ and $\ce{P},$ where "$\ce{P}$" is the protein. There are three "apparent $\mathrm{p}K_\mathrm{a}$'s" none of which represents a true $\mathrm{p}K_\mathrm{a}$ of a specific site. They instead correlate to the combination of possible reactions that give a change in the overall protonation state. For example, the first $\mathrm{p}K_\mathrm{a}$ represents the change in state from $\ce{H3P}$ to $\ce{H2P},$ regardless of which of the three sites loses a proton.

We can then write four equations starting with mass balance:

$$
\begin{align}
[\text{total protein}] &= [\ce{H3P}] + [\ce{H2P}] + [\ce{HP}] + [\ce{P}]\tag{1}\\
K_\mathrm{a1} &= \frac{[\ce{H}][\ce{H2P}]}{[\ce{H3P}]}\tag{2}\\
K_\mathrm{a2} &= \frac{[\ce{H}][\ce{HP}]}{[\ce{H2P}]}\tag{3}\\
K_\mathrm{a3} &= \frac{[\ce{H}][\ce{P}]}{[\ce{HP}]}\tag{4}
\end{align}
$$

We can also define the fraction of sites unprotonated as

$$\phi = \frac{3[\ce{P}]+2[\ce{HP}]+[\ce{H2P}]}{3[\text{total protein}]}\tag{5}$$

It's relatively straightforward to use the four above equations to get expressions for each of the species concentrations in terms of $[\ce{H}]$ and $[\text{total protein}]$ only. Substituting those expressions into the bottom equation will give you an equation to which to fit your data.

If you post your data in tabular form, others can play around with different fitting approaches and give feedback.


Why would the pH of a solution not be 7 at the equivalence point of an acid-base titration? Is it because the new substance formed might be acidic/basic ?
The internet defines this point as when the number of moles of the acid equals the number of moles of the base. But if we have for example 1 mole of NaOH and 1 mole of H2SO4 (both in dilute solutions) mixed together, the resulting solution will still be acidic because 1 mole of H2SO4 releases 2 moles of H+ right ? So we will have an extra mole of H+ which will make the resulting mixture acidic. So is this still the equivalence point even though it isnt really equal (bcz we have 1 extra mole of H+) or is it 1 mole of NaOH + half a mole of H2SO4 ?

The equivalence point is defined as when you have an equal amount of acid and base. It you have a strong acid and a strong base, the pH at the equivalence point should be pretty close to 7, but if you have a weak acid or a weak base involved, there is not reason why the equivalence point should be at pH 7. You would have a conjugate base or acid that is reasonably basic or acidic at the equivalence point.


Is the change in enthalpy (ΔH) for dissolution of urea in water positive or negative? To test the properties of a fertilizer, $\pu{15.0g}$ of urea, $\ce{NH2CONH2_{(s)}}$, is dissolved in $\pu{150 mL}$ of water in a simple calorimeter. A temperature change from $\pu{20.6^\circ C}$ to $\pu{17.8^\circ C}$ is measured. Calculate the molar enthalpy of solution for the fertilizer urea

I worked through this question by finding $Q = mc\Delta T$, and then dividing $Q$ by the moles of urea present. I can tell the process is endothermic because $\Delta T$ is negative, however my answer for $\Delta H$ comes out as negative, which would only make sense if this was an exothermic reaction. I'm not sure where I am wrong to be honest.
Here is my work:
$$
\begin{align}
\Delta H &= \frac{(\pu{150ml}) \times (\pu{1g mL^{-1}}) \times (\pu{4.18J g^{-1} K ^{-1}}) \times (\pu{-2.8 K})} {(\pu{15g}/\pu{60.07g})}\\
&= \pu{-7030.59J/mol}\\
&= \pu{-7.03kJ/mol}
\end{align}
$$
TL;DR - question asks for $\Delta H$ of an endothermic process, not sure if my answer should be positive or negative

The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.

In your answer, you are missing a negative sign in $\Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.


what's the mass of the original sample (g)? here is a question that i don't know how it is solved :

A sample that is 75 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that forms is 2.013 g, what was the mass of the original sample?

I hope you give me the the proper way of solving it

It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.

The molecular mass of AgCl = (107.87+35.45)=143.32g

Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473

Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.

If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g


What is the approximate lifetime of an SN2 transition state complex? Imagine some SN2 reaction involving an aliphatic sp3 carbon with a halide leaving group (e.g. iodide or bromide). Given the rise of femtosecond spectroscopy (e.g. using a mode-locked Ti:sapphire laser) and even attosecond spectroscopy, I'm curious if there are any studies that have measured the properties and approximate lifetime of the SN2 transition state complex? I guess that this is in the $\approx$ picosecond regime alongside the timescales for a bond stretching dynamics (e.g. between oxygen and hydrogen on a water molecule), however, is there a more precise value that is now known?

I am be curious about results from quantum mechanical simulations (using e.g. Jaguar from the Schrodinger computational chemistry suite) for SN2 reactions?

I'll specify that I'm referring to a "concerted" SN2 reaction, not, for example, the kind of reaction described in: [Chandrasekhar, J., Jorgensen, W. L. Energy profile for a non-concerted SN2 reaction in solution. J. Am. Chem. Soc. 107, pp. 2974 - 2975 (1985).].

Also, provided that non-concerted SN2 reactions exist (example) perhaps my question is underspecified. Perhaps, I should ask instead if there are any interesting long-lived SN2 interaction complexes, or tables for transition state lifetimes from ultrafast spectroscopy studies?

26-60 femtoseconds depending upon impact angle for $\ce{F- + CH3Cl}$ according to Mugnai et al. (Table IV.)


How do you calculate the dipole moment of larger molecules? I know that the dipole moment is obtained by:

$$
\vec{\mu} = Q \cdot \vec{d}
$$

with the condition that charges of a pair of atoms are equal and opposite, i.e. $\ce{NaCl}$.

But, in order to obtain the molecular dipole moment of water, how is the $\ce{-OH}$ group handled? What about even larger systems such as $\ce{NH3}$?

For example, the water molecular dipole moment is 1.85 Debye. How was this value was obtained?

I have read 1, 2, 3, and chemistry textbooks.

I am interested in a more mathematical or physical description.

