π Bonding in Carbon Dioxide I am fairly sure the first diagram I drew for carbon dioxide is wrong in terms of showing π bonding. This is because we use a π orbital twice, which isn't possible. The second diagram corrects this by realizing there are two unused p orbitals on the carbon.

EDIT: This is what I got for allene. Again we have an central $\ce{sp}$ carbon and this implies that two p-orbitals are free to form π bonds.

In terms of a simple bonding concept the answer of ron is correct, however due to symmetry constraints it is not possible to distinguish the two $\pi$ systems. The reason for this is, that these $\pi$ orbitals are degenerated and perpendicular to each other.

Carbon dioxide

The molecule is very symmetric and one of only a few with the point group $D_{\infty h}$. That also imposes constraints on how the spatial distribution looks like.

Along the molecular axis ($\ce{O-C-O}$) is a rotational axis of infinite fold, $C_\infty$, as the molecule is linear. Perpendicular to that axis there is a horizontal mirror plane $\sigma_h$ passing through the centre of mass, i.e. the carbon. Additionally there is an infinite number of two-fold symmetry axis $C_2$ lying in that plane, which is the criterium for $D_{xh}$ point group. In line with the main axis there is also an infinite number of vertical mirror planes $\sigma_v$ (which is implicit due to the horizontal mirror plane).

For the sake of convenience we can define an upper and a lower half of the molecule (taking the cartesian $z$ axis as the main symmetry axis). These two halfs are indistinguishable, hence their orbitals must reflect that.

Having a look at the $\pi$-orbitals, clearly shows that (enlarge):

So your first drawing was not too wrong after all. The only thing you forgot is, that there is an identical orbital perpendicular to that.

$\ce{CO2}$ is also a nice example to see the boundaries of the hybridisation concept. In this example, due to symmetry, all atoms have to be $\ce{sp}$ hybridised. And that matches all the $\sigma$ orbitals (enlarge).

Allene

The allene on the other hand is not linear and has a much lower symmetry. There is the main symmetry $C_2$ axis (for convenience) along the $\ce{C-C-C}$ axis, accompanied be two $\sigma_v$ mirror planes. There are also two diagonal symmetry planes $\sigma_d$ right between them. Perpendicular to the main symmetry axis we have two more $C_2$ axis (also perpendicular to each other). The point group is therefore $D_{2d}$ and has a lot less symmetry constraints.

However, the orbitals are actually (surprisingly) quite similar though. We can define $\sigma$ and $\pi$ orbitals accordingly. The biggest distinction is still in the $\pi$-orbitals (enlarge).

The Lack of symmetry and therefore less constraints make is possible, that the orbitals combine in a different manner. The top and the lower half of the molecule does not have to be the same anymore, hence the orbitals are more localised to one bond in particular. A notably thing is, that hydrogens take part in $\pi$ bonding (Hyperconjugation).

The sigma orbitals are quite the same as in $\ce{CO2}$ (enlarge).

However, the traditional bonding picture of localised orbitals is quite valid here. The central carbon is still being $\ce{sp}$ hybridised and the moieties are $\ce{sp^2}$ hybridised.

Ketene is a combination of the two above, with even lower symmetry ($C_{2v}$). Therefore the conventional bonding picture should hold.

For Oxygen it is usually always good to check, if hybridisation is a good concept to describe the bond. When bonded to more than one partner, the answer is usually yes. If less, then there is a good chance, that hyperconjugative effects outweigh the information gained from the hybridisation concept.

Additional fun note

One could argue that the carbon suboxide ($\ce{C3O2}$) is not too different, but here again you will be surprised, as it has a very shallow bending potential. (crystal structure, vibrations)


Polymerisation of methyl methacrylate - can the C=O group get involved? When methyl methacrylate undergoes free radical addition polymerisation, the C=C pi bond is broken during the process:

I'm wondering why the C=O pi bond doesn't get broken during the polymerisation. I suspect it may be because the C=O bond is stronger than the C=C bond, but are there any other ways to rationalise this?

C=O bond is strongly polar, and the cleavage of the double bond would rather involve heterolytic breaking with ions formed than homolytic with radicals.


Can hyperconjugation take place between lone pairs of an atom attached to an sp2 carbon? When $\ce{HBr}$ is added to 2-chloro-2-butene, the proton preferentially bonds to the $sp^2$ carbon that's not bonded to chlorine. Does this mean that hyperconjugation takes place between the lone pairs of chlorine and the positively charged $C3$? I would assume that since there is still a free rotation about the $\ce{C-Cl}$ bond and that chlorine is in a tetrahedral formation this would be the reason behind the regioselectivity.

Hyperconjuation is the interaction of electrons in sigma orbitals (in $C-H$ or $C-C$) with an adjacent empty p orbital or $\pi$ orbital. What you are describing is not hyperconjugation, but resonance; more specifically, the chlorine shows the $+R$ effect.

Now if you draw the resonance structures of this molecule you will realize that the $C_3$ has a negative charge on it in one canonical form. This means, in the actual molecule, that $C_3$ carbon has a partial negative charge, which is why the hydrogen from $HBr$ preferentially bonds to that carbon.

A more acceptable explanation would be that, when the hydrogen bonds to $C_3$, the resultant intermediate carbocation (positive charge on $C_2$) is stabilized by the $+R$ effect of chlorine, resulting in a racemic mixture of $2-bromo-2-chlorobutane$ to form as the major product.


Rank the following protons in order of acidity I have an acid and base organic chem quiz tomorrow and I need help determining how acidic protons are. I understand the concept of atoms, resonance, induction, and orbital when considering the acidity of protons. However, when I'm given these molecules, it overwhelms me so I don't understand how to break them down. I believe that the first step is that I have to find the conjugate base for each one and then just compare? Below is photo of my worksheet. I have already filled in the correct answers. Thanks in advance for your help.

the c hydrogen is more acidic because it is stacked between two electron withdrawing carbonly groups, and after the loss of hydrogen the carbanion is more stabilized, where as the same thing is not observed in case of d hydrogen the carbanion is in conjugation but only with one of carbonly group followed by b and c hydrogen.

now in case of second example, e is more acidic because it is allylic as well as next to the carbonyl group which means the anion after loss of hydrogen is more stabilized as come to f and g.


What can possibly turn litmus paper white if there is even no chlorine present? I did a hydrocarbon test on cyclohexane with bromine solution. The mixture was put inside a dark place. To be exact, a cupboard filled with laboratory equipments. I also put a piece of wet blue litmus paper on top of the tube.
It should've stays blue because there was no reaction occured, right? But the litmus paper turned white. What can possibly cause it?

Chlorine turns litmus paper white by oxidizing it.  There are, of course, other oxidizing agents (like bromine) that can do the same thing; chlorine is just the best known because we associate it with commercial bleaches.


Applying Bohr's model for Helium atom Consider the following atomic model for Helium. Helium contains 2 electrons revolving in circular orbit around the nucleus,in such a way that there is minimum possible repulsion between them.Calculate the first ionisation energy of this Helium.Will there be any difference in the ionisation energies calculated and experimentally found?

What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.
Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...
The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).
The experimental value is 1.81 (24.6 eV), which is significantly lower.
Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.


Does esterification require reactants to be present in the same plane? I believe that the first step is the Cannizzaro reaction which will give (c) as a product.

But since we are also adding $\ce{H+}$ in the second step, shouldn't an esterification reaction take place? That would give (d) as a final product.

However the answer given is (c).
My only guess as to why esterification wouldn't take place is the fact that the phenyl groups are in perpendicular planes, therefore esterification is difficult?
Is it necessary to have all the reactants in a plane for esterification?

The two commenters are collectively correct. But to clarify and augment their answers I would say that structure (d) cannot be correct because it is missing 2 carbon atoms relative to (c), and if you tried to cyclize (c) by forming esters, the result would be highly strained. Try making a molecular model of it and see how it looks.

Below is the product you would get if (c) cyclized. What is the difference between this product and (d)?


Color of Anhydrous Transition Metal Salt Why are some anhydrous transition metal salts colored?

Here are some examples I found.

Anhydrous cobalt(II) chloride ($\ce{CoCl2}$), with sky-blue color.
Anhydrous chromium(III) chloride ($\ce{CrCl3}$), with purple color.
Anhydrous copper(II) chloride ($\ce{CuCl2}$), with yellow-brown color.

I already know that complex ion could undergo d-d transition from its unfilled d orbitals, and that's what makes them pretty-colored. (The wavelength of absorbed light corresponds to the gap in energies between d orbitals.)

But in anhydrous salt, which I think isn't a coordination compound, the d orbitals are degenerate. Thus, no energy gap and excitation could occur, no light is absorbed, and the compound would be colorless / white-colored as it reflects all wavelength.

I summarize my understanding in this problem as below.

Solution of complex ion could give color (with exception of $\ce{[Zn(H2O)6]^2+}$ and some other complex ions). e.g. $\ce{CuSO4 (aq)}$, which form $\ce{[Cu(H2O)6]^2+}$ complex ion with blue color.
Hydrated salt is indeed a coordination compound (found related question here), so it could give color. e.g. $\ce{CuSO4.5H2O (s)}$ with blue color.
Anhydrous salt, because it doesn't have any ligands coordinating with the metal ion, it's not a coordination compound. It's a mere ionic salt instead. Shouldn't it be colorless?

Is there any explanation to this case?

Note that in all compounds mentioned above, the metal ions are transition metal. We don't consider s-block metal ions such as $\ce{NaCl}$ and $\ce{MgSO4}$. (Additional question: They should be colorless, am I right?)

A ‘mere ionic salt’, when inspected properly, turns out to be nothing different from a coordination compound. One side (the cation) is electron deficient while the other side (the anion) has spare electrons, so there will be some kind of coordination happening from one to the other. Indeed, all salts adopt crystal structures so that the favourable coordination interactions are maximised.

For anhydrous cobalt(II) chloride, I would immediately expect a closest packing with cobalt ions sitting in tetrahedral voids, occuping ¼ of them to give $\ce{[CoCl4]}$ tetrahedrons where each chloride coordinates two neighbouring cobalt ions. Analogously for chromium(III) chloride, I would expect chromium to occupy one-third of the anion structure’s octahedral voids, giving rise to $\ce{[CrCl6]}$ octahedrons where again each chloride coordinates two chromiums. Both are thus clearly coordination compounds. And going by what I know about $\ce{Cu^{II}}$, that salt will probably be a set of square-planar $\ce{[CuCl4]}$ fragments, again where each chloride coordinates two coppers.

Note that all of these are rough guesses (I wouldn’t even call them educated) since you can only predict solid-state (ionic) structures so far a priori. Oftentimes you can arrive at multiple possibilities and even more times you are surprised at what is actually adopted but then retrospectively you can usually explain the structure somehow with coordination.


