How is energy not stored in bonds? Consider the paraphrase from Ebbing:

  It takes energy to break bonds & energy is released when bonds are formed

Let's look at the first statement  It takes energy to break bonds.

Is this true of spontaneous processes?
If yes, is this true only in a closed-environment sense, where the energy required for a reaction to occur between two molecules is the energy contained within those molecules; i.e. no outside energy is necessary?
If yes to 2), is this energy, outside the kinetic energy of the molecules themselves, potential energy (PE) stored in the bonds?

Then for the second statement, energy is released when bonds are formed.
When the PE of the reactants is greater than the PE of the products, heat is released and we have an exothermic process. Consider the conversion of ethane to ethene. This is an endothermic reaction requiring approximately 220 kJ/mol of energy (bonds broken (2X C-H) - bonds formed (C=C).

As the PE of ethene is greater than that of ethane, why is this not reflected when considering the sum of bond energies in each molecule? For example, for ethane we have 2,812 kJ/mol and for ethene we have 2,246 kJ/mol.
Given the bond energy of C=C (602 kJ/mol) is higher than C-C (346 kJ/mol), why would it be incorrect to assume the energy necessary to drive the reaction is stored in the C=C bond, given the PE of ethene is greater than that of ethane?

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

Is this line of thinking correct, if not why?
If that line of thinking is correct, how do we reconcile this fact with "Energy is released when bonds are formed?
Would it be appropriate to say energy is both stored and released as bonds are formed?

I am a tutor hoping to master this content to best help my students. Please lend a hand to whichever questions you feel most confident in answering.

Let's start with the specific examples. For the conversion of ethene to ethane, you need a reaction partner to provide the extra hydrogen, e.g. dihydrogen:
$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$
In the reaction, a single bond and the pi-bond of the double bond are broken, and two single bonds are formed. If you look up the four bond dissociation energies (two broken and two formed), you get a fairly good estimate of the reaction enthalpy.
$$\ce{ADP + phosphate -> ATP + H2O}$$
I'm leaving out the charges and the protonation state of phosphate to make things simpler. If we look a bit closer, we can write:
$$\ce{RO3P-OH + H-OPO3R' -> RO3P-OPO3R' + H-OH}$$
So when ATP is formed, we are not only making the phospho-anhydride bond (and a bond within water), but also breaking two bonds.

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

Is this line of thinking correct, if not why?

You could say the energy is stored in the reaction, and when the reaction runs in the other direction (hydrolysis), it gets released again. Similarly, glucose is a source of free energy only if oxygen is available as reaction partner. Breaking glucose into atoms is not a source of free energy.

Let's look at the first statement It takes energy to break bonds.

Is this true of spontaneous processes?

Here is the reaction this refers to:
$$\ce{molecule -> free atoms}$$
The entropy of reaction is expected to be positive (more particles on product side). If the reaction were also exothermic, these molecules would not be stable under any condition (Gibbs energy of reaction would be negative at any temperature). So bonds of molecules that actually exist are expected to have a positive bond dissociation energy.
The other questions
The answers to the rest of the questions follow from the information above.


precipitation reaction - cellulose in schweizer reagent reacts with acid In the regeneration of dissolved cellulose (in a cuprammonium solution) in an acid bath, say sulphuric acid, what is the chemical reaction?

There is a lot about the chemistry of the production of the copper ammonia complex but nothing about the precipitation step.

Is the precipitation possible with any acid?

Apologies about the tag - can't find any existing ones that are appropriate

It's possible to regenerate cellulose from Schweizer's reagent by dilution in plain water, though neutralization of the high pH+ cuprammonium solution with an acid is more efficient. $\ce{HCl}$ works as well as $\ce{H2SO4}$. Economics probably drives the choice.

The chemistry is based on breaking cellulose, a polysaccharide (polymer)  β(1→4) linked D-glucose, into smaller, soluble, segments, and reconstituting the longer polysaccharide through neutralization.


Polarity of the phosphate ion? I understand that phosphate is bonded covalently, however, it is characterized as a polyatomic ion. This is making me confused on whether it's polar or non-polar.

There are multiple different things to consider for phosphate $\ce{PO4^3-}$:

The bonds themselves. Each of the four $\ce{P-O}$ single bonds is polarised towards the oxygen. Each bond is polar.
The molecule as a whole. It is of near-perfect tetrahedral shape thus having the point group $T_\mathrm{d}$. This point group includes the inversion symmetry operation $i$. Therefore, macroscopic properties of a vectorial type (like the overall dipole moment which is a vector more or less pointing from one pole to the other) must, by definition, be $0$. Thus, the entire molecule is not a dipole.
The molecule’s total charge. It has an overall charge of $-3$. This charge is located on the surrounding oxygens but the entire thing can be simplified from far away to a ball with the charge in the centre, i.e. the phosphorus atom. We would consider this molecular ion a monopole.


How is chirality center defined for an atom with 5 substituents? I understand that carbon atoms normally only have up to 4 substituents and the R and S chirality centers are defined for 4 substituents. Is there a similar convention of 'chirality' for an atom with 5 substituents?
The SMILES website (section 3.3.4 General Chiral Specification) shows that general 'chirality' for 5 and 6 substituents exists. But the text there focuses on how to convert them to SMILES strings. Is there any references about the classifications and nomenclatures?
https://www.daylight.com/dayhtml/doc/theory/theory.smiles.html
Sorry I am not a chemistry major and my nomenclature may not be correct.

Chirality depends not so much on coordination number, but on the symmetry of the substituents.  Meaning you have to pay attention to coordination geometry.  In the case of four-coordination you can have four different substituents and yet no chirality... if you have square planar instead of tetrahedron coordination.
For five-coordination you lack a mirror plane and thus have a chiral center in the following cases:

Square pyramidal geometry: if all four substituents on the base are different, or if just one pair of these basal substituents are identical and the identical pair is on adjacent vertices.  The axial substituent does not have any impact because all possible mirror planes pass through it.

Triangular bipyramid geometry: if all three equatorial substituents are different and the two axial substituents are different from each other.  You may still have chirality if one axial substituent happens to be the same as one equatorial one.


NMR spectrum chemical shift confusion Recently started on NMR spectrum, and I'm slightly confused about chemical shifts and their connection to the resonance frequency and magnetic field.
From what I know, the "new" way of doing NMR means keeping the radio-frequency constant and increasing the external magnetic field in small increments.
Now, I believe a greater electron density around the protons (more shielded) will protect the positive nucleus from the external magnetic field and thus, the positive nucleus experiences a lesser net magnetic field. Hence, there is a lower resonance frequency which should result in a smaller chemical shift (the magnetic field will determine the energy differences between the spins).
From what I understand, because of experiencing a lesser net magnetic field, we increase the external magnetic field so that we can get a reading (this apparently gives rise to the idea of high field on the more shielded side). However by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered a more modern way of doing NMR as it was developed after the CW technique. The pulsed technique was also accompanied by the development of FT processing which resulted in a Nobel price being awarded to Richard Ernst.
In CW NMR you usually "sweep" the field - increase the external field $B_0$ in small increments, as you explain - while keeping constant the frequency generated by a "read" RF coil. Coincidence between the read frequency and the Larmor frequency $\omega = \gamma \left(1-\sigma \right)B_0$ - the resonance condition - results in a signal registered with a pickup coil. Here $\sigma$ is the chemical shielding which reduces the effective magnitude of the applied field $B_0$  at the nucleus.

by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

This is exactly the point of increasing the field. The Larmor frequency is related to the splitting in the energy levels, and you need a greater field to bring about a splitting in a more shielded nucleus, so that the resonance condition is satisfied.
The resonance frequency in such a CW experiment is always $\omega = -\gamma B_1$, but as the field at the shielded nucleus is $\left(1-\sigma \right)B_0$ and the shielding opposes the applied field, it is necessary to apply a larger field ("go upfield") to excite a more shielded nucleus.


What is proximity effect? From Klein's Organic chemistry [1, p. 780]:

The 1,2-adduct is believed to form more rapidly as a result of a proximity effect. Specifically, the carbocation and the bromide ion are initially very close to each other immediately after their formation in the first step of the mechanism. The bromide ion is simply closer in proximity to C2 than C4, so attack at C2 occurs more rapidly.

Similarly, from Vollhardt's Organic Chemistry: Structure and Function [2, pp. 345–346]:

Ring size controls the speed of cyclic ether formation
A comparison of the relative rates of cyclic ether formation reveals a surprising fact: Three-membered rings form quickly, about as fast as five-membered rings. Six-membered ring systems, four-membered rings, and the larger oxacycloalkanes are generated more slowly.
Relative Rates of Cyclic Ether Formation
$$k_3 ≥ k_5 > k_6 > k_4 ≥ k_7 > k_8$$
$k_n =$ reaction rate, $n =$ ring size
[…]
A second enthalpic phenomenon, which has been called the proximity effect, operates, especially in 2-haloalkoxides.

The proximity effect was used to explain these two reactions. Is it applicable to every reaction?
What is the proximity effect?
References

Klein, D. R. Organic Chemistry, 1st ed.; John Wiley: Hoboken, NJ, 2012. ISBN 978-0-471-75614-9.
Vollhardt, K. P. C.; Schore, N. E. Organic Chemistry: Structure and Function, 7th ed.; W. H. Freeman: New York, 2014. ISBN 978-1-4641-2027-5.