The total dipole of a molecule can be thought of as the sum of dipoles of individual functional groups:

$$
\vec{\mu}_{\text{total}} = \sum_{i}^{N_\text{groups}} \vec{\mu}_{i}
$$

Because each dipole is represented as a vector with both magnitude and direction, this amounts to vector addition.

In the case of water, consider the dipole of each $\ce{-OH}$ group. The above reference gives it as 1.53 Debye, and the $\ce{H-O-H}$ angle is $105^{\circ}$. Because coordinates aren't given, the angle has to be used instead:

$$
\vec{P} + \vec{Q} = \sqrt{\lVert \vec{P} \rVert^{2} + \lVert \vec{Q} \rVert^{2} + 2\lVert\vec{P}\rVert\lVert\vec{Q}\rVert\cos{\theta_{PQ}}}
$$

which gives an estimate of 1.64 Debye. Not very good compared to 1.85 Debye! The group addition approximation completely neglects how the electronic structure of the molecule changes upon going from (using water as an example)

$$
\ce{2OH^{.}} \rightarrow \ce{H2O}
$$

The issue is probably that the two individual $\ce{-OH}$ groups interact strongly and favorably when combined together in water, compared to when they are apart; that is, the electron density changes quite a bit due to the bonding.

I've also neglected to explain how these groups can be defined; there is no unique definition of how to partition a molecule. For example, take a carboxyl group:

Are all four atoms treated together as a single group, or are the $\ce{C=O}$ and $\ce{C-O}$ and $\ce{-OH}$ treated separately? Maybe this is a simple case but it becomes more unclear for something like a protein or a coordination complex.

The molecular dipole can also be calculated using electronic structure theory, which removes the ambiguity surrounding how to partition a molecule and incorporates those "non-additive" effects.

As Geoff mentioned in the comments, the definition of the dipole moment created by a set of charges can also be used:

$$
\vec{\mu} = \sum_{a}^{\text{charges}} \vec{r}_a \times q_a
$$

where $q_a$ is the charge, and

$$
\vec{r}_a = \vec{R}_a - \vec{O} = (R_{ax} - O_x, R_{ay} - O_y, R_{az} - O_z),
$$

where $\vec{R}$ is the position of the point charge, and $\vec{O}$ is a common origin, usually taken to be either $(0,0,0)$, the center of mass, or the center of nuclear charge. It doesn't matter what this point is for uncharged (neutral) systems, but there is an origin dependence for anything with a non-zero total charge. The $\{\vec{R}\}$ are taken to be atomic positions, and the charges $\{q\}$ can come from simpler Mulliken or Löwdin population analyses, or more complicated schemes of which many are designed to reproduce total dipole moments. Care must be taken to deal with units properly, specifically when converting: $0.393430307 \, ea_0 = 1 \, \mathrm{Debye}$.


difference between glyceraldehyde-3-phosphate and 3-phosphoglycerate In the photosynthetic dark reaction there are two types of reactions: light-dependent and light-independent.  In the light-independent reactions during reduction 3-phosphoglycerate is converted to glyceraldehyde-3-phosphate.  I don't find any difference between them because both have a 3 carbon chain attached to a phosphate group. Can you please clear it up for me because my book says that the latter one is a sugar.  Why isn't the first one sugar too?

The difference between the two molecules is highlighted in red.

The functional group in 3-phosphoglycerate is a carboxylic acid. That in glyceraldehyde-3-phosphate is an aldehyde.

Sugars have the general formula $\ce{C_$n$H_{$2n$}O_$n$}$. To accomplish this specific ratio of carbon to hydrogen to oxygen, the requirement is that one of the carbons must have a carbonyl group (i.e. forms an aldehyde or ketone), whereas the remainder of the carbons have one hydroxyl group each. For example, the open-chain structure of glucose is:

The carbonyl group is in red and the hydroxyl groups are in blue.

Therefore, glyceraldehyde-3-phosphate is simply a phosphorylated sugar: C-1 is the aldehydic carbon and C-2 and C-3 have hydroxyl groups, the latter of which is phosphorylated.

3-Phosphoglycerate is one oxidation state too high.


How Can I Test Something for the Presence of Iodine? I want to test a supplement for the presence of iodine.
Iodine should turn starch blue-black. So, can I powderize the supplement tablet, add water to it, drop it on a piece of bread, and if it turns blue-black, that would be an accurate way to test?

Just keep in mind that only elemental iodine will cause a blue-black coloration. Nobody add elemental iodine to a supplement just like nobody can eat chlorine out of table salt (NaCl). The typical compound they add is KIO3, which will not color starch right away, until and unless you reduce it to KI and then oxidize KI to free I2. In any case, search natural foods which contain iodine.


Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from? Trial #1:

$$\ce{Fe^3+(aq) + SCN^- (aq)  ->   FeSCN^2+(aq)}$$

\begin{array}{|r|c|c|c|}\hline
\mathrm{Initial~}(M)&\phantom{-}6.00 &\phantom{-}10.00&\phantom{-}0.00\\\hline
\mathrm{Change}~(M) & –4.00 & –4.00 &+4.00  \\\hline
\mathrm{Equilibrium}~(M)&\phantom{-}2.00&\phantom{-}6.00&\phantom{-}4.00\\\hline
\end{array}
$$ K = 0.333$$

Consider the reaction represented by the equation:
$$\ce{Fe^3+(aq) + SCN- (aq) \longrightarrow  [Fe(SCN)]^2+(aq)}$$

Trial #2:

Initial: $10.0 ~M\ce{~Fe^3+(aq)}$ and $8.00~M\ce{ ~SCN- (aq)}$ (same temperature as Trial #1)

Equilibrium: $x~M\ce{~FeSCN^2+(aq)}$

I have tried putting $10.0$ and $8.00$ in an ice table, then when I have finished I start solving to get my equation as $$0=\frac{0.33x}{(10.0-x)(8.00-x)}$$ when I'm done I come out with $$0.33=\frac{x}{(x^2-18x+80)}$$ then I plug in for quadratic equation,
$$x=\frac{-6.94\pm\sqrt{6.94^2-4\times0.33\times26.4}}{2\times0.33}$$
my answers came out to be $-5$ and $-16.03$.  What am I doing wrong?