Basicity comparison of ammonia derivatives (and guanidine) Arrange the following in decreasing order of basicity:

  $\ce{N(CH3)3}$
  Guanidine
  $\ce{N(C6H5)3}$
  $\ce{NH(CH3)2}$

My answer - 2 > 1 > 4 > 3

Option '2' due to equivalent resonance structures on protonation. Then option
 '1' due to thrice +I effect. Followed by option '4' having +I effect twice which causes the lone pair electrons to be repelled from the nitrogen atom. Finally option '3', because lone pair electrons are delocalised on three rings.

Is my logic for the question correct or not? I am not sure if my answer is correct.

In triphenyl amine (3), the lone pair of nitrogen is in conjugation with phenyl groups and thus would be least available for protonation.

Due to resonance structures of guanidine (2), it should be the strongest base.

Although number 4 doesn't exist as it is written. Let's suppose it is $\ce{N(CH3)2}$ with a negative charge (the only possible way to have two methyl groups on a nitrogen).

If so, it is more basic than guanidine, because charged compounds are generally more reactive than neutral ones and will attract $\ce{H+}$ more quickly.

So if number 4 is charged, then the basicity will decrease in order
4 > 2 > 1 > 3


How to make hydrogen peroxide into gel How to make hydrogen peroxide into a gel or just simply really thick and it has to be able to be in your mouth?

The only thing i have came up with is (Gelatins).
What should i use?

There are a number of vegetable gums, such as carageenan, pectin and gum arabic, and man-made methyl cellulose which might be compatible with low concentrations of $\ce{H2O2}$, which you can determine easily. Any impurities, particularly transition metals, may make $\ce{H2O2}$ decompose, so it might be best to keep the gelling agent and peroxide separate until just before use.

If for oral use, this site suggests 0.75% to 1.5% solutions, one quarter and one half the strength of the common antiseptic N.B. $\ce{H2O2}$ bleach is much stronger and is not for oral use. Note that there are dangers with use of peroxide as a mouthwash.


Ionic bond and ionic radii in FeS Which of the following statements is true regarding an ionic bond in $\ce{FeS}$?

  Iron atoms gain electrons and sulfur atoms lose electrons.
  Iron and sulfur ions bond by electrostatic attraction.
  The ionic radius of an iron ion is equal to its atomic radius.
  Breaking an ionic bond between $\ce{Fe^{2+}}$ and $\ce{S^2-}$ releases energy.

My attempt:

I know metals lose electrons. I thought the answer was iron and sulfur ions bond by electrostatic attraction, but a classmate and I have a disagreement, so now I am not sure. I don't think the ionic radius of an iron (II) ion is equal to its atomic radius.

I think it is either 2 or 4.

Iron and sulfur ions bond by electrostatic attraction.

This is the right answer. I see a hard time maneuvering around it.

  Breaking an ionic bond between $\ce{Fe^2+}$ and $\ce{S^2-}$ releases energy.

Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right.

The misconception: Some students fall in the trap of believing that bonds are like boxes of energy. So, breaking these boxes of energy should release energy, right? No! In reality, ionic bonds are nothing but forces. This is like imagining that when you pull a box, you're applying or somehow putting some kinda a box.

The reality: Ionic bonds are just forces that are a result of attraction between two species (such as two atoms). Why do these happen in the first place? Because for instance in \ce{NaCl} the sodium atom really likes to lose an electron and chlorine wants to earn one. This love of giving and/or taking an electron is because of reaching the stability. Stability in this case means lesser energy. So, ionic bonds happen because atoms want to have lesser energy.

Finally, what would breaking this bond result in? If the atoms bonded together to lower their energies, then what would removing the bond result in?

  I don't think the ionic radius of an iron ion is equal to its atomic radius.

You're right! Think of the sodium chloride example again. The new sodium cation used to have 11 protons and 11 electrons. Now, after the lattice builds up, sodium crudely has only 10 electrons, with the same 11 protons. The electrons are outnumbered! They can't overpower protons anymore. So, what will happen to the distance electrons were able to travel to? (The distance they could "travel" further from the nucleus was called the atomic radius, and it's the ionic radius now)
So, is the atomic radius of iron equal to its ionic radius?

Rather pedantic and irrelevant, but there's no clear distinction between an ionic bond and a covalent bond. There are no "pure" ionic bonds.


Kolmogorov backward equations in active biological systems I am trying to understand the following equations from a paper by Wang et al. [1,  p. 5 at the SI]:

Michaelis–Menten Representation of the Kinesin Cycle
We assumed that there is a strong coupling between ATP turnover rate of kinesin and kinesin stepping on an MT. Each chemical step in the chemomechanical cycle of kinesin was assumed to be irreversible, except ATP binding and releasing (Fig. 1, step i). Consider the following scheme:
$$\large\ce{0 <->[$k_\mathrm{ATP}\lbrack\ce{S}\rbrack$][$k_\mathrm{-ATP}$] 1 ->[$k_\mathrm{s}$] 2 ->[$k_\mathrm{attach}$] 3 ->[$k_\mathrm{ADP}$] 4 ->[$k_\mathrm{h}$] 5 ->[$k_\mathrm{-P}$] 6}$$
Assume that $F_i(t)$ is the probability for the system to reach state $6$ if, at time $t = 0$, the system is in state $i.$ This function is governed by the backward master equations:
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)\label{eqn:1}\tag{S1}$$

Why does equation \eqref{eqn:1} read as
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)$$
and not
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{-ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)?$$
I would like to change/modulate $S(t)=\cos(t)$ in time and see how it changes the states $F_i(t)$
References
1 Wang, Q.; Diehl, M. R.; Jana, B.; Cheung, M. S.; Kolomeisky, A. B.; Onuchic, J. N. Molecular Origin of the Weak Susceptibility of Kinesin Velocity to Loads and Its Relation to the Collective Behavior of Kinesins. PNAS 2017, 114 (41), E8611–E8617. https://doi.org/10/gdj4h5.

UPDATED: I wrote a first answer assuming that $t>0$ which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that $t<0$ in these "backward equations". Here's a corrected explanation:

The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some time $t<0$, we want to know something about the probability of a system in each state reaching state 6 by time $t=0$, which is expressed as $F_i(t)$ where $i$ indicates the state number.

We start with a system in state 6 at time $t$. Since reversal back to state 5 is not permitted and there is no state beyond state 6, any system that is in state 6 will remain in state 6 up to (and past) time $t=0$. Thus the probability $F_6(t)=1$. In the paper, they use a Dirac $\delta$ function to indicate this, because that is necessary when $F_6(t)$ is used in the other equations.

For the other five states, the equations do not give explicit expressions for the probabilities, but instead express the change in probability over time as a function of the probability in a differential equation.

Let's consider a system that at time $t$ is in state $5$. During a time interval $\Delta t$, two things can happen: Either (1) the system remains in state $5$ or (2) the system switches to state $6$.

One assumption they seem to make is that the only way the probability can change is if the system changes state. So if the system remains in state 5, its probability $F_5(t+\Delta t)$ remains equal to $F_5(t)$. If it changes to state 6, then its probability increases to $F_6(t)$. The change is thus a weighted combination of these two.

In more detail, if we consider the probability of switching from state 5 to state 6 during the interval $\Delta t$ to be $P_{56}$, we can say that our new probability is the weighted sum

$F_5(t+\Delta t)=P_{56}F_6(t)+(1-P_{56})F_5(t)$,

so our change in probability is

$\Delta F_5=P_{56}F_6(t)+(1-P_{56})F_5(t)-F_5(t)=P_{56}[F_6(t)-F_5(t)]$.

Our last step is to figure out what $P_{56}$ is, and it is given by the rate constant for the change from state $5$ to state $6$. The rate constant is the average frequency of the reaction, so the probability of the reaction occurring in a given time interval is the rate constant times the length of time. So we replace $P_{56}$ with $k_{-P}\Delta t$ and get

$\Delta F_5(t)=k_{-P}\Delta t[F_6(t)-F_5(t)]$.

Dividing by $\Delta t$, we have

$\dfrac{\Delta F_5(t)}{\Delta t}=k_{-P}[F_6(t)-F_5(t)]$.

As $\Delta t$ goes to $0$, this becomes the derivative:

$\dfrac{dF_5}{dt}=k_{-P}[F_6(t)-F_5(t)]$.

You can repeat this for the other five states to generate the other equations in the paper. The key point is that even when back reactions are permitted (such as from state 1 to state 0), the probability function is only for the systems that began in the indicated state, so the rate constant for a system coming back to that state does not enter into the equation. An implicit assumption is that the system does not make two moves during the time interval $dt$. That is, it cannot move from state 0 to state 1 and back to state 0.


Can fluorescent & phosphorescent molecules still be fluorescent or phosphorescent as ions and not as salts? The way my pchem professor explained it is that fluorescence /phosphorescence is caused by collision excitation. During this collision the spin of the promoted electron is flipped to a parallel spin of the orbit it left. The Pauli Exclusion Principle prevents relaxation/emission until the electron is some how able to "flip" its spin.

Most of the phosphorescent compounds I've found online have been salts which makes a bit of sense, (I imagine that the oscillation of the ionic bond is what causes the collision excitation). However, not all of them are salts. In lab previously our solution of sodium fluorescein exhibited a small degree of phosphorescence even when in basic conditions. I'm curious how a molecule can promote to a triplet spin multiplicity internally.

My professor says that the ionic component is not necessary and these compounds are typically ionic for the purpose of stability and longevity.

If the ionic bond isn't contributing to the triplet spin multiplicity then what is causing it?

Fluorescence derives from an electronically excited state after the molecule has been excited by absorbing a photon, usually in the visible or uv parts of the spectrum. Almost all organic molecules fluoresce to a greater of lesser extent, most common are aromatics for example, benzene and its derivatives as well as most dye molecules such rhodamines or coumarins. They can be ions or neutral.

When a molecule is electronically excited an electron is transferred from the highest occupied molecular orbital (HOMO) into an anti-bonding orbital usually the lowest one (LUMO). From here the molecule, which now has excess energy, can emit a photon and return to the ground state. The lifetime of the singlet excited state  varies between different types of molecules but is generally speaking around a few nanoseconds and not longer than microseconds.

By definition the excited state produced has the same spin as the ground state, which is usually means spin paired electrons $\ce{^ (v)}$. They remain paired even in the excited state (called a singlet excited state). Fluorescence is emission from a molecule with spin paired electrons, the transition is called 'allowed' and happens rapidly.