I will not attempt to define "proximity effects" in my answer. If you are looking for a definition, this is probably not the best answer. I think the term itself is quite vague and rather qualitative if you will. I've personally seen the concept of "proximity effects" being used more in chemical biology / enzymes chemistry context, than in organic chemistry.

In the first example, I am not even sure if there is indeed any proximity effect in operation. There could be; but it really depends on the solvent effects. How does one really separate the different factors in play? Doing gas phase simulations could give you an answer. The first example you have cited, can indeed be more easily be explained in terms of relative stabilities of the "carbocation" resonance forms: it is more favourable to have the positive charge at position 2 (allylic, secondary), than to have the positive charge at position 4 (allylic, primary). In terms of resonance or MO theory, this picture translates into position 2 being more electron deficient than position 4. A nucleophile would attack at the most electron deficient position (ceteris paribus), and this would be the fastest reaction path. Attack at position 2 leads to the kinetic product, as shown in the book you've cited. Whether there is any additional "proximity effects" - such as the nucleophile being "closer" to position 2 than position 4, is up for debate and cannot be concluded from this result. On the grounds of carbocation stabilisation alone, the observed product can be explained. Note however, that under the conditions, this reaction could be reversible and eventually generate the more stable thermodynamic product.

The second example, is perhaps a more appropriate application of 'proximity effects'. However, I'm not sure if it is appropriate to call this an 'enthalpic phenomenon'. I would have guessed a more probabilistic/entropic origin of this phenomenon. In this example, I would interpret proximity effect as simply the fact that if you are doing a 1-3 ring closure (to form a 3 membered ring), your two reaction centres are closer in space, and hence the reaction is probabilistically more likely than a 1-6 ring closure. I don't know how using the term 'proximity effect' helps give a better explanation, though. If you're interested, you should read up on rates of ring closures in a book of your choice, that discusses both enthalpic and entropic contributions to analyse the relative rates of ring closure.


How are the chemical reactions of the table of standard reduction potentials experimentally determined? My background: I've read several books and articles on electrochemistry and even do photoelectrochemical research in a lab, so I'm somewhat familiar with the concepts and techniques.

In texts on electrochemistry, the standard reduction potential and its meaning is usually axiomatically presented early on and everything follows from there. What I would like to see in a treatment of the subject is an explanation of how the chemical equations of the reduction half reactions are actually determined. For example, let's take the equation $\ce{Cu^{2+} + 2e- <-> Cu}$. How did they figure out that it is in fact two electrons reducing the $\ce{Cu^{2+}}$ ion?

One could do open circuit potential measurements to determine the standard reduction potential, but it would seem to me that you would have to know the ionic state of the element a priori to use this technique. One could also apply a voltage until the reaction "turns on" then approximate that onset voltage as the standard reduction potential, but this wouldn't be strictly true as it would be a non-equilibrium measurement and would also be subject to the kinetics of the surface. Does anyone know of a book or other treatment showing how the chemical equations are determined or better yet has experience doing this?

The standard reduction potentials were mostly discovered by carefully measuring the reduction and oxidation of solid metals (or dissolved ions) in electrochemical cells, and monitoring the electrical current through the wire connecting the two electrodes. (The image below should have an ammeter instead of of a voltmeter, but you get the idea)

Take, for example, an electrochemical cell of zinc and copper.

At the anode, the solid zinc will dissolve to become Zn2+. At the cathode, the Cu2+ ions will be reduced, and become solid copper. During the redox reaction, the scientist measures the current across the two electrodes.

After some time, the scientist could pull the two electrodes out of solution, dry them off and mass them. They could then calculate the moles of zinc that dissolved from the anode, and the moles of copper added to the cathode. In addition, by measuring the current over the period of the experiment, they would also be able to calculate the number of electrons that passed from the zinc anode to the copper cathode.

In this instance, they would calculate that for every mole of zinc that dissolved, 2 moles of electrons were produced. Concurrently, 2 electrons were consumed for each mole of copper reduced! These measurements, over time, made it possible to determine the oxidation states of many metals and soluble ions.

(image from moodle2.rockyview.ab.ca/)


Electrolytes are ionic compounds that conduct electricity Electrolytes are ionic compounds that conduct electricity, or the flow of charged particles, when dissolved in water. This occurs via the separated ions acting as those charged particles, or charge carriers, when a voltage is applied. Which compound would not create an electrolyte solution when dissolved in water?

A) ammonium acetate, $\ce{NH4C2H3O2}$

B) sucrose, $\ce{C12H22O11}$

c) sodium hydroxide, $\ce{NaOH}$

D) potassium nitrate, $\ce{KNO3}$

Sucrose is an covalent compound and it does not ionize.
$\ce{KNO3}$, $\ce{NaOH}$ are good ionic compounds hence the can form electolytic solutions.
Ammonium acetate also can ionize.
Hence the answer is 2


Interpreting straight lines in a graph of isotherms of carbon dioxide I am having difficulty in understanding pressure-volume graphs of $\ce{CO2}$ at different temperatures which describes liquefaction of $\ce{CO2}$.

I know that at the critical temperature of $\pu{31.1^\circ C}$ it is possible to liquefy the gas at the critical pressure of $\pu{73atm}$.

But I am having difficulty in understanding some statements about the graph.

  (i)At temperatures below $\pu{30.98^\circ C}$ the behaviour of $\ce{CO2}$ towards compression is quite different.For example at $\pu{21.5^\circ C}$,  $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point $\mathrm B$ a liquid of certain volume appears.Further compression do not change pressure but volume decreases till point $\mathrm C$ is reached. In fact, the gas and liquid $\ce{CO2}$ coexist along the horizontal line BC.The decreases of volume represents condensation of more and more $\ce{CO2}$ gas till the point C is reached.

I know that if we are compressing a gas  its pressure should increase but why its pressure is constant along the line $\mathrm{BC}$?

  (ii)At point C all the gases has been condensed and further application of pressure simply compresses the liquid .The line $\mathrm{CD}$ represents the compressibility of liquid Carbon Dioxide.

I have studied in my earlier classes that liquids cannot be compressed but how can the liquid $\ce{CO2}$ has a tendency of compression here?

I will expand on the key points made in the comments. There are two issues that have been brought up:

Pressure is constant along line segment BC.

This is most easily intuited from a direct mechanical example. Consider a cylinder in which one face is a piston at vapor-liquid coexistence. Line segment BC then represents isothermal compression or expansion of the piston. As we push on the piston, the system volume decreases, and vapor condenses into liquid. As we pull on the piston, the system volume increases, and liquid evaporates into vapor.

That these processes occur just enough to maintain line segment BC as an isobar can be understood more quantitatively from thermodynamics. Phase coexistence actually places significant constraints on the system, as is quantified by Gibbs' phase rule---a system of a single chemical species at two-phase coexistence is specified by a single parameter, here the temperature $T$, because we are considering an isotherm. This implies that the pressure is known given $T$; that is, we have a relationship $P = P(T)$, and of course this is just the liquid-vapor coexistence line on a phase diagram. This pressure is, in particular, independent of the number of moles of substance in each phase, and so we must have the same pressure at phase coexistence; i.e., over the line segment BC.

Liquids cannot be compressed. (False!)

Liquids cannot be compressed easily. They are effectively incompressible for small changes in pressure (hence the steepness of the curve WX), but can certainly be compressed when subject to large pressures.


Lennard Jones liquid vapor phase transition I am trying to generate a phase diagram of pressure versus temperature in a NPT ensemble using molecular dynamics method. I am using the Lennard Jones potential with 108 particles and Berendsen thermostat and barostat to fix the parameter.

In order to observe phase transition I plot the radial distribution function (RDF) and also density versus temperature graph for various fixed temperature. However it is hard to differentiate liquid and gaseous phase from RDF.

Another peculiar thing is that there should be a discontinuity in density as the transition temperature is reached, which I observe only for solid liquid transition but not for the liquid vapor transition. The change in density should be huge for liquid vapor transition. Any idea on what possibly goes wrong?

The parameters:

Periodic boundary
cut off radius = 2.5
Truncated and shifted Lennard Jones potential
number of particle = 108
number of steps = 2000
timestep = 0.004
Berendsen barostat of scale coeefficient of 0.1 is used to scale the size of simulation box

The density versus temperature graph for pressure of 1.0MPa:

The solid liquid transition can be seen but not the liquid vapor transition.

I think your transition, if present, may be drowned out by noise, as your RDF is only produced by 108 particles. As the Lennard-Jones interaction is quite simple and you are using cutoffs (good idea!), is it possible to increase the number of particles in your simulation somewhat, or else take the mean of some large number of trials?

This said, it does not take much imagination to see a continuous PV plot from about 0.7 onwards, which might mean you have something analogous to a supercritical fluid. Maybe try a number of trials at different pressures?


What is the significance of 273.16 K? Today, at my chemistry class, we were learning how to convert Celsius to Kelvin with exact values. I said we add 273.15 K to given Celsius but my colleague said we add 273.16 K.

Yes, I did check on internet, we use 273.15 K for conversion, but I also found some relation with 273.16 K and I don't understand the concept of that. It's something to do with thermodynamics and triple point of water.

The Celcius scale was originally based on the freezing and boiling points of water, so 0 °C was chosen as the freezing point until 1954. But now, the size of one degree on both Celcius and Kelvin scales is defined as 1/273.16 of the difference between absolute zero and the triple point of VSMOW (for various reasons relating to the actual measurement), but the 0 point of the Celcius scale was left as the freezing point of water which means that even though both scales are defined at 0 K and 273.16 K (the triple point), the Celcius scale still starts at the freezing point and 0.01 °C was defined as the triple point of VSMOW, which means 0 °C = 273.15 K.

