Would the lone pair be in the equatorial plane or the axial plane for bromine pentafluoride? I've been learning chemical bonding and I'm confused about the structure of $\ce{BrF5}$.
It has a square bipyramidal geometry but a square pyramidal shape - I got that.
I don't  get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable.

And why does the angle change a lot from 90 degrees to 84 degrees? Similarly for $\ce{ICl4-}$- why do the lone pairs go to the axial plane?

As @IvanNeretin said in the comments, "There are no axial and equatorial positions here". This is a symmetric 3-D structure. Think of this structure as two planes cutting each other perpendicularly and you can rotate it any way you want. This is well reprsented by this image.

So your line "I don't get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable." does not make any sense. No matter where you place your lone pair it will still have no effect on the stability of the compound because everytime you think you are shifting your lone pair to a different position, it is forming the same molecule again.

Angle changes from 90 degrees to 84.8 degrees due to the lone-pair present. This creates a lone pair bond-pair repulsion and therefore slightly pushes the 4 Floruines, perpendicular to the lone pair, away from it making an 84.8 angle and not a 90.

Again in the case for $\ce{ICl4-}$ there is no axial or equatorial plane. Here's a diagram that should be helpful:

Keep in mind that (a), (b), (c), (d) can form a plane as well and according to your theory, it will become the equatorial plane. The main key point here is that the lone pairs must be furthest away from each other to be symmetric.


Understanding water models? I generally work on fluid flow, so some basic concepts in chemistry are new to me.
 Work I'm doing requires use of molecular dynamics (MD) to do classical potential simulations of water–NaCl systems but this question is about water models more than simulation and software details.

The simulations require definition of a "water model."  There's apparently one already implemented in an example/tutorial I'm using (below).  I want to compare the results of simulations based on different water models.  My problem is that I'm having a lot of trouble understanding how to change my water model (which I think is SPC/E) because I don't really understand what it is.

I have found tables that provide parameters for different water models, like this table at Wikipedia:
https://en.wikipedia.org/wiki/Water_model

However, when I look at my software, I can't relate what I'm seeing to the table.
 I can't find an example of where someone has taken information from this sort of table and explained how to implement it in a simulation.  So my first request is, please point me to anything out there!  Here's the part of my software that I think defines my water model:

&FORCEFIELD
  &BEND
    ATOMS H O H
    K 0.
    THETA0 1.8
  &END BEND
  &BEND
    ATOMS O H H
    K 0.
    THETA0 1.8
  &END BEND
  &BOND
    ATOMS O H
    K 0.
    R0 1.8
  &END BOND
  &BOND
    ATOMS H H
    K 0.
    R0 1.8
  &END BOND
  &CHARGE
    ATOM O
    CHARGE -0.8476
  &END CHARGE
  &CHARGE
    ATOM H
    CHARGE 0.4238
  &END CHARGE
  &NONBONDED
    &LENNARD-JONES
      ATOMS O O
      EPSILON 78.198 ! this is K, = 0.155 kcal/mol = 0.650 kJ/mol
      SIGMA 3.166
      RCUT 11.4
    &END LENNARD-JONES
  &LENNARD-JONES
      ATOMS O H
      EPSILON 0.0
      SIGMA 3.6705
      RCUT 11.4
    &END LENNARD-JONES
    &LENNARD-JONES
      ATOMS H H
      EPSILON 0.0
      SIGMA 3.30523
      RCUT 11.4
    &END LENNARD-JONES
  &END NONBONDED
&END FORCEFIELD

I can generally see that everything up to &NONBONDED relates to water somehow because it's about charge, H, and O (I have no idea what "K" is).  But what about the "LENNARD-JONES" stuff?  I know in general what a Lennard Jones model is, but cannot see how that would be related to water models.  If it is, how? Thank you for any advice on how to understand this.

I can explain some of this, and actually some of it is explained on the Wikipedia page you referenced.

Most water models represent the oxygen using a Lennard-Jones potential, which encapsulates the repulsive interaction that stops the molecules getting too close (this is physically because the electron shells don't want to overlap each other) and a long-range attraction due to Van der Waals forces. The Lennard-Jones potential can be parameterized in several ways, two of which are:
$$
v_{\text{LJ}}(r)
= 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} -
\left(\frac{\sigma}{r}\right)^{6} \right]
=
\frac{A}{r^{12}} - \frac{B}{r^6}
$$
The first of these appears in your force field specification file.
The $\epsilon$ parameter is actually given in Kelvin, which is $\epsilon/k_{\text{B}}$, where $k_{\text{B}}$ is Boltzmann's constant.
Hence the note "this is K".
This may seem an odd choice, but there are some historical reasons.
However, you'll see on the same line,
in a comment for the user,
that this number is translated into SI units and also into old-time cgs units of kcal/mol.
The $\sigma$ parameter is also given in your file, in old-style Angstroms.
Just to be clear, this is the interaction between oxygen atoms,
hence it is described as the O-O Lennard-Jones term.

The second form appears on the Wikipedia page. The values of $A$ and $B$ appear in the table, in old-time units. If you do the necessary multiplications
you can check that (to within a few significant figures, anyway),
$4\epsilon\sigma^{12}$ and $4\epsilon\sigma^{6}$ do indeed agree with the
tabulated $A$ and $B$, taking $\epsilon$ in kcal/mol and $\sigma$ in Angstroms,
for the SPC/E model.

There are some redundant entries in your file: BEND and BOND. For a molecule like this,
one could allow the OH bonds to vibrate, and the HOH bond angle to vary,
and these terms would need some force constants and equilibrium bond lengths and
angles.
But SPC/E is a rigid model: the bond lengths and angles are fixed,
so it looks like the force constants ($K$) have been set to zero.

One might imagine that the H atoms deserve also to be represented as Lennard-Jones atoms,
but they are quite small, and in this model this aspect is neglected.
Hence we see that $\epsilon$ is zero for the H-O and H-H interactions.

So then, all that is needed to complete the model is the charge distribution,
and it looks like you've identified the three charges,
one on each atom,
characterizing SPC/E.

This isn't the most sophisticated model,
as you can see if you scroll down that page.

For more details concerning the input file,
I would hope that your software package and/or exercise instructions,
give enough information.
Hope this helps at least to understand the correspondence between the
numbers in your file,
and the numbers on the Wikipedia page.


How can I work out what reaction will happen? I have recently started learning chemistry, and I've been reading a GCSE (UK high school level) textbook on the stuff :D

Much of it has been fairly straightforward, though there has been a reoccurring theme that has been bothering me.

I've encountered many instances where the author simply presents a reaction with little explanation, as if to suggest that the reason as to why it would occur is obvious.

Let me give you an example I have come across just today. Ethene in bromine water:

I think to myself, 'why would this happen?'. There are no 'free electrons' that I can see in ethene. OK, bromine is highly reactive, but what will it do? Will it react with H, or C, and how? What about the bromine bond? It's all very uncertain in my mind.

I always try to reduce it to an energy problem. We just want the most stable state, right? But, there are many possible outcomes. Am I supposed to apply some logic, follow some flowchart, or work out all possible outcomes and calculate which has the lowest energy? Do chemists just remember all simple reactions, and reduce problems to those? Or, am I just overthinking it?

A side problem to this is why, if we place an alkane in bromine water, we don't get the same thing happening with ethane.

There are no 'free electrons' that I can see in ethene

That's correct, but...  In ethene there are two bonds between the two carbons that hold the two carbons together.  A chemist calls this a "double" bond.  The important thing about a double bond (that's not obvious from the drawing) is that the two bonds are not the same.  One is a strong sigma bond, just like in ethane; the other is a weaker pi bond.  While the electrons in the pi bond are not free, they are not held as tightly

as the electrons in the sigma bond and are, therefore, more available for reaction.  This is why ethene is so much more reactive than ethane.

  Am I supposed to apply some logic... Or, am I just overthinking it?