You are just making a small sign error:
\begin{align}
ax^2 +bx +c &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a}
\end{align}

Taking $K=\frac13$:
\begin{align}
\frac13x^2 - \frac{21}{3}x +\frac{80}{3} &= 0 &
 x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\
&& x_{(1;2)} &= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\
&&           &= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\
&& x_{(1;2)} &= 16; 5
\end{align}

The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.


How does electrolytic conduction depends on the size of ions & their solvation? The textbook says,

  The smaller the effective size of the ion after solvation, the greater will be its conducting power."

But, on the Internet, it is explained as follows:

  Radius decreases, smaller ions carrying current in small area, more associations with anions & due to this, the ions become heavy & they move with reduced speed as compared to less associated ones, & thus conductivity decreases.

The explanation is different to the what the book says.
Could someone please explain this?

Small and/or highly charged ions tend to be highly solvated in water. These solvated ions may still be smaller than the solvated ions of larger ions.
However, water molecules are not as free as the molecules of a gas. Water has a much higher viscosity, for instance. So think of water as a loose gel. A large (unsolvated, or poorly solvated) ion has an easier time slipping between gaps in this gel. A smaller, more highly solvated ion is, by definition, more tightly bound to the gel network, and moves more sluggishly, as if it has to carry more of the network with it. This approach is more explanatory than predictive (see figure).

There are lots of numbers, some agree, some disagree with any simple explanation. But a clean example is sodium vs potassium: sodium bare ion is smaller than potassium bare ion, but their hydrated ions reverse that (hydrated potassium ion is smaller than hydrated sodium ion - and check out lithium!). So potassium has a greater conductivity, for the same number of ions.


How to rationalise the coordination number of CsCl versus NaCl? Question:

  Suggest an explanation for the coordination number in the $\ce{CsCl}$ lattice being different from those in $\ce{NaCl}$.

Standard answer:

  Caesium ion is larger than sodium ion, hence it can be surrounded by greater number of chloride ions.

My questions:

The question didn’t specify whether it is asking for the coordination number of $\ce{Cs+}$ or $\ce{Cl-}$... Is it because the coordination numbers of both ions are the same?
The given answer sounds like each $\ce{Cl-}$ ion will be surrounded by less $\ce{Cs+}$ ions, how is it possible that the ratio of $\ce{Cs:Cl}$ is 1:1 (as from its empirical formula)?

Since both compounds are of the form $\ce{AB}$ (and not $\ce{AB2}$ or $\ce{A2B3}$/etc), the coordination numbers will be equal for the ions. Similarly, for $\ce{AB2}$, the coordination number for $\ce{A}$ will be twice that of $\ce{B}$.

What happens is that each ion is connected to more ions of the opposite type in $\ce{CsCl}$. Each $\ce{Cs+}$ ion is connected directly to 8 $\ce{Cl-}$ ions, and vice versa. The net effect is that their ratio stays the same, they just have more connections.

Look at it this way: There are four men and four women in a room. First, each man holds hands with only one woman and vice versa. Now, we tell each man to hold the hands of two women, and we tell each women to do the opposite. Now, we shift it up to 3 hands (with the reasonable approximation that humans have an infinite supply of hands), and finally 4. Try this on a piece of paper -- you'll see that the man:woman ratio doesn't need to change as long as the "handholding number" (coordination number) for each person is the same.

Taking a look at the actual lattices:

For $\ce{CsCl}$ (body-centered cubic):

Note that each $\ce{Cs+}$ ion touches 8 $\ce{Cl-}$ ions and vice versa. Also note that the $\ce{Cs:Cl}$ ratio is indeed $1:1$. For each $\ce{Cs+}$ ion, there is a corresponding $\ce{Cl-}$ ion on its NE diagonal tilting into the plane of paper, and for each $\ce{Cl-}$ ion there is a corresponding $\ce{Cs+}$ ion on the SW diagonal tilting out of the plane of the paper.

Here's $\ce{NaCl}$ (face centered cubic, close packing):

Here each $\ce{Na+}$ ion touches 6 $\ce{Cl-}$ ions and vice versa. We can define a similar pairing as before -- each $\ce{Na+}$ ion has a corresponding $\ce{Cl-}$ ion to its east, and each $\ce{Cl-}$ ion has a corresponding $\ce{Na+}$ ion to its west.


Does solute add to volume of solution? I have to calculate the concentration of sucrose in water just by knowing their masses (and their temperature). Now, if the volume doesn't change when the sugar is being dissolved, I can do that quite simply over density. But I am not sure if it changes or not. Does it? And is that generally true (e.g. for electrolytes)?

Solute almost always changes the volume of final solution. The change (volume decrease or volume increase of final solution) of volume depends of solvent and solute chemicophysical properties, e.g. the full charge of each or the polar properties of each, or their molecular orbital structure and electron bond donors or acceptors. Two very non-interactive materials would not even mix, e.g. oil and water (H2O), though tiny tiny amounts of oil will "dissolve" in water as micelles, and tiny tiny amount of water will infiltrate oil, humidity of oil, just like humidity of breathable air.
If solute is not charged and does not have hydrogen-bonding-ready open orbitals or lone pairs of electrons, then the volume increases, both solute and solvent usually do not mix well.
If the solute is a ionic compound (e.g. salt, NaCl), then up to a certain amount of table salt the volume will decrease, the density will increase and reach the maximal amount of salt to reach the minimum amount of solvent, H2O, at a specific temperature and pressure.
If you mix a polar compound ready to accept or donate hydrogen bonds to H2O, then you get a similar picture as for NaCl in H2O, but not so drastic http://butane.chem.uiuc.edu/pshapley/genchem1/l21/vol.png , http://butane.chem.uiuc.edu/pshapley/genchem1/l21/1.html.


