Predict the number and nature of bonds using Molecular Orbital Theory I found the following question in a high school chemistry problem book:

According to Molecular Orbital Theory which of the following is true:
(A) In $C_2$ molecule both the bonds are $\pi$ bonds.
(B) In $C_2^{2-}$ ion there is one $\sigma$ and one $\pi$ bond.

Both of these are true. But how can one know this using Molecular Orbital Theory?
P.S.: I don't know anything about Quantum Mechanics as I am a high school student. My textbooks only provide a result-oriented explanation. I would appreciate if the answers keep this in mind.

I presume you can draw the homonuclear diatomic MO diagram for these two species.

As you probably already know, the s orbitals of the carbons form bonding and antibonding orbitals, both of which are filled. (The net effect of this is the “lone pairs” on each carbon.)

The p orbitals for 2 pi bonds and a sigma bond. Due to s-p mixing, the pi bonds are lower in energy and are filled first, and these 2 bonds (with their 4 electrons) are the 2 bonds we see in C2.

Add another 2 electrons to the system and you realise they fill the sigma bonding orbital, which means that the C2 2- anion has 2 pi bonds and a sigma bond (for a total bond order of 3).

Hope this helped.


Numbering in phenolphthalein versus ascorbic acid According to Wikipedia, the recommended IUPAC name for phenolphthalein is 3,3-bis(4-hydroxyphenyl)-2-benzofuran-1(3H)-one, and for ascorbic acid is (5R)-5-[(1S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one.

Why does the numbering in phenolphthalein start from the $\ce{O}$ atom next carbon atom, while in ascorbic acid it starts on the $\ce{O}$ atom? If we want to prioritize the keto group, why doesn't the numbering start on the carbonyl carbon? And other groups attached to the furan on both molecules don't seem prioritize any of the 2 counting.

The subsituents do not take the highest priority here, rather, it is the parent molecule that determines the numbering. According IUPAC's organic chemistry book for nomenclature, the Blue Book, for heteromonocyclic components fused to a benzene ring:

  P-25.2.2.4 (...) The presence of a single heteroatom determines the numbering in a monocyclic compound; the heteroatom has the locant ‘1’.

This gives the following numbering for the isomers of benzofuran:

The parent molecule of ascorbic acid is a furan ring, for which the numbering is dictated by skeletal replacement ('a') nomenclature:

  P-15.4.1.2 For naming and numbering purposes, the following decreasing order of seniority is followed: $\ce{F}$ > $\ce{Cl}$ > $\ce{Br}$ > $\ce{I}$ > $\ce{At}$ > $\ce{O}$ > $\ce{S}$ > $\ce{Se}$ > $\ce{Te}$ > $\ce{Po}$ > $\ce{N}$ > $\ce{P}$ > $\ce{As}$ > $\ce{Sb}$ > $\ce{Bi}$ > $\ce{C}$ > $\ce{Si}$ > $\ce{Ge}$ > $\ce{Sn}$ > $\ce{Pb}$ > $\ce{B}$ > $\ce{Al}$ > $\ce{Ga}$ > $\ce{In}$ > $\ce{Tl}$.


Why is the polar-nonpolar electronegativity difference defined to be 0.4? In introductory college chemistry, bonds between elements that have an electronegativity difference $\Delta EN > 0.4$ are generally consider to be polar (and thus have a dipole moment), while $\Delta EN < 0.4$ bonds are considered to be nonpolar. I think more advanced chemistry classes generally don't adhere strictly to this threshold, but I'm still curious if this 0.4 threshold has any meaning (or if it's just arbitrary).

For example, if there's an experiment plotting $\Delta EN$ vs. the intermolecular force strength, and there's a big jump on the graph at 0.4, that would be convincing evidence. Does anyone have an explanation along this line of reasoning?

(My question could also apply to the $\Delta EN > 1.7$ threshold for ionic vs. polar covalent. I saw a question on this before here, but I think it's worth further discussion.)

Both the $0.4$ threshold for nonpolar/polar bonds and the $1.7$ threshold for polar/ionic bonds are arbitrary choices. When defining nonpolar, polar and ionic bonds, it is important to allow $\ce{C-H}$ bonds ($\Delta \chi = 0.4$) to be nonpolar — it is the best description of alkanes’ low polarities and it facilitates discussions so much. Likewise, a few compounds commonly known as salts should fall just within the ionic boundary for ease of discussion. Teaching introductory levels is just so much easier if you can draw things black and white.

At higher levels, it should always be remembered that not only the thresholds are arbitrary but even electronegativity as a concept follows quite arbitrary definitions.

Except for one electronegativity scale that links electronegativity to ionisation enthalpy and electron affinity, all definitions and all defined reference points (typically fluorine having $\chi = 4.0$) are arbitrary.
Choosing a definite value for a specific element is also arbitrary since the physical implication of electronegativity depends on:

bonding partners
structure (i.e. hybridisation)


How can I get self-diffusion coefficient from MSD (mean sqaure displacement) I run some molecular dynamic simulations for polymer system (liquid), and I have data of MSD vs t(time). MSD-t relation looks linear proportional. Now, which one is the proper way to calculate the self diffusion coefficient?

1) D = MSD / 6t (for 3D system) and get the asymptotic value at large t.

2) Slope of normal plot of MSD-t

3) Slope of log-log plot of log(MSD)-log(t)

4) Slope of semi-log plot of MSD-log(t)

I searched for literature's but there are too many different methods, not sure what to choose here.

I would usually recommend the second option (fitting to the slope of the MSD at long times) over the first (dividing MSD by the time). If you compute $\langle \Delta r^2(t)\rangle/6t$, there is a residual effect of the velocity correlations at short time.

Suppose the velocity correlation function decays exponentially in time
$$
\langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle = \langle |\mathbf{v}|^2\rangle \exp(-t/\tau) .
$$
This is related to the diffusion coefficient by
$$D= \frac{1}{3} \int_0^\infty \langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle \, dt
=\frac{1}{3}\langle |\mathbf{v}|^2\rangle\tau .
$$
Since
$$
\Delta \mathbf{r}(t)=\mathbf{r}(t)-\mathbf{r}(0)=\int_0^t \mathbf{v}(t') \, dt',
$$
it follows that
$$
\frac{d}{dt} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, [1-\exp(-t/\tau)] ,
$$
and this converges towards the correct answer, $6D$, exponentially fast in the time $t$.
On the other hand
$$
\frac{1}{t} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, \left[ 1 - \frac{\tau}{t} + \frac{\tau}{t}\exp(-t/\tau)\right]  .
$$
This converges more slowly, proportional to $t^{-1}$.

Looking at a log-log plot may be useful, as it may highlight changes between different regimes of behaviour (for example, at long times) that you would miss otherwise. This might be especially true of a polymer system like yours. But your option (3) "slope of log-log plot"  does not give you the diffusion coefficient! It just gives you an idea of whether the behaviour is diffusive (slope $=1$) or not (slope $\neq1$). And option (4) "slope of semi-log plot" does not give you $D$ either.


Is there a relation between transition density and density differences? When I have an excited state, is there a relation between the transition density associated with this electronic transition and the density difference between the excited state density and the ground state density? Can the transition density be equal to the difference density?

By definition, the transition density is not the same as the difference density. In the following derivation, the transition density is $\mathbf{T}^{\mathrm{CIS}}$ and the difference density is $\mathbf{P}^{\Delta}$. Although it may seem like they can be formed from the same quantities, the true difference density contains orbital relaxation terms and can be used to build a relaxed density, which is the correct excited state density. Neglecting those terms (the $P_{ia}^{\Delta}$) will give an unrelaxed density. I will also try to show that the unrelaxed difference density is still not equivalent to the transition density.

Derivation of individual CIS densities

Here is the derivation of the transition density for doubly-occupied spin-restricted MOs, with quoted blocks and equations directly taken from the original CIS paper.

The CIS approximation is identical to the TDA approximation in TD-DFT. Full TD-HF or TD-DFT follows the RPA equations, which include deexcitation terms in addition to excitation terms. The equations would be different for RPA, but conclusion would be identical.

The standard indexing convention is followed: $\mu, \nu, \lambda, \sigma$ for AOs, $i,j,k,l$ for occupied MOs, $a,b,c,d$ for virtual MOs, and $p,q,r,s$ for any MO. Define the Hartree-Fock (or SCF-type, valid for DFT as well) density as a sum over occupied MOs:

$$
P_{\mu\nu}^{\mathrm{HF}} = \sum_{i=1}^{n} c_{\mu i} c_{\nu i}
$$

  The two-particle CI-singles density matrix, $\mathbf{\Gamma}^{\mathrm{CIS}}$, can be written in terms of the HF ground-state density matrix and the ground-to-excited-state transition density matrix, $\mathbf{T}^{\mathrm{CIS}}$:

  $$
\Gamma_{\mu\nu\lambda\sigma}^{\mathrm{CIS}} = \frac{1}{2} \left[ P_{\mu\nu}^{\mathrm{HF}} P_{\lambda\sigma}^{\mathrm{HF}} + 2 T_{\mu\nu}^{\mathrm{CIS}} T_{\lambda\sigma}^{\mathrm{CIS}} - P_{\mu\sigma}^{\mathrm{HF}} P_{\lambda\nu}^{\mathrm{HF}} - 2 T_{\mu\sigma}^{\mathrm{CIS}} T_{\lambda\nu}^{\mathrm{CIS}} \right]
$$

  where $\mathbf{P}^{\mathrm{HF}}$ is given (above) while $\mathbf{T}^{\mathrm{CIS}}$

  $$
T_{\mu\nu}^{\mathrm{CIS}} = \sum_{ia} a_{ia} c_{\mu i} c_{\nu a}
$$

I can't find the definition of $\{a\}$ in the paper, but it is the set of coefficients from a converged CIS calculation that describe the single-particle excitation from every occupied orbital to every unoccupied orbital. As an aside, they are the elements of the $\mathbf{X}$ vector in the RPA eigenvalue problem:

$$
\begin{pmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{B} & \mathbf{A}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
=
\omega
\begin{pmatrix}
\mathbf{1} & \mathbf{0} \\
\mathbf{0} & \mathbf{-1}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
$$

$\mathbf{A}$ and $\mathbf{B}$ are parts of the orbital Hessian. If you diagonalize the whole thing, you get the set $\{\omega\}$ of all excitation energies back as eigenvalues. As mentioned above, setting $\mathbf{B} = \mathbf{0}$ gives the TDA or CIS approximation, so there is no longer a deexcitation vector $\mathbf{Y}$.

Importantly, we always start from the set of ground-state MO coefficients $\{c\}$ and describe excitations on top of them.