Yes, there is logic and that body of logic is what organic chemistry is all about.  In the case of bromine reacting with ethene, the reaction belongs to a class of reactions known as electrophilic additions.  Most reactions in organic chemistry involve a molecule that has electrons available to share (a nucleophile) and a molecule that would like more electrons (an electrophile).  These two molecules react (form a chemical bond) by the one donating\sharing its electrons and the other molecule accepting\sharing those electrons.  Here, the pi bond in ethene has electrons that it is willing to share (they're not free, but they're not tightly held either), while the bromine molecule is highly polarizable, much like $\ce{Br^{\delta +}-Br^{\delta -}}$, with the positively polarized bromine atom wanting electrons.  The reaction proceeds as pictured below, to yield 1,2-dibromoethane.

  A side problem to this is why, if we place an alkene in bromine water,
  we don't get the same thing happening with ethane.

I bet you can answer that one now.  Ethene has a pi bond with electrons loosely held and available for reaction.  Ethane, on the other hand, only has a sigma bond where the electrons are more tightly held and less available for reaction.


Why doesn't oxygen replace chlorine? If oxygen is more electronegative why doesn't it replace chlorine in compounds?

Example:

$\ce{2NaBr + Cl2 -> 2NaCl + Br2}$

works, while

$\ce{4NaCl + O2 -> 2Na2O + 2Cl2}$

doesn't.

You can displace chlorine from hydrogen chloride.

With sodium chloride, if you did form sodium oxide it would be a strong base.  As such it extracts a $\ce{Cl^+}$ moiety from any potential chlorine molecules.  Sodium chloride is oxidized by oxygen, but the oxidized chlorine ends up as hypochlorite oxyanions such hypochlorite (i.e., oxide coordinated to chlorine).


Are there attempts to investigate phase states at negative pressure? Phase diagrams of various compounds and systems usually include only positive pressure. But I wonder whether it is possible to somehow investigate them under negative pressure?

Are there materials that change phase if mechanically stretched? For instance given that metallic form is sometimes achieved at high pressure, can a negative pressure applied to a metal produce a non-metalic form?

Or maybe the effect of negative pressure may be achieved by electrically uniformly charging the material?

I think there is a little misconception here.
There are no negative absolute pressures. When someone talks about negative pressures, they are referring to values under atmospherical pressure, this is,between 0 and 1 atm.
There are many ways of creating this vacuum, and of course, that matter has been studied under these conditions, for example, this kind of negative pressures are applied to get the boiling point of substances reduced, so thermolabile compounds can be vaporized (or the solvent which contains them) under its decomposition temperature.


What is inside nanobubbles? A recent article (ACS Omega 2021, 6, 8021−8027) confirms the stability of nanobubbles with theoretical principles. Nanobubbles, having a size in the range of 50-500 nm, are reported to be metastable, i.e., stable in various liquids for times ranging up to weeks. Is the inside of the bubble filled with a gaseous form of the liquid surrounding the bubble, or with some other gas? Are there repeatable observations of nanobubbles? Is the theoretical derivation even correct?

Inside nanobubbles would probably be "a high-pressure gas". Whether it be an inert gas or vaporized surrounding liquid depends on the temperature, pressure, components, and many other experimental variables.
Regarding theory on nanobubbles, and starting from a classical, macro-scale standpoint, the Laplace pressure describes the pressure inside a bubble due to surface tension on the curved vapor-liquid surface. The pressure will be necessarily higher inside the bubble in order that it doesn't collapse. The back-of-the-envelope calculation on that Wikipedia page indicates around 10 bar for a 300 nm bubble in water.
For smaller nanobubbles, things get more interesting. The surface tension is not constant with curvature and at the extreme curvatures necessary for nanobubbles this will change things. A first-order approximation to surface tension corrections is the Tolman length. Also, at the higher range of pressures involved gases would not be ideal, so some equation of state is likely necessary to describe the interior phase. As things get even smaller, then you get into the details where the continuum approximations like surface tension break down entirely.
Song et al. comes to my mind of a paper where they look at nanobubbles in an actual experiment. I'm sure there are many other experiments in this area.


Need help explaining OH-peaks in IR I have synthesized (1S, 2S)-1,2-diphenylethene-1,2-diol, and I'm wondering why the $\ce{OH}$-peaks in the IR-spectrum looks the way it does. I would expect a broad signal (because the solution wasn't diluted, and because free $\ce{OH}$-stretch happens at $\pu{3580-3700 cm^{-1}}$). So why these two separate, distinct, signals at $\pu{3394}$ and $\pu{ 3499 cm^{-1}}$?

(the IR-spectrum)

Off top of my head, I would think this is due intramolecular hydrogen bonding (a 5 membered ring), giving rise to a sharp peak at $\pu{3499.14 cm^-1}$, and intermolecular bonding, giving a slightly broader peak at $\pu{3394.45 cm^-1}$. But I don't have any references, this is just a hypothesis.


How to identify if a material is perovskite or not? This is more of a materials science question, but I believe many of you have referred NIST Inorganic Crystal Structure Database (ICSD) database.

I am collecting a database of perovskite materials and got a little bit confused on how the structure type is defined in the ICSD website.

Please find a screen shot of ICSD results on lead zirconate titanate (PZT) material.

Here, when mole fraction $\ce{Zr}=0.52$, the 'Struct. Type' is defined as $\ce{FeNbPb2O6}$. Does it mean that PZT material when $\ce{Zr}=0.52$ is not perovskite? And in general if the column 'Struct. Type' does not start with 'Perovskite,' does that mean the material is not perovskite?

Finally, sometimes I see the structure type named 'Elpasolie' or $\ce{LiNbO3}$ (disordered). Are these also perovskites? (or double perovskites?)

As MaxW indicated in the comment, perovskites are a huge family of natural and synthetic compounds. Following is a breif description of the group of compounds named perovskites:

The original Perovskite is a mineral of formula $\ce{CaTiO3}$. It was first discovered in 1839 by the Prussian minaralogist Gustav Rose in the Ural Mountains, and is named after Russian mineralogist Count Lev Aleksevich von Perovski (1792–1856). The natural crystals have a hardness of 5.5-6 and a density of $\pu{4000-4300 kgm-3}$. The crystal structure of this mineral was initially thought to be cubic, but was later  shown to be orthorhombic (Ref.1).

As with many minerals, Perovskite has given its name to a family of compounds with the same type of crystal structure, which are called perovskites that have a general formula close to or derived from the composition $\ce{ABX3}$. In these mineral type with formula $\ce{ABX3}$, $\ce{A}$ and $\ce{B}$ are usually two different sized cations, while $\ce{X}$ is an anion that bonds to both (e.g., $\ce{KNbO3}$). For instance, $\ce{PbTiO3}$) is a perovskite, crystals of which are Tetragonal (space group: $P_4mm$) with $a=\pu{0.3902 nm}$ and $c=\pu{0.4143 nm}$ of its unit cell. In contrast, the original Perovskite, $\ce{CaTiO3}$ crystals are Orthorhombic (space group: $Pbmn$) with $a=\pu{0.54035 nm}$, $b=\pu{0.54878 nm}$ and $c=\pu{0.76626 nm}$ of its unit cell (Ref.1).

To some extent, the multiplicities of phases that belong to the perovskites family can be rationalized by assuming that perovskites  are simple ionic compounds where $\ce{A}$ is usually a large cation and $\ce{B}$ is usually a medium-sized cations, and $\ce{X}$ is an anion (if the charges of the ions are written as $q_\ce{A}$, $q_\ce{B}$, $q_\ce{X}$, then: $q_\ce{A} + q_\ce{B} = 3q_\ce{X}$). At present, many hundreds of compounds are known that adapt the perovskite structure. Few examples are $\ce{NaMgF3}$ (Orthorhombic; space group: $Pbnm$), $\ce{CsPbI3}$ (Cubic; space group: $Pm \overline{3} m$),  $\ce{KCuF3}$ (Tetragonal; space group: $I4/mcm$), $\ce{LaAlO3}$ (Trigonal; space group: $R3c$), $\ce{KNbO3}$, $\ce{KNbO3}$ (Orthorhombic; space group: $Amm2$), and $\ce{(Fe,Mg)SiO3}$ (Orthorhombic; space group: $Pnma$). In fact, a perovskite structure mineral, Bridgmanite $\left(\ce{(Fe,Mg)SiO3}\right)$ is the most abundant solid phase in the Earth’s interior, making up $38\%$ of the total (Ref.1).