Occasionally the molecule can suffer a perturbation which causes one of the spins to be flipped to produce 'parallel' spins $\ce{^ ^}$ (heavy atoms or paramagnetic species can cause this). In this case a triplet state is produced and now the transition back to the ground state is far more difficult as the spin-flip has to be undone. The emission from such a state is called phosphorescence, it is long lived vs fluorescence, microseconds and longer, and as the triplet state is lower in energy than the singlet phosphorescence occurs at a longer wavelength in the same molecule, as fluorescence.

The figure  shows the energy levels. Different process compete with one another, thus the singlet can loose energy by fluorescence (with rate constant $k_f$) or by non-radiatively crossing to a triplet state (intersystem crossing) or non-radiatively to the ground state (internal conversion).

continued after question below.

The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule. Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed,  can only be produced by a spin forbidden process, singlet to triplet, and just as in phosphorescence, is a spin forbidden process and becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metal the forbidden transition is in absorption.

The general treatment starts with allowing the two states to interact via some  operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. If the operator is $H_0$, the rate of the transition from one state to another is

$$W_{if}^0 \approx \int \psi_f H_0 \psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau $$

where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one. As the operator does not depend on spin then
$$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i dtau = 0$$
and the  result is zero because the spin wavefunctions belonging to i and f are orthogonal (as singlet & triplet) and so the last integral is zero.

If we suppose that it is possible to mix a little of the initial state wavefunction into the final one and generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on. Now using this wavefunction as the final state and repeating the calculation above and separating out terms then as orthogonal spin terms are zero the only term left is

$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau$$
in which case the integral is no longer zero and evaluates to a number.(Note that all the indices are the same unlike the equation for $W_{if}^0$)  This equation shows that there is a finite rate for the process with operator $H_0$. In the case in point, spin-orbit coupling, the operator $\Omega$ depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why it is also called a ‘heavy atom effect’ when applied to emission processes.

The constant b is deribed using perturbation theory and turns out to have the form
$$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$
where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes

$$ W_{if} \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i} $$

and we can use symmetry to decide if this integral is zero or is finite. The dependence on energy gap between states is clear, rapid rate constants for small energy gap. (The degenerate case when the energy gap is zero needs special and a more complex treatment)


Kipp's apparatus with powders A Kipp's apparatus has three distinct bulbs. The topmost one contains a liquid and a tube to the bottommost one. The middle one contains a solid to react with the liquid.

This is very understandable when large chunks of solid are used. However, what do people do when they need to react a powder (e.g. sodium sulfite)? It seems that the powder would just fall through and keep reacting when the valve is closed.

According to the Wikipedia page on the Kipp's apparatus, if the reactant is a powder, and soluble in water, you can soak pieces of pumice in the solution, dry the pumice, and use it in a Kipp's apparatus. There is also a variant of the apparatus that allows two liquids to be used together rather than a liquid and a solid reactant. In that case, you would pre-dissolve the sodium sulfite in water.


Why does wood burn but not sugar? Fundamentally, they're both carbohydrates, although the cellulose in wood is essentially polymerized glucose, which combined with its isomer fructose forms sucrose.

So why does wood readily burn while table sugar chars?

Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started.

Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction going.

However sugar dehydrates and emits water when you heat it. Water isn't flammable (obviously) so there's no way to get combustion started. Dehydration leaves behind pure carbon and that is non-volatile so again there's no way to get this to burn. Carbon will burn of course, but you need a high temperature to get it going.


What is meant by viscous liquid? I think all the liquids are viscous. So I was wondering what is meant by a viscous liquid as it is written in the first line of page on Oil at Wikipedia.

  An oil is any neutral, nonpolar chemical substance that is a viscous liquid at ambient temperatures and is both hydrophobic (immiscible with water, literally "water fearing") and lipophilic (miscible with other oils, literally "fat loving"). ...

This is somewhat an elementary discussion, but viscosity is typically measured in some sort of viscometer. A typical one is the Oswald viscometer shown below.

Fluid is poured into the apparatus through tube at $f$ to fill neck and bulb at the bottom. The fluid is then sucked up above line $c$. The fluid is then released and drains down the tube to line $c$ at which time the timing for the flow rate starts. When the fluid falls to the line at $d$ is reached then a flow rate can be measured. By using samples of known viscosity, the flow time can be calibrated to the viscosity of the sample.

Now such measurements are in units of pascal-seconds. For reference a short list will be provided.

     Substance           mPa-s
     acetone                0.0306
     water                  0.0894
     sulfuric acid          0.242
     olive oil             81.
     motor oil SAE 40     319.
     glycerol           1,200
     corn syrup         1,380.6
     pitch              2.3×10^8

Wikipedia also lists tables for Viscosity of selected substances

Now let's defined three ranges of viscosity:

Low Viscosity -    0.1 mPa-s >= sample
Medium Viscosity - 0.1 mPa-s <= sample <=1.0 mPa-s
High Viscosity - sample > 1.0 mPa-s

Of course the "exact" ranges are arbitrary. So a viscous liquid is a High Viscosity liquid on our scale.


Is there a relationship between an electron's energy and its distance from the nucleus? I have read that in Bohr's model of the atom electrons in the same atomic orbital have the same energy, and that this energy (as a result of the electron existing in a fixed atomic orbital) is related to the distance of the electron from the nucleus. However, using the quantum mechanical model of the atom, in which an orbital is the area of space around an atom where the probability of interacting with an electron is greatest, it is implied that there is no relationship between an electron's energy and its distance from the nucleus. Is there no relationship, or have I misunderstood something else?

Also, it is often mentioned that an atom is 99.99% empty space etc., but since the probability of interacting with an electron is described by its wave function (which suggests it is possible, though unlikely, to interact with an electron very close to the nucleus), this suggests to me that this isn't necessarily true. What am I misunderstanding?

I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum mechanical description of the electron, but when you are studying it you absolutely have to remember that all of what he said was effectively rubbish.

Now on to your questions.

  Is there a relationship between an electron's energy and its distance from the nucleus?

The first thing that should be said is that in QM, unlike in the Bohr model, there is no well-defined "distance from the nucleus". Electrons exist everywhere (except at nodes) and the probability density is given by $\lvert\psi\rvert^2$. However, it is possible to calculate the "most probable distance from the nucleus" using the radial distribution function, $r^2 R^2$, where $R(r)$ is the radial part of the wavefunction. Essentially, you take the derivative of this with respect to $r$ and set it to 0 - you can read more here. For a 1s electron in hydrogen, you will find that the most probable radius is approximately $52.9 \text{ pm}$, a quantity called the Bohr radius and denoted $a_0$.

(As an aside: oddly enough, this correlates with the flawed Bohr model, which said that the radius of the 1s electron in hydrogen was $52.9 \text{ pm}$. However, what Bohr said was that the electron was always at this distance, which is wrong. QM says that the electron is most likely to be found at this distance.)

Anyway, you will find that there is some degree of correlation between the most probable radius and the energy of the electron. If we consider the 2s orbital of hydrogen, you can perform exactly the same calculation as you did for the 1s orbital, and you will find that the most probable radius is $4a_0$.

  Also, it is often mentioned that an atom is 99.99% empty space etc...

That's just a simplified view presented in popular science (or elementary chemistry classes) that ignores the quantum mechanical model. No serious chemistry book would write something like that. I don't know exactly how it's calculated, but I can make a good guess. You take any atom you like, let's say hydrogen. You look up the atomic radius - Google tells you it's $52.9 \text{ pm}$. (No surprises there.) So you can think of the atom as a sphere with that radius, and calculate the volume of the sphere: $V = \frac{4}{3}\pi r^3$. Then you go and look up the volume of an electron and a proton (a physicist would get a headache from just reading the phrase "volume of an electron"), and add it up, and you find that the "atom" is "99.99% empty space".

As Ivan succinctly said - with the QM description of the atom, there is no longer such a thing as free space since the electron "exists everywhere". Apart from that, just like how the distance from the nucleus is no longer well-defined in QM, the "volume" of an atom is also no longer well-defined. We pretty much threw that view out of the window.


Which silver halide is the most soluble? Halides are group 7 elements. They generally have a low melting and boiling point. They are highly reactive. Silver chloride and silver bromide are not soluble in water. So they will be precipitates. However, I do know that they have varying solubilities. So, they will form different amounts of precipitates. I want to know why they have different solubilities and which is the most soluble halide.

The solubilities of silver halides decreases down the periodic table:

$\ce{AgF}\ : K_{sp} = 205$
$\ce{AgCl}: K_{sp} = 1.8\times10^{-10}$
$\ce{AgBr}: K_{sp} =5.2\times10^{-13}$
$\ce{AgI}\ \ \ : K_{sp} =8.3\times10^{-17}$

The rationale for this trend is typically described using the concept of hard/soft acids and bases.  (See J. Chem. Educ., 1968, 45, page 581 and also page 643 for early discussions on HSAB theory.)  Briefly, ions with similar size and high polarizability will lend themselves to more covalent interactions, and therefore will have decreased solubility in water.  HSAB classifies ions as either hard or soft, with soft being large and polarizable and hard being small and highly charged.  H-H and S-S interactions are strong while H-S or S-H interactions are weaker.  Ag is classified as soft and the halides go from hard (F-) to soft (I-).


Gilmans and carbon dioxide I know that Grignards react with carbon dioxide to give carboxylate anions.

But will Gilmans do the same? Will Gilmans react with carbon dioxide in a productive fashion?

Yes, cuprates do react with carbon dioxide to give access to carboxylates / carboxylic acids.

Running a SciFinder search, I do not see any examples of traditional Gilman reagents (alkyl cuprates) reacting directly with carbon dioxide. This shouldn't be surprising though, since the alkyl cuprates are generated from the corresponding alkyl lithium compound. If you have the alkyl lithium compound, then you can just react that with carbon dioxide directly.

A more synthetically useful situation is when the alkyl cuprate is reacted with an alkyne, resulting in a syn-addition of the alkyl group and copper across the alkyne. The alkenyl (vinyl) cuprate can then be trapped with carbon dioxide to give a carboxylate.

A couple examples from the literature:

Z,Z-Dienes via acetylene carbocupration:  synthesis of navel orangeworm pheromone

The silylcupration of acetylenes: a synthesis of vinylsilanes

Also check the write up in Organic Reactions by Lipshutz (the world's expert in organocopper chemistry).

More complicated (non-Gilman) organocopper compounds have been reacted with carbon dioxide to give copper carboxylates:

Carbon-carbon coupling of activated alkanes with carbon dioxide by [(phen)(Ph3P)Cu(O2COH)]

Synthesis, Structure, and CO2 Reactivity of a Two-Coordinate (Carbene)copper(I) Methyl Complex


Safety concerns for radical halogenations of methane in sunlight? I wanted to know about the substitution reaction of alkanes particularly interested in the reaction of methane with fluorine and chlorine.

Let's have a look at some fundamental data and the chemistry involved.

