Calculating the Ionic strength of sodium acetylsalicylate With the addition of aqueous NaOH to an aspirin solutions in water the reaction will yield products of sodium acetylsalicylate and water. I would like to measure the amount of sodium apsirin formed that is disolved in water right after the NaOH reaction.

Attempt for the solution:
Given that sodium acetylsalicylate dissociates in water to form sodium ions and salicylate ions, I can consider finding the ionic strength i.e. ion concentration. However in order to calculate the ionic strength I need the molality which require the amount of solutes disolved, which is basically what I am trying to find in the first place.

Is there a recorded value for the Ionic strength of of sodium acetylsalicylate, which would intern allow me to calculate the mass of the solute or I'm going the wrong way about this?

Is there any other way I can use to algebraic calculate the mass of the sodium acetylsalicylate dissolved?

Any help will be greatly appreciated!

Aspirin has a solubility of ~3 g/L. In addition from a patent by Galat [1]:

  Sodium aspirin is almost 1000 times as soluble in water as aspirin itself.

All of the $\ce{NaOH}$  can be assumed to react with acetylsalicylic acid (aspirin) to form the sodium salt.  This means the amount of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$ (up to the equivalence point).

When computing the ionic strength you can therefore assume that the concentration of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$.

References

Galat, A. Stable Sodium Acetylsalicylate and Method for Its Manufacture. US3985792A, October 12, 1976. (PDF)


Initial guess for Unrestricted Hartree-Fock calculation I am implementing an Unrestricted Hartree-Fock (UHF) calculation, following Ref. [1]. It works fine for a odd number of electrons ($N_\alpha\neq N_\beta$) since, despite the same initial guess for $\alpha$ and $\beta$ spins, the density matrices at the second step will be different and I get different orbital energies for different spin orbitals. The problem arises when I have an even number of electrons ($N_\alpha=N_\beta$) . Authors of Ref. [1] state that (p. 216)

  When $N_\alpha=N_\beta$, a restricted solution of Roothaan equations is a solution to the unrestricted Pople-Nesbet equations. This restricted solution always exists and necessarily results if an initial guess $P_\alpha=P_\beta$ is used. [...] In seeking this second solution [the unrestricted one], it is imperative that an initial guess $P_\alpha \neq P_\beta$ be used or the equations will necessarily yield the restricted solution.

Unfortunately there is no clue on which initial guess should be taken. In fact in this book, for the restricted case, the Fock matrix is initially approximated by the core Hamiltonian (Guess=Core in Gaussian09), which consists in taking $P_\alpha=P_\beta=0$.

Since the core Hamiltonian guess cannot be differentiated for $\alpha$ and $\beta$ spins, I implemented Huckel guess (Guess=Huckel in Gaussian09) following Ref. [2] which consists in approximating the initial Fock matrix with the following:
$$
F_{ij} = KS_{ij}\frac{H_{ii}+H_{jj}}{2}
$$
$H$ is the core Hamiltonian, $S$ is the overlap matrix and $K$ is a constant. I tried to slightly change the coefficient $K$ (the suggested value is 1.75) for different spins in order to get a different initial guess. However, the SCF procedure converges towards a value that is higher in energy than the restricted solution. This should not be the case, in particular for a dissociating $\ce{H2}$ molecule (i.e. distance $R=4.4$ Bohr).

How I can properly differentiate the initial guess for $\alpha$ and $\beta$ spins?

EDIT

The reference of Gaussian09 states the following:

  Requests that the HOMO and LUMO be mixed so as to destroy α-β and spatial symmetries. This is useful in producing UHF wavefunctions for singlet states. [...]

I belive that this is the solution to my problem, but what does actually means HOMO and LUMO be mixed in practice (i.e. before starting the calculation)?

[1] Szabo and Ostlund, Modern Quantum Chemistry, Dover, 1996.

[2] M. Lee et al., Extended Hückel and Slater’s rule initial guess for real space grid-based density functional theory, Computational and Theoretical Chemistry, 2015.

When $N_{\alpha} = N_{\beta}$, a restricted solution of Roothaan equations is a solution to the unrestricted Pople-Nesbet equations. This restricted solution always exists and necessarily results if an initial guess $P_{\alpha} = P_{\beta}$ is used. [...] In seeking this second solution [the unrestricted one], it is imperative that an initial guess $P_{\alpha} ≠ P_{\beta}$ be used or the equations will necessarily yield the restricted solution.

The quote you state from Szabo and Ostlund is not wrong, but isn't a practical quote in the sense that it helps to think about the different spin cases that one would want to handle. You already know how RHF/closed-shell calculations work in theory and in practice. I won't talk about restricted open-shell Hartree-Fock (ROHF), because it isn't relevant to this problem.

There are three UHF/open-shell scenarios, which will be considered in order:

Odd number of electrons
Even number of electrons, high spin
Even number of electrons, low spin

Odd number of electrons

For this case, take the copper(II) atom, which has a 3d$^{9}$ valence electron configuration; it is a doublet, so it has 1 more $\alpha$ electron than $\beta$ electron. There's no trouble here, because the indices for occupied orbitals run over different ranges; in the simplest case, the range for $\alpha$ spin will be 1 larger than for $\beta$ spin.

Even number of electrons, high spin

For this case, let's take triplet oxygen. In practice, this is actually identical to the above case, except we have 2 more $\alpha$ electrons than $\beta$ electrons. Again, we have differently-sized occupied blocks for each set of MO coefficients which will lead to different densities.

Even number of electrons, low spin

For this case, let's take singlet oxygen.

This is the difficult case, because the orbital degeneracy means for a proper qualitative description of the molecular orbitals, two determinants must be used, requiring a CAS(2,2) (at a minimum). Because this valence electron degeneracy isn't possible in a single determinant picture, the degenerate MOs will "split" into one lower in energy, the other higher. Requesting a singlet spin multiplicity means the two electrons will now occupy the same orbital; there's no reason for them to occupy different orbitals, at least for the initial guess. Hopefully it's (visually) clear that if the spatial components of $\alpha$ and $\beta$ spins are the same in this case, our guess is equivalent to the RHF guess. This means that for an open-shell singlet, one must break spatial symmetry in the initial guess, just as Szabo and Ostlund states. Since the RHF solution always exists as a local minimum on the "determinental" potential energy surface, whatever you choose as your SCF converger is most likely going to get stuck in this minimum.

This is a little hand-wavy, since we know that UHF calculations approximate two real determinants at the cost of spin contamination, but that isn't our goal.

Breaking spatial symmetry

How do we break spatial symmetry? Consider the initial steps of a typical RHF calculation that diagonalizes the one-electron (core) Hamiltonian:

$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{0'} &= \mathbf{\epsilon}^{0'} \mathbf{C}^{0'} \\
\mathbf{C}^{0} &= \mathbf{S}^{-1/2} \mathbf{C}^{0'} \\
P_{\mu\nu} &= \sum_{i}^{\text{occ}} C_{\mu i} C_{\nu i}
\end{align}$$

where the tick mark signifies a quantity is in the orthogonal AO basis, as opposed to the original (non-orthogonal) AO basis. Now assume we're doing a UHF calculation:

$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{\alpha, 0'} &= \mathbf{\epsilon}^{\alpha, 0'} \mathbf{C}^{\alpha, 0'} \\
\mathbf{F}^{0'} \mathbf{C}^{\beta, 0'} &= \mathbf{\epsilon}^{\beta, 0'} \mathbf{C}^{\beta, 0'} \\
\mathbf{C}^{\alpha, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\alpha, 0'} \\
\mathbf{C}^{\beta, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\beta, 0'} \\
P_{\mu\nu}^{\alpha} &= \sum_{i}^{\text{occ }\alpha} C_{\mu i}^{\alpha} C_{\nu i}^{\alpha} \\
P_{\mu\nu}^{\beta} &= \sum_{i}^{\text{occ }\beta} C_{\mu i}^{\beta} C_{\nu i}^{\beta}
\end{align}$$

This is misleading! Because we only have one initial Fock matrix, it will only have one set of eigenvectors, so $\mathbf{C}^{\alpha, 0'} \equiv \mathbf{C}^{\beta, 0'}$! If the ranges for forming the two densities are different (as it would be when $N_{\alpha} \neq N_{\beta}$), then having an initial pair of identical MO coefficient matrices is not an issue. For your Huckel guess (called the Generalized Wolfsberg-Helmholtz or GWH guess in Q-Chem, scf_guess = gwh), the initial MO coefficient matrices would still be identical, because you modify a quantity they have in common (the core Hamiltonian).

So, we must modify the MO coefficients (coefficient matrices) themselves. There are three options that I know of:

mixing
rotation
switching

These are variants of the same thing; rotation is the most general, since you can define switching or mixing of two orbitals via rotations.

Mixing

The most straightforward approach of the three that requires no prior knowledge of your system is mixing. The usual choice is to "mix" some amount of the HOMO with the LUMO of the same spin:

$$\begin{align}
C_{\mu,\text{HOMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( C_{\mu,\text{HOMO}}^{\text{old}} + k C_{\mu,\text{LUMO}}^{\text{old}} \right) \\
C_{\mu,\text{LUMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( -k C_{\mu,\text{HOMO}}^{\text{old}} + C_{\mu,\text{LUMO}}^{\text{old}} \right)
\end{align}$$

where $k = [0, 1]$. If $k = 0$, nothing happens; if $k = 1$, the HOMO and LUMO become the symmetric and antisymmetric linear combinations of the two, respectively. In theory, this doesn't need to be the HOMO and LUMO; any two MO indices could be chosen, as long as they will result in different densities. This is also only for a single $\mu$; because we want to work with matrices, we would be working with whole rows of the MO coefficient matrix. For Q-Chem, this is $\alpha$ spin only (scf_guess_mix = k); the Gaussian manual doesn't say which spin it takes.

Rotation

The most general form for operating on pairs of MOs is rotation. Recall the rotation matrix in two dimensions:

$$
R(\theta) = \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
$$

This can be used to "rotate" two arbitrary molecular orbitals $i,j$ by an arbitrary angle:

$$
\begin{bmatrix}
C_{i}^{'} \\
C_{j}^{'}
\end{bmatrix}
= \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
C_{i} \\
C_{j}
\end{bmatrix}$$

where again each $C$ is a row vector. This can be expended into a set of 2 equations:

$$\begin{align}
C_{i}^{'} &= C_{i} \cos(\theta) - C_{j} \sin(\theta) \\
C_{j}^{'} &= C_{i} \sin(\theta) + C_{j} \cos(\theta)
\end{align}$$

There are a few interesting cases.

180 degree rotation: change phase

$$
R(180^{\circ}) = \begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= -C_{i} \\
C_{j}^{'} &= -C_{j}
\end{align}$$

This is kind of pointless, since the energy should be invariant to a phase swap.

90 degree rotation: interchange

$$
R(90^{\circ}) = \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= - C_{j} \\
C_{j}^{'} &= C_{i}
\end{align}$$

This is the usual case when "rotate orbitals" is seen. This is equivalent to switching orbitals. I don't know about other programs, but if you do this in ORCA, the orbital energies will be swapped as well.

45 degree rotation: symmetric and antisymmetric linear combinations

$$
R(45^{\circ}) = \begin{bmatrix}
1/\sqrt{2} & -1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} - C_{j} \right) \\
C_{j}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} + C_{j} \right)
\end{align}$$

This last example is identical to the $k = 1$ (100%) case from the mixing example shown earlier.