Figure A in following diagram illustrates an ideal cubic perovskite, $\ce{SrTiO3}$ (space group: $Pm \overline{3} m$):

The traditional view of the perovskite lattice is that it consists of smaller $\ce{B}$ cations within oxygen octahedra, and larger $\ce{A}$ cations which are nine fold coordinated by oxygen. According to this Princeton website where I abstracted the structure A (Note: I have modified to some extent):

  The easiest way to visualize the structure ($\ce{ABO3}$) is in terms of the $\ce{BO6}$ octahedra (e.g., $\ce{TiO6}$), which share corners infinitely in all 3 dimensions, making for a very nice and symmetric structure. The $\ce{A}$ cations ($\ce{Sr}$) occupy every hole which is created by 8 $\ce{BO6}$ octahedra, giving the $\ce{A}$ cation a 12-fold oxygen coordination, and the $\ce{B}$-cation a 6-fold oxygen coordination. In the example shown below, ($\ce{SrTiO3}$) the $\ce{Sr}$ atoms sit in the 12 coordinate $\ce{A}$ site, while the $\ce{Ti}$ atoms occupy the 6 coordinate $\ce{B}$ site. Note that there are many $\ce{ABO3}$ compounds for which the ideal cubic structure is distorted to a lower symmetry (e.g., tetragonal, orthorhombic, etc.).

Based on these facts, your last two entries, $\ce{Pb(Zr_{0.45}Ti_{0.55})O3}$ and $\ce{Pb(Zr_{0.871}Ti_{0.129})O3}$, are of course belong to perovskite family. The first one have adapted to the structure of $\ce{PbTiO3}$ (Tetragonal; space group: $P_4mm$), which is a perovskite (Ref.1). The second one have adapted to the structure of $\ce{LiNbO3}$ (Trigonal; space group: $R3c$), which is a synthetic perovskite (Wikipedia).

There is another family of compounds with formula $\ce{A2BB'O6}$, which is called double perovskites. The crystal structures of double perovskites feature an ordered rock-salt like arrangement of corner-sharing $\ce{BO6}$ and $\ce{B'O6}$ units in the crystal structure (Figure B). This is what researchers in Max Plank Institute where I have extracted the Figure B have to say about double perovskites:

  Proper choice of the metal ions $\ce{B}$ and $\ce{B'}$ and of the alkali, alkaline earth, or rare earth ions $\ce{A}$ allows to realize a large variety of physical properties in double perovskites. Prototype compounds which have found much interest are $\ce{Sr2FeMoO6}$ and $\ce{Sr2FeReO6}$ as they are half-metallic ferromagnets with Curie temperatures above room temperature and large magnetoresistance effects at ambient conditions.

The double perovskite structure is so named because the unit cell of double perovskite is twice that of usual perovskite. An example of crystal structure of a synthetic double perovskite, $\ce{Sr2FeMoO6}$, is shown in Figure C, which has a Orthorhombic (Phase Prototype: $\ce{Sr2WNiO6}$; space group: $Cmme$) unit cell (Ref.2). It has the same architecture of 12 coordinate $\ce{A}$ sites and 6 coordinate $\ce{B}$ sites, but two cations are ordered on the $\ce{B}$ site. The structure of $\ce{Sr2FeMoO6}$ shown here,the $\ce{Fe}$ and $\ce{Mo}$ atoms have ordered in a 3D chessboard type fashion.

Hence, it is safe to suggest that the compound used for first three entries, $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$, is an example of double perovskites, solely because it has a prototype structure of $\ce{Pb2FeNbO6}$ (Monoclinic; space group: $Cm$), which resembles a double perovskite ($\ce{A2BB'O6}$ type). The crystal structure of $\ce{PbFe_{0.5}Nb_{0.5}O3}$ is slightly distorted cubic ($a = b = c = \pu{4.0056 \mathring A}$; and $\alpha = \beta = \gamma = \pu{89.664 ^\circ C}$) and declaired to be a perovskite (Ref.3). On the hind site, $\ce{PbFe_{0.5}Nb_{0.5}O3}$ resembles more of $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$ than that of $\ce{Pb2FeNbO6}$.

Note: The best advice to OP is start with the book (Ref.1), which is written in basic language, understandable for any novice student learning about Perovskite materials. In addition, since OP is conveniently in England, I believe he can consult the author of the book, Prof. Richard J. D. Tilley, Ph.D., the Emeritus Professor in the School of Engineering at the University of Cardiff, Wales, United Kingdom, if OP has additional questions on Perovskite materials.

References:

Richard J. D. Tilley, In Perovskites: Structure-Property Relationships; John Wiley & Sons Ltd.: Chichester, West Sussex, United Kingdom, 2016 (ISBN: 978-1-118-93566-8).
Y. C. Hu, J. J. Ge, Q. Ji, B. Lv, X. S. Wu, G. F. Cheng, "Synthesis and crystal structure of double-perovskite compound $\ce{Sr2FeMoO6}$," Powder Diffraction 2010, 25(S1), S17-S21 (https://doi.org/10.1154/1.3478711).
Renbing Sun,  Wei Tan,  Bijun Fang, "Perovskite phase formation and electrical properties of $\ce{Pb(Fe_{1/2}Nb_{1/2})O3}$ ferroelectric ceramicse," Physica Status Solidi A: Applications and Meterial Science 2009, 206(2), 326-331 (https://doi.org/10.1002/pssa.200824432).


Interpretation of 1H-NMR spectra for Z-ochtodenol I'm struggling to understand the 1H-NMR data for this compound. I managed to find the compound using the data I had, but at the end I can't assign the peaks properly.

My confusions are:

Shouldn't the $1.96~\mathrm{ppm}$ (2H, s) protons couple with the proton from the blue carbon and give a (2H, d, $J~2~\mathrm{Hz}$) by allylic coupling? [Same issue with the protons from the blue carbon.]
What's up with the broad multiplet?
Counting the protons from the data gives us 15 instead of 18; why's that?

Let’s get the obvious out of the way first: Yes, the light blue protons should probably be the allylic ones since their shift is so similar to the other allylic protons’ shifts. (The green ones, that is; the red ones also experience deshielding from their neighbour oxygen.)

Then also, the report of the spectrum has been done badly. It should have included the field used. It makes a big difference if a spectrum is measured at $90~\mathrm{MHz}$ or at $700~\mathrm{MHz}$ — it may well be that weaker couplings are unobservable in lower field spectra but clear in higher ones. If I had to guess, I would assume a rather weak magnet since the smallest coupling reported is $6~\mathrm{Hz}$ — and they don’t even include a further significant figure.

Therefore, concerning your first bullet point: It is entirely possible that those protons do couple only the spectrum had too low a resolution to show it. However, remember that coupling should always be considered a possibility. If you check out the Karplus curve, you will see that for an angle of $\approx 90^\circ$ vicinial coupling effectively becomes zero. Same things may happen with long-range couplings: They may be absent with no apparant reason to the organic chemist since we cannot see the molecule’s exact orientation.

A ‘broad multiplet’ makes no sense to me whatsoever. In my opinion, multiplet alone should be enough. And I also strongly suspect an error in the book here: That is the shift where the four missing question-marked hydrogens should appear at. They should have badly overlapping, possibly even complex spins since each one is magnetically unequivalent and we are dealing with a cyclohexane system. This line should be, in my opinion:

  $1.35-1.70$ (4H, m)

And that also explains why three protons are missing. The pink ones are the only ones that would allow more protons in the same spot.

All in all that last bit is an error that really shouldn’t have happened in a textbook exercise. I do hope there is an erratum somewhere.


In antacids like Eno that fizz up when mixed in water, why is the citric acid important? I know that you can get rid of the acidity if you just take some baking soda added in water. I learned that these antacids that fizz up also have some citric acid and they react when added in water and produce $\ce{CO2}$ along with some other compound.
I like to know the use of citric acid in an antacid. Is it just for the feel of the fizz that subliminally reminds us of fixing the acidity in stomach?