Determination of percentage of Methanol in an Alcohol based hand sanitiser Note: I am looking for advice on a chemical test - not a sermon.
I expect that this is entirely on topic in this forum, downvotes and sermon comment notwithstanding.
More than about 5% of Methanol in alcohol based hand sanitisers indicates that corners have been cut during manufacture and that the product should be avoided.
I'm looking for an easy test to determine the approximate percentage of Methanol in an Ethanol / Methanol hand sanitiser mix. Many hand sanitisers may contain small percentages of Methanol (typically in the 1% - 5% range) and this is acceptable. Anything over 5% I'd like to avoid. Much over 5% (say 10% +) I'd like to be very clearly distinguished. If actual approximate percentage can be indicated, so much the better.
Ideally I'd like a test with either low cost, simple to use, readily available non-hazardous reagents (one can hope) or failing that, something that can be prepackaged for easy use by the unskilled.
In web searches I found a range of tests for Methanol but most were not very quantitative and most were not intended to work in an ethanol-methanol mixture.

Background: Added:
Some guy y'know said:

Using disinfectants outside of a heathcare environment is still nonsense in the year 2020. Using disinfectants of questionable provenience is simply hazardous. Don´t do it.

The use of hand sanitisers is well covered in recommendation documents from regulatory authorities in many administrations and by internationally relevant bodies. Sanitisers are sold under regulatory guidelines which if followed would guarantee their safety when used appropriately. In the US many out of specification brands are being sold which are indistinguishable from those which comply. Several of the largest US sellers have been selling out of spec product. There is no way to determine provenance based on labelling.
While the use of soap and water is often a good or even superior solution it is incorrect to suggest that the use of sanitisers in general applications is nonsense.
In this instance I am interested NOT in their efficacy or application but re their specific bonafides with respect to a crucial component which is known to be used in excessive and illegal amounts in some cases.
I'm an engineer. Sanitisers are sold by a wide range of manufacturers and repackagers. I read labels, note sources, note claimed contents and viscosities and residence times and more.

Russell, agree with your down voting frustration but it is a common problem of any public forum. The bad news is that "simple" distinguishing test do exist, but they generate more toxic products (including a carcinogen) in a so-called iodoform test. Good for school labs only. The problem is not in distinguishing methanol versus ethanol, the problem is the matrix (as analytical chemists call it), i.e., these alcohols are present in a complex mixture.
The only reliable test test to determine methanol in a complex mixture like a hand sanitizer is gas chromatography. You need to consult a professional certified service to ask them to analyze the sanitizer. There is no home-kit or simple test to differentiate the two in a mixture.


What reaction occurs when etching copper to create circuit boards? The components used for etching copper are acetic acid, hydrogen peroxide and sodium chloride. I think the equation may be similar to this:

$$\ce{2H3C2O2H + H2O2 + 2NaCl + Cu -> CuCl2 + 2H3C2O2Na + 2H2O}$$

Is this equation correct? There are a lot of bubbles created. What is the gas that is given off?

Well your equation balances but that isn't the right reaction.

First lets break this down into half cell reactions.

Oxidation reaction

  $\ce{Cu_{(s)} <-> Cu^{2+}_{(aq)} + 2e^{-}}$

Reduction reaction

  $\ce{H2O2 + 2 H^+ + 2e^{-} ->  2 H2O_{(l)}}$

Decomposition reaction

Hydrogen peroxide spontaneously decomposes in acid solution.

  $\ce{2H2O2 ->  2H2O_{(l)} + O2_{(g)}}$

Stoichiometric reaction between copper and hydrogen peroxide

So the stoichiometric reaction between copper and hydrogen peroxide is:

  $\ce{H2O2 + 2 H^+ + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2 H2O_{(l)}}$

the $\ce{2 H^+}$ ions come from the vinegar. (Let's use $\ce{HOAc}$ for vinegar).

  $\ce{H2O2 + 2 HOAc + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2OAc^{-} + 2H2O_{(l)}}$

Sodium chloride's Role

Finally the role of the $\ce{NaCl}$ plays in the etch. The $\ce{NaCl}$ must be fairly high to drive the $\ce{Cu^{2+}}$ to a copper chloride complex. This thus reduces the "free" $\ce{Cu^{2+}}$ in solution which keeps the electrochemical potential of copper high so that the solution keeps dissolving copper. (Think of the half-cell copper reaction as if it is in a battery. The voltage of the half-cell would drop as the copper builds up.)


O-carboxylation in the Kolbe-Schmitt reaction In the Kolbe-Schmitt reaction, we know that the pi electrons of the benzene ring attack the carbon in carbon dioxide, but why doesn't the reaction occur on oxygen instead (as shown in the diagram below)?

The resulting product would be a carbonate hemiester monoanion, specifically monophenyl carbonate.  Similar species have been observed in carbonated alcoholic beverages.  (Instead of phenol, the alcohol involved is ethanol so the carbonate hemiester is monoethyl carbonate.)  There is an equilibrium between bicarbonate and the hemiester:

$\ce{HCO3- + ROH <=> RCO3- + H2O}$

According to the paper I linked the equilibrium constant is about 2.0, i.e. $K_{eq} = \frac{[\ce{H2O}][\ce{RCO3-}]}{[\ce{HCO3-}][\ce{ROH}]} \approx 2$.  Under aqueous conditions, the concentration of water is very large and the concentration of bicarbonate is much lower, so usually the equilibrium concentration of the carbonate hemiesters is very low, on the order of a few mM.

In contrast, under the conditions of the Kolbe-Schmitt reaction, the carbon-carbon bond formation is effectively irreversible.  There is no equilibrium, and the desired product can be obtained in high yield.


Can you decay to half an atom? In school we are learning about half-life, and I was wondering if it is possible to decay to half an atom?

To clear up what are becoming confusing comments.

An atom is a nucleus of protons and neutrons surrounded by electrons.

An atom can decay by fission to make two or more atoms with a smaller number of protons and neutrons in each new atom.  The electrons get distributed between them.  Now in physics we'd normally describe this as a nuclear decay, rather than thinking about electrons around the nucleus which, to a certain extent, are irrelevant to the underlying process.

Now depending on how you want to think about this you're either increasing the number of atoms by splitting one into two, or you're creating two (or more) smaller sized atoms which you might loosely describe as halves of the original.

Fission products (the results of fission) do not generally contain exactly half the number of protons and neutrons and electrons in the original atom, even when those numbers are even, so we can't say they're generally exact halves.

Fission can also release neutrons - so a nucleus looses e.g. one neutron but still has the same number of protons left.  Even when the split is closer to 50-50, the process can still release some neutrons, and in fact that's exactly what is used in nuclear weapons and reactors.