Switching

We could also alter occupation numbers, such as by placing the $\alpha$ (or $\beta$) HOMO electron into the same-spin LUMO, or some other occupied orbital, but this isn't an automated procedure like orbital rotation, because you need to know your occupied/virtual orbital indices prior to starting the calculation. This is equivalent to switching orbitals.

A short summary of program capabilities (to the best of my knowledge):

Q-Chem: mix, switch
Gaussian: mix, switch
GAMESS: mix, switch
DALTON: switch
MOLPRO: rotate
ORCA: rotate

An important note about mixing: it doesn't matter that choosing the HOMO and LUMO to permute may be incorrect. Programs that have baked-in support for symmetry (DALTON, MOLPRO, unsure about GAMESS) will only mix/switch/rotate orbitals within a symmetry unless symmetry is disabled, but the point is only to break symmetry in the initial guess so that the local RHF minimum is escaped. If you know beforehand that your system has particular symmetries that need to be preserved, then you would probably be doing something more complex than just mixing the HOMO and LUMO anyway.


Mass Spectroscopy peak assignments in but-3-en-1-ol i have the mass spectrum for the molecule but-3-en-1-ol. The Mr of this is 72

There are four peaks that are labelled which I have to name the species responsible for forming that peak.

The peaks are:
m/z = 66
m/z = 67
m/z = 71
m/z = 73 - this peak is labelled as (M + H)^+

My answer so far...
The peak at m/z=73 is the but-3-en-1-ol with one added hydrogen, somewhere to the double bond?
For the other peaks am I just removing hydrogens as the mass only decreases by 8 in total so i couldn't remove a carbon or the oxygen.

Does this mean that the peaks are:
m/z=71 is the removal of 1H
m/z=67 is the removal of another 4H (5H in total)
m/z=66 is the removal of another 1H (6H in total)

is this is correct? Which hydrogens are removed first and what does each species look like for each peak, and are there any charges on the species produced. Also is there a special name for this sort of fragmentation, e.g alpha cleavage?

As you have pointed out, the peak at M = 73 is M+H, protonation is however more likely to occur at the oxygen atom.

In my opinion, peaks at M = 66 and 67, i.e. with a rather small mass difference to $\mathrm{M^+}$, are negligible.

Common sense and the link in permeakra's comment suggest that the major fragmentation of but-3-en-1-ol occurs along the red dotted line, yielding an allylic and a hydroxymethyl fragment.


Why do Fullerenes have even number of carbon atoms? My teacher started a chapter called "Carbon and its compounds", Which is just an intro to organic chemistry . In which she taught about fullerenes structure and examples and also told that fullerenes have even number of carbon atoms. I got curious and asked why but she didn't answer me. So I went and googled it but still did not get any answer. Is there any exception or any reason behind it?

The structure of (closed) fullerenes is governed by Euler’s polyhedron formula: vertices (carbon atoms) + edges (bonds) - faces = 2. Geometric constraints determine that at least some of the faces must be pentagons or heptagons, and that the number of atoms must be even.


How do chemists measure purity in high purity metals and other reagents? I have noticed that gold bars are often marked "999.9" fine:

Whenever I have metallurgical assays done, they always seem to do them with ISP or DSP or whatever, some spectrometer-based system. These systems have errors of 1% to 2%. When I complain about the high error, they say "it's good enough" and "nobody does the old fashioned stuff anymore". So, if ± 2% is the new standard for assayists, how are these banks determining that the gold is pure to a 9999 out of 10,000 standard?

As the comments alluded, to analyze purity at high purity, you analyze the impurities. I'll use XRF for this example, but ICP, MS, HPLC, or other methods could work in the same way.

With XRF's $1-2\%$ precision directly measuring the gold would give you a maximum supportable purity of $98-99 \%$. To verify a metal as $99.99\%$ the impurities are measured which are usually in the parts per million range, added up and subtracted from $\pu{100\%}$.

For example, let's say you have a gold bar with $\pu{35ppm}$ silver, $\pu{20ppm}$ copper, $\pu{10ppm}$ lead, and $\pu{5ppm}$ mercury with all other elements below a , $\pu{0.5ppm}$ detection limit (also called a threshold limit). The total impurity would be the total of the measured impurities and the detection limit of the unmeasured impurities. Note: most analysis is by metal basis which excludes the other elements that may be present.

$$\mathrm{Impurity = 35 + 20. + 10. + 5 + [0.5 \times (59-5)] = \pu{97 ppm} }$$

The $59$ is for the $59$ naturally occurring metal elements and the $5$ is for the five elements we measured ($\ce{Au, Ag, Cu, Pb, Hg}$). $\pu 2 \pu\%$ of $\pu{97ppm}$ is $\pu{1.94ppm}$ and the measurement plus the maximum error is $\pu{99\!.ppm}$. Taking $\pu{99\!.ppm} = \pu{0.0099\%}$ and subtracting from $\pu{100.00\%}$, we get an inferred purity of $\pu{99.991\%}$ Gold ($\mathrm{100\% - 0.01\% = 99.991\%}$). The gold would still be sold at $\pu{99.99\%}$ regardless of it being purer than needed as selling the gold by lots would be unnecessarily complicated and time intensive.


Why is the resonance concept not required in molecular orbital theory? In valence bond theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why is it told that "MO theory provides a global, delocalized perspective on chemical bonding"?

(This is a follow up to What is actually the difference between valence bond theory and molecular orbital theory?)

Because MO is closer to reality.

Resonance is a convenient fiction used to simplify the visualization of some molecules, and rationalize their actions in the frame work of Lewis structures. But every resonance structure you have ever seen in your life is a lie.

Well, this is not exactly true. It's more precise to say that resonance structures are like blind men trying to describe an elephant: they each give a partial snapshot of truth while being wholly inadequate to describing it all. The "true" structure is a weighted average of all the resonance structures based on how stable they would be if they were the actual molecule.

MO cuts to the chase and attempts to directly describe the molecule in terms of its molecular orbitals: another convenient fiction, mind, but one that is closer to the truth. The reason you don't need "resonance" MO structures is because molecular orbitals are inherently a probabilistic construct: they already take into account the uncertainty in the electrons' positions.


What's up with SiO₂ being tetrahedral? If silicon dioxide tends to form a crystal lattice with four $\ce{O}$'s around a central $\ce{Si}$, why isn't the molecular formula $\ce{SiO4}$ then? I'm confused why it's unique in that its molecular formula doesn't match up with its geometry.

Every $\boxed{^{14}\textrm{Si}}$ atom is surrounded by 4 $\boxed{^8\textrm{O}}$ atoms.   On the other hand, every $\boxed{^8\textrm{O}}$ atom is surrounded by 2 $\boxed{^{14}\textrm{Si}}$ atoms.

Suppose that the number of $\textrm{Si-O}$ bonds is $n$, then the number of $\boxed{^{14}\textrm{Si}}$ atoms is $n/4$, while the number of $\boxed{^8\textrm{O}}$ atoms is $n/2$.  Therefore you arrive at the formula $\textrm{Si}\textrm{O}_2$.


What happens when 0.02 mole NaOH is added to a buffer solution? I am asking this question for my son.

(This question is related but we still  have our confusions.)

We have one liter buffer solution of 4.74 pH which contains 0.1 mole $\ce{CH_3COOH}$ and
0.1 mole $\ce{CH_3COONa}$.

What will be the resulting pH when 0.02 mole $\ce{NaOH}$ is added to this solution? (Given $\ce{Ka} = 1.8 \times 10^{-5}$.)

As the definition of a buffer solution [A, B] goes, pH value should not change when small quantity of acid or base is added to it.

So, one way of answering the question  will be to simply comment that since
a small quantity of base has been added, the pH value will not change.

But, we find another solution (presented and accepted in some
moderately acceptable study-sheet) in the following line.

We will have a reaction like,

$\ce{CH_3COOH} + \ce{NaOH} = \ce{CH_3COONa} + \ce{H_2O}$

So, after the reaction, we have,

$\ce{CH_3COONa}$: $(0.1 + 0.02)\ \text{mole} = 0.12\ \text{mole}$

$\ce{CH_3COOH}$: $(0.1 - 0.02)\ \text{mole} = 0.08\ \text{mole}$

Now, $pH = pKa + \log{\text{[Salt]}\over\text{[Acid]}}$

Hence,
$pH = pKa + \log{{{n_\text{Salt}}\over V}
\over
{{n_\text{Acid}}\over V}}$

Which after plugging in the values gives,
$pH = 4.92$.

We are  confused about which of the above solutions is
correct.

The second solution using the Henderson-Hasselbach equation is correct. While buffer solutions change pH very little near the pka of their constituent weak acid, it would not be correct to say that they do not change at all. Also, an addition of 20 mmol of NaOH is not insignificant to a 100 mM buffer—a 20% change in concentration. The Henderson-Hasselback equation in the form you've shown is an acceptable estimation of the buffer pH in a moderately concentrated buffer with an acid/base ratio close to 1. Deviations from it in that form can occur for very dilute or concentrated solutions or if it's well away from the pka.


Melting of ionic compound hydrates My professor asked us the reason of the melting point of potash alum being so low, despite it being an ionic salt. He later went on to answer it by saying that melting point is a misnomer for the phenomenon that is happening here.

  What's happening here is that the alum is losing its water of hydration and is then dissolving in that water itself. So it is actually a dissolution point rather than melting point.

I couldn't find any sources for what he said. Though what he said makes logical sense, I would like to read a bit more about it. Can you please provide some sources confirming what my professor said?

Your professor's premise of dissolution is only true if the process is incongruent melting which by the way is still melting. Whereby the process proceeds as:

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][T1] KAl(SO4)2~_{(s)} + 12H2O_{(l)}\tag{$\beta$ to $\alpha$ + liquid}}$$
Then
$$\ce{KAl(SO4)2~_{(s)} ->[\Delta][T_2 > T_1] KAl(SO4)2~_{(aq.)}\tag{$\alpha$ to liquid}}$$

This concept is outlined in the generic binary phase diagram below where $\alpha$ is anhydrous potash alum, $\beta$ is the hydrated form, and liquid is potash alum solution. Above a certain temperature $\beta$ will decompose into $\alpha$ and liquid. As the temperature rises more potash alum will dissolve until eventually it will all be liquid again.

Figure 1. Binary Phase Diagram with Peritectic melting

However if it is a direct conversion from liquid to solid then it is congruent melting and your professor is wrong. This can occur either as melting an eutectic (shown in Figure 2) or congruent melting of a compound (shown in Figure 2) and proceeds directly as follows.

$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][92.5^\circ C] KAl(SO4)2~_{(aq.)} + 12H2O_{(l)}\tag{Interionic Melting}}$$
$$\ce{Sn_{(s)} + Pb_{(s)} ->[\Delta][183^\circ C] Sn_{(l)} + Pb_{(l)}\tag{Eutectic}}$$
$$\ce{Ti5Si3_{(s)} ->[\Delta][2130^\circ C] 5Ti_{(l)} + 3Si_{(l)}\tag{Intermetallic Melting}}$$

Figure 2. Binary Pahse Diagram with Eutectic.

Figure 3. $\ce{Si\! -Ti}$ Binary Bhase Diagram with Compound at $\ce{Ti5Si3}$ That Undergoes Congruent Melting.