Absolute zero is the coldest possible thermodynamic temperature, 0 K. The triple point is the temperature and pressure at which solid, liquid, and gas phases can exist in a stable equilibrium. If you look at the phase diagram below, it's the point at which the solid, liquid, and gas regions all meet. For water, the temperature of the triple point (273.16 K) is remarkably close to its freezing point at atmospheric pressure (273.15 K). Because 0 °C was originally defined as the freezing point of water, instead of changing the zero point when they redefined the scale based on absolute zero and the triple point, they decided to define the triple point as 0.01 °C.

We add 273.15 to °C to get kelvins because that's the difference between 0 K and 0 °C. We don't add 273.16 because the triple point is not 0 °C.

Source


Difference between oxidation and formal charge What is the difference between oxidation number and formal charge? I am confused about the two.

This is best explained in an example. Let’s take methyl isocyanide:

$$\ce{\overset{-}{C}#\overset{+}{N}-CH3}$$

This structural formula already includes the formal charges. How do we arrive at these? First, we draw a Lewis structure with the correct, required connectivity.

In this case, we have to know that a $\ce{CN}$ fragment is connected to the methyl group via the nitrogen. We can also find out that we need to form a total of seven bonds, three of which must be $\ce{C-H}$ bonds in the methyl group. Meaning that from the intermediate of $\ce{C-N-CH3}$ we need to add two more bonds but the carbon on the right-hand side already has a completed octet so these can only be between nitrogen and the left-hand carbon.

We also know that we should have a single lone pair; the only atom that can accomodate this is the left-hand carbon. All atoms now have a completed octet, our new intermediate solution is $\ce{C#N-CH3}$.

To determine formal charges, each bond is split homolytically, meaning that each bonding electron pair is split apart giving each atom one electron. Lone pairs are left alone, since they are, well, lone. The total we get is compared to the number of electrons an atom should have according to its position in the periodic table. This gives us:

$$\begin{array}{lcccc}\hline \text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{charge}\\ \hline
\ce{C} & 5 & 4 & 1 & -1 \\
\ce{N} & 4 & 5 & -1 & +1 \\
\ce{C(H3)} & 4 & 4 & 0 & \pm0 \\ \hline\end{array}$$

This concludes the determination of formal charges; the left-hand carbon gets a single formal negative charge while the nitrogen gets a single formal positive charge.

For oxidation states, a different method is employed. Here, all bonds are split heterolytically in such a manner that the more electronegative partner gets both electrons. Nitrogen is more electronegative than carbon which in turn is more electronegative than hydrogen. The resulting number is again compared to the number an atom should have; it will likely be different than before. This method will give us the following:

$$\begin{array}{lcccc}\hline
\text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{oxidation state} \\ \hline
\ce{C} & 2 & 4 & -2 & \mathrm{+II} \\
\ce{N} & 8 & 5 & 3 & \mathrm{-III} \\
\ce{C(H3)} & 6 & 4 & 2 & \mathrm{-II} \\ \hline \end{array}$$

If we add the three $\mathrm{+I}$ of the three hydrogens, we arrive at a neutral $\pm0$ again.

It is important to note that both resuts are quite different and the only thing in common is that they both add up to zero for a neutral molecule or the ionic charge of a molecular ion (unless you did it wrong). The formal charge rarely has any non-formal meaning except in ammonium derivatives where it denotes the actual centre of positive charge. The oxidation states corresponds to the actually observed electron density in an atom somewhat.

Both however are formal concepts and you need to know how to cope with them to use them to your advantage.


Identifying an extra NMR peak in silyl protection reaction I protected a phenol with the TIPSCl/triethylamine procedure. The product was isolated by column (hexanes/EtOAc) and shows an extra singlet at 1.05 ppm in $\ce{CDCl3}$, integrating to ~10 H. Whatever this is I cannot find in under UV or various TLC stains ($\ce{KMnO4}$, vanilin, PMA). The integration of the peak goes smaller after one more column. Any suggestions? Thanks!

I believe some of your triisopropylsilyl chloride may be hydrolyzed triisopropylsilylol, $\ce{((CH3)2CH)3SiOH}$. That may gives you $\delta$ 1.04 and 12.4 (for $\ce{C}$ attached to $\ce{Si}$) in $\mathrm{^1H}$- and $\mathrm{^{13}C}$-NMR spectra, respectively. Compare the spectral values in Ref.1 and 2 (see below; corresponding $\ce{^{13}C}$ values are in red):

References:

A. Goti, M. Cacciarini, F. Cardona, A. Brandi, “A convenient access to (3S)-3-(triisopropylsilyl)oxy-1-pyrroline N-oxide, a useful intermediate for polyfunctionalized enantiopure indolizidine and pyrrolizidine synthesis,” Tetrahedron Letters 1999, 40(14), 2853–2856 (https://doi.org/10.1016/S0040-4039(99)00310-X).
R. F. Cunico, L. Bedell, “The triisopropylsilyl group as a hydroxyl-protecting function,” J. Org. Chem. 1980, 45(23), 4797–4798 (https://doi.org/10.1021/jo01311a058).


Why do aromatic compounds have an upfield shift upon coordination to metals? Why do aromatic compounds have an upfield shift upon coordination to metals e.g. ferrocene? The 1H NMR shows a single environment at ~4.1 ppm. This is in contrast to the usual aromatic shift of between 6-9 ppm. At the moment, I have the following rationale, although I am not convinced about 2/3:

1) Coordination of the aromatic group to the Fe essentially localises the double bonds, as Fe will backbond into the aromatic $\pi$*-orbitals.

2) Bonding of the aromatic group to the Fe will disrupt the ring current which would also cause an upfield shift. However, there seem to be several papers online which suggest the presence of ring currents even in sandwich complexes?

3) There are 6 $\pi$-electrons for only 5 carbons in Cp, therefore there are essentially 1.2 $\pi$-electrons per Cp carbon whereas there is 1 $\pi$-electron for each carbon in Cp. Therefore Cp is more shielded than benzene naturally and will have a lower shift.

Chemical shift data dug up for a number of iron compounds in the website of Hans Reich reveals that the shift is not unique to aromatic compounds, but rather appears due to Fe(2+):

The base shift for a vinyl proton (C=C-H) is 5.25 ppm.

The following is a series of spectra from magritek, illustrating the effect of Fe(2+) complexation on the shift of the cyclopentadienyl ligand.

Sandwich compounds based on magnesium do not show such large upfield shifts:

This seems to eliminate your conjecture 3 as an explanation, since cyclopentadienyl is aromatic and addition of Fe(2+) causes an upfield shift, whereas Mg(2+) does not cause such a large one.

Now Fe(2+) complexes may be paramagnetic or diamagnetic depending on the the ligand (low/high field). Paramagnetic metals have unpaired electrons that can induce strong dipolar magnetic fields. In porphin, for instance, paramagnetic Fe(3+) results in strong shielding in the aromatic ligand, with one proton shifted upfield ~20 ppm. A diamagnetic Zn(porphin) complex does not show such strong shielding. Ferrocene is diamagnetic.

The quick and sloppy answer therefore appears to be that the shift observed in ferrocene is mainly due to Fe(2+), even though the compound as a whole is diamagnetic.

Regarding the bonding arrangement, the wikipedia has this to say:

  In terms of bonding, the iron center in ferrocene is usually assigned
  to the +2 oxidation state, consistent with measurements using
  Mössbauer spectroscopy. Each cyclopentadienyl (Cp) ring is then
  allocated a single negative charge, bringing the number of π-electrons
  on each ring to six, and thus making them aromatic. These twelve
  electrons (six from each ring) are then shared with the metal via
  covalent bonding. When combined with the six d-electrons on Fe2+, the
  complex attains an 18-electron configuration.

To add a little more detail, Mulay and Fox hypothesize on the nature of metal-ligand bonding in ferrocene based on the magnetic anisotropy:

  The anisotropy between K3 and the average of K1 and K2 was found to be
  $\pu{49.5×10—6 cgs units}$, and is evidence in favor of a single dπ—pπ bond between each of the rings and the iron atom. Pauling's "resonance
  structures" also seem plausible; the studies do not support Fischer's
  "complete π‐electron donation" structure. Theoretical calculations
  based on Langevin's formula for diamagnetism show a close agreement
  with the experimental value.

A more detailed account of bonding in ferrocene can be found in Miessler and Tarr's textbook.

[6]: Magnetic Anisotropy of Ferrocene and Chemical Bonding, J. Chem. Phys. 38, 760 (1963); https://doi.org/10.1063/1.1733734, L. N. Mulay and S.N.D. Mary Eleanor Fox


Total pressure in a chamber calculation I have a chamber containing two different gases and a liquid which needs to be heated at a certain temperature to bring it to vapor state.  I have worked out that we can calculate the total pressure of a mixture of gas using Dalton's law using partial pressure of each gas. To calculate total vapor pressure we have Raoult's law. If I have both gases and vapors in the medium can we calculate the total pressure by adding them both, as in a "collecting gases over water experiment"?

Dalton's Law is only true for ideal compounds, which theoretically have no intermolecular forces between their molecules. Those forces become more and more significant when increasing the concentration of any compound, because all molecules become packed up closer.

This will lower the pressure, because if any molecule is about to hit the container wall, it gets pulled backwards $-$ and thus slowed $-$ due to the attraction of the other molecules behind it. But for high temperature this effect is negligible, because they can't be slowed down so easily.