Another typical decay type is by alpha particle emission.  An alpha particle is, in fact, a Helium nucleus, so again that's a split to two different nuclei.


first-order binding kinetics with multiple identical binding sites I have beads in solution at concentration $B$ which have $N$ independent binding sites for a single target molecule at concentration $T$. $T$ is small compared to $B$, so most beads will probably not have anything bound at all. The on and off rates for a single target binding a single site on the bead are $k_{on}$ and $k_{off}$.

I want to determine the equilibrium distribution of the number of bound targets per bead. I expect it to be a Poisson distribution (right?) with an average binding count that depends on the parameters above.

Define $B_n$ to be the concentration of beads with $n$ bound targets. First order kinetics suggest we should have the coupled system of equations:

$\frac{dB_n}{dt}=k_{on}T(B_{n-1}-B_n)+k_{off}(B_{n+1}-B_n)$,

$\frac{dT}{dt}=\sum_{n=0}^{N-1}\left(-k_{on}TB_n+k_{off}B_{n+1}\right)$

with $B_{-1}\equiv 0\equiv B_{N+1}$

First question: is there an analytical solution for this system of equations? I am perfectly happy to simulate it, but if there is a way to avoid that, it would be nice.

Second question: the wikipedia page for Dissociation Constant has some ambiguous wording suggesting that even if the receptors are all identical, the rate constants will be dependent on $n$ as well, via

$K_d(n) =K_d \frac{n}{N-n+1}$

where $K_d=\frac{k_{off}}{k_{on}}$

How does that play here? Are my equations in need of modification, to something like:

$\frac{1}{k_{on}}\frac{dB_n}{dt}=T(B_{n-1}-B_n)+K_d \frac{n}{N-n+1}(B_{n+1}-B_n)$.

If so, could someone explain where that modification comes from?

EDIT: it seems that this is probably due to the fact that there is a higher change of dissociation if a bead has many targets bound since there is an equal probability for each target to unbind. However, it seems that the fraction presented is 1 for $n=N$, which suggests that the dissociation constant is defined in terms of a fully bound bead and not on a per-site basis. Could someone show me the derivation?

The scheme you need seems to be of the form
$$\ce{S + M_i  <=>[k_2][k_{-1}] M_{i+1}    }$$
where S is the molecule in solution and M the bead with i sites filled.
If there is no restriction on how many S may be associated with any bead and as the  S are indistinguishable molecules the number of ways of arranging the S among a total of M beads is
$$ W=\frac{M!}{\prod M_i!}$$
and the constrains are $\Sigma_0^N M_i = M$ and $\Sigma iM_i = N$ where N is the total number of S molecules.
Solving with these constraints is done using Lagrange multipliers, in the same manner as the Statistical Mechanics calculation of Boltzmann distribution, and leads to the distribution of finding i molecules of type S in a bead as,
$$p_i = \frac{\bar n ^i}{(1+\bar n)^{i+1}}$$
where $\bar n$ is the average occupancy of a bead. This distribution, which is clearly not Poissonian, has the form that it decreases as i increases even for moderate $\bar n$.

If the kinetic scheme is changed slightly to
$$\ce{S + M_i  <=>[k_2][(i + 1)k_{-1}] M_{i+1}    }$$
where the ‘off’ rate constant now depends on how many S are on a bead then the probability becomes Poissonian. This change/trick effectively makes the molecules distinguishable and now
$$p_i = \frac{\bar n^i}{i!}e^{-\bar n}$$
with $\bar n = k_2\ce{[S]}/k_{-1}$.

Which of these or other models applies will have to be determined by experiment.

If you want to solve the equations numerically then it is best to try a Master equation approach.

The set of equations for the population of M is written as
$$d\vec M/dt = \vec k \vec M$$
and has the formal solution $\vec M=\vec M_0e^{\vec kt}$ where $\vec M$ is a vector of concentrations of species i and $\vec k$ a matrix of rate constants such as

$$ \bar k=\left [\begin {matrix}
-Sk_2 & 2k_{-1} & 0 & 0 & 0 & \cdots\\
Sk_2 & -Sk_2 -2k_{-1} & 3k_{-1} & 0 & 0 & \cdots \\
\vdots& \vdots&\vdots & \vdots& \vdots& \\
\end {matrix}\, \right ]$$

where S is the concentration of S and is assumed to be in excess so that its value is constant. (If you want S to vary then you will need to do a numerical integration on all N equations instead of this method).

To solve, the secular determinant is set up and the eigenvalues $\lambda$ obtained. The populations at time t are given by
$$M(t)= x[e^{\lambda t}]x^{-1}M_0$$
where x is the matrix of eigenvectors corresponding to eigenvalues $\lambda$ and $[e^{\lambda t}] $ is a diagonal matrix of $e^{\lambda_1t}, e^{\lambda_2 t} \dots $ etc.


C=O bond length comparison among cyclohex‐2‐en‐1‐one and cyclohex‐3‐en‐1‐one I can see that in the first case a resonance would occur, and in the second case, there would be no resonance effect. I have read the fact that resonance reduces bond length due to partial double bond character.

According to that, shouldn't the $\ce{C=O}$ bond length in 1 be lesser than that in 2? The answer in my book is that the $\ce{C=O}$ bond in 1 is longer (reason is simply given as: "Resonance"). Could someone help me understand what's going on here?

The case 1 has partial double bond character (due to resonance) while case 2 has total double bond character of the C-O bond. So, relative to case 2, the C-O bond length in case 1 is longer.


Why is mercury much more toxic when inhaled than injected? Following wikipedia:

  Cases of systemic toxicity from accidental swallowing are rare, and attempted suicide via intravenous injection does not appear to result in systemic toxicity,[1] though it is still causes damage by physicaly blocking blood vessels both at the site of injection and the lungs.

Shouldn't both inhalation and IV injection of elemental mercury produce the same result, namely, mercury in the bloodstream?

I guess the answer is that mercury atoms stick with each other very well (e.g. high surface tension in comparison to most other liquids $\gamma_{293K}^{\ce{Hg}} = 0.47 \text{ N/m}$, where  $\gamma_{293K}^{\ce{H2O}} =  0.073 \text{ N/m}$ [1]) and that this continues to happen in the blood stream. Such clumps of mercury are not able to pass the blood brain barrier (BBB) effectively, especially in comparison to finely divided dissolved vapor, which is transported trough the BBB by the formation of a complex with cysteine.