  The CI-singles excited-state density matrix, $\mathbf{P}^{\mathrm{CIS}}$, is also constructed as a sum of HF and excited-state terms:

  $$
P_{\mu\nu}^{\mathrm{CIS}} = P_{\mu\nu}^{\mathrm{HF}} + P_{\mu\nu}^{\Delta}
$$

  Here we have introduced $\mathbf{P}^{\Delta}$, the CI-singles $\Delta$ density matrix. This can also be called a "difference density matrix", since it represents the changes in electronic distribution upon excitation. It is not, however, the same as the transition density matrix $\mathbf{T}^{\mathrm{CIS}}$ defined above. As we shall demonstrate, it is the use of the true CI-singles density matrix required by (the CIS gradient expression that isn't important here) and not the simple one-particle density matrix (1PDM) which allows the realistic computation of charge distributions, orbital populations, and electronic moments of the excited state. To see this distinction, first consider the $\Delta$ density matrix which would be added to the HF density matrix in order to generate the 1PDM for an excited state. In the MO basis, it is a symmetric matrix with both occupied-occupied (OO) and virtual-virtual contributions

  $$
\begin{align}
P_{ij}^{\Delta} &= -\sum_{ab} a_{ia} a_{jb} \\
P_{ab}^{\Delta} &= +\sum_{ij} a_{ia} a_{jb}
\end{align}
$$

  with the occupied-virtual (OV) elements all zero. The true CI-singles density matrix required in (the CIS gradient expression) will have exactly the same OO and VV contributions, but the OV terms are not all zero. The appearance of these off-diagonal block elements in the excited-state density matrix can be interpreted as orbital relaxation following the initial gross charge rearrangement due to excitation. That is to say, the CI coefficients will by themselves describe some of the gross features of charge redistribution in the excited state, but the response of the wave function to an external perturbation will account for further refinement in electronic properties. These OV terms can be found by solving a single set of CPHF equations:

  $$
L_{ai} = \sum_{bj} \left[ (ij||ab) - (ib||ja) \right] P_{bj}^{\Delta} + (\epsilon_a - \epsilon_i) P_{ai}^{\Delta}
$$

  where the $\mathbf{L}$ vector is the CI-singles Lagrangian:

  $$
\begin{align}
L_{ai} &= C1_{ai} - C2_{ai} + \sum_{kl} P_{kl}^{\Delta} (al||ik) + \sum_{bc} P_{bc}^{\Delta} (ab||ic) \\
C1_{ci} &= -2 \sum_{jab} a_{ia} a_{jb} (cb||ja) \\
C2_{bk} &= -2 \sum_{ija} a_{ia} a_{jb} (ik||ja)
\end{align}
$$

  The total CI-singles $\Delta$ density matrix required for $\mathbf{P}^{\mathrm{CIS}}$ can be generated by backtransforming the entire MO basis $\Delta$ density matrix defined by $P_{ij}^{\Delta}$, $P_{ab}^{\Delta}$, and $P_{ia}^{\Delta}$:

  $$
P_{\mu\nu}^{\Delta} = \sum_{pq} P_{pq}^{\Delta} c_{\mu p} c_{\nu q}
$$

Show that the transition density and unrelaxed difference density are not equivalent

We want to compare $T_{\mu\mu}^\mathrm{CIS}$ and $P_{pq}^{\Delta}$, but they are not in the same basis, so one must be transformed. Because I don't want to mess with the block-diagonal structure of $\mathbf{P}_{\mathrm{MO}}^{\Delta}$, I will transform $\mathbf{T}$ to the MO basis:

$$
\begin{align}
T_{pq}^{\mathrm{CIS}} &= \sum_{\mu\nu} c_{\mu p} T_{\mu\nu}^{\mathrm{CIS}} c_{\nu q} \\
&= \sum_{\mu\nu} \sum_{ia} c_{\mu p} c_{\mu i} a_{ia} c_{\nu a} c_{\nu q}.
\end{align}
$$

Now consider the unrelaxed difference density, meaning the ov/vo blocks are zero due to neglecting orbital relaxation effects:

$$
P_{pq}^{\Delta} =
\left(
\begin{array}{c|c}
-\sum_{ab} a_{ia} a_{jb} & \mathbf{0} \\
\hline
\mathbf{0} & +\sum_{ij} a_{ia} a_{jb}
\end{array}
\right)_{pq}.
$$

Here is a slightly hand-waving proof. I am not sure if the MO coefficients in the double sum can be simplified, but it doesn't matter; assume they are unity. $T_{pq}^{\mathrm{CIS}}$ contains terms with $a$, while $P_{pq}^{\Delta}$ contains terms with $|a|^2$. Unless there is an extra transformation that can be done, I don't see how they can be equivalent.

If you want a little test, generate some fake matrices and see for yourself. $\texttt{x} \equiv X_{ia} \equiv a_{ia}$.

#!/usr/bin/env python

from __future__ import print_function

import numpy as np

np.random.seed(42)

nocc = 2
nvirt = 2
norb = nocc + nvirt
nov = nocc * nvirt
nbasis = norb

x = np.random.rand(nocc, nvirt)
c = np.random.rand(nbasis, norb)
cocc = c[:, :nocc]
cvirt = c[:, nocc:]

p_delta_oo = -np.einsum('ia,jb->ij', x, x)
p_delta_vv = np.einsum('ia,jb->ab', x, x)

p_delta_ov = np.zeros(shape=(nocc, nvirt))
p_delta_vo = p_delta_ov.T

p_delta = np.asarray(np.bmat([[p_delta_oo, p_delta_ov],
                              [p_delta_vo, p_delta_vv]]))

# same as np.dot(cocc, np.dot(x, cvirt.T))
t_ao = np.einsum('ia,mi,na->mn', x, cocc, cvirt)
# same as np.dot(c.T, np.dot(t_ao, c))
t_mo = np.einsum('mp,mi,ia,na,nq->pq', c, cocc, x, cvirt, c)

print(t_mo)
print(p_delta)

which gives

[[ 1.82856579  1.89892758  0.59008502  3.09931192]
 [ 1.48134568  1.53535055  0.49193325  2.47904162]
 [ 0.54242862  0.56250364  0.17874155  0.9109361 ]
 [ 1.79242654  1.85791848  0.59456093  3.00118592]]
[[-1.75629929 -1.76345302  0.          0.        ]
 [-1.76345302 -1.77063588  0.          0.        ]
 [ 0.          0.          1.22441763  1.71443377]
 [ 0.          0.          1.71443377  2.40055604]]

Final comment

My original understanding of the term "difference density" comes from a coworker who was taking the real-space representation (cube files) of two state densities, say the ground state $0$ and an excited state $n$, and subtracting them.

$$
\Delta \mathbf{P}^{0\rightarrow n} = \mathbf{P}^{(n)} - \mathbf{P}^{(0)}
$$

Is this equivalent to the definition of the difference density from above?

$$
\begin{align}
\mathbf{P}^{(n)} - \mathbf{P}^{(0)} &= \mathbf{P}^{\mathrm{CIS}(n)} - \mathbf{P}^{\mathrm{HF}} \\
&= \left( \mathbf{P}^{\mathrm{HF}} + \mathbf{P}^{\Delta(n)} \right) - \mathbf{P}^{\mathrm{HF}} \\
&= \mathbf{P}^{\Delta(n)}
\end{align}
$$

Looks good!

References

James B. Foresman, Martin Head-Gordon, John A. Pople, and Michael J. Frisch. Toward a systematic molecular orbital theory for excited states. The Journal of Physical Chemistry 1992 96 (1), 135-149, DOI: 10.1021/j100180a030
Frank Neese. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009 253 (5-6), 526-563, DOI: 10.1016/j.ccr.2008.05.014


What exactly does the term "gram atoms" mean? The Wikipedia article on the mole states:

   The term gram-atom (abbreviated gat.) has been used for a related but distinct concept, namely a quantity of a substance that contains Avogadro's number of atoms, whether isolated or combined in molecules. Thus, for example, 1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.

Can anybody explain the difference between moles and gram atoms, and how is it that 1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$?

$\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.

Why is this so? Let us consider this problem in the familiar mole:

If we were asked how many moles of the substance (molecule) were in this, we would state 1 mole, if there were $\pu{45.3 g}$ of substance. However, if we were asked how many moles of atoms were in the substance, we would accordingly state 3. Thus, in one molecule, there are 3 atoms, so if we simply consider the number of atoms, there would be the necessity of counting the number. of atoms in a molecule, as a molecule is comprised of more than a single atom

* This is purely hypothetical. One would probably not consider  $\ce{MgB2}$, as J. LS pointed out, it decomposes . However, the concept can be applied to other compounds


How can hydrogen and oxygen coexist in high temperature? Where did the oxygen come from? A gasification worked with the combustion of bio pallets with production of syn gases. Pure oxygen and $\ce{H2O}$ were fed into the combustion reactor to combust the pallets with no opening of any syn gas outlets. Once the pallets got ignited enough, then there was no more oxygen input. A kind of gas reformer (catalyst) was equipped in the reformer. The temperature in the reactor reached more than $\pu{1000 ^{\circ}C}$. Then all the syn gases were collected and analyzed. What I found was that the composition of the gases were methane (43%), hydrogen (23%), oxygen (20%), and other gases (carbon dioxide and monoxide). I don’t understand how such a high volume of oxygen coexisted with hydrogen together in that high temperature reactor. Is it possible that those gases stay together without an oxidation reaction occurring? Where did the oxygen come from?

I'll answer the question in general of how Hydrogen and Oxygen can coexist in metastable mixtures at higher temperatures, though without hazarding a guess for your specific case. Like you and others, I'd be very surprised if your bulk gas is actually well mixed at $\pu{900 ^{\circ}C}$ without exploding! It's far enough beyond the autoignition temperature, that an explosive reaction seems inevitable. However, metastable combinations of methane, hydrogen and oxygen in the $\pu{550 ^{\circ}C}$ range have been reported in the literature.

Hydrogen can be a surprisingly picky burner. I was told by a colleague who specialized in rocket propulsion that in between Hydrogen's upper and lower explosive limits are a host of regions where it strongly resists ignition, even above its reported autoignition temperature of ~$\pu{570 ^{\circ}C}$. According to her, the reaction between hydrogen and oxygen is quite complex with several important intermediates. Having concentrations in these non-explosive regions apparently allows the hydrogen-oxygen mix to be metastable for long periods. This would of course not extend to long time periods at $\pu{900 ^{\circ}C}$!

More in my territory, there's also many chemicals which inhibit the slow combustion (non-explosive) of hydrogen and oxygen up through the $\pu{500 ^{\circ}C}$s. The most common is $\ce{KCl}$, which strongly inhibits the reaction of oxygen and hydrogen in non-explosive conditions allowing a metastable mixture to exist. Walls of vessels can also similarly inhibit reaction. The reaction mechanism apparently stems from $\ce{KCl}$ and other surfaces' abilities to consume reaction intermediates of the combustion such as monatomic Hydrogen:

$$\ce{KCl + H^. -> K^. + HCl}$$
$$\ce{SiO2 + H^. -> SiO(OH)}$$
or direct adsorption on metal surfaces like nickel, palladium, etc.