Now lets come to the calculation using Kay's rule:

Since you only have one liquid component, you don't need Roult's Law to calculate the vapor pressure for a mixture of liquids.

1.) You need to find the values for the critical temperatures and critical pressures for your components (gases and water vapor).

2.) You need to know the relative amounts of each substance:
$$y_i = \frac{n_i}{n_m}$$

$n_i$... amount of a specific component [mol]

$n_m$... total amount of substance of the mixture [mol]

3.) Calculate the pseudo-critical temperature and pressure:
$$T_c' = y_1 \times T_{c1} + y_2 \times T_{c2} + y_3 \times T_{c3} $$

$$P_c' = y_1\times P_{c1} +y_2\times P_{c2} +y_3\times P_{c3}$$

$T_c$... critical temperature of a specific component [K]

$P_c$... critical pressure of a specific component [Pa]

4.) calculate the pseudo-reduced values for T and the volume:
$$T_r = T / T_c'$$
$$v_r' = V / (R \times T_c' / P_c')$$

$T$... temperature in the container

$V$... volume of the gas phase [m$^3$]

$R$... universal gas constant = 8.314 J/(K mol)

5.) Now you need to look up $v_r$ and $T_r$ in a compressibility chart to get the value for a specific $Z$ factor, where their lines intersect.

6.) And finally calculate the pressure: $$P = ZRT / V$$

This will be more accurate than if you would use the standard way for ideal gases, especially for low temperatures

References:
real gases

calculation

compressibility chart examples


How does one determine the relative basicity of a nitrogen in a molecule? I'm trying to solve this problem:

Specifically, how does one know which nitrogen is more basic? I'm assuming that for the first molecule, that it's the nitrogen without the proton, since it seems like the other nitrogen has already been protonated (and already has acted as a base.)

Is my reasoning correct? How does one solve the other molecules?

Use a combination of

the exclusion principle: sort out the unlikely centres with low electron density and
paper work to figure out those nitrogen atoms for which the additional positive charge can be distributed over a larger part of the molecule without moving the proton.

In the case of caffeine (1) the amide and the imide are rather unlikely centres for protonation! You have probably already ruled out the amide nitrogen in the benzamide case (your second molecule) in favour of the aniline site.

From an initial number of four possible centres in 1, only two remain.

Now, let's see whether the positive charge resulting from protonation can be stabilized.

This looks fair, but creates a quarternary ammonium ion for which no obvious charge distribution seems to be available.

Let's examine the last possibility:

Would you agree that this looks much more promising?


Does the Battery Bounce Test actually work? I heard that a good way to test if a battery is dead or alive is to see if it bounces. Supposedly, dead batteries bounce higher than living ones. Can someone tell me if this is a legitimate claim, and if it is, an explanation would be nice.

Yes.

The battery bounce test kind of works. You can tell whether the battery still has a charge greater than 90%, but bounce hight levels off at about 50% and doesn't go higher. This means that it is not clear whether the battery is completely discharged when you observe a bounce.

Why?

Steingart et al. just recently published a paper on the subject, and through scientific rigor could confirm this claim. The reason they give has  to do with how an alkaline battery is built:

(Source)

As you can see, a large part of the inside is devoted to the zinc anode. During the discharge, a redox reaction takes place to produce the current:
$$ \ce{Zn + 2MnO2 -> ZnO + Mn2O3} $$
The zinc (usually in some gel form at first) is slowly oxidized. As a result of this oxidization, crosslinks between the small zinc granules are formed, which make the whole center more rigid. As such, its "bounce" (or coefficient of restitution) becomes greater, because the downward kinetic energy cannot be fully absorbed by the gel anymore.

If you are interested in more details, check out the article they published. If you don't have access you can read a summary given on the Princeton University News web page, where you also can see an animation of a series of discharged batteries bouncing to different heights.


Finding the number of cyclic isomers algorithmically Is there an algorithm for finding the number of cyclic isomers a given compound might have?

I've found a general algorithm for all isomers: link, but I'm not sure how to apply it for cyclic isomers only.

If I remember correctly, several years ago I stumbled over the SMOG program addressing this question (description here).  The software requestested sum formula and a selection of allowed structural elements, cyclic substructures were among them.  Maybe the algorithms implemented are not the first ones to answer such a question, yet publications like this showcase continuing interest in this field, too.


However what is so acidic about CaO and basic(no pun intended) about SiO2 while calculating basicity of slag? I have seen Basicity to be calculated as $\mathrm{B} = wt\%\:\ce{CaO}/wt\%\:\ce{SiO2}$ particularly in slag bascity/acidity calcuations. Now I do not think that $\ce{CaO}$ is the most basic oxide that we have or $\ce{SiO2}$ is the most acidic oxide then why are they calculated accordingly? On a lighter note in my geology classes, I have been taught to classify minerals as acidic on the basis of Silica content. Are they related? Please explain with close reference to context.

What is so acidic...and basic...?

Nothing.

This derives from the historical perception that silica in geological systems and in melts was in the form of silicic acid ($\ce{H2SiO4}$) and the alkali and alkali earth elements were considered as bases.

We now know that in high temperature silicate liquids (that eventually solidifies into slag, or rocks, or glass) there is no acid–base chemistry, at least in the form that we know it from low-temperature aqueous acid–base chemistry. However, the usage of the terms acid and base in relation to rocks and slags still persists, primarily with older professors (who still teach) or in certain parts of the world, such as Russia. Nowadays geologists usually prefer the term "felsic" over the term "acidic" for silica-rich rocks.

Nonetheless, the measure of $\ce{CaO}$ and $\ce{SiO2}$ contents in slags, glasses and rocks is still useful for various reasons. Therefore, the ratio is still calculated and used, and the old name persists. A slightly better parameter, called optical basicity, takes into account the compositional variability other than $\ce{CaO}$ and $\ce{SiO2}$, by adding other oxides such as $\ce{Na2O}$ and $\ce{K2O}$ into the calculation. It is widely used and has many applications.


Aluminum powder reaction with air I've made aluminum powder in a ball mill a few times, but my last batch began to smoke and caught on fire soon after I poured it from the tumbler. Nothing unusual was afoot. It was poured into a paper plate. The ball mill aggregate had steel bearings and a few rounded rocks, same as I had used before.
The only difference was I didn't place any charcoal along with the aluminum during the milling? So why would there be spontaneous combustion?

It rapidly oxidized. If the ball mill is sealed for long periods of time the oxygen inside the ball mill is consumed coverting aluminum to aluminum oxide. When the lid is opened the aluminum is rapidly oxidized with a exothermic reaction. If the temperature gets hot enough it can spontaneously ignite.
The reason for the carbon is it coats the aluminum particles to prevent them from oxidizing.


Are the lone pairs in water equivalent? I've read that the oxygen atom in water is $\mathrm{sp^2}$ hybridized, such that one of the oxygen lone pairs should be in an $\mathrm{sp^2}$ orbital and the other should be in a pure p atomic orbital.

First, am I correct about the lone pairs being non-equivalent?

Second, if so, does this have any significance in actual physical systems (i.e. is it a measurable phenomenon), and what is the approximate energy difference between the pairs of electrons?

Lastly, if it turns out the lone pairs are actually inequivalent, can this be reconciled with the traditional explanation (due to VSEPR theory) that oxygen is $\mathrm{sp^3}$ and the lone pairs are equivalent?

Water, as simple as it might appear, has quite a few extraordinary things to offer. Most does not seem to be as it appears.

Before diving deeper, a few cautionary words about hybridisation. Hybridisation is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will result from various factors, such as steric and electronic interactions, and furthermore interactions with the surroundings like a solvent or external field. The geometric arrangement will not be formed because a molecule is hybridised in a certain way, it is the other way around, i.e. a result of the geometry or more precise and interpretation of the wave function for the given molecular arrangement.

In molecular orbital theory linear combinations of all available (atomic) orbitals will form molecular orbitals (MO). These are spread over the whole molecule, or delocalised, and in a quantum chemical interpretation they are called canonical orbitals. Such a solution (approximation) of the wave function can be unitary transformed form localised molecular orbitals (LMO). The solution (the energy) does not change due to this transformation. These can then be used to interpret a bonding situation in a simpler theory.

Each LMO can be expressed as a linear combination of the atomic orbitals, hence it is possible to determine the coefficients of the atomic orbitals and describe these also as hybrid orbitals. It is absolutely wrong to assume that there are only three types of spx hybrid orbitals.
Therefore it is very well possible, that there are multiple different types of orbitals involved in bonding for a certain atom. For more on this, read about Bent's rule on the network.[1]

Let's look at water, Wikipedia is so kind to provide us with a schematic drawing:

The bonding angle is quite close to the ideal tetrahedral angle, so one would assume, that the involved orbitals are sp3 hybridised. There is also a connection between bond angle and hybridisation, called Coulson's theorem, which lets you approximate hybridisation.[2] In this case the orbitals involved in the bonds would be sp4 hybridised. (Close enough.)

Let us also consider the symmetry of the molecule. The point group of water is C2v. Because there are mirror planes, in the canonical bonding picture π-type orbitals[3] are necessary. We have an orbital with appropriate symmetry, which is the p-orbital sticking out of the bonding plane. This interpretation is not only valid it is one that comes as the solution of the Schrödinger equation.[4] That leaves for the other orbital a hybridisation of sp(2/3).
If we make the reasonable assumption, that the oxygen hydrogen bonds are sp3 hybridised, and the out-of-plane lone pair is a p orbital, then the maths is a bit easier and the in-plane lone pair is sp hybridised.[5]

A calculation on the MO6/def2-QZVPP level of theory gives us the following canonical molecular orbitals:

(Orbital symmetries: $2\mathrm{A}_1$, $1\mathrm{B}_2$, $3\mathrm{A}_1$, $1\mathrm{B}_1$)[6,7]

Since the interpretation with hybrid orbitals is equivalent, I used the natural bond orbital theory to interpret the results. This method transforms the canonical orbitals into localised orbitals for easier interpretation.