What does the charge distribution around a Stone-Wales defect look like? I'm reading a book on carbon nanotubes (P.J.F. Harris, Carbon Nanotube Science, to be specific) at the moment and puzzling about Stone-Wales or 5775 defects. This is where two adjacent carbons undergo a 90-degree rotation about their mutual centroid and generate a defect consisting of two five membered rings and two seven membered rings, as I've illustrated below (This is only supposed to be qualitative, and is on a finite PAH fragment rather than a nanotube, but you should get the idea):

I note that tropylium ($\ce{[C7H7]^+}$) is aromatic as a cation, and the ever famous cyclopentadienide ($\ce{[C5H5]^-}$) is aromatic as an anion, so what I'm wondering is: do the 5 and 7-membered rings in a Stone-Wales defect pick up either formal or partial negative and positive charges to maintain aromaticity in a fashion consistent with the rest of the structure?

P.S.

Big props to F'x, who attempted electronic structure calculations on the PAHs depicted, regrettably without SCF convergence.

FWIW, here are some Hückel calculations on the 5775 PAH.

I'm showing the first 7 energy levels. Each molecular graph presents the contribution of each atom to the electron density ($\mathrm{Abs}[c_i]^2$). This is not normalized. I do not use wavefunctions, only the coefficients multiplying them, assuming

$$\Psi_{\mathrm{mol}} = \sum_i^n c_i\psi_i$$

In this approximation (HMO), the first 7 energy levels will be occupied (14 pi electrons).

I will first give you pictures of each energy level, and then a linear combination of each. Take it for what it is, as there is much information lost in the process.

To me, this simply suggests that indeed, the system will behave as if composed of tropylium and cyclopentadienyl ions. But this is not publication quality calculation either =)

Here are compared the linear combination of the first 7 energy levels of the 5775 and 6666 systems. It is clear that electron distribution is not the same in both PAHs, which hints that my calculations are not so far fetched.

5775

6666


Lithium-ion batteries I understand what occurs on a regular Li-ion battery - the positive Lithium ion moves from one electrode to the other of the battery. But what is exactly the physical mechanism for this intercalation to occur?
Is it that the Lithium in the LiCoO2 electrode lowers the chemical potential between electrodes? I am really curious about the process - any reference would also be helpful!

Yes, lithium ion batteries works with the principle of intercalation and deintercalation of ions from cathode to anode and anode to cathode respectively. To be specific for intercalation, lithium (in cathode) having high oxidation potential oxidize, loosing an electron which goes to the current collector to anode. So lithium ion get positive charge. Since the battery source you use also has positive terminal hence repulsion between ion and positive terminal occurs. Thus positive ion of lithium goes towards anode, which has layered structure thus result in intercalation. Another way of seeing it is because of difference in the concentration the diffusion of ions occur or you can also consider the fact charge neutrality.


Is hydrolysis of polypeptides and polysaccharides "anabolic" or "catabolic" When a polysaccharide or polypeptide is hydrolyzed into mono-saccharides or amino acids, the building blocks can be oxidized to release energy.  The oxidation is considered to be catabolic since it reduces the building blocks to simple compounds: carbon dioxide, water, ammonia, and releases energy.

Is the process of hydrolysis that breaks up polypeptides and polysaccharides a net endothermic or exothermic process?
Do the free amino acids and monosaccharides have more or less stored energy than the polypeptide or polysaccharide that they were broken down from?
Is it proper to call the isolated process of "hydrolysis" of proteins and polysaccharides "catabolic"?
Are protein synthesis, glycogen synthesis, (and triglyceride formation), by dehydration synthesis processes that require energy or release energy.  I think that they release energy which is semantically interesting since protein and glycogen synthesis are the main examples of anabolism in the body but may actually release energy which is a key component of the definition of catabolism.  Even if the energy released from protein synthesis is not generating ATP directly, wouldn't the heat produced conserve ATP in the long run.

1) Is the process of hydrolysis that breaks up polypeptides and polysaccharides a net endothermic or exothermic process?

Under physiological conditions, it is a process that goes forward, i.e. the Gibbs energy is negative. As a consequence, it can happen outside of cells in the absence of ATP. When we eat, the hydrolysis of polysaccharides starts in our mouths, while the hydrolysis of proteins occurs under harsher (acidic) conditions in the stomach and continues in the intestine (slightly basic conditions).

For warmblooded animals like us, exothermic or endothermic is less important, but you could look it up.

  2) Do the free amino acids and monosaccharides have more or less stored energy than the polypeptide or polysaccharide that they were broken down from?

Less Gibbs energy, as the hydrolysis is spontaneous. However, this Gibbs energy is not captured (different from downstream oxidation of building blocks to $\ce{CO2}$, which is coupled eventually to synthesis of ATP).

  3) Is it proper to call the isolated process of "hydrolysis" of proteins and polysaccharides "catabolic"?

If your definition is just breaking down complicated molecules into simple ones, then you would call it catabolic. If you also require that Gibbs energy is captured in a coupled reaction, then you would classify these reactions as digestion, and the downstream oxidation to carbon dioxide as catabolism.


Why are the concentrations in the equilibrium constant multiplied? $$\alpha\,\mathrm{A}+\beta\,\mathrm{B}+\cdots\rightleftharpoons \rho\,\mathrm{R}+\sigma\,\mathrm{S}+\cdots$$
The equilibrium constant for this reaction can be calculated using the following formula:

$$\mathrm K_c = \frac{{{[\mathrm{R}]}}^\rho {{[\mathrm{S}]}}^\sigma ... } {{{[\mathrm{A}]}}^\alpha {{[\mathrm{B}]}}^\beta ...}$$

In this formula for the equilibrium constant of a reaction, why are the quantities R and S (and likewise A and B) multiplied?

I understand the equilibrium constant to be a ratio of the amounts of the products and the amounts of the reactants. But if this is the case, wouldn't a formula like $\mathrm K_c = \frac{[\mathrm{R}] + [\mathrm{S}] ... }{[\mathrm{A}] + [\mathrm{B}] ...}$ make more sense? I think this question is really a result of not understanding the meaning of the equilibrium constant.