After doing a bit of digging, methane is apparently a superb inhibitor of hydrogen and oxygen combustion. The effect is explained in a 1959 paper by R. Baldwin, et al. (see following reference). The paper explains that when methane is above a sharp critical concentration it drastically raises the autoignition temperature of hydrogen's explosive reaction and almost completely eliminates slow combustion in non-explosive conditions. In other words, when enough methane is present, the hydrogen and oxygen mixture either does nothing or explodes completely. From data taken in systems between $\pu{450\! -\! 550 ^{\circ}C}$, there was a very clear relationship of methane concentration vs. explosive limit taking the form of (values approximate):

Explosive Limit (in $\pu{K}$) = $\frac{\pu{3350K}}{2.3-\log_{10}(X_\ce{CH4})}$

where $X_\ce{CH4}$ is the mole fraction of methane, studied in the 0.01 - 0.1 range (significantly lower than yours). For this particular heuristic, the mole fraction of hydrogen was 0.28, oxygen 0.14, and the balance was nitrogen. The paper suggests the mechanism is non-equilibrium formaldehyde formation.

As a final note, please be very cautious when handling or recreating these mixtures. Just because the hydrogen isn't burning, doesn't mean it can't explode. A lab down the hall from mine was fortunate to loose only a wall to a hydrogen evolving system which ran fine for months and then, quite suddenly, decided to explode.

Best of luck learning more about your system! It sounds fascinating.

Reference:

R. R. Baldwin, N. S. Corney, R. W. Walker, “The inhibition of the hydrogen + oxygen reaction by methane,” Trans. Faraday Soc. 1960, 56, 802–814 (DOI:10.1039/TF9605600802).


Methanol vs Ethanol Substrate cleaning We have to clean substrates, often made of STO or MgO, before PLD for thin film growth. Previously we would use ethanol for sonication to clean the substrates. We recently ran out of ethanol but have a lot a methanol. In looking on the web I have only found conflicting viewpoints on which to use, or how corrosive methanol is. Are the two interchangeable for cleaning substrates? From where do you know?

Both $\ce{SrTiO3}$ and $\ce{MgO}$ can safely be washed with methanol. These compounds are insoluble in most any solvent other than certain acids. I suspect that any online reference you've found saying methanol is corrosive is with respect to organic compounds, nothing like the composition of your substrates.

One (somewhat handwaving) way to convince yourself that this is the case is that we know the compounds are insoluble in water (see their Wikipedia and PubChem pages) and ethanol (from your experience, also confirmed for $\ce{MgO}$ in it's Wikipedia and PubChem pages). In terms of solvent polarity, methanol lies between water and ethanol. It would be extremely unlikely for an ionic solid to be completely insoluble in water and ethanol yet have some significant solubility in methanol.

Although we can be confident that the substrate materials will not be destroyed by methanol, only trial will let you know how effective it is for cleaning whatever contamination may be present in your situation. Methanol is fairly, though not completely, similar to ethanol in it's solvent properties, and there is a good chance that it will be as effective as ethanol. Again though, only trial will verify this in your situation.


Lab test to compare the reactivity of liquid 1-heptene and liquid heptane Describe how you would carry out a lab test to compare the reactivity of liquid 1-heptene and liquid heptane. State clearly what the possible results of the test are and what they would indicate.

Would you carry out the lab test using bromine?
Add about 1 mL (about twenty drops) of 1-heptene to one
container and about 1 mL of heptane to the other container.
Add an equal volume of bromine water to each of the liquids then
shake each container to the same extent.
Leave the containers to settle, then observe.

Would this process work for heptane and 1-heptene? Would the colour of the solution change?

According to this Wikipedia article:

  "In electrophilic halogenation the addition of elemental bromine or
  chlorine to alkenes yields vicinal dibromo- and dichloroalkanes
  (1,2-dihalides or ethylene dihalides), respectively. The decoloration
  of a solution of bromine in water is an analytical test for the
  presence of alkenes."

Alkanes tend to be much less reactive toward bromine, so yes, your bromine water test should allow you to distinguish between heptane and 1-heptene.


Designing a mathematics course for 1st year chemistry students Background:
I am a mathematician with very little knowledge in chemistry (I do know some
physics). I will probably teach a course "mathematics for (1st year) chemistry
students" in the future (1st semester 3h + 1h exercises per week, 2nd semester
2+1h). This will be the only (mandatory) mathematics course in the chemistry Bachelor curriculum (and even the subsequent Master curriculum, as far as I know). The students should have high school mathematics background, such as onedimensional derivates and integration (at least of polynomials), but probably no complex numbers. I will have a lot of freedom in the selection and presentation of topics.
Questions:

Which topics are commonly covered in a "mathematics for chemistry
students" courses?
Closely related: Which textbook are most commonly used?

In more detail:

Can I safely assume that mathematical proofs and even rigorous definitions can/should mostly be skipped?

How black-boxy is Quantum physics used in the typical chemistry curriculum?
Even the most basic Quantum physics will need, as minimum, some linear algebra (Hermitian, normal and unitary operators, exponention of operators, etc.) as well as Fourier transformation and a little bit of differential equations. Should these topics be coverd, or is it enough to say "one can show that in an H atom, orbitals look like this", and just talk a bit about complex valued functions in R^3?
Do you ever do calculations with spins? (Mostly spin 1/2, I would assume? That might be a motivation to show some calculations with complex 2x2 Matrices?)

Which computer mathematics software are most commonly used in a chemistry curriculum? (Mathematica? Maple?) I assume it would make sense to do some calculations and exercises using such software?

Update answering some comments:

I am in Austria (sorry, do not know how to add an according tag), but I would be interested in how these things are handled in other countries as well.
This course has been held before at our department (not by me), but I would like to get "unbiased" input from people from chemistry, and therefore did not list the traditional topics / lecture plan.
Of course I will speak to people in the Chemistry department, but I would be happy to get some ideas here first (I do not know the Chemistry department people, and it could be that a randomly selected person who happens to be willing to indulge me has some unorthodox views).

At least in the US, most undergraduate chemistry curricula do not go deep enough into the math of QM that you need to spend a great deal of time on it. Students that do take more advanced QM classes can learn the math at that time. I also am not sure it's safe to assume that all incoming students at that level have a good understanding of 1D calculus, though most will have had some exposure.
With that in mind, here are topics that I'd include (not in any specific order), structured around the relevant applications:

basic calculus review of integration and differentiation in 1D - main intro level application is chemical kinetics. Also applies to intro quantum.

setting up and solving systems of equations, especially including iterative solving methods - intro level application is solving systems of multiple equilibria by mass and charge balance

a module on statistics with a focus on dealing with experimental error and error propagation. Include methods of curve fitting and associated pitfalls. - applies to lab component of curriculum. Probably one of the more commonly overlooked aspects of math in chemistry

Fourier transform and other techniques involved in spectroscopy

Basics of symmetry operations and group theory


C6H5CO2H dimerization in Benzene Why does $\ce{C6H5COOH}$ in benzene form dimers? Also why only in benzene, why not in other solvents such as water?

In general, what are the properties of compounds which form dimers or polymers in benzene?

In polar solvents such as water, the water molecules will solvate the carboxylic acid. Thus, hydrogen bonds will form between the 2 molecules and the carboxylic acid will dissolve.

The water is in large excess as it is the solvent so it will solvate most of the carboxylic acid molecules.

In a non-polar solvent such as benzene, no such hydrogen bonds can take place with the carboxylic acid and benzene molecules. As a result, the carboxylic acids simply interact with themselves and form dimers.


Why is fresh ice sticky? Fresh ice cubes are almost instantly sticky and easily cling on to fabric and other similarly rough surfaces.

A few minutes later, however, the effect almost completely disappears.

What is the cause of this adhesion?
Why does the effect vanish after some time?

Whether or not the ice cube will adhere to something is dependent on the temperature of the ice, the state of the surface of the ice, and the moisture content of the material.

The temperature of an ice cube can range from the freezing point of water ($0\ \mathrm{^\circ C}$) to well below this temperature. Typically most materials have a small degree of moisture regardless of whether they are “wet” or not. If the ice cube is cold enough, it will freeze the water present in the material. The giant covalent network structure of the ice extends into the material, accounting for why it sticks to the ice—the moisture in the material has become part of the ice cube.

If the surface of the ice cube is has melted, it will not be able to freeze the water in the material any more than it will be able to freeze the ice already melted on its surface. No giant covalent network solid will form, and the ice will not stick to the material.


Will the motion of particles always increase with the rise of temperature? I came across this following question while poring over my textbook.

  Predict the change in the motion of particles you would expect in the following if the temperature was increased:
  a) Solid
  b) Liquid

I think if the temperature increases, solid or even liquid particles will receive more energy and therefore their motion will speed up. I'm not sure why there are two parts for this question. Aren't they supposed to be similar?

Thank you so much for helping! Since this is my first post, please pardon me if I accidentally violate any rules of the forum.

Indeed they are similar, but there is a difference. For solids, particles can only vibrate about their own position. Heating results in a stronger vibration. For liquids, particles can wander around the whole volume of the liquid. Heating increases the speed of the particles so that they can move more quickly inside the liquid, hence a higher diffusion constant.


Are common names of substituents accepted by IUPAC? Are names of substituents like isopropyl and sec-butyl accepted by IUPAC?

Or do we need to name them according to standard IUPAC conventions?
(Ex- 1-methylethyl for isopropyl)

My book uses common names for substituents at a lot of places but I am doubtful about this.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general (IUPAC) nomenclature. However, many prefixes are no longer recommended.

  Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

The prefix “isopropyl” for $\ce{(CH3)2CH-{}}$ is retained only for use in general (IUPAC) nomenclature.
For the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”.
The prefix “1-methylethyl” may be used in general (IUPAC) nomenclature.

The prefix “sec-butyl” for $\ce{CH3-CH2-CH(CH3)-{}}$ was still contained in the 1993 recommendations but is no longer recommended as approved prefix.
For the preferred IUPAC name (PIN), the preferred prefix is “butan-2-yl”.
The prefix “1-methylpropyl” may be used in general (IUPAC) nomenclature.


Separating iron from blood How can I separate iron from blood, without doing any damage to the blood/haemoglobin?

I have thought about using a magnet, but apparently iron in blood is not magnetic, and I have also looked at filtration techniques but not sure if that will affect the haemoglobin.

Updated Dec 2017: Let me add some context, I was/am looking for an alternative for chelation therapy such as Desferaloxamine infusions to remove iron in patients with iron overload as a result of repeated transfusions due to anemia

It is not possible to separate iron from blood without damaging the haemoglobin. Iron is a part of haemoglobin, and if iron is removed, the haemoglobin becomes something else. The blood then becomes incapable of transferring oxygen.


Which software to use to output molecular orbital diagrams? I want to output a molecular orbital diagram. Can anyone recommend a software to do this? Also I searched for a python module, but didn't found a pure solution. Maybe I am using the wrong keywords.
I want to connect two atoms with a specific electron configuration, to see at which energy levels they have pi- and sigma-bindings.