Here is an excerpt of the output (core orbital and polarisation functions omitted) giving us the calculated hybridisations:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   2. (1.99797) LP ( 1) O  1            s( 53.05%)p 0.88( 46.76%)d 0.00(  0.19%)
   3. (1.99770) LP ( 2) O  1            s(  0.00%)p 1.00( 99.69%)d 0.00(  0.28%)
   4. (1.99953) BD ( 1) O  1- H  2
               ( 73.49%)   0.8573* O  1 s( 23.41%)p 3.26( 76.25%)d 0.01(  0.31%)
               ( 26.51%)   0.5149* H  2 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
   5. (1.99955) BD ( 1) O  1- H  3
               ( 73.48%)   0.8572* O  1 s( 23.41%)p 3.26( 76.27%)d 0.01(  0.30%)
               ( 26.52%)   0.5150* H  3 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
 -------------------------------------------------------------------------------

As we can see, that pretty much matches the assumption of sp3 oxygen hydrogen bonds, a p lone pair, and a sp lone pair.

Does that mean that the lone pairs are non-equivalent?
Well, that is at least one interpretation. And we only deduced all that from a gas phase point of view. When we go towards condensed phase, things will certainly change. Hydrogen bonds will break the symmetry, dynamics will play an important role and in the end, both will probably behave quite similarly or even identical.

Now let's get to the juicy part:

  Second, if so, does this have any significance in actual physical systems (i.e. is it a measurable phenomenon), and what is the approximate energy difference between the pairs of electrons?

Well the first part is a bit tricky to answer, because that is dependent on a lot more conditions. But the part in parentheses is easy. It is measurable with photoelectron spectroscopy. There is a nice orbital scheme correlated to the orbital ionisation potential on the homepage of Michael K. Denk for water.[8] Unfortunately I cannot find license information, or a reference to reproduce, hence I am hesitant to post it here.

However, I found a nice little publication on the photoelectron spectroscopy of water in the bonding region.[9] I'll quote some relevant data from the article.

  $\ce{H2O}$ is a non-linear, triatomic molecule consisting of an oxygen atom  covalently bonded to two hydrogen atoms. The ground state of the $\ce{H2O}$ molecule is classified as belonging to the $C_\mathrm{2v}$ point group and so the electronic states of water are described using the irreducible representations $\mathrm{A}_1$, $\mathrm{A}_2$, $\mathrm{B}_1$, $\mathrm{B}_2$. The electronic configuration of the ground state of the $\ce{H2O}$ molecule is described by five doubly occupied molecular orbitals:
  $$\begin{align}
\underbrace{(1\mathrm{a}_1)^2}_{\text{core}}&&
\underbrace{(2\mathrm{a}_1)^2}_{\text{inner-valence orbital}}&&
\underbrace{
  (1\mathrm{b}_2)^2 (3\mathrm{a}_1)^2 (1\mathrm{b}_1)^2
       }_{\text{outer-valence orbital}}&&
\mathrm{X~^1A_1}
\end{align}$$

  [..]

  In addition to the three band systems observed in HeI PES of $\ce{H2O}$, a fourth band system in the TPE spectrum close to 32 eV is also observed. As indicated in Fig. 1, these band systems correspond to the removal of a valence electron from each of the molecular orbitals $(1\mathrm{b}_1)^{-1}$, $(3\mathrm{a}_1)^{-1}$, $(1\mathrm{b}_2)^{-1}$ and $(2\mathrm{a}_1)^{-1}$ of $\ce{H2O}$.

As you can see, it fits quite nicely with the calculated data. From the image I would say that the difference between $(1\mathrm{b}_1)^{-1}$ and $(3\mathrm{a}_1)^{-1}$ is about 1-2 eV.

TL;DR
As you see your hunch paid off quite well. Photoelectron spectroscopy of water in the gas phase confirms that the lone pairs are non-equivalent. Conclusions for condensed phases might be different, but that is a story for another day.

Notes and References

What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

Formal theory of Bent's rule, derivation of Coulson's theorem (Wikipedia).
Worked example for cyclo propane by ron.

A π orbital has one nodal plane collinear with the bonding axis, it is asymmetric with respect to this plane. A bit more explanation in my question What would follow in the series sigma, pi and delta bonds?
With in the approximation that molecular orbitals are a linear combination of atomic orbitals (MO = LCAO).
The terminology we use for hybridisation actually is just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$
In theory $x$ can have any value; since it is just a unitary transformation the representation does not change, hence
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 4\times\mathrm{sp}^3 \\
   &\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \\
   &\leadsto \text{etc. pp.}\\
   &\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)}
\end{align}
There are virtually infinite possibilities of combination.
This and the next footnote address a couple of points that were raised in a comment by DavePhD. While I already extensively answered that there, I want to include a few more clarifying points here. (If I do it right, the comments become obsolete.)

  What is the reason for concluding 2 lone pairs versus 1 or 3? For example Mulliken has in table V the b1 orbital being a definite lone pair (no H population) but the two a1 orbitals both have about 0.3e population on H. Would it be wrong to say only one of the PES energy levels corresponds to a lone pair, and the other 3 has some significant population on hydrogen? Are Mulliken's calculations still valid? – DavePhD

The article Dave refers to is R. S. Mulliken, J. Chem. Phys. 1955, 23, 1833., which introduces Mulliken population analysis. In this paper Mulliken analyses wave functions on the SCF-LCAO-MO level of theory. This is essentially Hartree Fock with a minimal basis set. (I will address this in the next footnote.) We have to understand that this was state-of-the-art computational chemistry back then. What we take for granted nowadays, calculating the same thing in a few seconds, was revolutionary back then. Today we have a lot fancier methods. I used density functional theory with a very large basis set. The main difference between these approaches is that the level I use recovers a lot more of electron correlation than the method of Mulliken. However, if you look closely at the results it is quite impressive how well these early approximations perform.
On the M06/def2-QZVPP level of theory the geometry of the molecule is optimised to have an oxygen hydrogen distance of 95.61 pm and a bond angle of 105.003°. This is quite close to the experimental results.
The contribution to the orbitals are given as follows. I include the orbital energies (OE), too. The contributions of the atomic orbitals are given to 1.00 being the total for each molecular orbital. Because the basis set has polarisation functions the missing parts are attributed to this. The threshold for printing is 3%. (I also rearranged the Gaussian Output for better readability.)

Atomic contributions to molecular orbitals:
2:  2A1  OE=-1.039   is   O1-s=0.81 O1-p=0.03   H2-s=0.07   H3-s=0.07
3:  1B2  OE=-0.547   is   O1-p=0.63             H2-s=0.18   H3-s=0.18
4:  3A1  OE=-0.406   is   O1-s=0.12 O1-p=0.74   H2-s=0.06   H3-s=0.06
5:  1B1  OE=-0.332   is   O1-p=0.95

We can see that there is indeed some contribution by the hydrogens to the in-plane lone pair of oxygen. On the other hand we see that there is only one orbital where there is a large contribution by hydrogen. One could here easily come up with the theory of one or three lone pairs of oxygen, depending on your own point of view. Mulliken's analysis is based on the canonical orbitals, which are delocalised, so we will never have a pure lone pair orbital. When we refer to orbitals as being of a certain type, then we imply that this is the largest contribution. Often we also use visual aides like pictures of these orbitals to decide if they are of bonding or anti-bonding nature, or if their contribution is on the bonding axis.
All these analyses are highly biased by your point of view. There is no right or wrong when it comes to separation schemes. There is no hard evidence for any of these obtainable. These are mathematical interpretations that do in the best case help us understand bonding better. Thus deciding whether water has one, two or three (or even four) lone pairs is somewhat playing with numbers until something seems to fit. Bonding is too difficult to transform it in easy pictures. (That's why I am not an advocate for cautiously using Lewis structures.)
The NBO analysis is another separation scheme. One that aims to transform the obtained canonical orbitals into a Lewis like picture for a better understanding. This transformation does not change the wave function and in this way is as equally a representation as other approaches. What you loose by this approach are the orbital energies, since you break the symmetry of the wave function, but this is going much too far to explain. In a nutshell, the localisation scheme aims to transform the delocalised orbitals into orbitals that correspond to bonds.
From a quite general point of view, Mulliken's calculations (he actually only interpreted the results of others) and conclusion hold up to a certain point. Nowadays we know that his population analysis has severe problems, but within the minimal basis they still produce justifiable results. The popularity of this method comes mainly because it is very easy to perform. See also: Which one, Mulliken charge distribution and NBO, is more reliable?
Mulliken used a SCF-LCAO-MO calculation by Ellison and Shull and was so kind to include the main results into his paper. The oxygen hydrogen bond distance is 95.8 pm and the bond angle is 105°. I performed a calculation on the same geometry on the HF/STO-3G level of theory for comparison. It obviously does not match perfectly, but well enough for a little bit of further discussion.