Usually differential scanning calorimetry is used to determine which melting regime you are in indicated by a range of temperature causing a spike in heating rate corresponding to incongruent melting or a single temperature where heating spikes which would indicate congruent melting.

In this case, however, we can confidently assume that $\ce{KAl(SO4)2.12H2O}$ is under going incongruent melting since it melts above the temperature of water and below the temperature of the anhydrate and there are no other hydrates that it forms, the process is the peritectic melting thus the process your professor describes is correct, but it is still considered a melting process.


Calculate specific heat of a substance given other information What is the specific heat of a $\pu{20 g}$ substance that releases $\pu{979 J}$ of heat when changing from $\pu{70 ^\circ C}$ to $\pu{20 ^\circ C}$?

I started out with the following equation: $c = \frac{Q}{m\Delta T}$ and $Q = 979\ \mathrm{J}$ and $m = 20\ \mathrm{g}$ and $\Delta T = 45\ \mathrm{^\circ C}$.

My final answer was $2202.75\ \mathrm{J}$

Did I use the right equations?

Your equation is correct, as are most of your measurements.

The unit for $c$ is $\ce{Jg^{-1}K^{-1}}$ if using grams as mass, ($\ce{Jkg^{-1}K^{-1}}$ if you convert the mass to kilograms).

Using your formula, however, does not give the answer you gave. When calculating using a calculator, consider that a parenthesis needs to be put around the denominator, as such:

$$c = \frac{q}{(m\Delta T)}$$

or else the calculator takes it as being $\ce{979J / 20g \times 45K}$ which gives the value you got.

It needs to be $\ce{979J / (20g \times 45K)}$ which gives an answer that is much more reasonable.


I'm trying to make an optical dissolved oxygen sensor. I need help finding an appropriate oxygen sensitive dye I have been trying to make a dissolved oxygen sensor, the basics are described on this page.

I have a good understanding of the electronics behind this measurement, but I cannot find a suitable oxygen sensitive dye to use. I have read quite a bit about room temperature phosphorescence, and thought that erythrosin b might but a suitable candidate, it has an excitation wavelength of 525 nm, and radiates in near-IR. After more reading, I found that it needs to be in a sol-gel or similiar substrate. All of the reading I've done seems to require some sort of substrate and this isn't what commercial dissolved oxygen sensors are using. They appear as just a thin, hard disc. Additionally, I don't have any equipment or experience with sol-gels.

I'm looking for any information on what chemical and process commercial systems are using. Or, a simple way to get erythrosin b to work. My tests with it so far have not shown any phosphorescence at all.

Thanks for your time.

This review article might be useful for your purpose:

Title: Optical methods for sensing and imaging oxygen: materials, spectroscopies and applications
Chem. Soc. Rev., 2014, 43, 3666-3761 (DOI: 10.1039/C4CS00039K)

!http://pubs.rsc.org/en/content/articlehtml/2014/cs/c4cs00039k

This is a review article with 694 references (96 pages). The discussion focused on main sections such as direct spectroscopic sensing of oxygen, absorptiometric and luminescent probes, polymeric matrices and supports, additives and related materials, spectroscopic schemes for read-out and imaging, and sensing formats (e.g., waveguide sensing, sensor arrays, multiple sensors and nanosensors). They also discuss future trends and applications, and concluded it with summary of the properties of the most often used indicator probes and polymers.

There are several probes has been discussed. Some of them may be useful to develop oxygen sensors: Those include: a) Hemoglobin and myoglobin as optical probes for oxygen (the affinity of hemoglobin for oxygen is a function of pH and ionic strength); b)  Molecular absorptiometric probes for oxygen (e.g., several cobalt–organic compounds are capable of reversibly binding molecular oxygen); c) Methods based on the oxidative power of oxygen (the reduced or "leuco" form of Methylene Blue (MB) is colorless and quickly converted back
to blue MB by oxygen); etc.

PS: Selected irreversibly acting chromogenic probes for oxygen along with respective polymeric supports and color transitions, and Organic fluorescent probes for oxygen were also given in Table 2 & 3, respectively.


Software or library for creating images of ball and stick models? Is there any code library (in any programming language) or software capable of producing images of ball and stick diagrams?

Example of an image of a ball-and-stick model for Ammonia:

(Image source)

Preferences:

It should be capable of producing images programmatically through a code library. Any mainstream programming language (C, Java, Python, etc.) is acceptable.
It should be able to take chemical descriptions as input to produce images of ball and stick diagrams as output.
It should be free and open source.

I know that it is possible to create these models using Mathematica's ChemicalData built-in symbol, but I am looking for something that can integrate better into my existing code base (hence the preference for free and open source libraries in mainstream programming languages).

For general open source solutions:

Avogadro - Molecular editor GUI. Has a Python API.
Jmol - Chemical structure viewer. Fully scriptable using a custom scripting language. It also has a JavaScript-only version named 'JSmol', allowing models to be displayed in web browsers.
ghemical - Has a GUI for editing molecules, but no API.

For a $\LaTeX$ solution, see this question on the $\TeX$ Stack Exchange site: Help creating solid cylinders along a path in Tikz for crystal lattice diagram

Other solutions:

If you have access to Mathematica (e.g. on a Raspberry Pi running Raspbian [which comes with a free copy of Mathematica]), you can use Mathematica's ChemicalData built-in symbol to create ball and stick models.


Why isn't CH3OH basic if it contains OH? I was asked to find the most basic solution from the following:

  $\ce{K2SO4}$
  $\ce{CH3OH}$
  $\ce{Na2CO3}$
  $\ce{(NH4)2CrO4}$
  $\ce{BaCl2}$

I know that I have to identify the solution that contains the conjugate base of a weak acid.

Knowing this, I immediately eliminated $\ce{BaCl2}$ ($\ce{HCl}$), $\ce{K2SO4}$ ($\ce{H2SO4}$), and $\ce{(NH4)2CrO4}$ ($\ce{H2CrO4}$ — which I've heard is a strong acid).

This left $\ce{CH3OH}$ and $\ce{Na2CO3}$.

The correct answer is $\ce{Na2CO3}$ since $\ce{CO3^{2-}}$ is the conjugate base of $\ce{H2CO3}$.

However, why can the final answer not be $\ce{CH3OH}$, especially considering that $\ce{OH}$ is basic? Also, could $\ce{OH}$ be considered the conjugate base to a weak acid?

Unfortunately, chemistry is not as simple as going

  Oh, I spotted $\ce{OH}$, it must be a basic compound!

I would argue that the vast majority of all compounds one could write with $\ce{OH}$ inside are either neutral or acidic and only a minority is basic.

Take, for example, sulphuric acid. As Ivan correctly mentioned in a comment, rather than writing the traditional $\ce{H2SO4}$, you could also write it as $\ce{O2S(OH)2}$ which lays more emphasis on the molecular structure (to hydroxy groups and two oxy groups all bound to sulphur). We all know that $\ce{H2SO4}$ is acidic, not basic, even though it has $\ce{OH}$ groups.

Or take alcohols such as methanol ($\ce{H3C-OH}$) or ethanol ($\ce{H3C-CH2-OH}$). These also feature $\ce{OH}$ groups, they are even commonly written that way, but are about as acidic or basic as water ($\ce{HOH}$) is.

So the question may be why did you get taught that $\ce{OH}$ is basic? Well strictly speaking, it is not the $\ce{OH}$ hydroxy group that is basic but the $\ce{OH-}$ hydroxide ion. Only those compounds that dissociate into some cation and a hydroxide ion can be considered basic. For that to happen, ‘some cation’ typically needs to be metallic, at least at your level. So recognise a metal bonded to $\ce{OH-}$ and you should most likely be right calling the compound basic.

Some final notes:

You should always check the directly corresponding conjugate acid. So when analysing $\ce{K2SO4, Na2CO3}$ and $\ce{(NH4)2CrO4}$, trace them back to $\ce{HSO4- , HCO3-}$ and $\ce{HCrO4-}$, respectively. That still leaves us with two rather strong acids but the third ($\ce{HCO3-}$) is even weaker.
The conjugate acid to methanol $\ce{H3C-OH}$ would be the methyloxonium ion $\ce{H3C-OH2+}$, which is a strong acid. This goes for pretty much all organic hydroxy groups.
The $\mathrm{p}K_\mathrm{a}$ of water — hydroxide’s conjugate acid — is $15.7$, so yes, that is a very weak acid.


How can a vacancy defect be a stoichiometry defect? NCERT (India) explain stoichiometric defects as those defects which do not disturb the stoichiometry of the solid. Vacancy defects are stated to be under the category of stoichiometric defects. When some lattice sites are vacant, the crystal is said to have a vacancy defect.

My question is: if lattice points are missing in vacancy defects, shouldn't this disturb the stoichiometry of the compound? I think yes but the book states no. Why?

Yes, you are not completely wrong but try to answer these questions to check your understanding (mouse over for the answers):

First of all, what is a Non-stoichiometric compound?

   Non-stoichiometric compounds are chemical compounds, almost always
 solid inorganic compounds, having an elemental composition whose
 proportions cannot be represented by integers.

In which way a vacancy affects the ratio of the elemental composition?

   If the compound it's a "pure" metal the vacancy will not affect at all the ratio of the elemental composition. However, it could affect the ratio if, for example, we have $\ce{NaCl}$ crystal where a lot of $\ce{Na+}$ ions are missing.

How is likely to have an ionic compound where a lot of anion or cation are missing?

   Not at all, the ionic compound, usually, maintain an overall neutral charge see Schottky defect.


Why do more substituted alkenes undergo epoxidation, but the less substituted alkenes undergo hydrogenation? We know that, in case of a choice, catalytic hydrogenation will hydrogenate the less  substituted alkene, because the less substituted alkene is less stable. But, we also know that, in case of a choice, peroxyacids perform epoxidation of the more substituted alkenes, because the latter are more electron-rich (i.e. more nucleophilic) (details)

In one case, we are favoring the stability of the pi bond. But in another case, we are prioritizing the nucleophilicity of the pi bond. Why is this so?

I believe this definitely has something to do with their respective mechanisms, which I've studied from Libretexts (hydrogenation and epoxidation). The only difference I see is that in hydrogenation, the pi bond first attaches to the catalyst surface, and then a hydrogen atom (already attached to that surface) is transferred to the pi bond. But, in epoxidation, the alkene directly attacks the electrophilic $\ce{-OH}$ group of the peroxy acid. So, this explains the need for more nucleophilicity in the latter case, but how does this explain the need for less stability in the former case? Or is this direction of reasoning entirely wrong?

As you already know, the alkene has to approach the catalyst surface in the first step. For optimum adsorption, the molecule must be able to orient itself parallel to the surface of the metal. Less number of substituents promote this.

  The activated hydrogen is in the form of $\ce{Pt-H}$ or $\ce{Pd-H}$ bonds on the surface of metal particles, and hindered alkenes can't approach the $\ce{M-H}$ bonds easily.

Source

Epoxidation on the other hand is free from any such need for to attach to the surface. So we simply consider nucleophilicity.


Strange 1H-NMR signals/signal ratio Fedorov et al. assign the $^1$H-NMR ($\ce{CDCl3}$) chemical shifts of the $\ce{CH2}$ group in 2-(bromomethyl)phenylboronic acid

to three singlets at $4.75$, $5.09$ and $5.15~\ce{ppm}$ in a ratio of $3:3:14$.

Can someone explain their reasoning?