It's the citric acid that reacts with the sodium bicarbonate. When the tablet hits the water, the citric acid and bicarbonate react, forming $\ce{CO2}$ and sodium citrate. This causes the tablet to break up, thereby much speeding up its dissolution.

The table will contain a stoichiometric excess of bicarbonate, with respect to the citric acid, so it's anti-acid properties remain present, even after bubbling/dissolution. Dissolving a tablet that doesn't disintegrate due to the acid-base reaction and its fizz would take much longer.

The fizzing effect probably also has some aesthetic/marketing value, separate from anti-acid properties of the tablet.


Confusion in unit cells of crystal system Unit cells are divided into two main types

Primitive
Non-primitive

Primitive includes simple cubic lattice whereas non-primitive includes fcc bcc end centered..

Among the seven types of crystal systems orthorhomic crystal system posses both primitive as well as non-primitive unit cells i.e. face centred body centred rnd centred and simple cubic but the cubic system posses primitive , body centred and face centred arrangements only.

My question is if face centred is possible in cubic system why is not end centred type of unit cell arrangement possible in cubic system if there is no sort of hindrance in vacant space as in orthorhombic which posses all types of unit cell arrangements.

First, a note: the classification of crystal systems, or the reduction of possible lattice types into primitive Bravais lattices, has nothing to do with “space available” or “hindrance”. It is purely a mathematical property having to do with the symmetry of the lattices.

To answer your question, the base-centered cubic lattice is not a Bravais lattice, because it is equivalent to a simple tetragonal lattice:


Why is the auto ignition temperature of petrol higher than that of diesel? I can't find any reasonable explanation please help.

Auto ignition temperature is defined like this by Wikipedia:

  The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark. This temperature is required to supply the activation energy needed for combustion.

Gasoline (or petrol for non-americans) has an auto-ignition temperature about 50 °C higher than diesel.

The answer boils down to explaining why the typical chemicals making up the two fuels have different activation energy barriers to combustion in air. This reaction basically involves the constituent molecules getting enough energy to react with oxygen.

We can get a crude (and probably oversimplistic) understanding by looking at the typical components of the two fuels. Diesel tends to be made up from linear saturated hydrocarbons; petrol is typically a mixture of branched and unsaturated hydrocarbons. So, a little simplistically, the typical component if diesel will have a larger surface area than a typical component of petrol (branched chain hydrocarbons are more "bunched up" than their linear counterparts with the same number of carbons). To put it another way, in linear hydrocarbons each carbon is has, at least, two hydrogens between it and any surrounding oxygen molecules from the air but in unsaturated or branched hydrocarbons the component carbon atoms are often buried inside because they are surrounded by other carbons (consider the molecule shapes of linear hexane and 2,2 dimethyl butane which have the same number of carbons).

This is a very simplistic way of looking at the question but give a little insight. In typical diesel compound there are simply more opportunities for oxygen to attack the molecule and kick off a reaction because the amount of the molecule exposed to the outside is higher. By being more compact (lots of carbons buried deep inside the molecule) there is less opportunity for this in the components of petrol.

For the reaction to kick off some of the molecules in the mix have to have enough thermal energy to react when they meet. And oxygen has to hit the hydrocarbon hard enough or there is no reaction. Once an oxidation reaction starts, it releases energy increasing the likelihood of another molecular reaction. At a certain point when there is enough energy from new reactions and the mix doesn't spread this out too quickly, this becomes a runaway process and the mixture goes bang. This process is–at the molecular level–easier for diesel than petrol because the chances of an oxygen causing a chemical changes when hitting a hydrocarbon is higher.

Of course a lot of other factors are involved but this crude picture gives at least some useful insight. This was discovered experimentally in the early days of engine design as the designers observed that different extracts from oil had different combustion properties in engines and refined the way oil was distilled to give them the behaviours they wanted in their engines.


Ideal gasses reaction under const temp and pressure Conceptual problem:

Let $A + 2B \rightarrow C + 2D$ be a reaction under constant pressure ($P$) and temperature ($T$), that goes all the way towards the products. $\ce{A,B,C,D}$ are ideal gases. Let $\Delta H$ be the molar enthalpy change in the reaction.
What is the change in the internal energy?

The total amount of moles in the reaction is constant throughout it, can we say that $\Delta U=\alpha nR\Delta T$ is the term for the internal energy? (if so that will be 0).
Another idea is that $PV=nRT \rightarrow V=\textrm{const}$, and then the internal energy equals the sun of chemical work and heat which we know.
All help appreciated.

Got it!

as $H=U+PV$ and $PV = nRT=const$ we ger $\Delta(PV)=0 => \Delta U= \Delta H$


Can you purify a metal oxide with a laser? Is there any metal oxide (like aluminum oxide, titanium dioxide etc.), that can be put into a vacuum and hit with a heat source, like a laser and be converted into pure metal with a slightly weaker vacuum having oxygen in it?
From my understanding pure metals tend to oxidize at higher temperatures, however, since "nature hates a vacuum", is it possible to use heat to do the opposite process by providing enough energy to break it's chemical bonds and then the oxygen joins the vacuum leaving the metal behind?
If this is impossible. Is there another way to use a laser to purify metals? If heat doesn't do the trick even in a unique environment, perhaps is it possible to focus ionizing radiation to essentially perform electrolysis on it?
Again, this might not work. So one last idea. For example, aluminum oxide and titanium dioxide are dielectrics. If you were to hit one side with an electron beam from a cathode ray tube to negatively charge one surface and use ionizing radiation on the other side could you then create a chemical reaction to purify the metal?

this is an interesting question and I will answer it a bit differently.

There is a well known analysis technique in materials chemistry called ICP-MS (Inductively Coupled Plasma Mass Spectrometry) where you have a sample, and using an inductive electric field, the plasma of the sample containing ions of the elements present is created. This is then passed thorugh a mass spectrometer to know the elemental composition of a material synthesised. It is one of the best known techniques for elemental analysis.

However, the idea of using ICP-MS like technique for purification of metal oxides is novel to me. Instead of induction to introduce electric field and rip apart the electrons from the atoms, you can achieve the same using a laser. The rest idea is same.
However, you can try it, and in case you design an instrument and get a publication for the same, kindly let me know here. I will look forward to it.

References:

https://en.wikipedia.org/wiki/Inductively_coupled_plasma_mass_spectrometry

Kindly contact me in comments in case you have any questions.
Stay Safe. With greetings from India. जय हिन्द. Jai hind.


Identification of Sir Raman's handheld scope I read an article somewhere around 5-10 years ago that talked a little bit about the life of Sir C. V. Raman. The focus was on the effect of Raman spectroscopy but included an interesting little tidbit.

Sir Raman apparently used to carry a small handheld scope of some description which allowed him to examine an object and only see the scattered light. He would always keep it in his pocket so that he could pick a flower or look at an everyday object and see how it changed. The article included a photo of him using it. I think I even remember seeing a video of him using it. As I recall, it was a small tube, about a hands length in size, and I think it was made from brass.

What was that tube, how did it work, and is there a resource for me to get one of my own?

Sir Martyn Poliakoff did the video about the spectroscope on his Periodic Videos channel on youtube.

To my knowledge, he presented the original instrument but did not mention where to obtain a replica.


Toxicity of a sheet of lead? I just went camping and someone brought a game from Australia. There was a wooden box with the top made of a thin, soft sheet of lead with a hole in it. The participants then throw coins at the box, and getting the coin inside gives some points, while landing a coin on top gives a little less. I've been looking online, and I think the game is called "Toad in the hole".

I know that lead is toxic, but I don't know in what way. Is it toxic to touch this lead sheet? And to eat afterwards without washing hands?

Maybe lead is only toxic in some form?

No, it is not toxic to touch solid lead.  Lead poisoning results from

ingestion: paints used to contain lead-based materials, and kids would sometimes eat the peeling paint leading to health problems.
inhalation: if you were grinding or polishing a block of lead and fine particles were generated, you might inhale them if you weren't using the proper protective equipment and health problems could develop.
dermal contact: if you were working with organic lead compounds that were easily absorbed through the skin, serious health problems would result if you weren't using the proper protective equipment.  Gasoline used to contain lead compounds, but they were removed from gasoline in order to prevent problems.