If your interest is a qualitative reprepresentation, you may reach out for an editor like inkscape and use lines, squares and Unicode to create a picture like the following one

In text fields you may enter Ctrl + U followed by an unicode code like
2191 (for $\uparrow$), 2193 (for $\downarrow$), 21C5 (for $\uparrow\downarrow$).*)
If you have access to LaTeX, you may be interested in modiagram, especially if you are interested in an output as .[e]ps or .pdf.  Only a few parameters are needed for an illustration like

(credit)
(It is possible to construct more advanced orbital diagrams like this and this, too.)

For the quantitative, and more programmatic approach you seek, especially with Python, there are solutions with the matplotlib library which might require some adaptation by you (example).  One of them is PyEnergyDiagrams once shown here on ChemSE:

the other EnergyLeveller.py:

Addition:  With the special chars used here, maybe a couple like Ipe + LaTeX is more handy to use, than inkscape.


Does possibility of Tunneling influence Ground State Energy of PIB with and without finite energy barriers? Suppose there are two Particle in a Box models with the same initial energy and same size. Suppose model A has a finite wall or barrier in between the two infinite walls, and the system has sufficient energy for tunneling to occur.

Would the occurrence of Tunneling influence the ground state? If so, how? Which model would have a higher or lower state and why?

You are asking about a double-well potential where the two wells are separated by a finite square wall. This model is very well studied as it is the simplest model which involves tunneling through a barrier, so it provides a conceptual understanding to tunneling along a reaction coordinate.

The potential is visualized below along with the two lowest energy solutions:

Note, however, that in order to make a good comparison to the particle in a box, one must think about two times the particle in a box energy. This is because the double-well potential is really two infinite square-wells with a finite barrier between them. So, this tells you that in the limit that the barrier in the middle becomes infinitely large, there will be two ground state solutions with equal energy. Namely, the particle could be on the left or on the right.

So, as we lower the barrier and the particle is allowed to tunnel between the two wells, this degeneracy will be broken. These are the two wavefunctions which you see in the diagrams above. Namely, the solutions very closely resemble a symmetric and an antisymmetric linear combination of the ground state of a particle in an infinite box.

So, this has the effect of splitting the degeneracy into two states, one of which is higher in energy and one of which is lower in energy than particle in a box.

Why should the ground state of this double well be lower in energy than the ground state of a particle in a box where the particle stays on only one side of the barrier?

One way to think about this is that tunneling allows the particle to "leak" into the other box, which has the effect of making the effective box size of the particle larger.

This is all very qualitative, but the details of the math aren't really all that much more enlightening than simply looking at the qualitative features of the problem. The most important point I think is to recognize that tunneling through this barrier has the effect of splitting an otherwise degenerate state into a lower and higher energy state.


What is the difference between polyatomic ions and compounds? I don't understand why things like hydrogen phosphate are polyatomic ions but things like potassium nitrate are compounds. Can someone explain to me what the difference is, besides that ions are charged?

The ions, by themselves, are charged and thus cannot be complete compounds which are electrically neutral.

We sometimes do have names for polyatomic ions as If they are compounds, like cyanidin (which is a cation containing a pyrylium ion function).  But we understand that a "generic" counterion is needed to complete the compound.  For example, "cyanidin" (a polyatomic cation) can't be completely isolated by itself, but it could be incorporated into cyanidin chloride (a compound), which in principle could be isolated.


Michaelis Menten derivation for 2 enzyme substrates We know the Michaelis Menten derivation for the following reaction:

$$\ce{E + S <=> ES -> E + P}$$

However, what if the reaction took place in a different scenario whereby:

$$\ce{E + S <=> ES1 -> ES2 -> E + P}$$

What would the derived Michaelis Menten equation be now?

I will propose an outline for determining an expression for the rate of substrate conversion in the given case:

Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $$\ce{E + S <=>[k_1][k_{-1}] ES_1->[k_2] ES_2 ->[k_3] E + P}$$
Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.
Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES1 and ES2.
Make the pseudo-steady-state assumption (PSSA): assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e. $\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$.
Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.


Will heating of KClO3 be considered disproportionation? In all examples of disproportionation I have seen, there is one particular atom whose oxidation state is both increasing and decreasing to give rise to different products

For example, on heating phosphorous acid

$$\ce{4 H3PO3 -> 3 H3PO4 + PH3}$$

the oxidation state of phosphorous changes from +3 to +5 and -3.

However, in the case of heating $\ce{KClO3}$

$$\ce{2 KClO3 -> 2 KCl + 3 O2}$$

the oxidation state of oxygen changes from -2 to 0, while that of chlorine changes from +5 to -1.
So, will such a reaction be called a disproportionation since its the same molecule giving rise to two products, or will it be simply a redox reaction?

Thermal decomposition of potassium chlorate is not disproportionation, just a redox reaction.
Disproportionation refers to the same element acting both as oxidizing agent and a reducing agent, resulting in compounds which contain the same element in different oxidation states.

On the other hand, if you consider preliminary decomposition stage involving perchlorate formation, then it is a disproportionation reaction:

$$\ce{4 K\overset{+5}{Cl}O3 ->[\pu{400 °C}] 3 K\overset{+7}{Cl}O4 + K\overset{-1}{Cl}}$$

Also, synthesis of $\ce{KClO3}$ is a disproportionation reaction:

$$\ce{6 KOH + 3 \overset{0}{Cl}_2 -> 3 H2O + 5 K\overset{-1}{Cl} + K\overset{+5}{Cl}O3}$$


Selectivity of m-CPBA I have a question about the selectivity of m-CPBA in this reaction. In my experience, I have used m-CPBA for epoxidation of alkenes. It surprised me that in this reaction, the ketone reacts to give an ester and the alkene remains unchanged? Is it due to the presence of the overhanging ether group? Does this provide a sort of steric crowding so that the peroxyacid can't get in? Cheers.

The transformation taking place in your question is a name reaction known as a Baeyer-Villiger.

The Baeyer-Villager transforms ketones into esters/lactones (cyclic esters) via an alkyl migration in which the group most able to stabilise a positive charge is observed to migrate.

A general mechanism is presented below (taken from the wonderful name reactions book cited). The key steps are (1) attack of the peroxide into the carbonyl followed by (2) the alkyl migration to form the ester grouping.

Source: Strategic Applications of Name Reactions, 2005.


How does copper oxidise alcohols? Consider for example ethanol being heated in the presence of copper at 300 °C. How does it give ethanal as a product?
Copper being in its elemental state (oxidation state 0). Since negative oxidation state is not possible for Cu, which means that it should reduce ethanol to  ethane. Certainly, high temperature should have some role here. What could it be?

You likely had either or both of the following:

Copper had some oxide on it, and this oxidized the alcohol leaving the metal behind.

If there was air in your system, especially on heating copper could have reacted with it to generate the oxide (a relatively fast reaction with heating) and you're back to (1).  The metal would then, of course, be available to react with more air.  In effect this would be a catalytic oxidation.

Copper is actually known in negative oxidation states, but that does not happen or need to be invoked here.


can i calculate equilibrium constant of this reaction? Good day,
i had organic chemistry labs. And we had this reaction:
$$\ce{(CH3)3COH + HCl -> (CH3)3CCl + H2O}$$

My question is:
how do i explain my yield of reaction is only 40%?

My theory is:
reaction is reversible so the K isn't so high and that lead to yield 40%.
If my theory is right please explain me how to calculate K?

I had these informations:

V$\ce{((CH3)3COH)}$ = 10 ml

V$\ce{(HCl)}$ = 20 ml
HCl was concentrated.

Thank for answers.

$\ce{H2O}$ is a very poor nucleophile and we are dealing with a tertiary substituted carbon, so we can minimize the scope of possible reaction mechanisms to the $\ce{E 1}$ and $\ce{S_{n}1}$ type mechanisms. $\ce{E 1}$ will compete here with $\ce{S_{n}1}$, forming tert-Butyl chloride and isobutylene respectively. That is probably where most of your loss is coming from. Other reasons can be indeed that a bit is left in the equilibrium, but I don't think that would be much since $\ce{H3O+}$ is a much better leaving group than $\ce{Cl-}$. Note that is chemistry papers the yield obtained is almost never a 100% of theory.


What are the priorities for substituent groups on a cyclohexane? I was wondering about this specific scenario:
If you have a cyclohexane, for example, and the cyclohexane has two consecutive substituent groups, for example a methyl group and an ethyl group, how do you decide where the numbering starts?
Would it be 1-ethyl-2-methylcyclohexane or would it be 2-ethyl-1-methylcyclohexane?
Whoever answers this, could you just give some sort of general rules on what to do in this situation (e.g. does it follow alphabetical order, etc.)?

If the numerical sequence of locants is equivalent, the numbering should be assigned to correspond with alphabetical order. Hence, in your example, correct would be 1-ethyl-2-methylcyclohexane. Quoting the IUPAC Blue Book, Rule A-2.4:

  If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.


Kb values of N,N-dimethylaniline derivatives I need the $K_\mathrm{b}$ values for ortho-, meta- and para- positions of the following compounds:

I've looked and couldn't find them anywhere.

chemicalize.com is  a cheminformatics website which is the kind of resource you are looking for.

You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.

Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are calculated / estimated, not measured $\mathrm{p}K_\mathrm{a}$ values.

Table 1. $\mathrm{p}K_\mathrm{a}$ values (of the conjugate acid)

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & 3.71 & 5.19 & -0.98 & 1.44 & 2.66 \\
\textit{meta-} & 3.89 & 5.15 & 1.73 & 4.77 & 3.45 \\
\textit{para-} & 5.55 & 5.46 & 0.65 & 4.68 & 3.65 \\
\hline
\end{array}
$$

Since the question is asking for $K_\mathrm{b}$ values, I converted them too.

Table 2. $K_\mathrm{b}$ values

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & \pu{5.13e-11} & \pu{1.55e-9} & \pu{1.05e-15} & \pu{2.75e-13} & \pu{4.57e-12} \\
\textit{meta-} & \pu{7.76e-11} & \pu{1.41e-9} & \pu{5.37e-13} & \pu{5.89e-10} & \pu{2.82e-11} \\
\textit{para-} & \pu{3.54e-9} & \pu{2.88e-9} & \pu{4.47e-14} & \pu{4.79e-10} & \pu{4.47e-11} \\
\hline
\end{array}
$$


What is the difference between grafting and bonding in reference to polymers? I'm reading research papers and some keep mentioning grafting of polymers. Looked it up in wikipedia and it talks about adsorbing polymers, and im not sure what the difference is. Thanks for any help.

Grafting a polymer B on a polymer A is attaching  polymer B to the linear backbone of polymer A (co-polymer).
If A and B are identical, we obtain a branched polymer (Homopolymer).


Ring opening of aziridinylmethyl and oxiranylmethyl radicals When a radical is alpha to a three-membered ring, it is common for the ring to open via β-scission (to relieve ring strain):

When the ring contains a heteroatom there are a couple of different possibilities:

Ab initio studies of these reactions1 showed that for both X = O and X = NH, the C–C cleavage was thermodynamically favoured (presumably due to stabilisation of the subsequent radical by the adjacent lone pair). For X = O, the C–C cleavage was also found to be kinetically favoured, but for X = NH, the C–N cleavage is kinetically favoured!