NO  SYM    HF/STO-3G : N(O)     N(H2)      |      Mulliken : N(O)     N(H2)
1   1A1    -550.79     2.0014   -0.0014    |      -557.3     2.0007   -0.0005
2   2A1     -34.49     1.6113    0.3887    |       -36.2     1.688     0.309
3   1B2     -16.82     1.0700    0.9300    |       -18.6     0.918     1.080
4   3A1     -12.29     1.6837    0.3163    |       -13.2     1.743     0.257
5   1B1     -10.63     2.0000    0.0000    |       -11.8     2.000

As an off-side note: I completely was unable to read the Mulliken analysis by Gaussian. I used MultiWFN instead. It is also not an equivalent approach because they expressed the hydrogen atoms with group orbitals.
The results don't differ by much. The basic approach of Mulliken is to split the overlap population to the orbitals symmetric between the elements. That is a principal problem of the method as the contributions to that MO can be quite different. Resulting problematic points are occupation values larger than two or smaller than zero, which have clearly no physical meaning. The analysis is especially ruined for diffuse functions.
At the time Mulliken certainly did not know about anything we are able to do today, and under which conditions his approach will break down, it still is funny to read such sentences today.

  Actually, very small negative values occasionally occur [...]. [...] ideally to the population of the AO [...] should never exceed the number 2.00 of electrons in a closed atomic sub-shell. Actually, [the orbital population] in some instances does very slightly exceed 2.00 [...]. The reason why these slight but only slight imperfections exist is obscure. But since they are only slight, it appears that the gross atomic populations calculated using Eq. (6') may be taken as representing rather accurately the "true" populations in various AOs for an atom in a molecule. It should be realized, of course, that fundamentally there is no such thing as an atom in a molecule except in an approximate sense.

For much more on this I found an explanation of the Gaussian output along with the reference to F. Martin, H. Zipse, J. Comp. Chem. 2005, 26, 97 - 105, available as a copy. I have not read it though.
Scroll down until the bottom of the page for the image, read for more information: CHEM 2070, Michael K. Denk: UV-Vis & PES. (University of Guelph) If dead: Wayback Machine
S.Y. Truong, A.J. Yencha, A.M. Juarez, S.J. Cavanagh, P. Bolognesi, G.C. King, Chemical Physics 2009, 355 (2–3), 183-193. Or try this mirror.


Is peroxoborate and perborate the same thing? Recently I came across the following question:-

  Preparation of Sodium Peroxoborate- Equations

The term peroxoborate came from the fact that it has the peroxo bond $\ce{-O-O}-$. So, it should be strictly called peroxoborate. But some books says  peroxoborate while other books say perborate. Is it true?

I googled sodium perborate.  Wikipedia recognizes the compound as sodium perborate. Then, I googled sodium peroxoborate, the following message appear:-

So, is peroxoborate and perborate the same thing? If so, why some book says  peroxoborate while othe books perborate? It should be strictly called peroxoborate since it contains the peroxo bond $\ce{-O-O}-$.

Update

This is what written in Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

  Sodium perborate is used as a bleach  in laundry powder, automatic dishwasher powders, and whitening toothpastes. Although the formula is given as $\ce{NaBO3.H2O}$ or $\ce{NaBO3.4H2O}$, the compound contains th peroxide anion, $\ce{O2^{2-}}$ and is more accurately given by $\ce{Na2[B2(O2)2(OH)4].6H2O}$.

But this is what written in Lee, J.D. Inorganic chemistry

  $$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[B2(O2)2(OH)4].6H2O }$$
  Sodium peroxoborate is used as a brightener in washing powder. It is compatible with enzymes which are added to some biological powders.

So, what is the correct name:- sodium perborate or sodium peroxoborate?

This is necessarily  not  a chemistry answer but is a continuation of aventurin's answer.

Google ngram viewer is an online search engine h  that charts frequencies of any word or short sentence found in the sources printed between 1800 and 2012 in American English, British English etc.

So, here is the Google ngram for peroxoborate and perborate.

So, you can see that perborate is more often used than peroxoborate. Though, the correct name (IUPAC name) is tetrahydroxidodi-µ-peroxido-diborate, people often mention the compound sodium perborate. Sodium peroxoborate seems to be obsolete.


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

Lets take a case:

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

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

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

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

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

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

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

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

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

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

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

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


Semi-sandwich complex IUPAC name I've tried looking here, but I can't figure it out.

As a start, what is just the cobalt and cyclopentadienyl anion compound called.
How does that name get added to the trimethylphosphine and carbonyl groups.

I want to ask further questions about this compound on chemistry stack exchange, and I think it would be a good idea to be able to use the name in those questions.

The name of this compound is carbonyl(η5-cyclopentadienyl)(trimethylphosphane)cobalt. Note that, generally, the prefixes denoting the organic groups and any other ligands are placed in alphabetical order before the name of the metal.
The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) presents the rules for formulating and naming coordination compounds (Chapter 9) and organometallic compounds (Chapter 10). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
In particular, the ‘η5-cyclopentadienyl’ term used here is explained in Section IR-10.2.5.1. as follows:

The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons has led to the development of the ‘hapto’ nomenclature to designate unambiguously the unique bonding modes of the compounds so formed. (…) The Greek symbol η (eta) provides a topological description by indicating the connectivity between the ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc. The symbol η is added as a prefix to the ligand name, or to that portion of the ligand name most appropriate to indicate the connectivity, (…)
The ligand name η5-cyclopentadienyl, although strictly speaking ambiguous, is acceptable as a short form of η5-cyclopenta-2,4-dien-1-yl, due to common usage.
These ligand names are enclosed in parentheses in the full name of a complex. Note the importance of making rigorous use of enclosing marks, etc. to distinguish the above bonding modes from the other four cases below. (…)


Why potential energy is neglected in this problem for calculating the energy difference between two stationary Bohr Orbits? Question
An electron in a hydrogen like atom makes a transition from a state in which its de-Broglie wavelength is $\lambda_1$ to a state in which its de-Broglie wavelength is $\lambda_2$, then wavelength of photon generated due to this transition will be ___________.

The solution for this problem as given in my book is as follows:

In the first step, it is given that the energy difference is just equal to the kinetic energy difference. How is this possible? Why have they neglected the potential energy in calculating the energy difference since the two stationary orbits in Bohr's model have different potential as well as kinetic energies.
Please clarify my doubt.


Can someone explain the unintuitive ordering in the metallic activity series? I would understand it if the metallic activity series Li, Cs, Rb, ... Pt, Au corresponded to electronegativity or ionization energy. However, the way it is written, lithium will displace caesium, while caesium will displace rubidium (it goes from the top of family 1 to the bottom and then up again).

This order seems so random. Is there any order to it or is it just determined experimentally through trial and error? In addition, wouldn't different valences on the metals affect their placement in the series?

The metals in the activity series are ordered by their oxidation potentials; there is nothing random about it. The displacement reactions, though, are but one of the applications of these potentials, and not the most important one at that. Moreover, with active metals it does not make much sense anyway. The oxidation potentials are defined for water solutions; their significance ultimately stems from the fact that we live in a world dominated by water. Well, you can't react lithium or rubidium with anything in water, because they would rather react with water itself. And if you try a reaction in other media than water, the oxidation potentials do not apply and the outcome can't be predicted that easily. For example, magnesium would reduce sodium from its oxide, despite being nominally "less active".

Indeed, the very concept that some things are "more active" than others is just an approximation; admittedly, a pretty good one when we talk about metals, but still an approximation.


Addition reaction of hydroiodic acid to 2-bromo-3-chloro-2-butene How can the major product of the addition reaction of 2-bromo-3-chloro-2-butene with hydroiodic acid be predicted, since Markovnikov's rule fails to distinguish it?

$$\ce{H3C-C(Cl)=C(Br)-CH3~+~HI~->~?}$$

TL;DR - Sometimes you need to do the experiment.

Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:

  At C2 (Bromo)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$

The two issues to consider are resonance and induction.

Resonance

You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:

$$\ce{R2C+-Br <-> R2C=Br+}$$

However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.

Induction

The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.

  At C3 (chloro)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-C+(Cl)-CH(Br)-CH3}$$

We once again examine resonance and induction.

Resonance

$$\ce{R2C+-Cl <-> R2C=Cl+}$$

Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.

Induction

The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.

Compare:

                  Chlorine              Bromine
Resonance         Stronger donor        Weaker donor
Induction         Stronger withdrawer   Weaker withdrawer

In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.

Now... what else might happen.

Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a halonium ion. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better).

This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.

So in effect, you can get three products:

$$\ce{CH3-CHCl-CIBr-CH3}$$
$$\ce{CH3-CICl-CHBr-CH3}$$
$$\ce{CH3-CClBr-CHI-CH3}$$

You would then want to actually do the experiment to determine which of these is really the major product.


Absorption HOMO-LUMO, canonical? So I already asked yesterday about canonical and localized orbitals together with Koopman's theorem where we came to the conclusion that for for example ionization potentials localized orbitals cannot be used. So I assume a ionization would be from the HOMO which leaves me with the question, whenever we say an electron is excited from the HOMO into the LUMO of a system, are we talking about the canonical forms then? Because before I knew about those two I always imagined that there is a local position somewhere in the molecule like a double bond where one electron is exited into an antibonding MO creating a single bond or 1.5 bond and that you can really tell from where to where the transition is. But canonical orbitals do not really look like we imagine orbitals they are distributed over the whole molecule.

Short answer is yes, HOMO and LUMO refer to canonical orbitals, since only those have at least some physical meaning.

Two more remarks about that:

Koopmans' Theorem does only work if you add or remove one electron.