The paper specifically addresses this issue and ascribes it to the tendency for boronic acids to exist as solvent and concentration dependent oligomers:

  The dependence of the intensity ratio for these signals and the
  chemical shifts in the 1H NMR spectra on the concentration of the
  samples under study as well as on the nature of the solvents used
  indicates that anhydrides of boronic acids with different degrees of
  oligomerization can exist along with monomeric forms of these
  derivatives (ArB(OH)2) in solutions of substituted arylboronic acids.
  The conclusions about the existence of oligomeric forms of boronic
  acids and the absence of byproducts in the compounds isolated are
  supported by the results published earlier.

This type of equilibrium can be expressed like:

Higher order oligomers are also possible.


How does NMR spectrometer determine area under each peak? I am confused about exactly how does NMR machine figure our area under each peak, which then tells us how much protons present in the molecule. Does it have to do something with radio waves being emiitted after proton relaxes? Thanks.

The modern basic FT 1H NMR experiment involves irradiating and simultaneously exciting a wide bandwidth of proton frequencies. That is to say, all spin populations of the proton nuclei have their spin state reversed with a carefully calibrated Rf pulse that delivers the correct amount of power for the correct duration. As each spin relaxes back to equilibrium, it emits a photon of Rf energy. So, yes your assumption is pretty well correct. This energy packet has a specific frequency (a product of the chemical shift), and a specific intensity - it is a quantum packet of energy. If two nuclei of the same frequency each emit a photon of energy, the intensity of that frequency detected will be twice as large a single nucleus emitting a photon of energy.

Across the entire frequency range, the data collected in the free induction decay is the sum product of all frequencies,  and their intensities. The Fourier transform function then converts this time domain data into the preferred frequency domain spectrum.


Are glucose and galactose cis-trans isomers of each other? Glucose and galactose are diastereomers of each other. However, is it correct to say that they are cis-trans isomers of each other? Does it make a difference with regard to terminology if glucose and galactose are in cyclic form i.e. are glucopyranose and galactopyranose cis-trans-isomers of each other? And is it correct to say that glucopyranose has $2^5-1=31$ cis-trans isomers?

I would also appreciate some references to verify the correct use of terminology in this situation.

NOTE: I have edited this question to clarify that I'm interested in the cyclic form as well.

Glucose and Galactose are best described as epimers of one another.
The three structures below all have the same molecular formula and the same connectivity, but differ in the absolute configuration at a single stereocentre (this is the definition of an epimer).
You can see that glucose and galactose are epimeric at the $\mathrm{C4}$ position, whilst glucose and mannose are epimeric at the $\mathrm{C2}$ position.

Image taken from Principles of Biochemistry, Lehninger et. al. (5 ed.)

To comment on the use of cis and trans.
The cis/trans nomenclature is used to describe the geometric isomerism of alkenes - it is not correct nor appropriate to use it in this context.  What you may be referring to is whether two adjacent stereo centres are syn or anti to one another.


Why does it require so much pressure to create diamonds? Why does it require so much energy/pressure to create diamonds, yet with silicon it is very easy and (I believe) naturally occurs in a "diamond" structure, i.e. where each atom is covalently bonded to four adjacent atoms? I guess it has something to do with the fact that the outer shell is further away from the nuclear, since that is the only significant difference between the two atoms, but I can't work out why that should matter here.

There are two allotropes of carbon crystal, graphite and diamond.   Both
are stable, so the formation of diamond structure is competitive with
the other structure (or structures, if you include buckyballs
and tubes).   The important consideration, therefore, is crystal growth rate,
which will depend on solvent and seed material, as well
as temperature and
pressure.

For manufacture of large artificial diamonds, liquid iron 'solvent' is
used, which is why high temperature is important; carbon has to diffuse through the liquid solvent. At high temperature,
the (relative) instability of diamond means that
a pressurized environment is required: diamond has to expand (diamond is 3.53 g/cc, but graphite is 2.23 g/cc) in order to become graphite, and pressure prevents that.   Pressure alone might
become so high that graphite recrystallizes as denser diamond, in a shock wave (as long as you only want to make small diamonds) made with explosives.  At elevated temperatures and low pressure, diamond
recrystallizes as graphite, however.

It is
possible to grow diamond films (depending on a substrate that
seeds the crystal form of interest) at low pressures as well, from
vapor.   Pyrolyzing CH4 in the presence of a cold surface that doesn't
seed the graphite form is the general approach.

To summarize: in growth of a carbon crystal, the environment determines
the dominant crystal structure in a number of ways.  The Diamond Makers by Robert Hazen, is a good treatment of this subject.


Distinguishing stereoisomers using reagents Given the following possible reagents:

acetone,
$\ce{MnO2}$,
ozone, or
aluminium propoxide,

which one would distinguish between the cis and trans isomers of cyclopentan-1,2-diol?

The only reagent listed that will distinguish between cis and trans diols is acetone. In the presence of catalytic acid it will form an acetal with the cis diol. This cannot occur with the trans diol as the two OH groups are too far apart. Try drawing the cis and trans diol in 3D.


Complete combustion of mixtures I am working on my scholarship exam practice (around high school and first year university level). I could actually solve the problem below but my solution is pretty time consuming. It would be appreciated if I could find other aspects in solving this.

  Exactly $\pu{4.32 g}$ of oxygen gas was required to completely burn a $\pu{2.16 g}$ sample of a mixture of methanol and ethanol. How many moles of ethanol are contained in the sample? (Answer: $\pu{0.04 mol})$

What I did is setting $x$ and $y$ as the mass of methanol and ethanol, respectively. Then I formed 2 equations. First one was $(x + y = \pu{2.16 g})$ and another was the equation $(3x/64 + 3y/46 = \pu{0.135 mol})$ from the number of moles of oxygen gas, derived from two complete combustion equations. And then I solved simultaneous equations. I am wondering if this was the only way to solve this problem. If you think differently, please let me know.

Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in  $\pu{mol}$. Also,  you may not needed to do more calculations since the answer is in $\pu{mol}$. Thus, your two equations should be:
$$32x + 46y = 2.16 \;\text{.... (1), and}$$
$$\frac{3}{2}x + 3y = 0.135 \;\text{.... (2)}$$
Simplify equation (2) as:
$$x + 2y = 0.090 \;\text{ and }\therefore x=0.090-2y $$

Substitute $x$ in equation (1):
$$32(0.090-2y) + 46y = 2.16 $$
$$(64-46)y = 32 \times 0.090 - 2.16 = 0.72$$
$$\therefore y = \frac{0.72}{18} = 0.04$$


Brown stains on skin after working with Tollens' reagent I was working in the lab preparing Tollens’ reagent. When I came back home, I observed that my fingertips were brown, and I tried to wash it with soap several times but it didn't work.

It has been a day and they don't seem to go.

What may have been the reason? Is it permanent and how can I get rid of it?

The stains are due to silver nitrate, the reactive portion of Tollens’ reagent. It is very prone to reduction to metallic silver.

If you got it on your hands (a clear signal you weren’t wearing gloves when you should have been) it will create small silver particles in your skin — this is why it won’t wash off. Since the particles are small, they don’t exhibit the classical metallic characteristics (looking silvery, reflecting etc).

Don’t worry, they will grow out within a week’s time or so, due to your skin renewing itself.
And wear gloves next time.


Kinetic energy of molecules in liquid state? My book (book link) has this question:

  The kinetic energy of molecules at constant temperature in gaseous state is:

  more than those in the liquid state
  less than those in the liquid state
  equal to those in the liquid state
  None of these

The answer given is (3). It gives the reason as KE per molecule is still $=3/2\cdot k\cdot T$. So, the KE are both same!

But, I wonder, how can the kinetic energy of atoms in gaseous state be equal to that in the liquid state? Being in the gaseous state implies rapid and random motion of molecules, with little to no intermolecular attraction, spreading in all directions of any container. A liquid, however, is bound by intermolecular forces and is not so free to move.

In fact, I even wonder whether the formula $\text{KE per molecule }=3/2\cdot k\cdot T$ is valid in the liquid state, as it's derived from the kinetic theory of gases.

  What exactly is correct in this situation here?

The average translational kinetic energy of a molecule is $3kT/2$ irrespective of whether the molecule is in the gas, liquid, or solid phase.  In the liquid the motion giving rise to kinetic energy is restricted to a narrower range about the potential energy minimum  than it is in the gas phase. The equipartition theorem is quite general. A derivation is given here;  Derivation of mean kinetic energy

For ideal gases, the result may also be derived via kinetic theory. However, the most general derivation comes from the equipartition theorem.


Origin of Virtual Molecular Orbitals in Hartree-Fock equations Let's look at scheme
Do I understand correctly, the origin of Virtual Orbitals is a solution of (for, exampre, restricted) HF equation eigenfuncions?

Yes, they result from the HF equations just like the occupied orbitals. The diagram should also be show them as present in the initial set of equations, unless the basis  (used for the LCAO) was changed between the two states shown. The number of occupied orbitals plus the number of virtual orbitals is equal to the basis set size because they are the eigensolutions of
$$
 \mathbf{F}\mathbf{C} = \mathbf{\epsilon}\mathbf{S}\mathbf{C},
$$
where $\mathbf{F}$ is the Fock matrix, $\mathbf{C}$ is the matrix of orbital coefficients, $\mathbf{\epsilon}$ are the eigenvalues which are the orbital energies, and $\mathbf{S}$ is the atomic orbital overlap matrix (because the basis set used is typically not orthogonal). The dimension of all matrices involved is determined by the size of the basis set, which will be finite because otherwise the computational effort becomes infinite.


Solving carbon black I want to remove a layer of 2 mm thick carbon black on a copper bowl on the outside from hanging over wood and coal fire.
I have tried different solvents, but the carbon seems to be very resistive.

Which formula solves carbon black without solving copper? The bowl is made from pure copper. It will melt at roughly 1000 degree Celsius and has a high heat capacity.

By far the safest solution is to physically abrade the surface until you have worn away everything that isn't copper. This is slow and requires some sort of grinding agent but has the upside that you are not likely to injure yourself.

There is a chemical solution that will work but it requires access to some dangerous chemicals and is extremely hazardous even in a well-equipped lab. It is insane to try anywhere else as you won't have adequate safety equipment.

The best way to remove carbon black from anything is to use a mixture of concentrated sulfuric acid and hydrogen peroxide. This is often used (but also often highly discouraged because of the danger involved) in laboratories to clean sintered glass filters contaminated with things like activated carbon. In the case of a copper bowl, the solution will remove the carbon but will also etch the copper surface (it is sometimes used as an etchant to remove copper from printed circuit boards).

The mixture of sulfuric acid and hydrogen peroxide is sometimes known as Piranha solution of reasons that are obvious when you have seen it work.

I would not encourage anyone without good chemical safety training and access to very good safety equipment to work with the mixture. And even then they should work only with small amounts and in a fume hood behind a shield and while wearing eye protection, strong gloves, and a fireproof coat.


Inorganic IUPAC for Compounds with bridged ligands? Can we use  bis(tetraammine) instead of octaammine is the 2nd example ?

As in the 1st example they used  bis(pentaammine)  and also in the 3 one the used bis(carbonylcyclopentadienyliron)

Here is an example of a similar ligand and it can be named in three ways.
All you have to do is to put hydroxido/hydroxo in place of superoxido.