This regioselectivity is also backed up by experiment2 and also has potential for use in synthesis.3

Is there an explanation for why the kinetically favoured products are different for X = O and NH?

References

Pasto, D. J. Ab Initio Theoretical Studies on the Ring-Opening Modes of the Oxiranyl-, Aziridinyl-, Oxaziridinyl-, and Thiaranylmethyl Radical Systems. J. Org. Chem. 1996, 61 (1), 252–256. DOI: 10.1021/jo9505942.
Marples, B. A.; Toon, R. C. An investigation into the β-cleavage of aziridinylcarbinyl radicals. Tetrahedron Lett. 1999, 40 (26), 4873–4876. DOI: 10.1016/S0040-4039(99)00900-4.
Prévost, N.; Shipman, M. Intramolecular Radical Rearrangement Reactions of 2-Methyleneaziridines: Application to the Synthesis of Substituted Piperidines, Decahydroquinolines, and Octahydroindolizines. Org. Lett. 2001, 3 (15), 2383–2385. DOI: 10.1021/ol016194k.

The 2-methyl-3-phenyloxiranylcarbinyl radical has been shown EXPERIMENTALLY to cleave the C-O bond kinetically while the C-C bond is cleaved thermodynamically. This means that, in this example, The oxygen-centered radical forms faster than the highly stable benzyl radical. The formation of the oxygen radical is necessarily reversible. See ref. 11 in your ref. 2 (Marples paper)

Ziegler, F. E.; Petersen, A. K. Allyloxy Radicals are Formed Reversibly from Oxiranylcarbinyl Radicals: A Kinetic Study, J. Org. Chem. , 1995, 60, 2666.


Could thin layer chromatography separate enantiomers? My textbook states the thin layer chromatography could separate enantiomers only if both/or the mobile and solid phases are chiral?

Why so?

The reason is simple: the physical properties of chiral pairs are exactly the same.

So, for any given compound that is chiral, the melting point, the solubility, the partitioning between liquid and solid phase in a chromatography column etc. are all exactly the same for the pairs of chiral molecules. Simple physical processes can't distinguish the two molecules. The solvents and solid components in chromatography are rarely chiral, so they can't tell the enantiomers apart.

The only things than can are other chiral things. If the solid phase of a chromatography column were chiral (or the solvent, though that is a lot less common) then the column could separate the enantiomers.

Think of shaking hands. It is natural that a right handed person shakes hand with the right hand of another person but the interaction is radically different if they to shake hands with the other persons left hand. The interaction of chiral things (like hands) depends on chirality. But if you bump fists instead (fists being closer to a sphere which has no handedness) then it matters a lot less which hand your friend uses.


Derivation of an equality in Michaelis–Menten kinetics Enzymatic action may be described as follows:

$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_\mathrm{2}] Enzyme + Product}$$

The initial rate of enzyme-catalyzed reactions can be described by the Michaelis-Menten equation:

$$\mathrm{rate} = \frac{V_\mathrm{max}[\ce{S}]}{K_\mathrm{M} + [\ce{S}]} = \frac{k_\mathrm{cat}[\ce{E}][\ce{S}]}{K_\mathrm{M} + [\ce{S}]}$$

where $V_\mathrm{max}$ is the maximum rate, $[\ce{S}]$ the substrate concentration, $[\ce{E}]$ is the enzyme concentration, $K_\mathrm{M}$ is the Michaelis constant and $k_\mathrm{cat}$ is the number of catalytic cycles per second.

It is known that

$$V_\mathrm{max} = k_2[E]_0$$

and by inspecting the equation above we can deduce that

\begin{align}
V_\mathrm{max} &= k_\mathrm{cat}[\ce{E}]\\
\implies k_2[\ce{E}]_0 &= k_\mathrm{cat}[\ce{E}]
\end{align}

How, and why, does this equality hold?

How do we know when to use ${[\ce{E}]_0}$ and $[\ce{E}]$ in rate equations, and what are the implications of using either?

${[\ce{E}]_0}$ refers to the enzyme concentration at the start of the reaction, and [E] refers to the concentration of enzyme at any point in time during the course of the reaction.

Your equation

  $$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_2] Enzyme + Product}$$

contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.

The correct schema for the Michaelis-Menten kinetics would be

$$\ce{Enzyme + Substrate <=>[k_1][k_{-1}] ES complex ->[k_2 = k_{cat}] Enzyme + Product}$$

Note that $k_{cat} = k_2$ is the rate constant for the decay of the ES complex in the second step.

From this you can derive the Michaelis-Menten rate law as described in the Wikipedia article you refer to.

$[E]_0$ is the initial concentration of the enzyme. During the reaction the enzyme exists as free enzyme ($[E]$) and bound in the enzyme substrate complex $[ES]$. These concentrations are related by the equation $[E]_0 = [E] + [ES]$.


For bond block the numbering can be done at random or it has to follow some order

Bonds don't need to be in any particular order, although usually bonds are created first for atom 1, then atom 2, etc. However, the first bond could be between atom 101 and atom 100 if you wish.

  Also in the atom block the order in which atoms occur can it be random or again it has to follow any order

Atoms can be in any order. Some programs use "canonical ordering" so that the same molecule will always have the same atomic order, but this is rarely used because it can be time-consuming.

  Is order of numbering done in bond block and atom block related

As indicated above, both the atom and bond blocks could be in random order. The relation, obviously, is that for a given molecular structure, there will be bonds between bonded atom pairs.

  Is there any free software which converts mol files to structure

There are many, but you don't describe whether you want 3D (presumably) or 2D structures.

Personally, I suggest Avogadro which can create 3D coordinates for files.

If you're generating a large batch of 3D structures, you will likely want to use Open Babel or RDKit or something similar. Both are open source and free.


Why we don't talk about unit cells in hcp/fcc structures? I'm studying basics of Solid State Chemistry from this source.

So there are 14 types of Bravais crystal lattices (Primitive + Centered) in 3D. Every solid in this universe is made by the unit cells of these lattices. OK so far so good.

Later on there was the topic of Close packing and Structure of Metallic Crystals and we are told about hcp /fcc types of structures.

Problems:

Why do we remain silent about the unit cells and crystal system in this context? It seems like a huge disconnect.
Is there any relation between the Hexagonal Unit cell and hexagonal close packing?

We do not remain silent about the unit cells in this context. The hexagonal close packed structure is called so because it has hexagonal unit cell, and the other one is called fcc precisely because its unit cell is face-centered cubic.

(source)


Effect of drinking ultra-pure water What would be the effect if someone were to drink ultra-pure water with an electrical resistivity of $18 \, \mathrm{M} \Omega \! \cdot \! \text{cm}$?

Would they immediately die?  Would they just need to pee more?  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

What would be the threshold for bad things to happen — a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I'm not sure what the science here would be.

What would be the effect if someone were to drink ultra-pure, 18 M-ohm water?

Not much, although if they drank many gallons of the water it could be a problem.

  Would they immediately die?

No.

  Would they just need to pee more?

Probably.

  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices.  Same as regular water.

  What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I don't think the effect would be that different from drinking regular water.  All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.

Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia.  The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM.

A very rough approximation to understanding how toxic water can be would then be to compare the concentration difference of sodium (or potassium) between the ingested water and the blood.  A 25 micromolar -- i.e. 25 μM or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than "pure" 18.2 megohm water (0.055 microsiemens per cm).  According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM.  For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water.


Why is the principal quantum number irrational? My textbook says that for a $d$-electron, the orbital angular momentum is: $$ \frac{\sqrt{6}h}{2\pi} $$
But since:
$$ mvr = \frac{nh}{2\pi}$$
$$ n \in N$$
where $N$ is the set of natural numbers, how is the value of $n$ an irrational number?

So, first, as I said in my comment, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Both can take only integer values (positive and non-negative respectively), but for totally unrelated reasons.

Secondly, the orbital angular momentum itself is related to its quantum number $l$ by the following equation,
$$
    \hat{L}^2 \Psi = l (l + 1) \hbar^2 \Psi \, ,
$$
where $\hbar^2{l(l+1)}$ is the square of the magnitude of an electron orbital angular momentum. Thus, while the orbital angular momentum quantum number $l$ is always a non-negative integer, the magnitude of an electron orbital angular momentum $\sqrt{l(l+1)} \hbar$ can take irrational values.

For instance, for a d-electron, $l=2$, and thus, the magnitude of its orbital angular momentum is $\sqrt{2(2+1)} \hbar = \sqrt{6} \hbar = \sqrt{6} h / 2 \pi$.


Electronic model with highest prediction rate Amongst many models, including the valence bond model (VB) or the molecular orbital (MO) model, which are the ones with best predictive power? (e.g. the MO is thought to predict spectroscopic properties better than the VB)

The so called gold standard of quantum chemical calculations of ground state energies and properties is usually considered to be Coupled Cluster Singles Doubles (with perturbative) Triples as a method. This method is based on ab initio molecular orbital theory and it is size consistent, but not variational. A basis set of at least a triple zeta quality with additional valence and polarisation functions should be employed, good examples for this are Dunning's (augmented) correlation consistent basis sets, e.g. (aug-)cc-pV{T,Q,5,6}Z, or Weigend's def2-basis set family, def2-TZVPP, def2-QZVP, but there are various others. (If you like to have some citations, drop a note in the comments.) This level of theory is usually denoted as CCSD(T)/TZVP, where TZVP is a placeholder for one of the various basis sets.

In principle CC theory is exact at the complete basis set limit, but in practise, this is not solvable. Another theory that is in principle exact, because they become equivalent, is Full Configuration Interaction, F-CI. The truncated variants CISD, CISD(T), CISDT, ... are variational, but not size consistent. For chemical systems it is much more important, that the calculations are size consistent, therefore these methods lost their importance. An attempt to solve this was to introduce Quadratic Configuration Interaction, QCI, but this is almost the same as coupled cluster theory, so it was not successful either.

The importance of Many Body Perturbation Theory, MBPT, of high orders, like Møller Plesset Perturbation Theory of fourth order, MP4, has very much decreased because of it inefficiency.

There are some cases, which cannot be accurately described by these models. Those are cases, where the ground state cannot be well defined by one determinante, the initial guess wave function obtained from Hartree-Fock, HF. Here you have to use a different description. A Multi Configuration, MC, or Multi Reference, MR, ansatz is necessary. The qualitative equivalent to HF is Complete Active Space Self Consistent Field, CASSCF. Based on this you can use Perturbation Theory to include more correlation effects. The best possible description so far (judging accuracy versus computational cost) is second order perturbation theory on a CASSCF wave function, for short CASPT2. It is also very reliable for spectroscopic calculations. There are also various developments on Multi Reference Configuration Interaction as another theory. (I have limited expertise on that field.)