If you change two or more electrons of your system (e.g. exciting one electron is first removing one, then adding one in another orbital) you get additional terms in energy difference between both states. Those terms are coming from the electron-electron interaction of the two changed orbitals/electrons.

In case of changing only one orbital, those additional terms do not appear since there is no electron-electron self-interaction.

Furthermore, the approximation of Koopmans' Theorem, that orbitals do not change (do not relax, remain stationary) on removing/adding electrons, gets worse the more electrons change.

Orbitals itself are only an approximation. Factorizing the total electronic wave function into orbitals is mathematically only possible for non-interaction electrons, or like in Hartree-Fock if you assume some mean-field for the electron-electron interaction.
The resulting canonical orbitals do not always match the simplified picture based on our chemical intuition.


how to increase Keq(equilibrium) with temperatrue $$\ce{C_{(s)} + H_2O_{(g)} <=> CO_{(g)} + H_{2(g)}}$$

$$\Delta H = +131\ \mathrm{kJ}$$

If the reaction is in a state of equilibrium under standard conditions, what can be done to increase the value of $K_\text{eq}$?

My question is would you have to increase or decrease the temperature to increase $K_\text{eq}$?

I know the basic principle that you have to decrease the temperature to increase $K_\text{eq}$, but in my notes that was for $\ce{A +B<=>C}$ and the reaction above is not in that form, Instead it has two products. Would this still apply?

You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation:
$$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$

Rearranging the equation a little bit, we get that:
$$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$

Now we can integrate each side of this equation:
 $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$
Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to :
$$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$
Solving the integral we will get this :
$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

For $\Delta H^\circ > 0$ we have two cases: when $T_2>T_1$ and $T_2<T_1$.
If $T_2>T_1$ then $\frac{1}{T_2}<\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} < 0$ . Knowing that $\Delta H^\circ>0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive:
 $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)>0$$
And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)>0 \Leftrightarrow \frac{K_2}{K_1} > 1 \Leftrightarrow K_2 > K_1$$
i.e. The equilibrium constant increases as the temperature is increased.
Using the same principle for $T_2 < T_1$, we get that $K_2 < K_1$


When removing a reactant and equilibrium shifts left, doesn't this make an excess of the 2nd reactant, forcing equilibrium back to the right? In the reaction:$$\ce{Ca^{2+}(aq) + ^-OOC-COO^- <=> Ca(OOC-COO)(s)}$$

If $\ce{Ca^{2+}(aq)}$ is removed, the reaction shifts to the left to produce more $\ce{Ca^{2+}(aq)}$ to compensate.

But when the reaction shifts to the left, isn't a side effect that more $\ce{^-OOC-COO^-}$ is also produced (as when $\ce{Ca(OOC-COO)(s)}$ dissociates it produces BOTH $\ce{Ca^{2+}(aq)}$ and $\ce{ ^-OOC-COO^- }$, and not $\ce{Ca^{2+}(aq)}$ by itself?

So if there is more $\ce{^-OOC-COO^-}$, won't the reaction want to shift to the right?

So wouldn't these two (i.e. wanting to shift left and shift right) counteract and the reaction not shift anywhere? I think there is something about Le Chateliers's that I have misunderstood, so please point it out if you see it.

The thing to be careful about with Le Chatelier's principle is that it relates to equilibria :

  Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium.

Le Châtelier's Principle

Now when you say in the above "But when the reaction shifts to the left, isn't a side effect  ..." note you can't use Le Chatelier's principle here again so fast because the system has not reached an equilibrium yet, it is just moving towards one.

It is likely that the next thing you will study is how to actually calculate the new equilibrium conditions. This gives usually a deeper understanding of how changes in various conditions make the reaction move forward/backward establishing new equilibrium conditions.


Why can a diamond be broken using a hammer if it's the hardest natural substance known? I've heard that diamond is the hardest natural known material but, on Google search, I found that it can easily be broken by a hammer as it's not tough.

So, what is difference between hardness and toughness? According to me, both should be the same, and if both are different than what is the toughest known substance?

Hardness and toughness are not the same

Hardness and toughness are very different qualities in materials and are weakly related.

Hardness is strongly related to the more well-defined quantity of stiffness which measures how easily a compound can be deformed under stress. Glass and diamond are very stiff materials, for example. If you try to poke them with something they resist deforming to accommodate your poke. (Hardness is not perfectly aligned to stiffness because of small scale microstructures in many materials, but this is good enough for now).

Toughness is a vaguer term for materials and there isn't a simple way to measure it. This is partially because it varies depending on circumstances in a way that stiffness does not and it is a property of the overall structure and not just the materials that make up the structure. Understanding it requires some insight into why things break (and why other things don't). we need to know a lot about the small scale structure of materials not just the substances involved. For example, stainless steel (used in knives and forks) is a well-known tough material but cast iron is brittle. Both are mostly made of iron. The differences are in the crystalline structures.

One key property of tough structures is that cracks don't propagate. So a stick of glass will break easily as will a glass fibre. But a bundle of glass fibres embedded in an epoxy resin can be very tough (because the cracks in individual glass fibres are not propagated through the epoxy resin). Some tough metals can adjust the micro defects in their crystalline structures to absorb the strain that would otherwise propagate cracks. Some very soft compounds are very tough because the deform so easily that it is hard to start cracks, nylon rope for example.

Diamond is very very stiff. But it has no protection mechanism against cracks. So, like glass, once a crack has started it doesn't take a lot of energy to cause it to spread, so it may be stiff but it isn't tough.

A fuller explanation of this took material scientists and engineers a long time to work out would require a whole book to do it justice. Luckily that book has been written and is called The New Science of Strong Materials. It is well worth a read.


What is the chemistry behind amalgamation? Mercury can form amalgams with metals, such as gold, however how exactly does the "reaction" if you can call it that take place?

My understanding of metal bonds is that the atoms of a metal are surrounded by a "sea" of electrons, however I don't understand how mercury can "penetrate" metals and weaken their structure or "dissolve" them.

How does it do this? Also does this apply to Gallium?

An amalgam is simply an alloy of some metal with mercury.

Alloys are simply solid solutions with metals. Since the mixture of multiple elements disrupts the metal bonding, the properties are often quite different than the components. (In some cases, alloys form so-called intermetallic compounds with defined stoichiometry, but more often, the ratios are very flexible.)

I wouldn't describe the mixture as a reaction, in much the same way that I wouldn't really call dissolving sodium chloride in water as a reaction. You're forming a mixture. Water disrupts the ionic bonding in NaCl, and mercury and other metals can also accept other components or impurities.

Gallium does alloy with a lot of metals, and indium-gallium eutectic alloy is a liquid at room temperature, as is "galinstan".

As to your question about mercury dissolving the metal atoms, consider that mercury is also a metal, so it also participates in similar metal bonding.


Is enthalpy of activation the same as activation energy? Is there a purely mathematical way to find that value? I asked about the effect of pressure on a reaction, and since then I learned a lot more. One of the things I learned was the Eyring equation:

$$\kappa = \frac{\kappa_B T}{h} e^\frac{\Delta^\ddagger S°}{R} e^\frac{- \Delta^\ddagger H°}{R}$$

where:

$\begin{multline}
\begin{split}
\kappa &= \text{reaction rate} \\
\kappa_B &= \text{Boltzmann constant} \\
T &= \text{temperature} \\
h &= \text{Planck's constant} \\
R &= \text{gas constant} \\
\Delta^\ddagger S° &= \text{entropy of activation} \\
\Delta^\ddagger H° &= \text{enthalpy of activation} \\
\end{split}
\end{multline}$

My questions:

Is the enthalpy of activation the same thing as the activation
energy?
Is there a purely mathematical way to determine both $\Delta^\ddagger S°$ and $\Delta^\ddagger H°$? (without having a laboratory to use to observe the reaction)

(1) Enthalpy of activation is a component of activation energy. The other component is temperature * entropy of activation. Enthropy controls "heat difference" while entropy controls "measure of disorder difference".

$\Delta$G = $\Delta$H - T*$\Delta$S. This applies to activation energy too.

(2) DFT functionals rely on 4 parameters per atom make predictions. So, if you give 4*5=20 parameters, then you can in theory predict behaviour of any compound, build with only C, H, N, O, S atoms. This covers most of biochemistry, excluding nucleic acids (you would need additional 4 parameters to characterize P).

These 4 parameters are known for most of the atoms (excluding some elements, going after uranium). But those four parameters were experimentally measured by fitting behavior of modeled molecules to real molecules.

Should we count it as "purely theoretically predict?"

Also there are purely ab initio methods like Hartree-Fock with even less input required. You would still need some input. In math you can build many different plausible worlds. For example purely "newtonian" vs "newtonian + quantum mechanics" words. experiment is the only way to tell, which model is correct.


distilled water and copper(II)chloride In class we did a lab of adding blue crystals of copper(II)chloride into distilled water, the water turned blueish green,were the changes chemical or physical?

https://en.wikipedia.org/wiki/Copper%28II%29_chloride

  Aqueous solution prepared from copper(II) chloride contain a range of copper(II) complexes depending on concentration, temperature, and the presence of additional chloride ions. These species include blue color of $\ce{[Cu(H2O)6]^2+}$ and yellow or red color of the halide complexes of the formula $\ce{[CuCl_{2+x}]^{x−}}$

On the Wikipedia page, you can see a photo of 3 solutions with different colour.