So yes, bridged ligands can be named in a few ways, combining any amount of recurring atoms separately, but most common used would be (if the LHS and RHS are the same) μ-amido-μ-hydroxidobis(tetraammine)dicobalt.
Reference

Crystal structure of .mu.-amido-.mu.-superoxo-bis[tetraamminecobalt(III)] tetranitrate, $\ce{[(NH3)4Co(NH2)(O2)Co(NH3)4](NO3)4}$ by
Gary G. Christoph, Richard E. Marsh, William P. Schaefer, Inorg. Chem. 1969, 8, 2, 291-297, DOI: https://doi.org/10.1021/ic50072a022


Boiling points of v. similar organic molecules Given the similarity in formula between pentane ($\small\ce{CH3(CH2)3CH3}$, b.p. 36.2 °C) and neopentane ($\small\ce{C(CH3)4}$, b.p. 9.5 °C), why are their boiling points so different?

Solids and liquids are hold together by cohesion. Intermolecular interactions contributing to cohesion depend on the geometry of the molecules in interaction,too.  Here, n-pentane with an elongated molecular structure exhibits a much larger molecular surface as compared to neo-pentane of a more sphere-like structure*).  As a result, other criteria kept constant (like the molecular mass), the lesser the molecular surface area of molecules of a substance, the lower the boiling temperature of this substance.

*)True, neo-pentane's methyl groups point towards the vertices of a tetrahedron. Actually I recommend to build a molecular model of both compounds and place them side-by-side to see it.


Why do Grignard reagents add to propargyl alcohols? In Phil Baran's lectures on heterocyclic chemistry at Scripps, one of the lecture problems is concerned with this synthesis of Vioxx:

The first step supposedly involves an anti addition of the aryl Grignard reagent to the propargyl alcohol:

and this intermediate is trapped with $\ce{CO2}$ to form a lactone (a butenolide in this case), and mCPBA oxidises the sulfide to a sulfone.

Why does the aryl Grignard add to the alkyne in this case? I would expect it to deprotonate the alcohol, and it does seem to do so, but as far as I know, Grignards don't usually add to alkynes which aren't conjugated to electron-withdrawing groups. Why is this propargyl alcohol special?

On top of that, what is the rationale for the selectivity observed (both regio- and stereoselectivity)?

The addition of Grignards to propargyl alcohols occurs via the alkoxide salt, after deprotonation:[1]

Here:

the reactivity may be explained by the stabilising influence of complexation;
the regioselectivity may be explained by the preference of a five-membered chelate ring over a four-membered one; and
the stereoselectivity arises because syn addition requires a five-membered ring with a trans double bond and is therefore precluded.

In fact, propargylic alcohols also get reduced by $\ce{LiAlH4}$ (Hans Reich's website has a page on it), with the same trans stereochemistry observed.

Reference

(1) Forgione, P.; Fallis, A. G. Magnesium mediated carbometallation of propargyl alcohols: direct routes to dihydroxydienes and enediyne alcohols. Tetrahedron Lett. 2000, 41 (1), 11–15. DOI: 10.1016/S0040-4039(99)01994-2.


Is any element base or acid? I have seen classification of Mg+2 or any other cation or anion for that matter as an acid or a base, but I have never seen an element labelled as an acid or a base. Isn't there a possibility of Mg+2 being the conjugate acid of Mg a base? If not,why so?

Short answer is: Yes and no. It depends on what acid-base theory is applied.

Redox reactions like $\ce{Mg -> Mg^2+ + 2 e-}$  are not usually considered acid-base reactions, but  over-generalized acid-base theories like Usanovich acid-base theory treat them as a special case of acid-base reactions.
Within this theory, $\ce{Mg}$ is relatively strong Usanovich base, $\ce{Mg^2+}$ is very weak Usanovich acid.

If we consider typical reactions by classical Broensted-Lawry theory:
$\ce{H2SO4 + H2O -> HSO4- + H3O+},$
or $\ce{NH3 + H3O+ -> NH4+ +H2O}$,
or by another classic - Lewis theory:
$\ce{NH3 + BF3 -> H3N-BF3}$,
then atoms involved in acid-base interactions have the same oxidation number before and after the reaction.
$\ce{Mg^2+(aq)}$ is the conjugate acid to the conjugate base $\ce{MgOH+ (aq)}$.
$$\ce{[Mg(H2O)6]^2+ + H2O <=> [Mg(H2O)5(OH)]+ + H3O+}$$
$\ce{MgOH+ (aq)}$ is is then the conjugate acid to $\ce{Mg(OH)2 (aq)}$.
$$\ce{[Mg(H2O)5(OH)]+ + H2O <=> [Mg(H2O)4(OH)2] + H3O+}$$
$\ce{\pu{Mg}}$ has in all 3 species the oxidation number +2.


Why doesn't NASA or SpaceX use ozone as an oxidizer for rocket fuels? Presumably, it would be expensive to use ozone ($\ce{O3}$) as an oxidizer instead of $\ce{O2}$, but would the extra oomph be worth it? Does $\ce{O2}$ provide as much thrust/energy/heat as can be provided, given the liquid hydrogen propellant?

I figured Chemistry was better than Physics, Engineering or Astronomy for this question.

As is usual with rocket fuels, the problems of ozone are practicality not performance
Almost every answer for why a specific rocket fuel component is used or not will end up referring to John D Clarke's magnificent and sparklingly written book: Ignition: An informal history of liquid rocket propellants (a rare technical book worth reading for the brilliant and humorous style in addition to the technical content).
His summary of why ozone, despite its apparent functional advantages, is not more widely used is fairly simple: the practical problems outweigh the apparent advantages.
He points out the advantages:

What makes it attractive as a propellant is that (1) its liquid density is considerably higher than that of liquid oxygen, and (2) when a mole of it decomposes to oxygen during combustion it gives off 34 kilocalories of energy, which will boost your performance correspondingly. Sänger was interested in it in the 30’s, and the interest has endured to the present. In the face of considerable disillusionment.

But every available way of creating a liquid with a high proportion of ozone is dangerous. Ozone is extremely toxic and unstable:

For it has its drawbacks. The least of these is that it’s at least as toxic as fluorine. ... Much more important is the fact that it’s unstable—murderously so. At the slightest provocation and sometimes for no apparent reason, it may revert explosively to oxygen. And this reversion is catalyzed by water, chlorine, metal oxides, alkalis—and by, apparently, certain substances which have not been identified. Compared to ozone, hydrogen peroxide has the sensitivity of a heavyweight wrestler.

Work was done on solutions of ozone in liquid oxygen which is somehow more stable. But has the disadvantage that ozone/oxygen mixtures separate into two phases the ozone rich version of which is difficult to prevent in feed tubes after firing and is extremely unstable. Another mixture considered to make handling easier was with liquid fluorine (!!!).
Ultimately he concludes the known work on ozone mixtures of any sort:

For ozone still explodes. Some investigators believe that the explosions are initiated by traces of organic peroxides in the stuff, which come from traces, say, of oil in the oxygen it was made of. Other workers are convinced that it’s just the nature of ozone to explode, and still others are sure that original sin has something to do with it. So although ozone research has been continuing in a desultory fashion, there are very few true believers left, who are still convinced that ozone will somehow, someday, come into its own. I’m not one of them.

Maybe there are theoretical advantages, but they are outweighed by the practical and safety problems. In a profession used to testing things like FOOF (fluorine dioxide) and chlorine trifluoride, this is some admission to make.


Balanced equation for synthesis of p-nitrophenol from p-nitrodiaznoium chloride salt and sulfuric acid/water? I am trying to write a balanced equation for the above mentioned reaction but having some trouble figuring it out. So far, I have:

$$\ce{([C6H4N3O2]+ Cl^{-}) + H2SO4 + H2O -> C6H5NO3 + N2 + ?}$$

How do I "incorporate" sulfuric acid and water into the reaction? Thanks for any help.

The sulfuric acid is not consumed in the reaction pathway to product, so there's no reason for it to appear in the equation for the reaction.  Therefor, the overall equation for the reaction is $$\ce{([C6H4N3O2]+ Cl^{-}) + H2O -> C6H5NO3 + N2 + HCl}$$


Is a meta-sodium sulfonate group on a benzene ring electron withdrawing or donating? I am wondering if a sodium sulfinate group would decrease or increase the electron density of the aromatic ring

As $\ce{SO3H}$ is a strongly electron withdrawing group, I'd say that $\ce{SO3Na}$ is basically the same, but to a slightly less extent. The reason for this is that being a group of highly electronegative atoms both groups have strong inductive electron withdrawing effects but they are both at donating electrons via conjugation (only possible from $\ce{S\bond{2}O}$ bond). This means that the sodium sulfonate group will decrease the electron density on the aromatic ring.


Method to determine ibuprofen content in dosage forms Please suggest some methods for determining the ibuprofen content in tablets, capsules and liquid dosage forms.

Is there a method which this can be done using UV spectroscopy?

Common methods in quantitative analysis of ibuprofen are

titration with sodium hydroxide in methanol,
potentiometric titration,
chromatography combined with mass spectrometry,
IR spectroscopy[1], or
UV spectroscopy (273 $\pu{nm}$[1]).

Some of these methods, e.g. UV or IR spectroscopy require extraction of the active ingredient.

Ref.: [1] S. R. Matkovic, G. M. Valle and L. E. Briand, Quantitative analysis of ibuprofen in pharmaceutical formulations through FTIR spectroscopy


Does sulfoxide H₂SO exist? The question Does H2SO or H2CO have a higher dipole moment? left me wondering whether $\ce{H2SO}$ has already been synthesized.

There are experimental evidences that both hydrogen thioperoxide $\ce{HSOH}$ (1) and  thiooxonium ylide $\ce{H2OS}$ (3) exist, whereas sulfoxide $\ce{H2SO}$, despite possessing global energy level lower than (3), is not [1]:

  The global minimum corresponds to hydrogen thioperoxide $\ce{HSOH}$ (1). The sulfoxide $\ce{H2SO}$ (2) is predicted to be $> \pu{25 kcal/mol}$ higher in energy (relative to 1) and the thiooxonium ylide 3 is the least stable
  isomer $(> \pu{34 kcal/mol}$ relative to 1).

There were numerous theoretical investigations (ab initio) suggesting that (2) is possible, but one of the most recent ones I found [2] still claims there is no experimental evidences to support its existence.
One of the problems with its structure the paper [2] refers to is that predicted $\ce{S-O}$ bond is abnormally short (1.4749 Å) and the valence state of sulfur is too high.

It would be nice if someone could point me towards a reputable source proofing/refuting existence of (2).
Additionally, what would be the complete preferred name of $\ce{H2SO}$?
MarvinSketch generates quite an amusing one: oxidanethione.

References

Iraqi, M.; Schwarz, H. Experimental Evidence for the Gas Phase Existence of $\ce{HSOH}$ (Hydrogen Thioperoxide) and $\ce{SOH2}$ (Thiooxonium Ylide). Chemical Physics Letters 1994, 221 (5), 359–362. https://doi.org/10/ct4m2m.
Denis, P. Theoretical Characterization of the $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS}$ Isomers. Mol. Phys. 2008, 106 (21), 2557–2567. https://doi.org/10/dnrk2v.