Chlorine is an inflammable gas with a boiling point of -35 °C and a nice yellowish-green colour. It is heavier than air.

It is highly irritant and toxic. It is a strong oxidant. With water, it partly decomposes to hydrogen chloride and hypochlorous acid. Your mucous membranes are wet. Inhalation of chlorine results in severe corrosive injuries. Inhale a tad longer to face dyspnoea, coughing of blood and asphyxia. Did I mention the pulmonary oedema and fatality?

Methane is a flammable gas with a boiling point around -160 °C and a flash point around -180 °C. The lower explosion limit of methane in air is around 5%.

Open flames and sparks are not your friend here.

For the mixture of methane and chlorine, red light districts are safe.
The bond dissociation energy of chlorine is $243\, \mathrm{kJ \cdot mol^{-1}}$, you will need blue light ($\lambda$ < 490 nm) to initiate the self-propagating reaction:

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$

Then the chain reaction starts:

$$\ce{Cl* +\  CH4 -> HCl + CH3*}$$
$$\ce{CH3* +\  Cl2 -> CH3Cl + Cl*}$$

All that goes on and on; termination steps are radical recombinations:

$$\ce{2 Cl*  -> Cl2}$$

$$\ce{2 CH3*  ->  C2H6}$$
$$\ce{ CH3* +Cl* ->  CH3Cl}$$

Apparently hydrogen chloride gas is is produced in the reaction, together with methyl chloride.

Methyl chloride is a flammable gas (boiling point around -24 °C, flash point around -20 °C, lower explosion limit in air around 8%).
GHS hazard statements file it with H-351: suspected of causing cancer.

If you perform the reaction with a large excess of chlorine, $\ce{Cl*}$ radicals will attack methyl chloride, generate dichloromethane via the chloromethyl radical $\ce{*CH2Cl}$, etc. Tetrachloromethane ($\ce{CCl4}$) would be the final product then.

That was the healthy and boring part!

When fluorine talks to chlorine, it goes like in the musical "Annie, Get Your Gun":

Anything you can do I can do better!

Particularly when it comes to running havoc on your tissue!

As far as the photolysis of fluorine in the presence of methane is concerned: People have done that!

Typical conditions were:

Deposit methane in a fluorine matrix at 15 K
Co-deposit methane/argon and fluorine/argon samples at 12 K
Trap fluorine in a methane matrix at 15 K

Yes, these guys had no sense for adventure - but they weren't blown away either!


Why does fructose reduce Tollen's reagent and Fehling's solution? Even though fructose is a ketohexose (ketone-containing hexose, a six-carbon monosaccharide), it reduces Tollen's reagent and Fehling's solution. Generally, a ketone does not reduces Tollen's reagent and Fehling's solution. So, why is this possible in the case of fructose?

The presence of the hydrogen atom in the aldehyde group gives reducing properties to the aldehyde molecule. Aldehydes are easily oxidized by mild oxidizing agents such as Tollens’ reagent ($\ce{Ag+}$ in aqueous ammonia), Fehling’s reagent ($\ce{Cu^2+}$ complexed with tartrate), or Benedict’s reagent ($\ce{Cu^2+}$ complexed with citrate). Ketones do not have such reducing properties and are not oxidized under similar conditions. Therefore, Tollens’ reagent, Fehling’s reagent, or Benedict’s reagent can be used to distinguish aldehydes from ketones.

However, fructose reduces such reagents even though it contains no aldehyde group. Reduction occurs because the reagents are basic solutions and fructose is readily isomerized to a mixture of aldoses (glucose and mannose) under basic conditions.

Under basic conditions, the proton alpha to the carbonyl group (aldehyde or ketone) is reversibly removed. If the resulting enolate ion reprotonates on the oxygen, an enediol intermediate results. Thus, the base-catalyzed enediol rearrangemet can move the carbonyl group up and down the chain.

Furthermore, in the enolate ion, the affected carbon atom is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either configuration. Thus, the base-catalyzed epimerization results in a mixture of stereoisomers.

Therefore, under strongly basic conditions, the combination of enediol rearrangements and epimerization leads to a complex mixture of sugars (see also: Lobry de Bruyn–van Ekenstein transformation).


Hyper conjugation in acetic acid? While comparing the relative stabilities of acetic acid and formic acid why don't we consider the 3 possible hyper conjugative structures which would stabilize the positive charge on carbon thus making the conjugate of acetic acid more stable.

My professor told me hyper conjugation does not operate here. But what is the reason behind this?

Hyperconjugation does play a role in your example.  Acetic acid ($\mathrm{p}K_\mathrm{a}=4.75$) is a weaker acid than formic acid ($\mathrm{p}K_\mathrm{a}=3.75$).  To understand this we must consider the relative stabilities of both the ionized and unionized acids.

In acetic acid there is a partial positive charge on the carbonyl carbon.  Both the inductive effect and resonance effect (e.g. hyperconjugated resonance structures like B) of the methyl group act to stabilize this charge, thereby stabilizing the unionized form of acetic acid.  In formic acid, without a methyl group, similar stabilization of its unionized form is not possible.

Now let's look at the ionized acids.  The negative charge in the anion is delocalized as shown in resonance structure C. This negative charge on the two oxygens makes them much less electron withdrawing.  Consequently, there is much less positive charge on the "carbonyl" carbon in the ionized form.  In turn, there is less need for stabilization of positive charge by the methyl group.  The methyl group does not stabilize the ionized form as much as it stabilizes the unionized form.

Since the methyl group stabilizes the unionized form of acetic acid more than a hydrogen and since the methyl doesn't stabilize the ionized form much more than a hydrogen, there is a higher energy barrier for acetic acid to ionize compared to formic acid.  This means that acetic acid will be a weaker acid than formic acid.


What makes Gorilla Glass more durable with each generation? Gorilla Glass (GG) is one of the most recognizable brands of toughened glasses today, thanks to their marketing and widespread usage in various gadgets and automobile industry.

According to Wikipedia, it's basically an ion exchange ($\ce{Na+ <-> K+}$) within the glass (boro-/alumosilicate?) matrix which gives this material resistivity against scratches and various mechanical impacts:

(Source 1, Source 2)

Still, there is a new generation of GG released every few years, showing better properties with thinner samples, making me think it's all not just about ion exchange.

Are there other ways of "chemical strengthening" besides cationic exchange that can be utilized to strengthen the glass, keeping it flexible and less fragile at the same time?

Are there other ways of "chemical strengthening" besides cationic exchange that can be utilized to strengthen the glass, keeping it flexible and less fragile at the same time?

In a word: No.

To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder.  The process ion exchange hardening, its done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (below Tg). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium; this mismatch creates the compression that in turn gives the glass its scratch resistance. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.

Now an astute person might be thinking:

  Are there other ways of "chemical strengthening" besides cationic exchange... What about anion exchange?

To this yes in theory if you could substitute the anions without rearranging the structure then yes anion exchange would be another way. But as Yogi Berra once said "in theory there's no difference between theory and practice, but in practice there is." The problem with anion exchange is you would be breaking and thus relaxing the bonds of atoms that you need to remain in place, thus no strengthening would occur.

I suppose you could also modify the surface to make a layer of silicon nitride or carbide and call that anion exchange, but would it this strengthening glass or changing the material altogether?

One way in which you could increase glass strength though still predominately a chemical process is to substitute sodium for lithium (or hydrogen), anneal the glass to allow the bonds to relax to form sites sized for lithium then substitute potassium back in where the lithium is to make it even stronger.

As far as what make each generation of GG stronger, I think it is base glass improvement such as modifying the aluminum content or making the glass thinner or making the ion diffusion more local to the surface and preventing the potassium from penetration too far.


How to calculate an enthalpy change for this reaction I've got an assignment to solve problem number 4 in this worksheet, that is to calculate change in enthalpy for reaction:
$$\ce{CHCl3 + O -> COCl2 + HCl}$$

How do I solve it? Why is oxygen $\ce{O}$, and not $\ce{O2}$?

I figured out that, it is not a combustion reaction, as combustion of any hydrocarbon will lead to formation of $\ce{CO2}$ and $\ce{H2O}$, as per this page.

Well, what I have at the beginning is not exactly a hydrocarbon, but rather, a halogenated hydrocarbon. Yet, no water and no carbon dioxide is produced. So as I mentioned above, I consider it is not a combustion. Any thoughts on this? Could it still be a combustion? Why or why not?

When I calculate $\Delta H$ of this reaction, the total outcome depends on using reaction as it is shown in the worksheet, or correcting $\ce{O}$ to $\ce{O2}$ and balancing it. Which option, do you think, is preferable, and why?

To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible.

Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten:
$$\ce{CHCl3 + 1/2O2 -> COCl2 + HCl}$$

It is better to take it for what the question presents it as, an oxygen atom, so I would therefore recommend you to leave the equation as:

$$\ce{CHCl3 + O -> COCl2 + HCl}$$

Though $\ce{O2}$ is a pure element, and its $\Delta H^0_\mathrm f=0\:\mathrm{kJ\:mol^{-1}}$, for the atom $\ce{O}$, $\Delta H^0_\mathrm f=58.99\:\mathrm{kJ\:mol^{-1}}^{[1]}$ (provided thanks to user1420303).

To calculate $\Delta H$, the equation is:  $$\Delta H=\Delta H^0_\mathrm f(\ce{COCl2}) + \Delta H^0_\mathrm f(\ce{HCl})-\Delta H^0_\mathrm f(\ce{CHCl3})-\Delta H^0_\mathrm f(\ce{O})$$

Though enthalpy is a state function, this question is asking you to calculate $\Delta H$ from a specific set of reactants, and their individual energies must be considered. There is more than just one way to calculate $\Delta H$, and an analysis of the $\Delta H^0_\mathrm f$ (the enthalpy of formation of a compound from its constituent elements) for each compound accounts for which bonds are being broken and formed, and will yield (about) the same results as a bond by bond analysis. I find this method to be preferable because it bond by bond analyses are based on average bond enthalpies, while $\Delta H^0_\mathrm f$ is often painstakingly measured for each compound, and will yield the most accurate $\Delta H$ calculation.

[1] Curtiss, L. A.; Raghavachari, K.; Redfern, P. C.; Pople, J. A. Assessment Of Gaussian-2 and Density Functional Theories for the Computation of Enthalpies of Formation. The Journal of Chemical Physics J. Chem. Phys. 1997, 106, 1063.


13C-NMR of aniline: Why are carbons that are further from the amine group more downfield than carbons that are closer? A $\ce{^{13}C}$-NMR of aniline shows that the carbons further from the $\ce{-NH2}$ group are more downfield than the ones near the top of the benzene ring. If electronegative atoms like nitrogen have a deshielding effect on nearby Carbons, why is this the case? Is it because of dipole vectors?