Being around or touching solid lead would not lead to ingestion, inhalation or dermal contact issues.

Here's a link to Wikipedia article on lead, see the section on "Health effects" if you'd like to read more or read about the biochemical mechanism of action.


Using combined gas law to calculate the volume of dry hydrogen gas Volume of wet hydrogen collected at room temperature: $\pu{40.48 mL}$, ambient pressure: $\pu{29.7 inHg}$ ($\pu{754 torr}$), ambient temperature: $\pu{295 K}$ ($\pu{22^\circ C}$), vapor pressure of water at $\pu{22^\circ C}$: $\pu{19.83 torr}$. Partial pressure of dry hydrogen: $\pu{9.87 inHg}$.
Find the volume of hydrogen under standard conditions.

The equation of combined gas law is $p_1V_1/T_1 = p_2V_2/T_2$.
Say if the volume of wet hydrogen gas is $\pu{40.48 mL}$, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?
One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of wet hydrogen if we are trying to find the moles of dry hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.

Consider the composition of the gas in your eudiometer.  Most of it is hydrogen, but some of it is water vapour.  In other words, not all the pressure is exerted by hydrogen, part of it is due to water.  You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.

You should have been given vapour pressure of water in a table.  You should also have taken down value of air pressure and ambient temperature.


Number of stereoisomers of polysubstituted cubane The molecule given below is a substituted cubane.

a) How many stereoisomers exist for this molecule?
b) How many pairs of enantiomers are possible?

How do I calculate the number of stereoisomers? Since each carbon is linked to four different groups, can I claim that all are chiral and, therefore, $2^8$ stereoisomers are possible?
If that is true, since there is no element of symmetry, can I say that $2^4$ enantiomeric pairs exist?
Edit: Looking at the comments clears things up but brings forth another question. If, say, iodine was present at any other position that would not lead to a tetrahedron, what would happen?

I hope I am not too late.

All carbons are $\mathrm{sp^3}$-hybridised and three bonds of every carbon are fixed (one corner shares three bonds), so fourth one has only one place in space to maintain an $\mathrm{sp^3}$ tetrahedron. This one is optically active, so this and its enantiomer are the only two possible isomers.

Now you get why we don't use the $2^n$ formula. And similar thing happens in a bridgehead position.


How to calculate Debye length of a 2-1 electrolyte I am given the Debye length to be 10nm for 0.001M KCl at 298K. I want to know its Debye length in 0.01M CaCl2. How can this be calculated?

I know that for 1-1, the Debye length is inveresely proportional to the root of its concentration. But how can I work out for electrolytes other than 1-1?

If we follow the derivation of the Debye length model, we would arrive at the following adapted model (doesn't assume uniform charge distribution):
$$\lambda_{D}^{2}=\frac{\kappa k_{B}T}{4\pi Z_{1}Z_{2}e^{2}c_{0}k_{c}}$$

In our equation:

$\lambda_{D}$ is the Debye length;
$Z_{1}$ represents the charge on the cation;
$Z_{2}$ represents the charge on the anion;
$\kappa$ is the dielectric constant;
$k_{B}$ is the Boltzmann constant;
$T$ is the absolute temperature; and
$c_{0}$ represents the number of ions per a cubic metre of solution.

Note that the general equation assumes that $Z_{1}=Z_{2}$, hence why this is usually written as $Z^{2}$.

Keep in mind this is only applicable for calculating the Debye length for a single dissociated species in solution! Further alterations would need to be made in order to consider a mixture.


Ester foam insert and isopropyl alcohol I recently purchased a used set of lens for my telescope. They came in a nice aluminum case with foam inserts, but the foam has an odor. I am currently using just the sun to try to rid the foam of the musty odor. It's slow going. I was wondering if there would be any adverse interaction if I sprayed the ester foam inserts with isopropyl alcohol? I am hesitant to spray anything on the foam for odor removal, for fear any lingering residue may harm the lenses in the long term.

The foam looks like a gray ester poly foam. Bottom is solid, top is convoluted.

Thank you.

I would just try soap and water first. If you want to try any kind of solvent like alcohols, you really needn't be concerned with lingering smell, but I would be concerned with the (slight) possibility of destroying the foam.

If soap and water doesn't work, just try placing a drop of isopropanol on both the top and bottom of the insert and let it dry. If there is any change in the consistency or feel to it I would stop there. If not, then progress to cleaning a small area, and if that does no damage go ahead and soak the whole thing if needed.


With which group does the middle oxygen atom of an anhydride remain, on hydrolysis? With which group does the middle oxygen atom of an anhydride remain, on hydrolysis? As an example, let us say we have the following modification of ethanoic methanoic anhydride, with the middle oxygen atom being the $\ce{^18O}$ isotope:

On acid catalysed hydrolysis with normal water, is this isotope found in formic acid, or acetic acid, or in both (i.e., hydrolysis is random and hence the product contains both the acids with both types of oxygen atoms)?

What I think is, the marked oxygen atom will stay with acetic acid because the carbon on the right side of the oxygen isotope has a slightly more δ+ charge than the carbon on the left due to the +I effect of methyl group attached to it. So, the marked oxygen should go with acetic acid instead of formic acid in the major product.

  For the reaction the mechanism is usually tetrahedral. Only under acid catalysis does the SN1 mechanism occur and seldom even then.

The tetrahedral reaction mechanism of hydrolysis of anhydrides is similar to that for the hydrolysis of carboxylic esters which usually also proceeds through the tetrahedral mechanism.

  Of the eight mechanisms given by Ingold for ester hydrolysis, seven have actually been observed in hydrolysis of carboxylic esters. The one that has not been observed is the BAC1 mechanism. The most common mechanisms are the BAC2 for basic catalysis and the AAC2 for acid catalysis, that is, the two tetrahedral mechanisms. Both involve acyl-oxygen cleavage.

So considering the mechanisms for ester hydrolysis and anhydride hydrolysis to be similar, it can be proved that the extent of δ+ charge on the carbon being attacked does play a major role in deciding the rate of reaction.

For instance, the rate of hydrolysis for the following esters is in the order shown:

So I think this should be enough to prove what I stated in the very beginning of my answer.

Source: Jerry March Organic Chemistry 6th edition, Organic chemistry by Paula Yurkanis Bruice.


Integration of the first order rate equation Given the simple first order reaction $\ce{A -> P}$ derive the integrated rate law.

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k[\ce{A}]$$
Collect terms:
$$\frac{\mathrm d[\ce{A}]}{[\ce{A}]} = -k\,\mathrm dt$$

Now for the bit I need help with, the integration:
Apparently the integrated form of $\frac{\mathrm d[\ce{A}]}{[\ce{A}]}= \ln[\ce{A}]$ but I'm struggling to see exactly how.

The next step (usually omitted)I think should be to split the fraction before integration
$$\frac{\mathrm d[\ce{A}]}{1}\frac{1}{[\ce{A}]}=\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}=-kt$$

$$\int{\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}}=\int{-kt}$$

$$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=\int{-kt}$$

Now, to evaluate this I use the general integration law for a constant $\int{[\ce{A}]\,\mathrm dt} = [\ce{A}]t + C$ and reciprocal  $\int{\frac{1}{[\ce{A}]}=\ln[\ce{A}]+C}$. Therefore $$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=[\ce{A}]t + C_1+ \ln[\ce{A}]+C_2=\int{\frac{\mathrm d[\ce{A}]}{[\ce{A}]}}$$

I know this is wrong because the textbooks say something different but I don't know how to correct it. Can anyone help?

Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.

The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$

Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$

Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$

Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$

Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$

Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$

Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$


Thermal properties of 301 stainless steel So I know that 301 stainless steel was chosen for the SpaceX Starship rocket as it can handle high temperature differences.
What about this alloy makes the steel harden when it gets colder?