Another way of treating chemical systems is Valence Bond theory, or with the application to modern computational aspects. Here you propose the mixing of various Lewis like structures, to obtain the ground state energy and properties. The formalism can get quite complicated, maybe this is why its performance for excited states properties is not very accurate at the moment. It is still a very active field of research, by a few scientific groups. (I am not one of them.)

However, affordable spectroscopic calculations have still huge errors and sometimes the numbers can vary quite a bit. Currently most used is a variety of methods derived from Time Dependent Density Functional Theory, TDDFT. An ab initio method of great importance in this field is Configuration Interaction Singles. Also fairly well implemented are approximate coupled cluster techniques, e.g. CC2. The struggle for finding an accurate, robust and computationally cheap method are one of the main goals of modern quantum chemistry. (I can apply these methods fairly well, but I have no deeper insight how they work.)

For extended systems like surfaces and solids there are various approaches. Since they get very demanding very soon, the most commonly used practical approaches are based on plain or hybrid density functionals. They often also employ plane wave basis sets with periodic boundary condition. (Unfortunately I am not an expert on that field either.)

As you can see, there is no definite answer to your question and the development of new and efficient methods is still a very active field. Depending on what type of description you want or need, there are many possibilities. The most advanced models are obviously for ground state energies and properties, since they have been researched for about a century now.

I hope this helps you, I apologise for the chatty answer. If something remains unclear, drop me a note in the comments.


How can nucleophilic substitution happen at the sp²-hybridised carbon of 4-chloropyridine? Why can substitution of the C–Cl bond occur at the sp2-hybridized carbon in the following reaction (taken from Clayden et al., Organic Chemistry (2nd ed.), p 700)?

Ordinarily, SN1- and SN2-type reactions should not occur at sp2 centres, as described here: Why do SN1 and SN2 reaction not occur at sp2 centres?. What is the mechanism of this substitution?

The mechanism is indeed a nucleophilic substitution but it does not follow the bimolecular substitution paradigm ($\mathrm{S_N2}$) or the unimolecular one ($\mathrm{S_N1}$) that one learns about in basic organic chemistry classes — and it also does not follow the general electrophilic aromatic substitution that is also often presented in introductory courses.

$\mathrm{S_N2}$ can be ruled out because a rear-side attack would require an attack through the ring — a path which is blocked by the nitrogen atom (and the other bits of the ring).
$\mathrm{S_N1}$ is highly unlikely because we would need a phenyl cation — a highly unstable intermediate.

The aromatic ring of 4-chloropyridine can be understood as electron-deficient since pyridine’s nitrogen exercises a strong $-I$ and $-M$ effect. You can draw the $-M$ effect by pushing the electrons of the $\ce{C=N-C}$ double bond to nitrogen to give $\ce{\overset{+}{C}-\overset{-}{N}-C}$ — and this cation can be moved around the ring, most notably also to the 4-position where chlorine is located. See the figure below.

Figure 1: $-M$ effect of the pyridine nitrogen.

The nucleophilic attack is that of the amino nitrogen’s lone pair into this π system; the attack flips electrons so that the pyridine nitrogen has an additional lone pair, two $\ce{C=C}$ bonds are adjacent to this and a tetrahedral carbon bonded to both $\ce{Cl}$ and $\ce{\overset{+}{N}H2R}$ completes the ring.

This zwitterion can break down by reverse electron flipping which displace the chloride to give the skeleton of the reaction product. Additionally removing a proton from the amino nitrogen gives the neutral product. Considering that we have an acid (2-aminobenzoic acid) in the reaction mixture, I have drawn the pyridine nitrogen (the more basic one) to be protonated throughout the mechanism. This assists in the nucleophilic addition step by further reducing the ring’s electron density. See figure 2 for the mechanism.

Figure 2: Mechanism of the nucleophilic aromatic substitution reaction. Nitrogen being protonated due to the acidic effect of 2-aminobenzoic acid facilitates the reaction.

This mechanism is called nucleophilic aromatic substitution $\mathrm{S_NAr}$ and can only occur on very electron-deficient aromatic compounds: typically pyridines or pyrimidines with additional electron-withdrawing potential leaving groups such as chlorine.


Is it feasible to make dendritic crystals at home? I've become interested in the forms of dendritic metal crystals from an aesthetic perspective, and I'm interested in whether it is feasible (from a cost, safety, resources) perspective to create metal crystals of this sort at home?

Would the energy involved, or the cost of materials, the difficulty of controlling conditions sufficiently, or the general complexity of techniques be prohibitive to a non-expert/ someone outside an institution? Are there some metals, or metal compounds (e.g. those with lower melting points - lead?) for which this might be easier?

The single displacement reaction between silver nitrate and metallic copper will yield dendritic crystal growth of metallic silver. Normally, a copper wire is placed into silver nitrate solution. The reaction that occurs is:
$$2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2(aq) + Ag(s)$$


Rate equation variations I am having a bit of trouble understanding variations to rate equation problems.
What I know:

For $\ce{A + B -> AB}$, if the rate is proportional to the concentration of A and B, the rate equation is:

$\quad$ rate=$\ce{k*[A]^1*[B]^1}$

For $\ce{A + A -> 2B + C}$, if the rate  proportional to $\ce{[A]}$:-

$\quad$ rate=$\ce{k*[A]}$ , as per  Question 1c and Answer 1c.

Question 1 : Why is this the case?
E.g. when it was $\ce{1A + 1B}$ and the rate was proportional to the concentration of $\ce{A}$ and $\ce{B}$ our rate equation contains $\ce{[A]*[B]}$. But for $\ce{1A + 1A}$ why does the rate equation contain $\ce{[A]}$ instead of $\ce{[A]*[A]}$?
Question 2: Do molecules change anything (compared to single atoms)
Is $\ce{A + A}$ equivalent to $\ce{A2}$
E.g. will all of these have the same rate equation (if the rate is proportional to $\ce{[A]}$)

$\ce{A + A -> 2B + C}$
$\ce{A2 -> 2B + C}$
$\ce{2A -> 2B + C}$

I can't see why they wouldn't all have the same rate equation as it would just be different ways to represent the same total $\ce{[A]}$.

The issue is that the rate equation is determined not by the reaction equation, but rather by the actual mechanism that lies beneath it. It turns out that reactions that appear fairly simple, such as the burning of methane in oxygen in the flame of a Bunsen burner, do not necessarily take place in a simple manner, I understand that the even the reaction for the Bunsen flame can in real life contain a very large number of steps. Depending on the reaction, different steps will require different reactants, and will take place at different speeds. Maybe for the reaction $\ce{A + B->AB}$, some step that takes a long time involves $\ce{A}$ decomposing on heating, the products of which go on to quickly react further to eventually form $\ce{AB}$. This step that takes a long time is called the rate-determining step, and decides the form of the rate equation. It may be that this causes only $[\ce{A}]$ to feature in the rate equation.

In this case however, the rate equation is given to you by the comments to the right of the equation. Just because we write it as $\ce{2A}$ or $\ce{A+A}$ doesn't make a difference, it's just how someone decided to write the equation, which does not necessarily reflect the steps of the mechanism. In fact, if a mechanism involved a slow step $\ce{A + A -> B}$, the reaction would be second order with respect to $\ce{A}$.

Hopefully that helps.


Why does 1,2-dimethylcyclohexane only possess three stereoisomers? Why does 1,2-dimethylcyclohexane in chair form have only three stereoisomers?

I understand that the trans isomer doesn't possess plane of symmetry and thus it has two stereoisomers. But my book says that the cis isomer has does not have two  stereoisomers. Why is this so?

The structure of cis-1,2-dimethylcyclohexane is shown above. Its mirror image is non-superimposable and cannot be obtained by rotation about single single bonds or by a ring flip. Can anyone explain?Thanks

Quoting , ORGANIC CHEMISTRY second edition by JOSEPH M. HORNBACK, UNIVERSITY OF DENVER

1,2-dimethylcyclohexane has two chirality centers.

The cis-diastereomer has a plane of symmetry bisecting the ring bond between
the two methyl groups  and therefore a meso compound.It is not chiral and does not rotate plane-polarized light.The trans-diastereomer exists as a pair of enantiomers.

The chair conformation of cis-1,2-dimethylcyclohexane  shows that it is chiral.
But due to  ring-flip , another  conformation ,its  enantiomer of the original conformation is also formed.These conformers interconvert rapidly at room temperature, therefore cis-1,2-dimethylcyclohexane is not chiral.

Further I quote

"stereochemical analyses such as the preceding one can be done with drawings using planar rings" pg.235.

A molecule cannot be optically active if its chiral conformations are in
equilibrium with their mirror images. We consider such a molecule to
be achiral. PG- 193 - ORGANIC CHEMISTRY  .EIGHTH EDITION
L.G. WADE , JR . WHITMAN COLLEGE


Percentage purity of Ba(MnO4)2 sample that completely reacts with given H2O2? $\pu{40 g}$ $\ce{Ba(MnO4)2}$ (mol. wt. = 375) sample containing some inert impurities in acidic medium is completely reacted with $\pu{125 mL}$ of $\pu{3 M}$ $\ce{H2O2}$. What is the percentage purity of the sample?

I actually found this question in a book, I found its solution but I cannot properly understand it. This is the first equation given:

Since milliequivalents of $\ce{Ba(MnO4)2}$ = milliequivalents of $\ce{H2O2}$
$$(w/375)\times10\times1000 = 3\times125\times2 $$

I understand the RHS as $3\times125$ gives number of millimoles which multiplied by n-factor gives milliequivalents. But where does the LHS come from? And what is $w$? In the next line it is given

$$ \text{percentage purity} = (w/40)\times100 $$

where the value of $w$ is taken from the first equation. Would someone please explain this to me?

The question does not demand to solve it using "equivalents," I'm going to attempt to solve the problem in universal way, using moles. As OP correctly suggested, this reaction of $\ce{Ba(MnO4)2}$ and $\ce{H2O2}$ is a redox reaction. Since the reaction has taken place in acidic medium and has completely reacted (assuming observation is made by appearance so assuming it was judged by clear solution), the two half reactions should be:

$$
\begin{align}
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & E^\circ &= \pu{1.507 V} \\
\ce{H2O2  &-> O2(g) + 2H+ + 2e-} & E^\circ &= \pu{-0.695 V}
\end{align}
$$

Thus, total redox reaction can be written as:

$$\ce{2MnO4- + 6H+ + 5H2O2 -> 2Mn^2+ + 5O2(g) + 8H2O} \quad E^\circ_\mathrm{rxn} = \pu{0.812 V}$$

The positive $E^\circ_\mathrm{rxn}$ means the reaction is spontaneous. And it also shows that you need $\pu{5 mol}$ of $\ce{H2O2}$ to completely react with $\pu{2 mol}$ of $\ce{MnO4-}$. Since $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$ contains $\pu{2 mol}$ of $\ce{MnO4-}$, it is correct to say that $\pu{5 mol}$ of $\ce{H2O2}$ would completely react with $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$.