Although the nitrogen atom is more electronegative than carbon, nitrogen will readily share its electron pair, allowing one to draw a resonance structure where the nitrogen is donating electron density into the ring. Therefore, the donating of the lone pair of electrons overcomes the inductive effects arising from the difference in electronegativity.


Determination of initial excited state via luminescence A one electron species initially in the some excited state
  ($n_i$)
  is irradiated with a light of wavelength 121 nm when
  the electron is promoted to a further higher orbit ($n_f$).
  In
  returning back to the ground state, it gives an emission
  spectrum containing 15 bright lines. Out of these 15 lines,
  9 lines were found to have wavelengths smaller than
  121 nm while 5 lines have wavelengths greater than
  121 nm.
  What is the value of $n_i$?

I got the value of $n_f$ using this:

$$\frac{(n_f-1+1)(n_f-1)}{2}=15$$
$$\therefore n_f=6$$

We do not know the atomic number($Z$), nor $E_{ground state}$, nor $n_i$. What is the approach we should use to solve this problem?

Sometimes formulae are more hindrance than help, and perhaps there are some red herrings in your problem.

  A one electron species initially in some excited state $n_i$ is
  irradiated with a light of wavelength 121 nm when the electron is
  promoted to a further higher orbit $n_f$. In returning back to the
  ground state, it gives an emission spectrum containing 15 bright
  lines. Out of these 15 lines, 9 lines were found to have wavelengths
  smaller than 121 nm while 5 lines have wavelengths greater than 121
  nm. What is the value of $n_i$?

There being fifteen bright lines suggest that $n_f = 6$, since $15 = 5 + 4 + 3 + 2 + 1$, there being $5$ transitions from the $n = 6$ state to $n = 5,\,4,\,3,\,2,\,1$; $4$ transitions from the $n = 5$ state, and so on. This is the basis for the formula you've employed.

Long wavelengths correspond to low energy; we see that the transition from $n_i$ to $n = 6$ is the sixth lowest in energy. $E \propto n_i^{-2}-n_f^{-2}$, so we pull out our handy-dandy calculator and work out: $$ 6 \to 5: 0.012,\,5 \to 4: 0.023,\,6 \to 4: 0.035,\, 4 \to 3: 0.049,\, 5 \to 3: 0.071,\, 6 \to 3: 0.083.$$ Thus $n_i = 3$. We did not need to use the exact wavelength of light used for irradiation and we did not need to know the identity of the species, because this problem is only concerned with the relative energy differences between energy levels.


What is the simplest structural unit of an element or compound? From "Smarter than a 5th Grader":

  What is the simplest structural unit of an element or compound?

  molecule,
  atom,
  neutron,
  proton

For covalent compounds, the answer would definitely be molecule. But I'm unclear as to how to show that the answer for ionic compounds or elements is "molecule".

For example, https://sites.google.com/site/rccchem2a/chapter-4-chemical-bonding-the-i states that ionic compounds: "Do not have molecules as basic structural unit. Instead extended array of positively and negatively charged particles called ions." Is the term 'structural unit' defined across Science as only for covalent compounds?

For a solid lattice (metal, ionic solid, crystal) I would say the answer is unit cell, for a non-charged compound  that isn't a lattice (including solution/gas phase) I would say molecule, and for a charged one I would say ion-pair / ternary ion-associates / quaternary-ion associates for two / three / four ion clusters. For polymers, I would refer to the basic unit(s) being the component monomer(s).

There is, as has been identified in the comments, an issue with allotropes, Carbon (Diamond, Graphite etc) can be encompassed within 'unit cell', but for elements like phosphorus and sulfur, it's a little more complicated. There I would contend that the basic unit of e.g. P4 (white Phosphorus) is the molecule, and for the rest, I would return to referring to the unit cell, or 'monomer' in the case of amorphous networks.


Is there a difference in the reactions of a dilute acid and a concentrated acid? All of the exams that I've sat so far, and the homework questions I've had, always ask for the reaction between a substance and a dilute acid. Is there a difference in the reaction between a dilute acid and a concentrated acid when each is applied to a substance? If so, what is it, and how dilute is dilute?

While it would be easier with examples, I leave here some ideas.

This could be because the idea of concentration is an approximation of the concept of thermodynamic activity, which is frequently referred to as effective concentration. Depending on the question being asked, it might be important that the solute behaves as if it was in an infinite concentration, where there's no interaction between solute molecules/ions and the concentration equals the activity.

The mention of dilution might also be important if the reaction involves water as a reagent. Even though water usually does not receive much attention when discussing reactions, since it's the solvent with "constant" concentration, it is important to note that several reactions wouldn't happen in its absence. That would be the case if the acid is available in a purified form, such as a gas (like HCl) or as solid (such as benzoic acid). It's also worth mentioning that the concept of acidity is inherent to solutions, so the two above aren't acids unless diluted.

The above also highlights something that is noteworthy. While diluted is the opposite of concentrated, it may also be said as the opposite of "not-in-solution", meaning not mixtured with the solvent. This could also be a reason to mention dilute acids, to make it clear they are in solution instead of any other form.

Edit:

Given the example, I do believe it's because the reactions happen in the presence of water, but honestly, I'm not sure.

Edit 2:

Another possible reason to mention dilute acids would be in case it's necessary to consider the concentration of a medium-strength acid. In dilute solutions, those can be considered completely dissociated, which makes calculations a lot easier, if that's the case.


Confusion in finding electrophilic centre In the compound CH3-CH2-CN the electrophilic centre according to my NCERT TEXTBOOK(class 11) is the carbon of cyanide but I think it should be the the carbon adjacent to cyanide since, during bond cleavage between cyaninde and ethyl the electron pair will be taken up by the carbon of cyanide so in this way it is a neucleophilic centre . Hence I think that the carbon of ethyl should be electrophilic centre. Am I right or iam missing something or my textbook is wrong?
Any help is greatly appreciated.

In all of the compounds illustrated, the starred carbon is connected to a more electronegative element.

Let's examine this one by one- I shall begin with methyl iodide. Here, the starred carbon is connected to a Iodine which is more electronegative, thus naturally, we can guess that the electron density shifts towards the halogen. Thus, the carbon in question is slightly electron deficient (slightly positive or electrophilic)

Now, let's look at the examples with the multiple bonds.
Consider ethanal: the carbon is again connected to a more electronegative atom i.e oxygen. So, if you were to cleave the pi bond, the electron density would shift to the oxygen, and again you are left with an electrophilic carbon right next to the oxygen.

The same can be said for the acetonitrile: the electron density shifts to the more electronegative element i.e nitrogen and the carbon right next to is left "slightly positive" or electrophilic.

The electron density would always be slightly greater in the vicinity of a more electronegative element across a bond between two atoms differing in electronegativity (essentially, what we mean when we say a bond is polar)


What is the need for so many theories of acids and bases? What is the need for the  Brønsted-Lowry theory and the Lewis theory of acids and bases?Is there any problem with the Arrhenius theory?

First came the Arrhenius theory. It defines an acid as a species that dissociates to produce $\ce{H+}$ in solution, and a base as a species that dissociates to produce $\ce{OH-}$ in solution. (Examples: $\ce{HCl}$, $\ce{NaOH}$)

Next came the Brønsted-Lowry theory, which expands the notion of acid and base. It defines an acid as a species that loses $\ce{H+}$ in a reaction, and a base as a species that gains $\ce{H+}$ in a reaction. All Arrhenius acids and bases are Brønsted-Lowry acids and bases; hence the Brønsted-Lowry theory generalizes the concepts of Arrhenius acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{NH4+}$, $\ce{CH3COO-}$)

Finally came the Lewis theory, which further expands the notion of acid and base. It defines an acid as a species that accepts an electron pair, and a base as a species that donates an electron pair. All Brønsted-Lowry acids and bases are Lewis acids and bases; hence the Lewis theory generalizes the concepts of Brønsted-Lowry acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{Li+}$, $\ce{CO}$)

Remark: Hopefully the repetition and structure within my answer has been suggestive. The benefits of generalization are obvious: the theory can explain more, and more things can be explained; this is the natural progression of science.


How to select a schematic of the coordinate vs energy representing a rearrangement reaction? Here's the question

For a reaction to take place the net energy should go down, and thus option (d) can be discarded.
We are now left with options (a), (b), and (c).
A $3^{\circ}$ carbocation is more stable than a $2^{\circ}$ carbocation, therefore the 3rd dip in the graph should be lower than the 2nd dip as more stable means lower energy.
Thus now we are left with options (b) and (c).
How do you decide between (b) and (c)?

The answer is (b).  The initial step of the reaction should take energy.  Otherwise the reactant molecule itself is unstable.  The very initial portion of the curve in (c) is downward-sloped, which doesn't make sense.  If the reacting alkene could relax to a different conformation without any activation barrier, then it would do so.  But the fact that the reactant is a stable, isolable molecule with a well-defined chemical structure indicates that no such barrier-less relaxation is possible.

Thus, (b) is a better answer than (c).


What are the molecular orbitals shaped like in diberyllium (mono)cation? I think one way of describing the result is that the bond in $\ce{Be^+_2}$? cation is a resonance hybrid of two possible pi bonds. Each pi bond has a bond order of 1 so the resonance hybrid structure has two partial pi bonds. But then can it further be in a resonance hybrid of different rotations and so be an even cylinder?

I guess if you wanted to be silly you could say that that the electron configuration for each beryllium ion is like:

$$ \underset{2s}{[\uparrow \downarrow]} \underset{(2p)^2}{[\uparrow \vert \;]} \underset{2p}{[\;]} $$

Short answer: the molecular orbitals will be $\sigma$ and $\sigma^*$ and thus cylindrically symmetric.

First off, your electron configuration is wrong. $\ce{Be2+}$ will have 3 electrons, so the $\ce{2p}$ orbitals need not be considered.

Here's a quick MO diagram:

So there aren't an $\pi$ bonds at all. There's a $\sigma$ and a $\sigma^*$ orbital, for a bond order of 0.5.

If I generate orbitals (using Avogadro) then I see something like this for the $\sigma$

And this is the $\sigma^*$


Is it possible to combust a plasma? In what we usually consider combustion your fuel and oxidizer are always neutral, but is it possible to have a similar "explosive" combustion process where both the fuel and oxidizer have been singly ionized? If so what might some potential fuels be, and where might more information on the topic be found? A google search doesn't turn up much (other than injecting a small amount of plasma to catalyze an otherwise non-ionized reaction.)

If you specifically mean a hot plasma, I think it might be possible but it's a tough request. To get combustion, you'll need to start off with form of bonding present between atoms in the plasma. This requires finding a molecule with atoms so strongly bound that the bond dissociation energy is higher than the molecule's ionization energy.