Relation between number of conjugated double bonds and absorbance? I observed the absorbance on Phenolphthalein in acidic and basic medium, and I know that the absorbances differ due to the number of conjugated double bonds, but I am not sure exactly why this happens. I believe it has something to do with the delocalised electrons, which causes lower energy differences, but I am unable to establish a solid link between the concepts. Why does this happen?

When 2 p orbitals mix, a pi and a pi* orbital is formed, and the electrons occupy the lower energy pi orbital.  The UV-absorbance is the energy difference between these two orbitals such that on electron may be 'promoted' to the p* orbital.

When 3 p orbitals mix, three pi-type orbitals are formed.  One is obviously bonding and one is obviously antibonding.  There is a third one 'in the middle', frequently called non-bonding.  Because of this third orbital, there is a smaller difference between any two states, and as such absorbing a photon to promote an electron requires less energy, redshifting the absorbance.

Add a fourth p orbital and the energy differences get smaller, but with two bonding orbitals and two anti-bonding orbitals.  Add a 5th p orbital and now you have two bonding orbitals and two anti-bonding orbitals and a non-bonding orbital, with even smaller differences.  Repeat ad nauseum.

Smaller energy differences means longer wavelengths.

Check out this picture for a graphical summary:


Smell of the methanol When I first time smell the methanol, for me it was similar to the benzene (completely different from the ethanol). But almost everywhere information says, that it should me very similar to the ethanol. (I am talking about almost dry methanol, not dissolved in water or mixed with ethanol.)

I understand, that smelling chemicals is not healthy, but kills the dosage, not the poison. The only breath won't kill anybody. (Off-course, if you are working with methanol or other dangerous compounds you must wear gas-mask and/or work with good ventilation!)

How would you describe the smell of the methanol?

Its a qualitative answer, so you'll get a lot of odd things in these responses.

I find methanol to be like ethanol but shimmed by a plastic/wood-life taste.

More importantly, having a higher volatility, its taste will be more intense because there are more vapors to stimulate your odor receptors which frequently change or augment tastes.


Are dicarbonyl amides stable in water? Dicarbonyl amides, that is carboxylic acid anhydrides with their single-bonded oxygen between their carbonyl groups replaced with an amide group, I'm curious as to whether they're stable in water or unstable like carboxylic acid anhydrides.

This is what I mean by a dicarbonyl amide:

where R1 and R2 are organic substituents/hydrogen.

This functional group is an imide. Imides are stable to water and require significantly more effort to hydrolyze than anhydrides (from personal experience). The acyl nitrogen group is a better resonance donor and poorer leaving group than acyl oxygen groups.

From the wikipedia page:

  Being highly polar, imides exhibit good solubility in polar media. The N-H center for imides derived from ammonia is acidic and can participate in hydrogen bonding. Unlike the structurally related acid anhydrides, they resist hydrolysis and some can even be recrystallized from boiling water.

As an example, succinimide, a cyclic imide has a solubility in water of 1 g for every 3 mL - if succinimide reacted with water, such a high solubility would not be possible.

The related N-bromosuccinimide (show below) is also water stable despite that unfriendly looking $\ce{N-Br}$ bond. It is prepared from succinimide in water and an be recrystallized from near-boiling water.


Application of LeChatelier's principle for reactions with solid products I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.
Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.

I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.

Let's get back to the fundamentals of Le Chatelier's principle :

  When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract (partially) the effect of the applied change and a new equilibrium is established.

Let's assume you have a reaction $\ce{A(g) <=> 2B(g)}$. When lowering the pressure, the system will adjust itself to counteract this, i. e. increase the pressure. This can be achieved by displacing the equilibrium towards the right, as the direct reaction increases the amount of gas molecules.

In your case, however, there is no increase in the amount of gas compounds. Regarding the solid, as it does not participate to the overall pressure in the system (only gases do), it should not be taken into consideration here. Therefore, a change in pressure wouldn't alter the equilibrium state of your reaction.

That's for the math-less part.

You can also understand this concept using the equilibrium constant. For my example, it would be $K_c = \frac{[\ce{B}]^2}{[\ce{A}]}$. Lowering the pressure corresponds to an increase in volume, thus a decrease in both concentrations. This leads to a ratio that is different from the equilibrium constant, which will force the system to a new equilibrium.
$$K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{\left(\frac{n_\ce{B}}{V}\right)^2}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V}$$
When increasing $V$ to $V'$,
$$\frac{[\ce{B}]^2}{[\ce{A}]} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V'} < K_c$$
As this ratio is lower than $K_c$, the reaction will evolve towards the formation of products.

For your equation,
$$K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{\frac{n_\ce{B}}{V}}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}}{n_\ce{A}}$$
there is no dependance on the volume, thus none either on the pressure.
C is not included as it is a pure solid ($a(\ce{C}) = 1$).

  Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.

It is true that, if you find a way to remove products from the system, you can reach full conversion of your reagents. But that's actually the catch : it can be very hard to remove those products. Take as an example the Haber-Bosch process :
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
As it is a slow reaction, a high temperature is required to get the product in a reasonable time. How would you remove $\ce{NH3}$ from the system, leaving the reagents inside? We then use other ways of increasing the yield (pressure and temperature), but they do not allow a full conversion of the reagents.


What happens with ammonia ions in tap water after they are separated from chlorine? Tap water is increasingly treated with chloramine which is composition of chlorine and ammonia. Chlorine can be removed by catalytic activated carbon and ammonia ions are then left. What happens with them in the water? Do they bind with something or are they left in ionic form? How quickly they evaporate at room temperature?

Chloramines (mono-, di-, and tri-) are used as a disinfectant for drinking water treatment (the source water, not the tap water) - this process is called combined chlorine. Upon contact with pathogens, organics, etc., chloramine is reduced to chloride (not chlorine), ammonia, and hydroxide. However, this half reaction is for alkaline conditions. The pKa of ammonia is around 9.4, so around circumneutral pH, some ammonia will exist as ammonium, the protonated state.

So, will activated carbon (AC) remove the products? First, we need to clarify something about AC - it is not catalytic, it is a physical adsorption removal process. AC will remove chloride ions, but it will not remove ammonium. Ammonia is quite volatile and will move from the water to the gas phase if e.g., a pitcher of water is left at room temperature. If you wanted to remove ammonium as well you could use something called biologically activated carbon (BAC). One last note, chloride will not "bind" will AC, it will be adsorbed through electrostatic forces.


What makes alcohol a good conservation product? A common way of preserving food and organisms is immersion in alcohol. But what causes alcohol to preserve these items well?

Alcohol kills bacteria and fungus, put simply. It does this as its metabolism products cause damage directly to the DNA, and directly to the cell respiration process. It also happens to tend to dry it out.

We can drink alcohol safely because of our alcohol dehydrogenase enzyme, which breaks the toxic alcohol into useful byproducts, that we can use for energy.

Also, it's a matter of scale. Consider submerging yourself in alcohol for a period of time. Don't you think you'd probably die, too?

It's very important to note that alcohol strength matters in this case. It also depends strongly on type. For example, ethyl alcohol: we drink this kind. It's not as toxic as say, methyl alcohol, or isopropyl alcohol. Those last ones are used as preserving and disinfecting agents, respectively.


Why does Methyl Carbocation have p orbital For a carbocation to form, it needs to lose an electron. In the example below the leaving group is Br and the products are a methyl carbocation and a bromide ion.

What I am unsure about is why the carbocation needs an empty p orbital. For a carbon atom with 6 valence electrons, the molecular orbital diagram looks like the one below. So when an electron is lost in heterolysis the $\ce{2p}$ orbital will be empty and I assume that is why the p orbital is empty. However, why do you include the empty p orbital in the shape. Does it serve a purpose?

You might want to read up on hyperconjugation.

Basically, the R groups (if NOT a hydrogen) can donate some electron density into the carbocation’s empty p orbital, stabilising the positive charge and increasing the ion’s stability. This is the reason for Markovnikov’s Rule when it comes to Sn1 mechanisms; the tertiary carbocation is most stable, followed by a secondary one, while a primary/methyl carbocation hardly exists at all.


Aside from carbon, what other substances can be made "superhard"? So diamonds are an allotrope of carbon, that is formed when carbon is put under immense pressure, right? What other elements or compounds can you make into "diamonds", as in more durable, harder versions of themselves? What happens if salt or iron for example, is put under immense pressure, do they become harder too?

Interestingly, nobody addressed the reason why diamonds are hard in the first place. The pressure (and temperature) are not the reason why they're hard, only the reason why they are formed. The diamonds are hard because the carbon atoms are bonded together by sigma ($sp^3$) bonds, which are the strongest chemical bonds. Other materials exhibiting the same kind of bonding are potentially also very hard. But the small atomic number of carbon (Z=6) means that the bonds will be stronger than in higher atomic number atoms, such as silicon.

However, amorphous (non-crystalline, like glass) materials can also be extremely hard. There is even an example that is even stronger than diamond (although not naturally occurring). And this material is made of... carbon! Exactly like diamond. This material, dubbed Q-carbon, was discovered in 2015, and is even created at atmospheric pressure, and around 4000 K with the help of a laser pulse. And it is estimated to be 10-20 % harder than diamond.

Bottomline: gigantic pressures are not required to create extremely hard materials, and diamonds is no longer the strongest known material.


How would you name this organic compound? Because there are at most seven consecutive carbon atom, and four alkyl groups attached to the chain of carbon, I would name it 2,4-tetramethyl heptane.

Yes, it's tetramethyl heptane, where are the 4 methyls? 2,4,4,6-tetramethylheptane.