Suppose that $\pu{40 g}$ of impure sample contains $x~\pu{g}$ of $\ce{Ba(MnO4)2}$. Then, amount of of $\ce{Ba(MnO4)2}$ in the sample is

$$\frac{x~\pu{g}}{\pu{375 g\cdot mol-1}} = \frac{x}{\pu{375 mol}}.$$

To completely react with that amount, you need

$$\left(5\cdot\frac{x}{375}\pu{mol}\right)~\text{of}~\ce{H2O2}.$$

Thus,

$$5\times\frac{x}{375}\pu{mol} = \pu{3\frac{mol}{L}}\times\pu{125 mL}\times \pu{10^{-3} \frac{L}{mL}} = 3\times \pu{0.125 mol} \label{eq:1}\tag{1}$$

$$\therefore x = \pu{\frac{ 3\times 0.125 \times 375}{5} g} = \pu{28.1 g}$$

Thus,

$$\text{percentage of $\ce{Ba(MnO4)2}$ in the $\pu{40 g}$ of sample} = \frac{28.1}{40}\times 100 = 70.2$$

Note that, the equation $\eqref{eq:1}$ is exactly the same as yours (with minieq.)


Polyhedron with four capping ions Let's say I have a trigonal prism in my crystal structure and this is further capped by two ions it's called a bicapped trigonal prism. If I have three ions it's called a tricapped trigonal prism. But what about four? I cannot find anything like quadcapped, fourcapped, fourfold capped trigonal prism. How would you name that?

After asking more people who already wrote reports with similar structures and a paper I found on the net the most convincing answer I could find was the term 'tetracapped'.

Source: 10.1002/zaac.201400143


Enthalpy and heat relationship They say that enthalpy of reaction is the heat when at constant temperature and pressure. So if enthalpy is

$$\mathrm dH = \mathrm dU + p\,\mathrm dV = T\,\mathrm dS - p\,\mathrm dV + p\,\mathrm dV + μ\,\mathrm dN = T\,\mathrm dS + μ\,\mathrm dN$$

and heat is $T\,\mathrm dS,$ then what happens to the chemical potential term if enthalpy is heat?

They say that enthalpy of reaction is the heat when at constant temperature and pressure.

They should not say that. At least not without saying in the absence of non-PV work. And in a closed system.

  So if [...] heat is TdS [...]

You should not say that without specifying a reversible (near equilibrium) process in a closed system where dS refers to the surrounding:

$$q_\mathrm{rev} = T dS_\mathrm{sur}$$

The sign of $q_\mathrm{rev}$ is from the perspective of the system, i.e. if there is heat transfer from system to surrounding (positive value), the entropy of the surrounding will increase (we don't know what happens to the entropy of the system because there is more going on in the system whereas the only change in the surrounding is that it got heat transferred to it).

On the other hand, all the thermodynamic quantities in your formula refer to the system. By the way, the enthalpy and the entropy of a reaction are not linked, there is no general formula that lets you infer reaction enthalpy from reaction entropy without knowing other quantities.


Conception and manufacture of the Avogadro Project's (nearly) perfect silicon sphere (and, what is “atomic feeling”?) The Avogadro Project, an effort prior to the recent redefinition of Avogadro's constant to design a nearly perfect sphere of silicon-28, apparently relied on the fine fingers of optician Achim Leistner - at least according to the Wikipedia:

  Achim Leistner is an Australian optician of German origin.3 During
  his retirement, he was asked to join the Avogadro project to craft a
  silicon sphere with high smoothness, as automated machining does not
  match his precision.3

  In addition to precision instruments, Leistner uses his hands to feel
  for irregularities in the roundness of the sphere.3 The research
  team has called his extraordinary sense of touch "atomic feeling".3

I suppose this is largely a set of questions about machining and metrology, but materials science is important too. Also, because of the relevance to the redefinition of $N_\text{A}$ I thought it appropriate for this site. With all due respect for Achim Leistner's amazing fingers:

How was this sphere conceived and crafted? For instance: how was the size of the sphere decided? How was the material selected? What is the concentration of defects in the silicon that allows assumption that the lattice is perfect? How is this degree of quality guaranteed? Why can't "automated machining" match the precision of (assuming the alternative) "machining by hand"? What is this purported "atomic feeling"? Is this PR talk, to make the manufacture of the sphere seem more mysterious?

Short answer:  The success of the project benefitted from many contributors.

According to BIPM and its Avogadro project, there are two spheres (reference).  The page equally refers to a publiction about the 2011 $\ce{^{28}Si}$ atom count (paywall) and the report in 2017 (here, open access).  From the later paper, you may identify the origin of the two spheres in the Czochralski crystal obtained initially:

as well as where samples were taken to determine the amount of carbon, oxygen, nitrogen, and boron (loc. cit., table 1), the molar mass of silicon (table 2), concentrations of point defects in the lattice (table 3).  Equally, the analysis took into consideration the $\ce{SiO2}$ layer naturally formed if the sample is exposed to air, as well as adsorbed water; for both, the upper limit of layer width as well as the mass are provided.  As an example, the deposit of oxygen on the spheres was mapped by X-ray photon spectroscopy (XPS):

The article describes that the diameter of the spheres was assessed by a sphere interferometer

in vacuum at $20\,^\circ{}\mathrm{C}$ within a tolerance of $\pm 3\,\pu{mK}$ to allow the statement

  The measured diameter was the apparent diameter, which is not corrected for the phase shift due to the surface layers. The mean apparent diameter was 93 723.723 61(61) μm.

As for imperfections, this property equally was mapped:

With papers titled like The self-weight deformation of an x-ray interferometer the dedicated issue of Metrologia 48(2), April 2011 shed some light on monitoring the quality of the samples and the used tools for the analyses.

Additions:
A video of channel veritassium is dedicated to the reference spheres, too.  The section starting at 7:00 min focusses on carving and polishing them (8:11 min, at 8:19 min likely showing Leistner himself).

A now open-access publication by Fujii et al. equally briefly describes the stages to carve-out the spheres:

Starting from the ingot, a section is cut.  A lathe is used to obtain a sphere already with an shape accuracy of $1\,\pu{mm}$.  To smoothen the surface, these spheres are lapped with alumina particles in water with finer and finer grains; eventually (fine lapping) within a margin of $100\,\pu{nm}$.  To polish the surface, collodial alumina and eventually titanium dioxide yielded a shape deviation of less than $50\,\pu{nm}$ and a surface roughness below $0.2\,\pu{nm}$.  This publication explicitly refers to Leistner's work including publications authored by him (example, example) and then established protocols easing to manufacture such samples reliably.

PTB continues both further development of the interferometers (here) and the deployment of these spheres as substitute of the old standards (example).  (As of writing, the mass standard's group seeks collaborators for Si spheres (reference, perma).)  Comments by Maurice and GM suggest the advancement of mechanical engineering and this new definition of the kilogram might offer smaller mass references which do not need to be as large and heavy as $1\,\pu{kg}$.


Is the Gibbs standard free energy always constant? I am a biochemistry student and we are learning about thermodynamics. Is the Gibbs standard free energy for a reaction always constant? The equation below suggests that it changes with temperature:

$$\Delta G^{\circ\prime} = -RT\ln K'_\mathrm{eq}$$

It is important to note that we are using the biochemistry version of the Gibbs standard free energy equation. To answer my own equation, I believe that it is always constant. Changing the temperature would probably just change the equilibrium constant.

In short, no, the standard Gibbs free energy change is not constant; it is a function of temperature. The same is true for practically all other standard-state quantities.

This gets a little confusing because of how standard-state properties are often explained in lower-level (high-school and college-freshman level) textbooks. The standard Gibbs free energy change, $\Delta G^\circ$, is the value of $\Delta G$ at a certain constant pressure (usually 100 kPa or 1 atm) and constant concentration of solutions (usually 1 molal or 1 molar) and a specified temperature (usually $0~^\circ \text{C}$ or $25~^\circ \text{C}$). What is intended but not immediately obvious is that the pressure and concentration are constants but temperature is a variable; since the temperature is a variable, $\Delta G^\circ$ is a function of temperature. When the value of $\Delta G^\circ$ is given, the temperature must also be specified.

In your equation, then, $\Delta G^\circ$ is the standard Gibbs free energy change of the reaction at the specified temperature ($T$ in the equation). Change $T$, and the value of $\Delta G^\circ$ also changes; but it is still the standard Gibbs free energy change at the (new) specified temperature. (This equation cannot be used to calculate $\Delta G^\circ$ at a given temperature because  $K_\text{eq}$ is also a function of temperature.)

The IUPAC "Green Book" (ed. 3) has a good explanation of what is meant by the term "standard state"; the discussion starts on p. 61 (as printed on the page; it's p. 76 in the pdf file). It explicitly expresses the standard chemical potential ($\mu^\circ$), which is very closely related to $\Delta G^\circ$, as a function of temperature.


What makes ammonium nitrate soluble? Ammonium nitrate is one of the classic examples of an endothermic dissolution.  People will usually explain why it's soluble by stating that the spontaneous dissolution process of ammonium nitrate is what drives the reaction, meaning that the obvious answer is it's soluble because ΔG of dissolving it is negative.  Here is an example.

However, after thinking a bit more about it, this explanation seems very vague and doesn't explain it well.  Enthalpy usually has a stronger role in determining the solubility of salts, since their ΔH values are usually much larger than the TΔS values they would have, especially at room temperature.

Both calcium and aluminum fluoride are also salts that have an endothermic dissolution, but by the number of ions dissociated, one would think that ΔS of dissolving both of these salts would be much higher than that of ammonium nitrate.  Ammonium nitrate is still several times more soluble though.

My question is, what mechanisms allow ammonium nitrate to have such a high ΔS when dissolving in solution that allow it to still be very soluble, despite having an endothermic dissolution?

Here is some data on other well-soluble salts:

Comparing the values, both the anion and the cation contribute to magnitude of enthalpy and entropy. The larger cations seem to contribute to a larger entropy of dissolution, as do the larger anions. The entropy is concentration-dependent, so I assume the data is for a standard state of about 1 mol/L dissolved salt.

There is a nice discussion about the dissolution of ammonium chloride here. The enthalpy of dissolution (14.4 kJ/mol) and the entropy of dissolution (75.3 J/(K mol), source) are close to that of KCl, so there is nothing super-special about the ammonium ion.

  [...] since their ΔH values are usually much larger than the TΔS values they would have, especially at room temperature.

At the beginning of the dissolution process, the entropy dominates. At the solubility limit, entropy and enthalpy contributions are equal and add up to zero.

  What makes ammonium nitrate soluble?

I did not find a data table that includes entropy and enthalpy of dissolution for ammonium nitrate, but you could easily calculate these from entropy and enthalpy of formation of the reactants and products. On second view, I am having trouble finding the data for aqueous nitrate, but this textbook has some values. There is a report that nitrate comes in two forms in aqueous solution, perhaps complicating things.