This is not an answer to the question "can the sulfoxide be synthesized in significant yield" (and under which conditions). Instead, this is a "naive" prediction based on the data you provide and two assumptions:

The predicted thermodynamic values in the articles you cite are accurate at $\pu{298 K}.$
The heat and entropy of formation of (2) and (3) from (1) are constant over $T.$

From the references we can build a data table of energies and entropies relative to "the ground state" $\ce{HSOH}$ assuming the $\Delta G^\circ$ values correspond to $T = \pu{298 K}:$

$$\begin{array}{lccc}
\hline
\text{Compound} & \Delta G/\pu{kcal mol-1} & \Delta H/\pu{kcal mol-1}^* & \Delta S/\pu{cal mol-1 K-1} & K_\mathrm{eq}\\
\hline
\ce{H2SO} & 25 & 15.8 & -31 & 10^{-19} \\
\ce{H2OS} & 34 & 38.2 & 14 & 10^{-25} \\
\hline
\end{array}$$

I include the order of magnitude of the equilibrium constants $K_\mathrm{eq}$ only to illustrate their ludicrously small values (and these are upper estimates!).

Could you find a temperature that would make $\Delta G^\circ<0$ for the reaction to $\ce{H2SO}$? The answer is no.  To see why we turn to the enthalpies. Since the reaction is endothermic you might hope that an increase in $T$ would help, but the accompanying entropy change is negative, so there is no way you can sufficiently increase the temperature to drive the reaction forward. The highest equilibrium constant you could hope for is $K_\mathrm{eq} = \pu{1.7E-7}$.

Things are not so bleak for $\ce{H2OS}$: the reaction is also endothermic but the entropy change is positive. At about $\pu{2800 K}$ $K_\mathrm{eq} = 1,$ and if we increase $T$ sufficiently, we can crank up $K_\mathrm{eq}$ to any arbitrarily large value.

This is of course not remotely exhaustive (therefore "naive") but hopefully a start to answering your question.

* Quoting your second reference:

  At the CCSD(T)/CBS limit and including corrections for scalar relativistic, spin orbit and core-valence correlation effects, the estimated enthalpies of formation are $\pu{−28.1 ± 1},$ $\pu{−12.3 ± 1},$ and $\pu{10.1 ± 1 kcal/mol}$ for $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS},$ respectively.


Finding partial pressure of a gas in a mixture A mixture of argon and carbon dioxide was initially at a total pressure of $\pu{1.72 atm}$ in a volume of $\pu{1.72 L}$ at $\pu{17 ^\circ C}$. The carbon dioxide was then completely removed from the mixture by passing it through a solution of sodium hydroxide ($\ce{NaOH}$). The argon was collected in a $\pu{1.1 L}$ vessel and its pressure found to be $\pu{1.55 atm}$ at a temperature of $\pu{28 ^\circ C}$. What was the partial pressure of carbon dioxide in the original gas mixture?  Assume that argon and carbon dioxide behave as ideal gases.

So, initially I found the moles of argon in in a $\pu{1.1 L}$ vessel which is $\pu{0.06896 moles}$

Then, i subtracted $\pu{1.72 L}$ from $\pu{1.1 L}$ in order to get a volume of carbon dioxide and used it in $PV = nRT$ and i got $\pu{0.04474 moles}$ of carbon dioxide .

Finally, I used the formula for partial pressure

$\frac{\text{moles of carbon dioxide} \times\text{total pressure}}{\text{moles of argon} + \text{moles of carbon dioxide}}$ and got $\pu{0.6771 atm}$.

However, my answer is not correct according to the textbook. Where did i make a mistake?

Your intial step is correct, finding the number of moles of argon in the 1.1 vessel. The next step however is incorrect. Because the pressure and temperature are both in different conditions for argon in the end and argon and carbon dioxide together in the beginning, you cnnot simply subtract the volume 1.72 - 1.1 L and then find the number of moles of carbon dioxide there. Remember, subtracting the volume and finding the moles of co2 would only work if the conditions are the same, but in this case, the pressures and temepratures are different. Thus, you can just simply go about finding the total n moles of Argon and carbon dioxide through PV=nrt.

So, 1.72 (1.72L) / 0.08206 x (273 + 17) = 0.1243 total moles. Subtract 0.06869 moles from earlier from 0.1243. You will get 0.055356 moles of CO2. Now, you can use your partial pressure method.

Hope this helps.


Benzyl Carbocation Stability: Would electron-withdrawing group on the ring destabilize the carbocation? In the following question I need to decide whether the reaction would undergo SN1 reaction or SN2 reaction.

I initially thought it would be SN1 since the resulting benzyl carbocation can be stabilized by resonance. Would the strongly electron withdrawing NO2 group destabilize it though? What kind of effect would this NO2 group have on the reactivity?

The reaction shown here is a Finkelstein reaction and SN2 reaction mechanism is followed.is an organic reaction that uses an alkyl halide exchange into another alkyl halide through a reaction wherein the metal halide salt is used. This reaction takes place at an equilibrium process by taking the advantage of poor acetone solubility in metal halide salt that is newly formed. The mechanism of the Finkelstein reaction is single-step SN2 reaction with stereochemistry inversion.

The success of this reaction depends on the below conditions.

Nucleophilicity of your halogen
Nature of leaving group
Carbon-halogen bond
Alkyl Halide reactivity

Holtz and Stock showed that electron-withdrawing groups increase the rate of SN2 reactions. This can be ascribed to stabilization of the transition state by withdrawal of some of the electron density. So NO2 will make reaction faster.

Ref:ADVANCED ORGANIC CHEMISTRY: REACTIONS, MECHANISMS AND STRUCTURE, 4TH ED


What is molecular system in general terms? What is a supramolecular system? What is a molecular system in general terms?

Could the solution of sugar in water be called a molecular system, if we were not interested in sugar-water interaction?

Today Jean Marie Lehn told me that bulk water is classical supramolecular system, but I cannot understand why.

Molecular chemistry is based predominantly on interactions through covalent bonds. It looks at molecules as individual entities, and is limited in its scope of accomodating intermolecular interactions. Your solution of sugar in water could easily be studied as a molecular system, if all you are interested in is the sugar molecule itself. Subtle differences in molecular conformation of the sugar in different solvent systems, for instance, would simply be written off as 'solvent effects'.

Supramolecular chemistry is based on intermolecular interactions. Typically, these are based on weak interactions (non-covalent forces), and include:

ion-ion interactions
ion-dipole interactions
dipole-dipole interactions
hydrogen bonding
van der Waals forces
pi stacking
hydrophobic interactions
and probably some others

It is through these types of interactions of that we can explore phenomena such as host-guest association, self-assembly, metal-ligand chelation, micelles, and even the structure of DNA, as all of these systems rely on the weak intermolecular interactions outlined above. You could also study your aqueous sugar system as a supramolecular system, and here you could consider what predominant intermolecular interactions are at work, and how they affect the overall chemistry of the bulk system.

Water is supposed to exist as water clusters, weak oligomeric and polymeric forms of $\ce{(H2O)_{n}}$


Oxidation state of carbon How do we find the oxidation state of individual carbon in Mg₃C₂. Explain with the help of structure if possible. For the average oxidation number, we can exploit the fact that the molecule will be electrically neutral, but what about the individual atoms of carbon?

Method 1: Assume the compound is ionic

For binary compounds consisting of a metal (but not a transition metal) and a nonmetal, the best method of determining oxidation numbers is to assume that the compound is ionic and determine the charge on the corresponding ions. The oxidation number of a monatomic ion is equal to its charge. This a good assumption because most compounds of this type are ionic, and it will work on many compounds that might be covalent. To make this work, you need to memorize some typical charges for ions:

Alkali metals - Li, Na, K, Rb, Cs - are typically 1+
Halides - F, Cl, Br, I - are typically 1-
Alkaline earth metals - Be, Mg, Ca, Sr, Ba - are typically 2+
Chalcogenides - O, S, Se, Te - are typically 2-

These are typically sufficient to deduce charges for most binary ionic compounds.

Let's do some examples.

  Definitely ionic: $\ce{Li3P}$

First, we assume that this compound is comprised of lithium ions and phosphide ions. The lithium ion is $\ce{Li^+}$, and there are three of them for a total of $3+$. Thus the charge on phosphorous must be $3-$. The oxidation number of lithium is +1, and the oxidation number of phosphorous is -3.

  Definitely covalent: $\ce{BeH2}$

Beryllium hydride is covalent, but this approach still works. If we assume this compound is a beryllium cation, $\ce{Be^2+}$, and two hydride anions, we can work it out. The hydride anions must be $\ce{H^-}$ to make the compound neutral. Thus, the oxidation number of beryllium is +2, and the oxidation number of hydrogen is -1.

Method 2: Based on structure

If we know the structure, we can use an alternative method. This method works best for covalent compounds, and does not work well for ionic lattices or network compounds. For each atom, we can assign the oxidation number as follows:

Each bond to a less electronegative atom contributes -1 to the oxidation number
Each bond to a more electronegative atom contributes +1 to the oxidation number
Each bond between two atoms of the same electron contributes +0 to the oxidation number

Let's look at some examples:

  $\ce{H2O2}$

The structure of $\ce{H2O2}$ is $\ce{H-O-O-H}$. Each oxygen atom has one bond to hydrogen (-1) and one bond to oxygen (+0) for an oxidation number of -1. Each hydrogen has one bond to oxygen (+1) for an oxidation number of +1.

  $\ce{HCN}$

The structure of $\ce{HCN}$ is $\ce{H-C#N}$. The hydrogen atom has one bond to carbon (+1), for an oxidation number of +1. The carbon atom has one bond to hydrogen (-1) and three bonds to nitrogen (+3) for an oxidation number of +2. The nitrogen atom has three bonds to carbon (-3), for an oxidation number of -3.

  $\ce{NaC2H}$

This compound is an ionic compound with one of the ions having covalent bonds. The structure is $\ce{Na^+ ^-C#C-H}$. The charge on sodium is 1+, so its oxidation state is +1. The first carbon atom has a formal charge of 1- and three bonds to another carbon atom (+0). The formal charge is added to the contributions from the bonds to get an oxidation state of -1. The other carbon atom has an oxidation state of -1 because of one bond to hydrogen (-1) and three bonds to the first carbon (+0). The oxidation state of the hydrogen atom is +1.

Your compound $\ce{Mg3C2}$

Either approach will work for $\ce{Mg3C2}$. If we treat it as ionic (a valid assumption based on results reported in this paper by Hongzhe Pan, et al.), there are three $\ce{Mg^2+}$ cations in the formula unit, and carbon needs to have a corresponding charge to make the compound neutral.

In the same paper by Pan, et al., the structure of a $\ce{Mg3C2}$ monolayer is described as a honeycomb kagome lattice. Below is a representation of part of this lattice:

.

In this lattice, each magnesium is bonded to two carbon atoms, and carbon is more electronegative than magnesium. Each carbon atom is bonded to three magnesium atoms.

The similar compound $\ce{Mg2C3}$ is a little trickier. If we use the ionic method, we can get an average oxidation state for each carbon atom of $\dfrac{+4}{3} \approx +1.333$. An atom cannot have a fractional oxidation state, so the total +4 must be distributed unequally over the three carbon atoms. If we look at the structure, we can see that one carbon atom (the one in the middle) is different than the other two.

$$\ce{Mg^2+ ^2-C=C=C^2- Mg^2+}$$

Using the formal charges on the carbon atoms, you can determine the oxidation states.