Given that the lowest ionization energy known (3.51 eV or about 340 kJ/mol for ditungsten tetra(hpp)) is comparable to relatively sturdy covalent single bonds, there's not much room for negotiation. Ionization energies easily become larger than even strong covalent double bond dissociation energies, so there's a large tendency for any substance to be stripped into its constituent atoms before getting a chance to ionize.

The situation is easier if you allow the usage of "cold" plasmas. Plasma generators are commonly used to scour surfaces completely clean of any organic residues. I don't see why you couldn't just insert oxygen and some organic gas into such a generator, ionize them both with the strong and rapidly varying electric field inside, and cause the organic gas to combust.


Does a timed reaction "ready indicator" paper exist? I'm trying to see if such a thing exists in paper form that upon activation,  changes colour or "indicates" after a set configurable amount of time.

In other words a litmus or pH paper that can act as an egg timer. I'm by no means a chemist so please excuse my lack of protocol.

I'm not aware of any commercially available examples, but I know that the speed of a fluid wicking through a paper microfluidic device has been used for timing. I'm also reminded of pencil detonators, dating back to WWII, where the time taken to dissolve a lead wire is used for timing. Since electronic timing is now so cheap, easy, accurate, and reliable, I'm not surprised that chemical timers have pretty much died out. There are a number of chemical phenomena that could be exploited for timing, but unless you need a huge number of cheap, one time use timers with relatively poor accuracy, there are probably better choices.


Why are 2 ways to calculate the energy of the ground state? Setup

I found in Szabo, Quantum chemistry page 88, these two ways to calculate the energy of the ground state:

$$E_0= \sum_a^N \langle a|h|a\rangle + \frac{1}{2} \sum_a^N \sum_b^N \langle ab||ab\rangle $$

and

$$E_0= \sum_a^N \langle a|h|a\rangle + \sum_a^N \sum_{b>a}^N \langle ab||ab\rangle $$

Question

Why?

Both formulas have exactly the same meaning. They are just two different ways of not double counting the Coulomb and the exchange interactions. In principle, the goal can be achieved in two ways:

double counting in the first place and then dividing by two (first formula);
counting wisely so that there is no double counting in the first place (second formula).

Obviously, for any practical purposes when you one does not want to double count one would almost always choose the second procedure, but the first representation in some cases provides more apparent picture when dealing with the formulas analytically.

Where does all this come from?

The whole derivation of the expression for the ground state Hartree-Fock energy basically starts from writing down the expectation value of the ground state electronic energy for the ground state Slater determinant $\Phi_{0}$
$$
\langle \Phi_{0} \mid \hat{H}_{\mathrm{e}} \mid \Phi_{0} \rangle \, ,
$$
in terms of the individual spin-orbitals of which $\Phi_{0}$ is constructed.
And in doing so we use the so-called Slater rules and it happens so that the electronic Hamiltonian $\hat{H}_{\mathrm{e}}$ contains only two different kinds of operators:

one-electron operators of the form
$$
    \hat{F} = \sum\limits_{i=1}^{N}\ \hat{f}(i)\,,
$$
and two-electron operators of the form
$$
    \hat{G} = \frac 12 \sum\limits_{i=1}^{N} \sum\limits_{j=1\atop{j\neq i}}^{N}\ \hat{g}(i,j)\,.
$$

Already at this stage an expression for any two-electron operator $\hat{G}$ could as well be written as
$$
    \hat{G} = \sum\limits_{i=1}^{N} \sum\limits_{j>i}^{N}\ \hat{g}(i,j)\,.
$$
The mathematical meaning is exactly the same: we do not want to double count pairwise interactions, so that if, say $\hat{g}(1,2)$ is already counted, $\hat{g}(2,1)$ has to be excluded.

Note carefully that we also have to exclude physically meaningless self-interactions, such as, say, $\hat{g}(1,1)$, which is done in the first case by an additional restriction on summation $j\neq i$ and in the second case by the strict inequality in the lower limit of summation $j>i$.

With respect to the exclusion of this non-physical self-interactions note now that they it is missing from the HF energy expression being written in the first form. But that is perfectly fine since each and every Coulomb self-interaction is perfectly canceled by the corresponding exchange self-interaction. Thus, you do not necessarily need the above mentioned additional restriction of the form $j\neq i$ ($b \neq a$ this time), although you could include it if you would like to,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b=1\atop{b \neq a}}^N \langle ab||ab\rangle \, ,$$
or adopting a rather usual convention for such summations, as follows,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b \neq a}^N \langle ab||ab\rangle \, .$$


Trying to figure out heat of reaction A question in Thermodynamics of Natural Systems by G. M. Anderson is to find the "standard heat of reaction" of this reaction:
$$\ce{C8H16N2O3(aq) + H2O(l) -> C6H13NO2(aq) + C2H5NO2(aq)}$$

As far as I understand it, it should be a rather simple problem of finding $\Delta_\mathrm r H$.
So it should be:

$\Delta_\mathrm r H=\Delta_\mathrm fH (\ce{C6}) + \Delta_\mathrm fH (\ce{C2}) - \Delta_\mathrm fH (\ce{C8}) - \Delta_\mathrm fH (\ce{H2O})$ (note that I shortened the organic formulae)

According to the data from the same textbook, it's:
$(-632077~\pu{J/mol}-513988~\pu{J/mol})-(-847929~\pu{J/mol}-285830~\pu{J/mol})=-12306~\pu{J/mol}$

However, the answer given is $−89475~\pu{J/mol}$.

Am I missing something? Do I need to add some terms of pressure or volume expansion? Is this not the enthalpy that I'm looking for?

There is a "readme" note from the author that says:

  The answers are not guaranteed, in the sense that not only
  might I make a mistake here and there, but I might have used different sources of
  data, or the data in a database may have changed. I have not been able to check all
  the answers, or to make them all consistent with the data in Appendix B, although I
  believe that most are.

So I think that this problem is an example of one of those situations.

The book itself works out the problem on page 267 and gets the same answer as you: −12306 J/mol.


What to say 'one mole of C atoms or one mole of C molecules? What to say 'one mole of C atoms or one mole of C molecules in CO? Please point out if I am wrong here
As we know,
one molecule of CO = one mole of CO molecules = 6.223 X 10^23 number of CO molecules
If I split one molecule of CO, we get one C and one O. Here are my questions, What to say these ''one C and one O" one mole of C atoms and one mole of O atoms or ''one mole of C molecules and one mole of O molecules'' or just like above like ''one C and one O''?
Could you clarify it, please? I am confused

What you need to remember when using terminology involving moles is that moles are just a way of counting things but that the terminology only makes sense if you unambiguously specify the thing you are counting. It is entirely possible to use moles to count non chemical things (for an amusing example of the consequences of having a mole of moles (the small burrowing mammals) see this.) Most confusion arises because the way the thing is specified is ambiguous.

Some examples:

a mole of CO molecules is perfectly correct (the units being counted are single molecules of CO)
A mole of C atoms is perfectly correct (and, if you are counting C atoms in CO molecules that is also fine and the answer will be the same)
a mole of C molecules in CO isn't correct but only because the molecules is CO which consists of atoms of C and O.
if you were counting the atoms in CO2, then it would be correct to say there are twice as many moles of oxygen atoms than of carbon atoms or carbon dioxide molecules

The use of moles to count doesn't imply anything about what you are counting, you have to specify that as clearly as possible. When talking about the parts of a molecule we usually talk about the atoms that make up the molecule. CO2 doesn't consist of molecules of oxygen and carbon, but atoms of oxygen and carbon. But this is independent of the use of moles as a way of counting those components.


Red Cabbage Indicator colour change with Sodium Metabisulphite I made some pH indicator by boiling and straining cabbage juice and in an effort to keep it from spoiling for a while without having to store it at fridge temperatures I added some wine making stabiliser which consists primarily of Sodium Metabisulphite. I had previously tested some of it to check if adding it would affect the pH of the indicator since I wanted a pH neutral preservative. Well strangely after leaving it for a while I came back to find that the indicator had turned a weird shade of purple I had never seen the indicator take before. At the concentration it is at the original colour was a very dark purple - almost black. Now I know that the metabisulphite ion decomposes to produce SO2 gas which is acidic under aqueous conditions but I would expect that to change the indicator to its normal purple for a not very concentrated acid. But the colour it currently is is very pale and in fact when I pipette out some of the indicator it is almost totally clear. So I guess really my question is does Potassium Metabisulphite have some kind of bleaching effect I'm not aware of?

http://mylespower.co.uk/2012/04/06/homemade-ph-indicator/

Now we add the reducing agent (model compound).

That lessens conjugation and bleaches the anthocyanine.  "But the colour it currently is is very pale" If the new compounds' residual optical absorbances are in the green, the solution will be iodine purple by transmission.


Why are hollow cathode lamps used for atomic absorption spectroscopy? In 1955, it was proposed that hollow cathode lamps are good emission sources for monochromatic radiation, and thus for atomic absorption spectroscopy (AAS).  In looking into the topic, I've read that "continuum radiation is not preferred in AAS." Why not?

Atomic absorption lines are very narrow.
For the Beer-Lambert law to be applicable, the bandwidth of the source should be narrow in comparison with the width of the absorption peak. Otherwise, the signal-to-noise ratio and the slope of the calibration curve would be low; the resulting sensitivity would be poor.
Continuous radiation sources use monochromators; however, even good monochromators have effective bandwidths that are significantly greater than the width of atomic absorption lines.
Therefore, the use of line sources with bandwidths even narrower than the width of absorption peaks is preferred.


Methyl vs halogen : order of precedence How would you name this compound?

4-bromo-2-methylpentan-3-one
or
2-bromo-4-methylpentan-3-one

It should be named as 2-bromo-4-methylpentan-3-one

The alphabatical order should be followed while naming this compound. As b in bromo comes before m in methyl, it should be named as above.

See here and here for more information.


Boiling point of ethanamide vs propanamide I just have a question regarding the boiling points of some primary amides. Ethanamide has a boiling point of 222 °C, while propanamide has a lower boiling point of 213 °C. Both amides are capable of hydrogen bonding, however, I would have expected propanamide (the molecule with more electrons) to have a higher boiling due to stronger dispersion forces. The same applies to butanamide which has a boiling point of 216 °C.

Is there anyway to rationalise why propanamide and butanamide have a lower boiling point than ethanamide or why the carbon chain length doesn't seem to affect the boiling point of amides?

The relatively high boiling points for such low molecular weight compounds suggests a primary effect from hydrogen bonding.