As discussed in comments, you can find a good derivation in a standard university level physical chemistry or thermodynamics textbook.

Let's instead consider why it should be true.

From a microscopic perspective, what we're trying to balance, in essence, is the density of states on both sides of the equation, in essence, the count of ways to rearrange the various pieces. These are heavily dependent on the relative spacing of energy levels and degrees of freedom on both sides. Fortunately, this complex information is captured in the value of $K$.

But the point is that if you're trying to find the total number of permutations of two different collections, you multiple the values. For example, if there are two dice, the total number of outcomes is 36.

And concentrations essentially measure the number of things to rearrange, so the total number of possibilities should scale with the product, not the sum.

This is all a bit hand-wavy, but again, I'm not trying to derive the answer. I'm just trying to suggest that even without a formal derivation, it seems very reasonable that the correct operation to use is multiplication.

EDIT:

After reading this again, I realized I can extend this with another hand-wavy justification.

Remember that we have:

$$\Delta G = -RT \ln K$$

If you have two reactions that you combine together, you want the energies to sum. But that translates into products inside the log! Once again, the proper operation for equilibrium constants is multiplication.


Reactions which are impossible theoretically Take an example of reaction like :

Burning of carbon in air to form Carbon dioxide, say heat required is 100 joule (assume so)
And we provide the system say 20 joule heat
Theoretically it would be not possible. But the alone C atoms present near heat source will absorb their required heat (which would be very very small indeed) and get converted into co2
Hence, there will be small value of reactants which are converted into products then how can we say reaction is not possible. Thus at room temperature is it true that there is some concentration of products in any reaction.

If the $20$ J you mention is the average energy this will correspond to a certain temperature. However, molecules have a distribution of energies, caused by their random motion in solution or in the gas phase, so that some have far more energy than the average and some far less. The relative populations of vibrational and rotational energy levels a molecule has are determined by the Boltzmann (exponential) distribution. This means that at a given temperature some molecules are more likely to react than others. If the $100$ J you mention corresponds to a temperature at which the reaction proceeds with a reasonable rate constant then if you only supply $20$ J the rate constant will be $e^{-5} \approx 7.10^{-3}$ times smaller. This may be a small enough rate constant to suspect that the reaction is not proceeding, but its just a matter of waiting !

(If you have a temperature gradient across your sample then of course the reaction rate constant will also vary across the sample being greatest where the temperature is greatest.)


Mathematical form of four hybrid orbitals The mathematical form of the four $\ce{sp^3}$ hybrid orbitals are given below.
  \begin{align}
\ce{
\tag{a} sp^3 &= 1/2s + 1/2p_x + 1/2p_y + 1/2p_z\\
\tag{b} sp^3 &= 1/2s + 1/2p_x - 1/2p_y - 1/2p_z\\
\tag{c} sp^3 &= 1/2s - 1/2p_x + 1/2p_y - 1/2p_z\\
\tag{d} sp^3 &= 1/2s - 1/2p_x - 1/2p_y - 1/2p_z\\
}\end{align}
  For each hybrid orbital the number in front of the $\ce{s}$ and three $\ce{p}$ functions, called a coefficient, describes the contribution and relative ratio of each canonical orbital to the hybrid wave function. Add up the coefficients and prove to that these orbitals are $\ce{s^1p^3}$.

Yes, I know that the grammar is terrible on that last sentence, that is how it is written. So, I know that for an $\ce{sp^3}$ orbital it is one part $\ce{s}$ orbital and three parts $\ce{p}$ orbital. However, I have no idea what this question wants me to do. Adding up the coefficients gives me $2$, $0$, $0$, and $-1$, but I have no idea if that is how it wants me to add it up (you could ignore the sign and get $2$ every time, and show that $1/2$ is a quarter of $2$, thus proving that each component contributes a quarter to the hybridized orbital, but this seems too simple and incorrect).

If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination

$$\psi = \sum_n c_n \phi_n$$

then loosely speaking, the "amount" of $\phi_n$ in $\psi$ is given not by $c_n$, but rather by $\left|c_n\right|^2$. (See, for example, Griffiths, Introduction to Quantum Mechanics, 2nd ed., Section 3.4.)

Therefore, summing the four values of $c_n$ for each AO has no physical meaning.

Instead let's take wavefunction (b) for example. The coefficient in front of the s orbital is $1/2$, so the "amount" of s-orbital character is simply $\left|1/2\right|^2 = 1/4$. The coefficient in front of the $\mathrm{p}_y$ orbital is $-1/2$, so there is $\left|-1/2\right|^2 = 1/4$ $\mathrm{p}_y$-character.

Armed with this knowledge, it's incredibly easy to see that each orbital has exactly $1/4$ s-character and a total of $3/4$ p-character ($1/4$ from each p orbital), because none of the $\pm$ signs matter (the minus signs are removed by the absolute value).


How many significant digits are appropriate when one or more digits are eliminated (during subtraction)? Say, for example, $50.0-49.6=0.4$. Does this result have $1$ significant figure, or $3$ (as in the data: $50.0$ and $49.6$)?

Had it been $50.0-4.6$, it is understood that the answer is $45.4$, by the rules of significant figures. How do I apply them in the "$50.0-49.6$" case?

This question deals with $50-49.6$; so a related but not the same question.

There's an easy way to look at this.

Lets say the value $50.0$ refers to $\pu{50.0 cm}$ measured accurately to $\pu{0.1 cm}$, and that $49.6$ refers to $\pu{49.6 cm}$ measured accurately to $\pu{0.1 cm}$. The difference would be, as you've said, $\pu{0.4 cm}$ measured to $\pu{0.1 cm}$ accuracy.

So, yes, the answer has only one significant digit.

Your initial measurements aren't more accurate than $0.1$, so adding two extra significant digits is incorrect.

I hope this helps.


Why is it ill advised to eat the solid aspirine formed from salicylic acid + acetic anhydride In the laboratory notes it is indicated that it is ill advised to eat the solid formed from this experiment. An esterification reaction, in which salicylic acid is treated with acetic anhydride.

If it is for the acid added as catalyst either sulfuric acid or phosphoric acid. Why isn't it possible to neutralize it with a strong basic?