How can three p orbitals per atom bond in tin (II) sulfide? Tin (II) sulfide, a mainly covalent compound, bonds using its p orbitals only (the 5s orbitals are considered to be part of the core in divalent tin). Its structure is made up of Sn–S–Sn–S chains, which are further stabilized by Sn–S bonding across the chains:

How would VB or MO theory explain the fact that the tin molecule bonds three times, when only two of its p orbitals are half filled (Electron configuration: [Kr]5s24d105p2), while the third is empty? The observed bond angles of ca. 90° (96°/88°) would indeed suggest plain p orbital bonding, as suggested by this paper, but how is this possible?

And how could sulphur bond three times (Electron configuration: [Ne]3s23p4) with similar bond angles, when one p orbital is full?

The only thing that comes up to my mind is a dative covalent (co-ordinate) bond, in which sulphur provides both p electrons, thus permitting coordination. But then, the tin atoms would have a negative charge attached to them (due to having more electrons around than protons) and the sulphur atoms a positive one. Wouldn’t this be energetically unconvenient, considering that the compound has ca. 9% of ionic character, which depends on the tin ions being positive and the sulphur ions negative?

Addendum:

The dative bond assumption clashes against the bond lengths observed in tin (II) sulfide. The paper I cited above states that the Sn–S bond length on the chains are 2.68 Å long, while those between the chains are 6 pm shorter (2.62 Å). The IUPAC Gold Book says that dative bonds

  are distinguished by their significant polarity, lesser strength, and greater length

when compared to covalent bonds. This is exactly the opposite of what happens in tin (II) sulfide.

I think your conclusion of a dative bond between sulphide anions and tin cations (assuming pure ionic bonding, which is clearly not the case if you study the image closely) is correct.

Note that the charges you are observing are nothing but formal charges. In a macroscopic picture, tin is still less electronegative than sulphur and the electrons would be considered sulphur’s e.g. when determining oxidation states. The mere fact that an electron pair with correct symmetry is adjacent to an atom is enough to stabilise said atom’s empty orbital.


What are complete active space methods and how are such spaces defined for molecules? I am new to computational chemistry and would like to know about the active space and inactive space terminologies in MCSCF type calculations.

For understanding this I have four systems:

$\ce{H2O}$
$\ce{Be}$
$\ce{Be(OH)2}$
$\ce{NO3-}$

For above atom, molecules and ion what would be the CAS space configuration (which is typically written as eight digits) separately for inactive and active space. If you could give explanation then it will be useful.

Multiconfigurational self-consistent field (MCSCF) calculations can be further categorized. What you are referring to is a complete active space SCF (CASSCF) calculation.

In CASSCF, the 1-particle orbitals are partitioned into 3 groups, called spaces, that must be continuous and non-overlapping:

core or inactive orbitals; these must always hold two electrons
active orbitals; these are allowed to be partially occupied
virtual or external orbitals; these must always hold zero electrons

What "complete active space" means is that within the active orbitals, a full configuration interaction (FCI) calculations is performed, so electrons may occupy orbitals in any configuration as long as spin is conserved. The core/inactive orbitals are identical to occupied orbitals and the virtual/external orbitals are identical to unoccupied orbitals from Hartree-Fock (HF), by definition only; starting orbitals for any MCSCF theory don't necessarily need to be HF orbitals. You will see the notation CAS(n,m), where n is the number of active electrons and m is the number of active orbitals. Again, this is for a given total spin.

Here is a rather crude depiction of what a starting active space before optimization would look like; here, there are 8 (active) electrons in 8 (active) orbitals to give a CAS(8,8):

There is also restricted active space and generalized active space (RASSCF and GASSCF). For RASSCF, there are additional spaces, 1 below and 1 above the CAS, where electrons can only be doubly excited from the lower space to the upper space. By definition, RASSCF is more general than CASSCF. The Molcas manual has a good explanation. For GASSCF, all restrictions are lifted and an arbitrary number of active spaces can be chosen with arbitrary restrictions on excitations between them. I will defer mostly to the primary literature.

I'm not sure what is meant by "typically written as eight digits". If you're referring to the strings that are printed to the screen by programs, like 222ab0ba00a or 21102110 (those are made-up examples), they represent the orbital occupations (and potentially electron spins) of a single determinant that comprises a configuration state function (CSF). The length is entirely dependent on how large the chosen active space is in terms of number of orbitals, and all the numbers should add up to how many electrons have been chosen for the active space.

Choosing the active space for an atom or molecule is done so that the static or nondynamical correlation is qualitatively recovered in the wavefunction. In FCI, all possible electron correlation is captured within a given basis. As the excitation level is truncated from all possible excitations to all possible 1 through 6-fold excitations (CISDTQ56) to all possible single and double excitations (CISD) and so on, we make the artificial distinction between static and dynamic or dynamical correlation. Crudely put, static correlation is a result of near-degeneracies in the wavefunction, requiring a linear combination of Slater determinants for a proper qualitative representation. Dynamic correlation is a result of electrons interacting with each other instantaneously and explicitly, rather than in an average way such as the mean field SCF approaches. There are more detailed explanations here and here. Again, in the FCI limit, or even as you approach something like CISDTQ or CCSDTQ for simple systems, these are the same thing.

Because an FCI calculation is being performed within a subspace, and the computational cost of FCI in general grows factorially, this subspace is by necessity kept small. If no static correlation is present in a system, then it's rather pointless to perform any MCSCF calculations, and it's better to do a dynamical correlation calculation (perturbation theory, truncated CI, coupled cluster) on top of a single-configurational wavefunction from HF.

With regards to these specific cases,

there is no static correlation in the water molecule, so any energy lowering from MCSCF would actually be caused by dynamic correlation. Compare correlation energies (post-HF energy corrections) between a singlet CAS(2,6) calculation (just to pick an active space that will run quickly) with an MP2 calculation.
For the beryllium atom, it is well-known that the ground-state wavefunction from CASSCF should be ~90% $\mathrm{(1s)^{2}(2s)^{2}}$ and ~10% $\mathrm{(1s)^{2}(2p)^{2}}$ divided evenly over the 3 $p$-orbitals (they are degenerate in the absence of any fields), so a CAS(2,2) will qualitatively capture this effect.
My guess for $\ce{Be(OH)2}$ is that this effect is only qualitatively changed by the presence of two hydroxide groups breaking spatial symmetry, which will remove the $p$-orbital degeneracy and require more orbitals in the active space.
I am not familiar with nitrate, but looking at the Lewis structure, there are 3 resonance structures which clearly contribute equally, with 2 electrons moving between 3 bonds. Without looking at orbitals, I would try a CAS(2,3) and a CAS(2,6) to start.

I have neglected to discuss any of the subtleties of MCSCF calculations which are numerous and not always obvious. The one I most blatantly ignored above is I assume the starting orbitals are in the correct order. Unless you reorder the orbitals prior to starting MCSCF iterations, CAS(2,2) will pick the last 2 occupied orbitals and the first 2 unoccupied orbitals. These might not be the ones you want! Look at your starting orbitals! The other one I left out is technical, which is how the MO and CI coefficients are optimized during CASSCF iterations leading to final CSFs, and single-state versus state-averaged or multi-state MCSCF.

Image source: Chemgapedia


Why is magic acid a superacid? The above is a representation of magic acid. The left part is the acid; the right part is the solvent. Why would this combination make for a particularly good acid?

The left part looks like a sulfuric acid derivative, except one of the oxygens has been replaced by an electronegative fluorine. Why can't we replace all the oxygens with fluorines? Wouldn't that make for an even stronger acid through inductive stabilization of negative charge after proton loss?

Also, does the solvent play a role in magic acid's acidity? Does the highly partially positive Sb central atom play a role in stabilizing negative charge on the solute after the solute has been deprotonated?

The explanation is that sufficiently strong Lewis acids like $\ce{SbF5}$ will form adducts even with very weak, traditionally non-coordinating Lewis bases like $\ce{SO3F-}$. The resulting anionic complexes are extremely stable, with the negative charge distributed over a large, highly polarizable molecule and inductively stabilized by the fluorine atoms. The formation of this complex, in turn, promotes the autoprotolysis equilibrium:

$$\ce{HSO3F + HSO3F <=> SO3F- + H2SO3F+}$$

The protonated species above is the most active molecule in sufficiently concentrated solutions. (In reality, the composition of these mixtures may be much more complex, with dimers of the form $\ce{Sb2F10(SO3F)-}$ and possibly more complex oligomeric structures. See, for example, this book.)

While there are some more recent and exotic examples of such Brønsted/Lewis acid mixtures, the idea is actually a fairly old one. Perhaps the most common example: sulfur trioxide in sulfuric acid gives a solution known as oleum, which contains a proportion of disulfuric acid molecules. The resulting mixture is considerably more acidic than pure sulfuric acid alone and has various uses in industrial applications as well as laboratory-scale synthetic organic chemistry.

As for why the oxygens cannot be subsituted with fluorines to yield an even stronger acid system, I can only give a speculative answer. It seems to me that a hypothetical molecule such as, e.g., $\ce{SF3OH}$ (if this is what you're imagining) would probably be very liable to decompose:

$$\ce{SF3OH -> SOF2 + HF}$$

In any case, I can find absolutely no relevant literature, which in and of itself says something about the stability of such a molecule (or perhaps my ineptitude, as the case may be).


How does the position of an electron donating group affect acidity in benzoic acid? I  know that the closer the electron withdrawing group to the $\ce{-COOH}$, the stronger is the acid. I am interested as to how this relationship works out in the case of an electron donating group.

The strength of the acid is determined by strength of the positive charge that is on the carbonyl carbon so in case of the presence of an electron donating group you have also to check what is its inductive effect whether it has an electronegative part which would attract electrons towards it or not ....the best way is to draw the resonating structure for every case and you will reach how is the activity affected by the EDG in different positions..one last factor you have to bear in mind is that the presence of an OH group in ortho position to the COOH will form a hydrogen bond which makes it even more stable than other positions ..if we take salicylic acid as an example then the strength is ortho>meta>para


Predict which hydrogen is more acidic in the following organic compounds? Between 2-chloro-2-phenylpropionic acid and hydrogen phthalate, which is the more acidic compound?

I'd say the $\ce{H}$ on the second one is more acidic as the $\ce{-COOH}$ group is directly attached to the aromatic group. Is that correct?

In the second molecule, the charge on the carboxylate ion is not stabilised through resonance, so we can rule out any resonance stabilisation. Moreover, the fact that the benzene ring is in fact electron donating within the $\pi$ system doesn't help either.

More importantly, however, (as it was pointed out to me by @orthocresol and @hippalelectryon in chat) one can expect intramolecular hydrogen bonding.

Moreover, after the second deprotonation, the second molecule would have two negative charges in close proximity to each other, and this, again, is obviously destabilising.

A combination of these factors would make the second molecule a weaker acid.

In the first case, however, the inductive electron withdrawing effect would increase the acidity, (and the benzene ring is far enough to not have an effect).

Thus, I would say it is the first compound that is more acidic.

@Hippalectryon used a software to compute a pKa of 3.87 for the first one, and Wikipedia records 5.51 for the second one, which confirms my claim.