301 is the lowest alloy content of the "18-8" family. The slightly lower nickel ( compared to 304) makes the austenite less stable so when cold rolled it work hardens more than 304. It is still basically austenitic so does not become brittle at low temperatures. So it was chosen for the higher strength ( when cold worked)  at all temperatures , relative to 304. Austenite is non-magnetic but as 301 is cold rolled it becomes somewhat magnetic depending on the amount of cold work. Presumable caused by formation of tiny grains of martensite ( the magnetic , strong, crystal of the iron family). The small martensite grains block the slip planes in the austenite matric making it stronger.


What is nanoporous structure? I would be very grateful if you help me to understand what is nanoporous structure. Does this term relate only to the nanoporous materials (as I understand, materials with regular porous structure) or to the such structures as nanotubes, fullerenes, layered silicates and so on? My doubt arises from the fact that the last do not seem for me to have regular structure.

I believe this question is both simple and difficult -- especially for those who are not native English speakers. I'm going to suggest a straightforward answer, but first let me explain the difficulty.

Unfortunately, terminology related to the nanoscale is often misused. But if we can agree on what the nanoscale is, the answer to the question becomes much simpler. So I'd like to respond to this question in two parts: 1) What is nanoporous? and 2) What is a nanoporous structure?

1) What is nanoporous?
This question relates to a more general question, "What is the nanoscale?" In my experience authoring articles in nanotechnology and serving as a referee for ACS Nano, I've struggled with this question.  It seems no accepted authority has defined the term. The closest I've found is this website which says clearly, "...the nanoscale, which is about 1 to 100 nanometers". I realize that good arguments can be made for extending this range to larger values… but surely it must stop short of 1000 nm = 1 micron. If someone knows of a more authoratitive body making a definition, I’d love to hear of it.

Unfortunately, I’ve found a significant number of scientific experts publishing statements suggesting that the nanoscale extends up to the 1000 nm range. This does not make sense to me: that would more properly be called the "microscale".  As a referee I've been surprised at the number of articles submitted for publication which claim nanoscale work when actually the objects of study are at the microscale. The motivation for this is obvious: it’s a lot more impressive to say you’re working at the “nano” than the "micro" scale these days.

So, my suggestion is that if we accept nanoscale means 1-100 nm, then nanoporous means the presence of pores in the 1-100 nm range.

2) What is a nanoporous structure?
Strictly speaking then, a nanoporous structure would be any structure having pores in the 1-100 nm range. The term itself implies nothing about the uniformity of the pores, and that is why we refer to sponges as porous despite very different sized pores. It is true that uniformity of pore size is of great interest to scientists for important reasons, but the term itself does not include that meaning. It is a case of denotation verses connotation. Though scientists have a reputation for precise terminology, they can be just as guilty as others in applying terms imprecisely – and for the same reasons: convenience and self-interest.


Why is more of a catalyst added, when it is not consumed? … catalysis, the acceleration of chemical reactions by substances not consumed in the reactions themselves—substances known as catalysts.

(Source)

Now as I’ve understood, to keep a reaction going, you must keep the solution heated and sometimes add more of the catalyst. Why then, is the definition that the catalyst is not consumed? Something of the catalyst must surely be consumed.

In short, a catalyst does participate in a chemical reaction the rate of which it increases, it just do so in a somewhat tricky way so that it does not appear in the overall reaction equation.

Wikipedia article on catalysis provides a concise explanation of why this is the case:

  Catalysts generally react with one or more reactants to form
  intermediates that subsequently give the final reaction product, in
  the process regenerating the catalyst. The following is a typical
  reaction scheme, where $\ce{C}$ represents the catalyst, $\ce{X}$ and $\ce{Y}$ are
  reactants, and $\ce{Z}$ is the product of the reaction of $\ce{X}$ and $\ce{Y}$:
  $$
     \ce{X + C → XC} \tag{1}
$$
  $$
     \ce{Y + XC → XYC} \tag{2}
$$
  $$
     \ce{XYC → CZ} \tag{3}
$$
  $$
     \ce{CZ → C + Z} \tag{4}
$$
  Although the catalyst is consumed by reaction 1, it is subsequently
  produced by reaction 4, so it does not occur in the overall reaction
  equation:
  $$
    \ce{X + Y → Z} \, .
$$
  As a catalyst is regenerated in a reaction, often only small amounts
  are needed to increase the rate of the reaction. In practice, however,
  catalysts are sometimes consumed in secondary processes.


Is the polarisation of halogens of higher atomic number easier because the electrons are further away from its core? This question concerns the dihalo addition reactions. I've learned that a halogen molecule (which, of course, consists of two atoms of the respective halogen) is more easily polarised by the double bond present in an alkene if it has a larger atomic number.
My question is this: Is this because the electrons of the halogens of higher atomic number are, on average, further away from its nucleus, so that they are easier to dislocate because the force that the protons exert on them is weaker?
A follow-up question would be: Is it, for the same reason, easier (in terms of energy requirement) to ionise elements of higher atomic number?

Short answer: yes.
The Wikipedia article on Polarizability has a nice section on this, which I would like to supplement with further quotations from its cited source.$^1$
pg 24

Since electrons are charged particles they respond to electric fields.
...
The electron cloud is said to become polarized in response to the electric field. The ability of the electron cloud to distort in response to an external field is known as its polarizability.

pg 25

We define the polarizability of a molecule as the magnitude of the dipole induced by one unit of field gradient, which works out to be in units of volume. Often, the larger the volume occupied by the electrons, the more polarizable those electrons.

My understanding is that it's less that the electrons are farther away from the nucleus in higher elements, but rather that they occupy a larger volume and are therefore more greatly affected by the electric field.

There is probably a nice analytical expression for the relationship of the polarizability to the terms we're used to seeing in the Schrodinger equation, but I'm not aware of it. In my experience, it is typically calculated numerically using perturbation theory. I did find an example where they worked through the perturbation solution for a particle in a box being subjected to an electric field here, section 4.1.1.

$^1$ Anslyn, Eric V. and Dougherty, Dennis A. Modern Physical Organic Chemistry. University Science Books, California, 2006.


How can I draw alpha helices in ChemDraw? I am trying to draw a polyisocyanopeptide with different kinds of monomeric units that intramolecularly interact with one another. However, I wanted to show these connections with its carbon backbone in the helix configuration.
An example of what I mean is shown below, taken from Kouwer et al (Nature 2013, 493 (7434), 651–655). How can I draw the alpha-helix as shown in red in ChemDraw (or any other program)?

I use ChemDraw Professional 19.1.1.32. If you follow  File>Open Templates>Advanced BioDraw, you will find the black helix that I have reoriented from horizontal to vertical. The red helix was enlarged and colored red. I hope this is of help.


what the solution of this problem, negative absorbance values in UV-vis spectrum In my project, I take the UV-visible spectrum of reaction between pyridinium chlorochromate and N-acetylcysteine. In some wavelengths, the absorbance value is negative for example it is (-0.006, -0.002, -0.003){negative but very small}.

Also when I take the spectrum of pyridinium chlorochromate alone, there are absorbance values in negative {negative but very small}.

The blank is NaClO4 solution.

what are the reasons for these negative values? And can I negligible these values?

Thank you...

The negative values can arise due to two major reasons: You did not mention that whether you are using a single beam instrument or a double beam instrument.
In a double beam instrument, absorbance of the sample and the blank is continuously measured. In a single beam spectrophotometer, A(blank) is recorded first and later subtracted. In each case, the output value at each wavelength is:
Abs (measured) = A(sample)- A (blank)

If at any region of the spectrum A(blank) is larger you get slightly negative absorbance values. This could be due to several reasons including refractive index differences. Under ideal cases, the cuvet which contains the blank is same as cuvet containing the sample or both of them should be optically matched i.e. they should show identical absorbance readings at given wavelengths.

Another reason to see random negative values is that noise is a part of any instrumental analysis. Eliminating all sources of noises, you still get stuck with shot noise, which originates due to statistical counting of photons. That is the very nature of detection especially in spectroscopic measurements- perhaps the ultimate limit in analytical sciences.


How many hydride ions are available in NaBH4? Different sources have different answers. In the reduction of an aldehyde or ketone to an alcohol by Sodium Borohydride, how many moles of aldehyde are reduced by one mole of Sodium Borohydride?

While my teacher claims it is 1, Solomons-Fryhle claims it to be 4.