Using ethanamide as a reference point, propanamide has a higher molecular weight, which should raise the boiling point a little, but three relatively nonpolar carbons (plus hydrogens) vs two in ethanamide. A lower boiling point would then not be unexpected. Butanamide seems to support this, being a little higher in bp than propanamide by virtue of its higher mw, but still lower than ethanamide because of its lower overall polarity.

But to throw in some difficulty, formamide (methanamide) has a bp of only 210 °C, whereas its low content of nonpolar atoms suggest it should be much higher.

And then there are the melting points, which are difficult to relate to polarity: formamide 2 °C; ethanamide (acetamide) 80 °C; propanamide (propionamide) 80 °C; and butanamide (butyramide) 115 °C.

Formamide could be an outlier because of nitrogen flipping back and forth thru almost-planar nitrogen. Butanamide's high mp is a bit surprising. There could be packing factors in the solid that are favorable.

Maybe too many factors and too small overall effects to be able to make a lot of predictions, even tho we may be satisfied with some explanations after seeing the numbers.


How do you calculate the total heat capacity of a reaction I need help with this one homework question:

I am combining $100\,\text{mL}$ of $0.5~\text{M}~\ce{HCl}$ and $100~\text{mL}$ of $0.5~\text{M}~\ce{NaOH}$ to obtain $200~\text{mL}$ of $0.25~\text{M}~\ce{NaCl}$. The change in temperature of this reaction was $\Delta \vartheta  =3.38~\text{°C}$. I have to calculate the heat of reaction for 1 mole of the reactant. I also need to calculate the $c_\text{total}$ of this reaction. How do I go about doing this?

So the equation is $$\ce{HCl + NaOH -> H2O + NaCl}$$

Suppose the heats of formation of $\ce{HCl}$, $\ce{NaOH}$, $\ce{H2O}$, and $\ce{NaCl}$ are $-166.19$, $-469.07$, $-285.85$, and $-406.77$, respectively, all in $\text{kJ/mol}$.

I went about doing this calculating the heat $$q=\Delta H=[-(-166.19-469.07)+(-285.85-406.77)]~\text{kJ/mol} = -57.36~\text{kJ/mol}$$

I then found the mass of the liquids used for the reaction:

$$\begin{aligned}m &= 0.1~\text{L}\times 0.500 ~\text{M}\times 36.4611~\mathrm{g\,mol^{-1}} \\
&\;\;+0.1~\text{L} \times 0.500~\text{M} \times 39.99715 ~\mathrm{g\,mol^{-1}}\\
&=3.822~ \mathrm{g} \end{aligned}$$

I then solved for $c$ using $q=mc\Delta T$.

$0.05\times 57360~\mathrm{J} = 3.822\ \mathrm{g}\times c_\text{total}\times 3.38~\mathrm{K} \Rightarrow c_\text{total}=222.01 ~\mathrm{J \, g^{-1}\, K^{-1}}$

By the way, I multiplied the $q$ by $.05$ because that's how many moles of reactant I have (since we have the molarity and volume of the reactant).

I am pretty sure I am doing something wrong, but I am not sure what.

Two things are wrong in your calculation. Well, it's really just one thing, but the other one bugs me more.

The total mass of the solution.
Units.

1. Mass of the Solution

To calculate the mass of the total solution, I think it is fair to make the following assumption:
$$ \rho_\ce{HCl} \approx \rho_\ce{NaOH} \approx \rho_\ce{NaCl} \approx \rho_\mathrm{H_2O} \approx 1000~\mathrm{kg\, m^{-3}}$$
$$ V_\text{tot} = V_1 + V_2$$

It says that we assume the densities of all the solutions to be the same, and that the total volume is the same as the sum of the two initial volumes.

Using the above assumptions, we can calculate the total mass of the reaction mixture:
$$ m_\text{tot} = \rho V_\text{tot} = \rho (V_1 + V_2) = 1~\mathrm{g\, cm^{-3}} \times (200 ~\mathrm{cm^{3}}) = 200~\mathrm{g}$$

You have calculated the reaction enthalpy to be $\Delta H = -57.36~\mathrm{kJ\,mol^{-1}}$.

Now we have to calculate the total heat that was released during this reaction, which leads me to

2. Units

Using unit analysis (we have $\mathrm{kJ\,mol^{-1}}$ and need $\mathrm{J}$) we know that we need to multiply the reaction enthalpy with some amount of substance (unit: $\text{mol}$). Now that we know what we have to plug in, we need to know how much of it. Since the reaction enthalpy that you calculated is per $\text{mol}$ of reactant, and we have exactly $0.25~\mathrm{M} \times 0.2~\mathrm{L} = 0.05~\mathrm{mol}$ of reactants, we calculate the total heat as follows:
$$ q_\text{tot} = n_\text{tot} \times -\Delta H_\text{reaction} = 0.05~\mathrm{mol} \times -57.36~\mathrm{kJ\,mol^{-1}} = 2.868~\mathrm{kJ}$$
Why the $-\Delta H$? Well, the heat that goes out from the reaction goes into the mixture. That is why the sign has to change here.

We also need the temperature difference, which was measured to be $\Delta T = 3.38~\mathrm{K}$.

The final result (the heat capacity) is calculated via
$$ q_\text{tot} = m_\text{tot}c_\text{tot}\Delta T \Rightarrow c_\text{tot} = \frac{q_\text{tot}}{m_\text{tot} \Delta T} = \frac{2.868~\mathrm{kJ}}{200~\mathrm{g} \times 3.38~\mathrm{K}} = 4.243~\mathrm{J\, g^{-1}\, K^{-1}
}$$
This is a little bit more than the specific heat capacity of water, so we're good!

I hope my walk-through has demonstrated all crucial points of where you went wrong, and that it clarified things for you.


Choose the molecule with the strongest bond Choose the molecule with the strongest bond: $\ce{F2, Cl2, Br2, I2,}$ or all are equal.

All of them are nonpolar-covalently bonded, so I had thought that the answer would be "all are equal". However, apparently, $\ce{Cl2}$ has the strongest bond, and I couldn't figure out why. Could someone help me understand this? Any help would really be appreciated.

The Bond Dissociation Energies of these molecules are in the order
Cl2 > Br2 > F2 > I2

All these molecules involve a single covalent bond. Covalent bonds rely on sharing of a pair of electrons between two atoms, an electron each from both atoms involved. The better the sharing, the stronger the covalent bond formed. This sharing is reflected in the valence shell orbital overlap that occurs between the two atoms, when a more effective overlap occurs, a stronger covalent bond is formed.

Now, the radii of the halogen atoms increases in the order of writing them down the group, F < Cl < Br < I < At. As the atoms get larger, so do the valence shell orbitals. The increase in size of orbitals carrying the same number of electrons would mean that the electron density would be lower in larger orbitals. So, for the same proportional overlap between two similar orbitals of similar atoms, the sharing of electrons would be poorer and would consequently result in a weaker covalent bond.

This should explain the strength of bonds decreasing as we go down the group.

However, in this orbital overlap consideration gets forgotten another factor that becomes important when atoms are small in size, and that is of electronic repulsions.

Fluorine is extremely small in size and so the electron density is pretty high as well. This results in an unstable arrangement if the atoms come too close, since like charges repel.

The interplay of the above mentioned effects results in Chlorine having the highest bond strength.

I may also add that the same is the case for (1) N-N bond energy as against P-P bond energy and (2) O-O bond energy as against S-S bond energy. But, do keep in mind this stands true only for single bonds.


Benzilic acid rearrangement I'm in need for the exact mechanism of the above reaction. I attempted some tries(mentioned in the mechanism above) and I felt the yield quite unsatisfactory.
- the last step remains stuck in an equilibrium. Is the energy of the product sufficiently high to surpass the equilibrium created??
- the probability of the product formation (according to the mechanism I assumed) is too low as the pair of electron would hardly prefer an attack on the 5th carbon (as it doesn't possess any partial positive charge as the 6th carboxy carbon does)

Czakó/Kürti$^{[1]}$ suggest a [1,2] shift, which in your case would mean a shift of the 5-6-bond to the 1-carbon; in concert with a free electron pair of the anionic oxygen neutralizing the 6-carbon. The carbonyl oxygen becomes anionic and accepts a proton in the next step. So instead of the 1-6-bond breaking, the 5-6-bond breaks.

Responding to a comment regarding yield: Czakó/Kürti give several references and cite one example yielding 63% over the two steps (addition of base and acid).$^{[2]}$ Other examples are given, but without yield.

1) L Kürti, B Czakó: Strategic Applications of Named Reactions in Organic Synthesis, Elsevier (2005).

2) P A Grieco, J L Collins, J C Huffman, J. Org. Chem., 1998, 63 (25), pp 9576–9579, DOI: 10.1021/jo981419z (as given by Czakó/Kürti).


Angular momentum of an electron in an atom The Orbital Angular momentum of an electron in an atom is given by $\hbar \sqrt{l(l+1)}$ and the Spin Angular Momentum is $\hbar \sqrt{s(s+1)}$. So what is the resultant Angular momentum of an electron in an orbit in an atom?

I came across a question which asked the value of Angular momentum of an electron in its ground state in an atom. Does this refer to resultant of both kinds of Angular momentum, or am I misinterpreting the question?

Also, what exactly does ground state refer to in an atom? Is it the 1s orbital?

The question does not state what kind of atom, but I assume this is about the Hydrogen atom. In this case you can combine orbital and spin angular momentum according to Russell–Saunders coupling.

Since we are talking about the Hydrogen atom, the ground state is essentially just the $1s$ orbital. In general however, for many-electron systems, electronic states and orbitals are different things!


What happens when you mix milk and coke? Add small amounts of milk to a bottle of coke, and you end up with a white-ish precipitate with a clear liquid on top. "The Internet" seems to agree that phosphoric acid in coke reacts with something in the milk (possibly proteins, it seems to me). The Telegraph even wrote that it is the "milk molecules" that react.

So what is the reaction(s) that takes place?  I can't find any proper sources that explain this in chemical terms.

Telegraph's milk-molecule analysis:
 This is what happens when you add milk to coke
or via the Internet Archive (without pictures).

The phosphoric acid causes denaturation of milk proteins which precipitate out of the solution (milk is an emulsion, to be exact). The flock that is formed also adsorbs dye molecules (caramel) which is why the solution become clear.

There is nothing coke-specific about it. Any acid would denature milk just like that. And nearly any huge organic molecule would get absorbed, just like caramel dye.


Simulating H1 NMR spectra using chemical shift/multiplicity data? I've encountered a number of free programs/internet plugins that can output IR spectra and the corresponding chemical shift/multiplicity data if you input a chemical structure, but are there any where you can input chemical shift/multiplicity data and get an output of at least the H1 NMR spectra?

Note: by "chemical shift/multiplicity data", I am not referring to chemical shift tensors (σ). I mean simple data like "1.92 ppm (2 H, m)".