In reaction equations, why are j and k sometimes capitalized to J and K? From 'Leighton relationship' on Wikipedia:
through the following mechanism:[2]
($J_1$) $$\ce{NO2 + h\nu (\lambda < 420 nm) → NO + O (^3P)}$$
($k_2$) $$\ce{O (^3P) + O2 + M → O3 + M}$$
($k_3$) $$\ce{NO + O3 → NO2 + O2}$$
This series of reactions creates a null cycle, in which there is no net production or loss of any species involved.
And here:
Since $\ce{O (3P)}$ is very reactive and $\ce{O2}$ is abundant, $\ce{O (3P)}$ can be assumed to be in steady state, and thus an equation linking the concentrations of the species involved can be derived:
$$\ce{[O_3]} = {J_1\ce{[NO2]} \over k_3\ce{[NO]}}$$
$${\displaystyle [{\ce {O3}}]={\frac {J_{1}[{\ce {NO2}}]}{k_{3}[{\ce {NO}}]}}}$$
$${\displaystyle [{\ce {O3}}]={\frac {J_{1}[{\ce {NO2}}]}{k_{3}[{\ce {NO}}]}}}$$
The ${J}$ is capitalized, but the $k$ (or $k$s) are not....
Does this signify anything?

The entry in Wikipedia misses to define the entries in the formula.  But note equations 1 and 2 in a paper by Griffin et al. (doi 10.1029/2006JD007604, open access) in the Journal of Geopphysical Research, Atmospheres. To establish context of the paper by Griffin et al., the authors introduce the reader to the topic citing earlier work.

«In the troposphere, a molecule of $\ce{NO2}$ is degraded photochemically to form a molecule of $\ce{NO}$ and an oxygen atom ($\ce{O}$). It should be noted that this process depends on the sufficient flux of photons associated with light of the appropriate wavelength to photolyze $\ce{NO2}$. The $\ce{O}$ atom quickly reacts with molecular oxygen ($\ce{O2}$) in the presence of a third body to form $\ce{O3}$. However, the $\ce{O3}$ formed can react rapidly with the $\ce{NO}$ to reform $\ce{NO2}$ and $\ce{O2}$. The net result of these reactions is a null cycle, and when steady state is assumed, the $\ce{O3}$ concentration can be predicted as [Seinfeld and Pandis, 1998]
$$ [\ce{O3}] = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}]}$$
where the bracket notation represents a concentration or mixing ratio (here, parts per billion by volume, ppbv), $j_\ce{NO2}$ ($\pu{s^{-1}}$) is the photolysis rate of $\ce{NO2}$, and $k_1$ ($\pu{ppbv^{-1} s^{-1}}$) is the temperature-dependent rate constant for the reaction between $\ce{NO}$ and $\ce{O3}$.»

The article continues by

«This is referred to as the photostationary state for $\ce{O3}$. If each side of equation (1) is divided by $[\ce{O3}]$, the right hand side is termed the Leighton ratio, $\Phi$ [Leighton, 1961]:
$$\Phi = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}] [\ce{O3}]} $$
[...]»


Are there natural or artificial enzymes that break down metals? Do natural or human engineered enzymes exist that can speed up the break down metals (element, compounds & alloys)?

If no, then what biological or chemical construct can break down metals in the same way that enzymes break down food in a person's stomach?

Silly answer

There is no enzyme that grinds down an Airstream, but an alligator might do the job!

Serious (but disappointing) answer

In nature, most metals seem to exist as oxides, sulfides, carbonates, halides, etc. and not as bare metals. Consequently, there was/is probably little evolutionary benefit in tackling the bare metals.

Breaking down something to into very small particles, as you described it, is a mechanical process that leaves the oxidation state of the metal unchanged. It is not a chemical transformation.

Nothing cool with enzymes an metals?

There's a lot! Two topics come to my mind:

Chemosynthesis

There's a bunch of bacteria that can oxidize $\ce{Fe(II)}$ to $\ce{Fe(III)}$

at different pH levels
in the presence and absence of oxygen
in the presence and absence of light

Acidithiobacillus ferrooxidans and Leptospirillum ferrooxidans - look at their family names ;-) - operate at $pH\ < 4$, use the oxygen from the air and munch away pyrite.

Metal hyperaccumulation

Some plants and bacteria harvest enormous amounts of metal ions even from highly contaminates soils and store them. The metals are held by by the action of proteins. If you're interested, here is a freely available article on PLOS One Pathogens with lots of further references.


Antiseptic role of kerosene When I was kid and got a huge cut on my finger, I remember my grandmother putting lots of kerosene on my wound, as there was no health facility around.

I want to confirm. Do chemists agree on the antiseptic role of kerosene, or was grandma just using useless remedy?

Just as a note to anyone considering it as an antiseptic agent, it is toxic and probably absorbs through the skin.
Any compound that causes a change in osmotic pressure within a pathogen may kill it through plasmolysis, depending on how much of the compound is soluble in water.  There may be other mechanisms by which a hydrocarbon poisons pathogens like the interruption of metabolic pathways.
Anyhow petroleum distillates are dangerous to apply to the skin and nobody should ever do it on purpose.


Electronic strucure of cyclopropene radical cation Suppose I have a cyclopropene molecule with a + charge on the singly bonded carbon.I think the singly bonded carbon is $\mathrm{sp^2}$ hybridised, isn't it? To obtain this species from cyclopropene, one can consider removing two hydrogen atoms, one homolytically (the one bonded to pure p orbital of carbon) and the other heterolytically (the one bonded with $\mathrm{sp^2}$ orbital of carbon).

So, finally, there is one electron in the pure p orbital and one unit positive charge on one $\mathrm{sp^2}$ hybridised orbital, shouldn't the electron in the pure p shift to the $\mathrm{sp^2}$ orbital as $\mathrm{sp^2}$ has more s character (so it is near the nucleus), does this shift happens?

I am unclear about this scenario, can someone please explain where the electron and the positive charge reside in the C atom.

You do not really have a singly bonded carbon atom. When you remove the hydrogens from the originally saturated carbon the pi electrons delocalize to make an aromatic ring.  In effect you have a cyclopropenyl cation with one of the hydrogen atoms homolytically removed.  The positive charge is thus delocalized in the ring but the odd electron is localized to the carbon with no hydrogen, in the outward pointing $sp^2$ hybridized orbital.


Which definition of Acids and Bases conforms to the pH scale? It's all in the title. Is it the Arrhenius or the Brønsted-Lowry definition? Or is it some other definition?

Also, our teacher told us that $pH$ stands for "Power of Hydrogen", but some sources say it is a number. So which it is?

A few things first...
Arrhenius Theory:

An "acid" is a proton donor; a "base" is a hydroxide donor.

So for example, $\ce{HCl}$ and $\ce{NaOH}$ (both in water) ionize to produce $\ce{H+}$ and $\ce{OH-}$ respectively.
$$\ce{HCl -> H^+ + Cl^-}$$
$$\ce{ NaOH -> Na^+ + OH^-}$$
Which gives us the all-too-familiar neutralization reaction:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
Brønsted–Lowry Theory:

An "acid" is a proton donor; a "base" is a proton acceptor.

At first, the base being a "proton acceptor", may appear to be in contradiction with the Arrhenius Theory... but it isn't.

 The following's from Chemguide. It's been put so well, that I didn't even bother with paraphrasing it
The Brønsted–Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.
Hydroxide ions are still bases because they accept hydrogen ions from acids and form water.
An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.

When an acid in solution reacts with a base, what is actually functioning as the acid is the hydronium ion. For example, a proton is transferred from a hydronium ion to a hydroxide ion to make water.
$$\ce{H3O^+ + OH^- -> 2H2O}$$

 I'm assuming you already know what the hydronium ion is...so I won't get into that

Or, meh, I'll re-phrase that last bit. Remember that example/equation we first  brought up?
$$\ce{HCl + NaOH -> NaCl + H2O}$$
Looking at it from the Brønsted–Lowry POV, your "acid" is really just the hydronium ion, because that's the real proton-donor here; the "base" is really the hydroxide ion, because that's the real proton-acceptor here.
The hydronium ion (acid) passes on a proton to the hydroxide ion (base).
$$\ce{H3O^+ + OH^- -> 2H2O}$$

Back to your question:

Which definition of Acids and Bases conforms to the pH scale?

Both the Arrhenius and Brønsted–Lowry definitions are "compatible" with $\mathrm{pH}$ . They both point in the same direction ;)

Our teacher told us that pH stands for power of Hydrogen, but some sources say it is a number. Which is it?

$\mathrm{pH}$ is, by definition, (approximately) the negative logarithm (base 10) of the molar concentration of $\ce{H+}$ ions in a solution.
Therefore, $\mathrm{pH}$, is indicative of the $\ce{H+}$ strength, in a solution.
I've been led to believe that the $\mathrm{p}$ in $\mathrm{pH}$, comes from the German "Potenz", or the French "Puissance", both which mean the English "Power". I'm under the impression it was originally "Potenz" since German was the language of science well up to the end of the 19th century (and then the whole World War thing ruined that...); besides, the man who came up with the $\mathrm{pH}$ notation was a Dane, and Danish is a Germanic language with little semblance to French. However, Wikipedia suggests it's more likely "Puissance", since the laboratory the man worked for was French-speaking ...that's still just a guess though.
Your teacher isn't wrong, the $\mathrm{p}$ does mean "power" (or even "potential").
However, $\mathrm{pH}$ is a number (it is dimensionless/unit-less...but not without meaning either), so your "sources" are correct too ;)


Do all conversion factors have an infinite number of significant figures? I believe conversion factors have an infinite number of significant figures. They should be exact values; but not all of them are, are they?

I looked at a pdf document (via the Internet Archive) and it agreed with the fact that all conversion factors have an infinite number of significant figures.

I know this is true with a conversion factor such as $\frac{\pu{1 cal}}{\pu{g ^\circ C}}$, because this is how a calorie is defined. Similarly, it's the same for $\frac{\pu{1 kg}}{\pu{1000 g}}$, $\frac{\pu{4 quarts}}{\pu{gal}}$, etc.. We know these are exact, and in dimensional analysis, if you multiplied by the first conversion factor, you wouldn't end up with one significant figure because of the single significant figure in $\pu{1 cal}$.

However, what is the case with measured conversions, such as $\frac{\pu{4.184 J}}{\pu{cal}}$ or $\frac{\pu{1.60934 km}}{\pu{mi}}$? These were measured using significant figures, so it would make sense to round with significant figures when multiplying (dimensional analysis), but the above link says otherwise.

What should I do, especially with the below type of conversions?

Your intuition is correct. There are two types of conversion factor.
One kind of conversion factor is based on definitions. These conversions are exact (infinite significant figures).
In addition to your examples of the specific heat of water (in calories), the kilogram/gram conversion, and the quart/gallon conversion, other conversions are exact, including:

The speed of light - $c=299,792,458\ \mathrm{m/s}$ the length of the meter is then derived from this constant.
The price of gasoline/petrol.
The inch is defined as exactly $25.4\ \mathrm{mm}$, and thus the conversions from all imperial to metric lengths (including your  are exact as well.
The specific heat of water in joules - the relationship between the calorie and the joule is defined exactly as $\mathrm{1\ cal = 4.184\ J}$

Note that most conversions of this type are unit conversions. Many units have been defined exactly in terms of SI-units.
Other conversion factors are based on measurements. One common example is atomic mass. Atomic masses are valuable conversion factors that are based on measurements. However, some unit conversions live in this category.
The electronvolt is based on the measurement of the elementary charge of the electron (which is not exact). Thus the electronvolt is $\mathrm{1\ eV=1.6021766208(98)\times 10^{−19}\ J}$.