The addition of water should be enough to quench (destroy) any remaining acetic anhydride and turn it into acetic acid, and this latter one is more soluble than aspirin so it should be possible to crystallize aspirin without any acid.

The question appears in this lab notes http://www2.xula.edu/chemistry/crs-orgleclab/Organic_Lab_1_2230L_Web_Files/23_Aspirin_notes.pdf

I think it is self-evident that your basement shouldn't become a pharmacy. I think it has more to do with the acids used in the synthesis. Does the industrial process differ from the synthesis presented in the article with regards to this aspect (acid catalyst and their neutralization after the process)?

Unless you worked to Pharma industry Good Manufacturing Practice you shouldn't trust your product

The simple reason why you should not trust the product is that that you don't know it is safe unless you have followed rigorous quality control processes. The drug industry follows carefully regulated quality-control processes known, in Europe, as Good Manufacturing Practice. This is far from easy to meet and even major drug firms sometimes have their factories closed because they fail to ensure it.

For your particular reaction there are many possible things you would need to check before knowing the product was safe to consume. You really don't want to guess whether all the acetic anhydride or acetic acid has been removed: even a small amount will be bad for you (you can't rely on intuition that it should all have been removed). And are you sure there are no nasty side products? Have you checked? Have you checked the purity to make sure? Have you had a check by an independent party?

If you can't say yes to all of these (and more: go read the GMP standards for aspirin manufacture), don't take the product.


Why do ionic compounds dissociate whereas coordinate complexes won't? An ionic bond is the bonding between a non-metal and a metal, that occurs when charged atoms (ions) attract after one loses one or more of its electrons,and gives it to the other molecule, for example sodium and chlorine. This makes the bond stronger and harder to break.
  In other words, an ionic bond is the electrostatic force of attraction between two oppositely charged ions. The positive ion is called cation, and the negative ion is the anion. It is like the north and south poles of a magnet.
  A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. The stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding. For many molecules, the sharing of electrons allows each atom to attain the equivalent of a full outer shell, corresponding to a stable electronic configuration.
  A dipolar bond, also known as a dative covalent bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.
  NOTE: It is important to recognize that pure ionic bonding - in which one atom "steals" an electron from another - cannot exist: all ionic compounds have some degree of covalent bonding, or electron sharing. Thus, the term "ionic bond" is given to a bond in which the ionic character is greater than the covalent character - that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bond to be more polar (ionic) than other forms of covalent bonding where electrons are shared more equally. Bonds with partially ionic and partially covalent character are called polar covalent bonds. Nevertheless, ionic bonding is considered to be a form of noncovalent bonding.

$$\begin{equation}\tag{1}F=\frac{1}{4\pi\epsilon}\cdot\frac{q_1q_2}{r^2}\end{equation}  $$
 According to Coulomb's law equation $(1)$, the force between two charges depend on the medium in which they are placed. The value of $\epsilon_r$ (relative permitivity) of air is found to be approx $1$, whereas relative permittivity of water is found to approx $80$ at room temperature.

I thought Coulombs law could answer why ionic compounds dissociate into ions whereas other compounds of different bonds won't. I thought, ionic compounds are formed by the electrostatic interaction of two oppositely charged ions (I assumed these to be charges), they will be stable in air, as relative permittivity value is not much effective. When, the same is dissolved in water, I thought they will dissociate as relative permittivity value of water effects more. I found the explanation, so as to why ionic compounds dissociate into ions when dissolved in water. In case of coordinate bond, I thought an atom losses a pair of electron and thus can be called as cation, whereas the other becomes anion. So, even here we can expect electrostatic interaction, and as discussed above must dissociate into ions when dissolved in water. But coordination complex of coordinate compound won't dissociate? I think, I have misunderstood somewhere or else there must be some chemistry theory which could explain, why coordinate complex won't dissociate?

Even in case of covalent bond we can see the bond to be formed by stable electrostatic attraction and repulsion between electrons and protons, so I thought, even covalent bonds should break, when dissolved in water, as electrostatic interaction is going to be affected. I thought, there must be some other forces which must be keeping them bonded. If so what are they?

I don't know whether I am right or wrong, if any was the case please explain.

First of all, who says coordination complexes don't dissociate? The interesting chemistry happens right there, and without dissociation no catalytic coordination compounds would be possible!

Now, the way I see it, when you move your compound, be it ionic, covalent or coordination, from air to a condensed phase, it will automatically interact with said phase. I think that you assume this phase is water, but it might as well be dichloromethane or ammonia or diethyl ether or anything else that is liquid (basically).

What makes a bond happen (or not) is the overall energy of the system. See, in chemistry (as in physics) the energy minimum is the goal of the system, since it is the most stable. Using this line of reasoning we can derive that bonds are formed because the overall energy (of the bonding electrons) is lower than if they were not bonded. Note that it doesn't matter where these electrons come from (they might be shared by both bonding partners or made available by only one of them). A good and easy way to visualize this is if you scratch the surface of Molecular Orbital Theory.

So why do ionic compounds dissolve? Mostly due to the solvent-compound interactions. See more here.

(I also think that your approach using Coulomb theory doesn't work that well, since the change of the dielectric constant around the compounds has nothing to do with what is between the bonding partners, thus affecting attractive forces between the two partners.)


Reduction of hemiaminal by triethylsilane / boron trifluoride I was trying to understand how the following reaction occurs (taken from a synthesis of homopipecolic acid[1]):

I thought that triethylsilane, $\ce{Et3SiH}$, would reduce the ester to an aldehyde. However, it turns out that the hemiaminal carbon has also been reduced with cleavage of the $\ce{C-N}$ bond. How does this happen?
References:

Chiou, W.; Chen, G.; Kao, C.; Gao, Y. Syntheses of (−)-pelletierine and (−)-homopipecolic acid. Org. Biomol. Chem. 2012, 10 (13), 2518. DOI: 10.1039/C2OB06984A.

My organic chemistry is rusty by now, but the full mechanism for the reduction itself is likely something along these lines:

In a "typical" lactone / ester, the alkyl oxygen can't be lost so easily, but in this case there's a nitrogen lone pair that can assist in that. The resulting iminium ion is reduced by the silane. The acid at the end will remove the remaining boron trifluoride.