Practically speaking, a chemist will use one equivalent of sodium borohydride to reduce an aldehyde or ketone. However, theoretically only 0.25 equivalents are necessary. Borohydride donates a hydride to the carbonyl carbon, generating borane, which is a weaker reducing agent than borohydride. However, borane is a Lewis acid that can react with the alkoxide formed in the reduction. This gives a new borohydride, which can reduce a second molecule of the ketone. This process can continue until all of the hydrides are consumed from the boron center.

Keep in mind that sodium borohydride reductions are commonly performed in alcohol solvents, so the exact nature of these steps could be altered. The alkoxide could be protonated by solvent, and then the solvent's conjugate base could reactivate a neutral boron species.

Also keep in mind that while sodium borohydride isn't VERY basic, it is a little basic, so it will deprotonate alcohols slowly. It's that reaction that causes the practical requirement for more than the theoretical borohydride stoichiometry.


How to synthesize a material at 2000atmospheres? I am trying to combine various elments and at high pressure. How to achieve these pressures for a gas and in quantities of at least 1 cubic centimeter.

I will try to give some more information knowing what you added in the comments. As you didn't precise which temperature I can't answer you in a very precise way.

However as I already said you can try to figure out how a reactor for polyethylene's production is made. If you're able to have access to some engineering literature, then you should have a look to tubular and autoclave reactors. If not you will have to contact someone or a company to explain what you want and get maybe an answer. In any case I doubt that having a reactor able to handle both ultra high pressure and (ultra) high temperature could be cheaper. The main reason is the savety. I bet that if you need such high pressure and temperature, you might use some dangerous chemicals and if not you have the savety and security problems of high pressure and temperature.

I found something that may be interesting for you here. You can achieve a pressure of 2.000 bars at a maximum temperature of 427 °C. On these reactors you can go up to 10.000 bars but with a lower temperature.

I would advise you to contact them they might have some solutions for your problem.


How can I measure the internal temperature of a solid metal? I’m heating up several metal blocks as part of an experiment to determine their specific heat capacities. In a nutshell, I’m heating the blocks of known mass to 100 °C, then adding them to an insulated vessel containing water of known temperature and measuring the increase. Pretty basic stuff.

I’m trying to work out how I can measure the temperature of the blocks, so that I know they’re at 100 °C. Leaving them in the boiling water for a long period of time is not ideal – I want to be sure that they are at the desired temperature that I can record.

How can I do this effectively? I thought about using an infrared thermometer, but wouldn’t that just measure the temperature on the surface of the block?

I also thought about something to do with measuring the resistance of the metals, but is this actually accurate and easy to do? Any advice or suggestions are appreciated.

You can use ultrasound thermometry. It uses change in speed of sound under heat. AFAIK, there are industrial-grade solutions on the market.


How to determine which p-type GTO to choose in HF method? I am trying to understand the working principle of basis sets in quantum chemistry. As far as I understand in the Hartree-Fock method, the basis set is used to calculate the matrix representation of Fock operator in each iteration in the SCF procedure. Usually each basis function in a chosen basis set is represented as a linear combination of several primitive Gaussian type orbital (GTO). I want to use cc-PVDZ basis for C atom which I can take from the EMSL Basis Set Exchange Library. For example in the Gaussian format:

!  cc-pVDZ  EMSL  Basis Set Exchange Library   10/19/17 2:40 AM
! Elements                             References
! --------                             ----------
! H     : T.H. Dunning, Jr. J. Chem. Phys. 90, 1007 (1989).
! He    : D.E. Woon and T.H. Dunning, Jr. J. Chem. Phys. 100, 2975 (1994).
! Li - Ne: T.H. Dunning, Jr. J. Chem. Phys. 90, 1007 (1989).
! Na - Mg: D.E. Woon and T.H. Dunning, Jr.  (to be published)
! Al - Ar: D.E. Woon and T.H. Dunning, Jr.  J. Chem. Phys. 98, 1358 (1993).
! Sc - Zn: N.B. Balabanov and K.A. Peterson, J. Chem. Phys. 123, 064107 (2005),
! N.B. Balabanov and K.A. Peterson, J. Chem. Phys. 125, 074110 (2006)
! Ca     : J. Koput and K.A. Peterson, J. Phys. Chem. A, 106, 9595 (2002).
!

****
C     0
S   8   1.00
   6665.0000000              0.0006920
   1000.0000000              0.0053290
    228.0000000              0.0270770
     64.7100000              0.1017180
     21.0600000              0.2747400
      7.4950000              0.4485640
      2.7970000              0.2850740
      0.5215000              0.0152040
S   8   1.00
   6665.0000000             -0.0001460
   1000.0000000             -0.0011540
    228.0000000             -0.0057250
     64.7100000             -0.0233120
     21.0600000             -0.0639550
      7.4950000             -0.1499810
      2.7970000             -0.1272620
      0.5215000              0.5445290
S   1   1.00
      0.1596000              1.0000000
P   3   1.00
      9.4390000              0.0381090
      2.0020000              0.2094800
      0.5456000              0.5085570
P   1   1.00
      0.1517000              1.0000000
D   1   1.00
      0.5500000              1.0000000
****

In the lines such as S   8   1.00 I know that the letter S is the orbital type, 8 is the number of GTO in this orbital's expansion, but what is 1.00 for?
The GTO for a p-type orbital has the form $x^l y^m z^n \exp(-ar^2)$ where the powers satisfy $l+m+n=1$ and thus there are three possible sets of $\{l,m,n\}$. But for instance in that link for lines with p-type orbitals, which possibilities shall I pick? The same question for d-type orbitals.

Have a look in the Gaussian manual if something is unclear.

If you want to use an extern basis set in Gaussian you have to respect the gaussian input conventions for the basis set, which you can find here (gaussian.com/gen/). Since you are using the EMSL Basis Set Exchange the server gives you a file that satisfies these conventions. Like:

Type  NGauss  Sc
$α_1\quad d_{1μ}$
$α_2\quad d_{2μ}$
…
$α_N\quad d_{Nμ}$

Type stands for the shell, then you have the number of primitives with NGauss and scaling factor Sc. Afterwards you put your exponents and coeffients of the primitive gaussian functions which then will be contracted to the respective contracted gaussian functions. You might notice that the scaling factor is set to unity for almost all basis sets—I checked some Dunning, Pople and Ahlrichs basis sets and always found it set to unity, which seems resonable. In the TURBOMOLE format the scaling factor is omitted totally, so it seems to be a Gaussian specific hack.

Now to your second question. You are specifing the shell, not a single function, a shell contains always all spherical harmonics belonging to the given quantum number or all cartesian gaussian functions. The spherical harmonics are the solution of the Schrödinger equation for an electron on a surface. You have generally $2l+1$ spherical harmonics where $l$ is the azimudal quantum number specified as 0/s, 1/p, 2/d, 3/f, 4/g, 5/h and so on. There exists also the possibilty to use cartesian gaussians with $(l+1)(l+2)/2$ degenerated functions.

Why am I telling you that? You always choose a whole shell with a azimudal quantum number, how many function belong to this shell depends on the quantum chemistry code you are using. ASAIK, Gaussian supports both.

In case of a p-shell it does not matter, both expressions evaluate to three p-functions (as expected). So you get three p-functions with one p-shell.

The basis set you have chosen would be evaluted as:

(9s4p1d) → [3s2p1d]

Having nine primitive s type gaussians (note that Dunning basis sets use a general contraction scheme), four primitive p type gaussians and one primitive d type gaussian which form a basis set out of three s shells, two p shells and one d shell with 14 contracted spherical harmonic gaussian functions or 15 contracted cartesian gaussian functions for carbon.


Electrochemistry - Redox & Electroplating I've stumbled across an electrochemistry problem that I need help on:

$$\ce{Fe~|~Fe^{+2}~||~Cl^{-},Cl2~|~Pt}$$

I understood that iron is at the anode and is being oxidized, however I do not know what was going on with chlorine, it looks like it is being oxidized while it is supposed to be reduced.
Maybe this is a non-spontaneous reaction? Can this be possible, can chlorine be used as an oxidation reaction? And if so how can we calculate the voltage?