Some people seem to think that the dissolution of nitrates is special in terms of enthalpy (compare to e.g. carbonate) and others argue that the entropic contribution is special. It would be easier to answer the question, for example, "what makes ammonium nitrate more soluble than ammonium carbonate".


NMR of a bridging Methyl complex I was provided with the following structure of an inorganic metal complex:

And asked to give information on its NMR spectrum shown below

The highfield signal is caused by a singlet with Platinum ($I = \frac{1}{2}$) due to the terminal $\ce{CH3}$ groups.

As $\ce{^{195}Pt}$ has an abuance of $33.8\%$ this meant that in $66.2\%$ of cases, the singlet will not split, whereas in around $33.8\%$ of cases a doublet would be observed due to the satellites, which would have intensities of $\frac{33.8}{2} = 16.9\%$ hence the "Triplet" was in the ratio of around $16:67:16$.

The low-field signal however for me was easy to deduce but more difficult to calculate ratios for.
The Methyl's could couple to

No $\ce{^{195}Pt}$ resulting in a singlet
One $\ce{^{195}Pt}$ resulting in a doublet
Two $\ce{^{195}Pt}$ resulting in a singlet

The calculations given in the answer are shown:

  No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1

These calculations confused me.
Why would for a singlet, the intensity be $0.66 \times 0.66$ if both $\ce{CH3}$ give the same signal and hence for the doublet would I need to multiply $0.66 \times 0.33$ by $2$?

These calculations also seemed erroneous.
The calculation for the intensity of the doublet and singlet give the exact same answer and I could not reach the final ratio.

What is the logic behind these calculations?

Seems to me there is an error in the last line of this list:

  No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1

Simultaneous J-couplings to two $\ce{^{195}Pt}$ should result in a triplet, not a singlet, given that the J-couplings are identical in magnitude.

Therefore the expected relative intensities in the quintet are as follows:

$$\begin{align} &0.34^2\times \frac14 &= 0.029      \tag{1}
\\&0.66\cdot0.34\times1 &=0.224    \tag{2} \\& 0.34^2\times \frac24 + 0.66^2 \times 1 &=0.493   \tag{3} \\& 0.66\times0.34\times1 &=0.224   \tag{4}  \\& 0.34^2\times \frac14 &=0.029\ \tag{5} \end{align}$$

Note there is an implicit assumption that molecules with two $\ce{^{195}Pt}$ result in a 1:2:1 triplet contributing to peaks (1) (3) and (5) above, and that molecules with one $\ce{^{195}Pt}$ account entirely for peaks (2) and (4).

Normalized so that the central peak (3) has an integral of 18 the expected integrals become

$$\begin{align} & 1.0      \\&8.2     \\& 18.0    \\& 8.2     \\&1.0\end{align}$$

Separately I attempted digitizing and integrating the provided image of the spectrum. I obtained the following integrals:

$$\begin{align} & 0.25      \\&7.0     \\& 18.0    \\& 6.2     \\&0.7\end{align}$$

Given the resolution perhaps the disagreement is acceptable.


Calculating weight of soluble salt in a solution I got a solution which density is $\pu{1.035 g mL-1}$ and it contains soluble salt $\ce{Ba(NO3)2}$ at concentration of $\pu{0.250 M}$. The teacher is asking from me to "clear" the solution from the $\ce{Ba^2+}$ ions by using a sedimentation with $\ce{PO4^3-}$ ions that the concentration of $\ce{Ba^2+}$ would be less than $\pu{7.00 ppb}$.

Then she is asking if its possible (well yes it is, but I don't know how to explain this though)? And, if it is possible, what would be the weight of the soluble salt $\ce{Na3PO4}$ that we need to put into a $\pu{50.0 mL}$ of the solution?

And I got the $K_\mathrm{sp}(\ce{Ba3(PO4)2}) = \pu{4.03e-25}$.

Now I probably started by making the right ions equation, which is:

$$\ce{2 Na3PO4 + 3 Ba(NO3)2 -> 6 NaNO3 + Ba3(PO4)2}$$

Then, I calculated the net mass of the solution by multiplying the given density  $\pu{1035 g L-1}$ by given Volume $\pu{0.05 L}$.

And finally I calculated the amount of $\ce{Ba(NO3)2}$ and of $\ce{Na3PO4}$ which are $\pu{0.0125 mol}$ and $\pu{0.0083 mol}$, respectively.

That is the point where I don't know how to mix all this information and to get the right answer.

The final answer: It's possible. If we will put more than $\pu{1.80 g}$ of $\ce{Na3PO4}$ to $\pu{50.0 mL}$ of the solution, the ion concentration of $\ce{Ba^2+}$ will be less than $\pu{7.0 ppb}$.

Ok... It is hard to read you post since the details are scattered around.

Problem Statement

Given: 50 ml of a 0.250 molar solution of barium nitrate with a density of 1.035 g/ml  and that barium phosphate has a $K_\mathrm{sp} = 4.03 \times 10^{-25}$.

Question: How much $\ce{Na3PO4}$ must be added to reduce the concentration of barium to less than 7.00 ppb.

Complications with problem

Initial density

  The density of the initial solution, 1.035 g/ml, is useless.

Final density

  We need a density for the final solution to convert 7.00 ppb $\ce{Ba^{2+}}$ to a concentration of grams per liter. There is no density for the final solution, so the assumption will be that the density is 1.000 g/ml which means that a liter of the final solution will weigh 1000. grams. Since each $\ce{Ba^{2+}}$ cation will be replaced by 2 $\ce{Na^+}$ cations, the final solution will be at least 0.50 molar in $\ce{NaNO3}$. Thus the assumed density of 1.000 g/ml will be too low.

A pragmatic point here about the experimental procedure.

  You typically wouldn't want to add solid $\ce{Na3PO4}$ to a solution containing $\ce{Ba^{2+}}$ since $\ce{Ba3(PO4)2}$ would precipitate on the particles of $\ce{Na3PO4}$ sealing them from the solution and thus preventing complete dissolving of the $\ce{Na3PO4}$ particles. So experimentally you first need to dissolve the $\ce{Na3PO4}$ in a small amount of water and then add the solution of $\ce{Na3PO4}$ to the solution of $\ce{Ba(NO3)2})$. Thus the initial volume would really be over 50 ml.

Rounding errors

  The calculations will be carried to 5 significant figures to try to avoid rounding errors due to intermediate results. The final result will be properly rounded to 3 significant figures

Calculations

Moles of barium in the initial solution

In the initial solution $\ce{[Ba]} = 0.250\ \mathrm{m}$ so 50 ml contains

$\mathrm{m}\ \ce{Ba^{2+}} = 0.250\ \mathrm{m/l}\times 0.050\ \mathrm{l} = 0.0125\ \ \mathrm{m}$

Moles of barium in the final solution

In the final solution for Ba:

$\ce{Ba^{2+}} = 7.00\ \mathrm{ppb}$
$\quad\quad = (7.00\times 10^{-9})\times 1000\ \mathrm{g/l}=7.00\times 10^{-6}\ \mathrm{g/l}$
$\quad\quad  = \dfrac{7.00\times 10^{-6}\ \mathrm{g/l}}{137.33\ \mathrm{ g/m}} = 5.0972\times 10^{-8}\ \mathrm{m/l} $

For 50 ml of solution this is:
$\mathrm{m}\ \ce{Ba^{2+}} = 5.0972\times 10^{-8}\ \mathrm{m/l}\times 0.050\
\mathrm{l} = 2.5486\times10^{-9}\ \ \mathrm{m}$

$2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ so the residual amount of barium in solution can be ignored.

    (1) *In determining the amount of barium phosphate that will precipitate, the amount of barium left in solution is so small that it doesn't matter if the solution has a density of 1.000 g/ml or 1.05 g/ml. So the density of the sodium nitrate solution doesn't really matter for that part of the problem.

    (2) $2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ This is a significant figures argument. In other words for intermediate calculations I'm using only 5 significant figures.
    $$\begin{align}
0&.012\,500\,000\,000\,0 \\
-0&.000\,000\,002\,548\,6\\
\hline
0&.012\,499\,998\,551\,4
\end{align}$$ Now when trimmed to 5 significant figures the intermediate result is 0.012500. But the 0.0125 is exact, so I don't need the two extra zeros at the end.

Residual phosphate in the final solution

For the residual phosphate, in excess of that needed to precipitate the stoichometric amount of barium is:
$K_\mathrm{sp} = \ce{[Ba^{2+}]^3[PO4^{3-}]^2}$ or $\ce{[PO4^{3-}]} = \sqrt{\dfrac{K_\mathrm{sp}}
{[Ba^{2+}]^3}}$

$\ce{[PO4^{3-}]} = \sqrt{\dfrac{4.03\times10^{-25}}{(5.0972\times 10^{-8})^3}} = 5.5164\times10^{-2}\ \mathrm{m/l}$

For 50 ml this would be:

$\mathrm{g}\ \ce{PO4^{-3}}\ \mathrm{as }\ \ce{Na3PO4} = (5.5164\times10^{-2}\ \mathrm{m/l})\times 0.050\ \mathrm{l}\times 163.94\ \mathrm{g/m} = 0.45218\ \mathrm{g}$

  This is too great of an amount of residual phosphate in solution to be ignored. Thus we need to add this to how many grams of trisodium phosphate are needed to precipitate 0.0125 moles of $\ce{Ba^{2+}}$ as $\ce{Ba3(PO4)2}$.

Stoichiometric amount of phosphate to precipitate all the barium

Each mole of $\ce{Ba3(PO4)2}$ requires 3 moles of $\ce{Ba^{2+}}$. So for 0.0125 moles of $\ce{Ba^{2+}}$ there would be

$\mathrm{moles}\ \ce{Ba3(PO4)2} = \dfrac{\mathrm{moles}\ \ce{Ba^2+}}{3}= \dfrac{0.0125}{3} = 4.1667\times10^{-3}$

But each mole of $\ce{Ba3(PO4)2}$ requires two moles of $\ce{PO4^{3-}}$

$\mathrm{moles}\ \ce{PO4^{3-}} = 2\times(\mathrm{moles}\ \ce{Ba3(PO4)2})= 2\times( 4.1667\times10^{-3}) = 8.3333\times10^{-3}$

$\ce{Na_3PO4} = \mathrm{MW}\ 163.94\ \mathrm{g/m}\\
\quad\quad\quad\quad= 163.94\times8.3333\times10^{-3} = 1.3662\ \mathrm{g}$

Total amount of trisodium phosphate needed

$\mathrm{Total}\ \ce{Na_3PO4} = 1.3662 + 0.4522 = 1.8184\ \mathrm{g}$

Now rounding the final result to 3 significant figures we get 1.82 grams.

Now the other pertinent fact is how much $\ce{Na3PO4}$ can be dissolved in 50 ml at 25 °C? Wikipedia shows 14.5 g/100 mL (25 °C), so 7.25 g/50 mL (25 °C). Thus it should be possible to dissolve the $\ce{Na3PO4}$.