I am not aware of any software that can transform a spectrum into a molecule. But if you are asking for look up database have a look at Spectral Database for
Organic Compounds SDBS

Here you can input simple data as you asked for like ppm, and then get suggested moleulces as output.


Conversion of a value between g/gmol and lb/lbmol I am doing some air emissions calculations and I really just have a bad feeling about my results.  By definition, molecular weights are given in $\pu{g/gmol}$.  For the substance I have, I was told by a technical contact at Citgo that this material has a molecular weight of $\pu{140 g/mol}$ (which should be $\pu{g/gmol}$).  Whenever I convert this $\pu{lb/lbmol}$ (which is required for this particular equation), I'm getting a value of $\pu{0.3086 lb/lbmol}$.  There are 453 grams in a pound.  Is it not correct to divide 140 grams by 453 grams to get the total number of pounds?

My final result for annual emissions is pretty low.

A gram-mole ($\pu{gmol}$),often just called a mole ($\pu{mol}$), is the amount of a material whose mass in grams equal to the numerical value of the atomic or formula mass. For example, a gram-mole of water has a mass of 18.017 grams, this number being the formula mass for water.

A pound-mole ($\pu{lbmol}$) is the amount of a substance whose mass in pounds-mass equals the atomic or formula mass. For water, one pound mole has a mass of $18.017$ lb-mass.

The pound-mole is bigger than the gram-mole by the same factor as a pound is bigger than a gram - so pounds per pound-mole equals grams per gram-mole.  You do not divide by anything.


Is it considered safe to sonicate with flammable solvents? I use a sonicator at work in a cleaning process. It involves sonicating acetone, NMP, and IPA. We would like to move forward with other flammables as well. We currently do the work in a hood, but no other precautions are taken. OSHA has no guidance. What is your experience with this issue?

In general it should be relatively safe considering you take all the precautions required when working with volatile flammable liquids.
Specifically for sonication I can think of two safety aspects:

Sonification promotes formation of the aerosol/fog consisting of the atmospheric gases and solvent vapors.
Depending on its composition, it might or might not be fire hazardous or even explosive.
That's why keeping an ultrasonic setup in a ventilated area — a fume hood, ideally — is a good idea.
Keep the power supply and all controller circuits outside, if possible, in order to prevent spark vapors ignition.
Be careful when sonicating heterogeneous solutions.
Some materials would absorb the ultrasonic energy and this can result in local overheating.
To prevent this, use mechanical stirring.


Oxalic acid structure What is the reason why oxalic acid has its second $\ce{-COOH}$ group flipped?

I'm guessing it's something to do with steric repulsions between the $\ce{-COOH}$ groups but I'm not sure.

Like ron already stated, the reason for that arrangement is the possibility to form more intermolecular hydrogen bonds. For oxalic acid there are two modifications which depict this quite clearly.

From charge considerations itself it is not obvious, that the periplanar alignment should be better. However one can see, that there might be a little less repulsion. Given here are natural partial charges from NBO/BP86/cc-pVDZ.

A rotational scan reveals that the rotation barrier is just around $7~\mathrm{kJ/mol}$ and the difference between these are just around $2~\mathrm{kJ/mol}$ (BP86/cc-pVDZ). The dihedral angle of $0^\circ$ refers to the hydroxyl moieties being on the same side.

The barrier results mainly from breaking the partial $\pi$ bond between the two carbon centres.

The $\ce{O...H}$ intramolecular distance is around $230~\mathrm{pm}$ and in gas phase and higher temperatures hydrogen tunneling should be also considered.

In aqueous solution, exchange rates between water and oxalic acid are very fast, so that all oxygen atoms should be equal - there is no difference observable at this time scale.
\begin{aligned}\ce{
(COOH)2 + 2H2O &<=> HOOC-COO- + H3+O + H2O\\
               &<=> (COO^{-})2 + 2H3+O\\
}\end{aligned}


Balancing a redox reaction with only one product Consider the following comproportionation reaction:
$$\ce{NH4NO3 -> N2}$$
How would I go about balancing this using the half reaction/ion-electron method?

You should use the the same product species in both half -reactions. In the example you gave, this is $\ce{N2}$.
$$\ce{NH4NO3-> N2}$$
The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states:
\begin{align}
\ce{NH4+ &-> N2} \tag{1} \\
\ce{NO3- &-> N2} \tag{2}
\end{align}
Balancing the reduction half-equation $(1)$:
\begin{align}
\ce{2NO3- &-> N2} & & \text{(balance N)} \\
\ce{2NO3- &-> N2 \color{red}{+ 6H2O}} & & \text{(add } \ce{H2O} \text{ to balance O)} \\
\ce{2NO3- \color{red}{+ 12H+} &-> N2 + 6H2O} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NO3- + 12H+ \color{red}{+ 10e-} &-> N2 + 6H2O} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)}
\end{align}
And balancing the oxidation half-reaction:
\begin{align}
\ce{2NH4+ &-> N2} & & \text{(balance N)} \\
\ce{2NH4+ &-> N2 \color{red}{+ 8H+}} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NH4+ &-> N2 + 8H+ \color{red}{+ 8e-}} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)} \\
\end{align}
In order to combine the two balanced half-equations, we need to multiply the first by $4$ and the second by $5$:
\begin{align}
\ce{8NO3- + 48H+ + 40e- &-> 4N2 + 24H2O} \\
\ce{10NH4+ &-> 5N2 + 40H+ + e-} \\ \hline
\ce{8NO3- + 10NH4+ + 48H+ + 40e- + 10H2O &-> 9N2 + 24H2O + 40H+ + 40e-}
\end{align}
which finally reduces to
$$\ce{8NO3- + 10NH4+ + 8H+ -> 9N2 + 14H2O}.$$


Basic Understanding of Relative Atomic Mass I’m currently trying to learn some chemistry and have run into some trouble understanding some of the basics. The confusion stems from a passage in my textbook, so I will list the brief passage and then ask my question.

  In addition to predicting molecular formulas, Avogadro’s hypothesis gives correct results for the relative atomic masses of the elements. Analysis by chemists during the 18th century had revealed that 1g of hydrogen combines fully with 8g of oxygen to make 9g of water. If Dalton’s formula for water, HO, were correct, then an atom of oxygen would have to weigh eight times as much as an atom of hydrogen; that is, Dalton’s assumption requires the relative atomic mass of oxygen to be 8 on a scale where the relative atomic mass of hydrogen is set at 1.

So far so good, I believe that oxygen has a relative mass of 8 because there are 8g of oxygen in the original reaction and likewise with the 1g of hydrogen, but this next part is where it starts to get confusing.

  From Avogadro’s hypothesis, however, each water molecule contains twice as many atoms of hydrogen as oxygen, so to achieve the experimental mass relationship, each oxygen atom must have twice as large a relative atomic mass. This gives a relative atomic mass of 16 for oxygen, a result consistent with modern measurements.

Alright, so basically what I’m confused about is how they arrived at a relative atomic mass of 16 given that there is 8g of oxygen and 1g of H in the original experiment. What exactly are they observing in order to establish the relative atomic mass of oxygen as 16. Any help understanding this and relative mass in general would be greatly appreciated.

It's because there's two hydrogen atoms in water, not one. In order to show you, let's assume we're making HO instead of H2O. This would clearly only require half the amount of hydrogen, but still the same amount of oxygen. Thus we have .5 grams of hydrogen combining with 8 grams of oxygen in a 1:1 relationship. Thus the oxygen weighs 16 times more than the hydrogen. Do you understand?

In H2O they combine in a 2:1 relationship. We have 1 gram of hydrogen combining with 8 grams of oxygen, but because there's 2 hydrogen for every oxygen, we actually need TWICE as many oxygen atoms to have the same amount of atoms. In other words, there will be the same number of atoms in 1g of hydrogen as there is in 16g of oxygen.


Has the half-life of an isotope ever been accurately predicted before it could be measured? The half lives of many radioactive isotopes have been measured directly. But are their half-lives known to be a function of their composition?

In other words, can someone say "given an atom with this many protons, neutrons and electrons, and given the known strengths and ranges of the various forces at play inside an atom, we can quantify how unstable such an atom would be by saying that a collection of them would have a half-life of X"?

Has such a prediction ever been made about an isotope before people had encountered it, then later borne out by experiment?

A quick perusal of the post-WWII literature (particularly Glenn Seaborg) pulls up papers like The New Element Californium (Atomic Number 98) where comparisons of the observed alpha decay half life are said to be in line with predictions. Such predictions follow along from papers such as Systematics of Alpha-Radioactivity.

So, yes, it would seem that predictions of radioactive half lives were part and parcel of the creation of new elements in the laboratory.


What type of bond would be formed if dyz orbitals overlap along the z axis? I know that when two $\ce{p_z}$ orbitals overlap in a particular direction - along the z axis - they form a sigma bond. Now, if two $\ce{d_{yz}}$ orbitals overlap in same direction, which type of covalent bond would be formed?

Before I answer the question, you should understand that - since everything is symmetric - the answer would be the same if we took two $\ce{d_{yz}}$ orbitals along the x-axis, two $\ce{d_{xy}}$ orbitals along the y-axis, etc.

Have a look at this snippet I picked from Wikipedia Commons (by Jeremy77, public domain):

Here you can see the overlap of two $\ce{d_{xy}}$ orbitals along the y-axis constructively (in the same phase). There are four lobes overlapping, which would lead to a total of two nodal planes. Such a bond is a speciality of $\ce{d}$ orbitals and it is called the delta ($\delta$) bond.

Now, can you understand what the answer to your question would be?

See also: Which d orbitals can form Sigma, Pi, Delta bonds?


How to prepare KF/Al2O3 i.e. potassium fluoride on alumina (/on aluminum oxide) It's surprisingly hard to find an answer to this. One book I found referenced a method wherein one would simply mix (with a mortar I assume?) equal (another source said 40:60) portions of Al2O3 and KF and that's it.

In Bull. Chem. Soc. Jpn., 1983, 56, 1885-1886 (PDF), the authors (Takashi Ando and coworkers) give the following method:

  Alumina (30 g, neutral alumina, type 90 for column chromatography, activity I, 70-230 mesh, Merck Jpn.) was mixed with $\ce{KF}$ (20 g) in 200 mL of water and the water was removed at 50-60 °C in a rotary evaporator. The impregnated alumina was then dried in a vacuum drying oven at 75 °C for several hours.

In another publication, J. Chem. Soc., Perkin Trans 2, 1986, 1133-1139, by the same main author, which focuses on the properties of the $\ce{KF}/\ce{Al2O3}$ system, a "dry preparation" by grinding $\ce{KF}$ and $\ce{Al2O3}$ under a nitrogen atmosphere is mentioned.