Here the iron electrode is made by dipping an iron plate or wire in a $\ce{FeCl2}$ or $\ce{FeSO4}$ solution. It is an anode, where $\ce{Fe}$ is oxidized according to $$\ce{Fe -> Fe^{2+} + 2 e-}$$ The chlorine electrode is a cathode made by dipping a platinum electrode in a solution where some chlorine gas $\ce{Cl2}$ is arriving from under the platinum plate, so that a few bubbles remain adsorbed on its surface. This $\ce{Cl2}$ gets reduced to $\ce{Cl-}$ by absorbing the electrons coming from the iron anode according to the equation : $$\ce{Cl2 + 2 e- -> 2 Cl-}$$ The global effect of these reactions is equivalent to the following equation that describes the general change of composition in the cell : $$\ce{Fe + Cl2 -> Fe^{2+} + 2 Cl-}$$


Is chirality a property of only molecules with exactly 1 chiral centre? I have seen chirality being defined in a number of ways in terms of a molecule not being superimposable with its mirror image.
My syllabus has a statement: "know that optical isomerism is a result of of chirality in molecules with a single chiral centre". This seems odd because surely you can have a have optical isomerism between molecules with more than 1 chiral centre so why a single.
Overall, I am asking for either a confirmation that the statement is incorrect or an explanation of where my understanding is wrong.

I don't want to try to interpret the intent behind the wording of the phrase you quoted, so I'll just state the situation and you can read into it as you wish.

There is no truly general relationship between the number of chiral centres in a compound and whether it displays chirality overall.

However, if a compound has exactly one chiral centre, then it must be chiral. Compounds with more than one chiral centre may or may not be chiral, depending on what those chiral centres are and their relative stereochemical relationship. And as has already been rightly pointed out (4 seconds before me), compounds with zero chiral centres may still be chiral.


Brittle d-block metal trend I remember seeing a periodic table that had the top-left corner of the d-block shaded and marked as "brittle". If I recall correctly, the elements were $\ce{Sc,Ti,V,Cr,Mn,Y,Zr,Nb,La}$. I think (sorry, I can't find much on it on the net).

The triangular shaded area is a clear indicator of a groupwise and period wise trend (similar to how the non-metals form a triangle due to the electron affinity/Ionization potential trends).

So, I ask, what are the trends leading to the brittleness of a metal, and, more importantly, why are these trends there (usually trends can be explained in terms of simpler things like atomic size/shielding/etc)?

So, I ask, what are the trends leading to the brittleness of a metal

Brittle materials absorb very little energy before fracturing. The only trend that can be linked to this is the strength of bonding in the metal (Other factors like crystal structure and purity do not have any trends and so there are a lot of exceptions). This also means that the trend in brittleness is not very regular since it is complex and depends on too many other factors.

  why are these trends there? (usually trends can be explained in terms of simpler things like atomic size/sheilding/etc)

Since brittleness cannot be explained by a very simple trend, I will explain what reasons I have found.

The stronger the bonding in the metal, the more difficult it is to change its structure and thus it tends to break instead of deforming. Here's a graph (melting point vs. Group number) that gives a general measure of strength of metallic bonding in all the transition elements of the first three series.

According to the graph, we predict that Tungsten (Period 6 Group VIB) should the most brittle and this is true for impure tungsten. Again, this successfully predicts that Silver, Gold and Copper (Group IB) are least brittle (and so most malleable and ductile). Also Lanthanum (Period 6 Group IIIB) is predicted to be quite malleable and this is also true.

Note that this also explains the high malleability of alkali metals (the graph is not very clear at that point).

There may be much more exceptions to this trend when examined closely so take this only as an approximation.

Crystal structures also play an important role in brittleness but they cannot be explained by a trend.

Purity of a metal tends to increase malleability and decrease brittleness because there are less points at which a fracture can take place.


How to calculate crystal field stabilisation energy? How to calculate it? I know that it depends on the coordination compound and the number of electrons present in t2g and eg orbitals and the final answer is expressed in terms of crystal field splitting parameter. But, how to know how many electrons are there in t2g and eg orbitals?

I think rather than a broad answer and example would help. So, consider the compound $\ce{[Ni(CO)4]}$ and how to calculate CFSE for it?

Step 1: Look up Nickel Carbonyl and find out what geometry it has.

We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.

  Step 2: Find the appropriate crystal field splitting diagram for this geometry.

  Step 3: Figure out how many $d$ electrons there are.

Nickel is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations. For more information on $d$ electron count, go to this Wikipedia page.

  Step 4: Fill in the $d$ electrons.

You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?

  Step 5: Determine $\Delta E$.

$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This Wikipedia page walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:

$e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)
$t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)

Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.

$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$


Why are porphyrins colored when porphyrinogens are colorless? In heme synthesis, porphyrinogens are colorless but upon their oxidation (from removing 6 hydrogen atoms), they become colored porphyrins. Why does the change in color occur? Is red the only color it can turn into or are there others (if so, what does that depend on)?

The precursor porphyrinogen does not have a conjugated pi system that spans the macrocyclic ring. On oxidation to porphyrin, however, the macrocyclic ring becomes fully "conjugated", i.e. connected by sp2 hybridized atoms. The HOMO/LUMO spacing in such systems is much narrower than in non-conjugated systems, which means the photon energy required to excite electrons into the LUMO is lowered from "deep" UV to the visible region, resulting in strong absorbance of light in the visual frequency range.


Which rate, the forward or reverse rate of acid dissociation, is more strongly affected when diluting acetic acid in aqueous solution? Acetic acid is a weak acid. It is in equilibrium with acetate and hydronium ions in aqueous solution:

$$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}$$

Ostwald's law states that the degree of ionisation of a weak acid increases upon dilution (in quantitative terms: as the square root of the volume).

Consider two scenarios:

We start out with 1.001 moles of water and 1.000 moles of acetic acid (so water is the solvent and acetic acid is the solute), and dilute it with water to twice the volume.
We start out with 100 mM acetic acid, and dilute it with water to
twice the volume.

How do the rates of the forward and reverse reaction change in these two scenarios, and how do these changes help to explain Ostwald's dilution law? (I know that once equilibrium has been reached after dilution, the forward and reverse rates will be equal again, so I am asking about the rates after dilution, but before changes to concentrations by the reaction going towards equilibrium. If mixing is slower than the acid-base reactions and you can't do the actual experiment, treat it as a thought experiment.)

This question was inspired by Why degree of dissociation/ionisation affected by dilution? . I am specifically interested in addressing the following ideas from the OP of that question:

  as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

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  since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

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  since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

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  What motivates or pushes [the acid] to dissociate more?

If it helps, it would be fine to discuss what happens with a single acetic acid molecule and a single acetate anion upon dilution.

Let's consider case 2. To make the math easier, I'm going to assume that acetic acid and any mixture of acetic acid/water has the same density as water. Thus, the molarity of pure acetic acid is approximated as 16.7 mol/L. That isn't exactly right, but conceptually that doesn't change anything. Also, since the mole fraction of water is a bit less than 1, let's use its actual molarity of approx 55.6 M.

So, to start, we have 0.1 M of acetic acid in water, which means only ~6 mL/L  water, and we can assume the water is still 55.6 M. And assume a total volume of 1 L. At equilibrium, we know that the forward and reverse rates of ionization are the same, so

$k_f[\ce{H2O}][\ce{AcOH}]=k_r[\ce{AcO-}][\ce{H3O+}]$

Based on the published $K_a$, we can calculate that [AcOH]~0.1 M and [AcO-]=[H3O+]=0.0013 M.

If we instantly add an additional 1 L water to double the volume, the AcOH, AcO- and H3O+ concentrations all drop by half, but the water concentration stays at about 55.6 M. Intuitively, it should be clear that the result is a much greater drop in the rate of reverse reaction than the rate of the forward reaction. More rigorously, we relative rate of the forward reaction is (55.6x0.05)/(55.6x0.1)=0.5, while that of the reverse reaction is (0.00065)^2/(0.0013)^2=0.25. Since the rates were equal before dilution, the forward rate is now approx twice that of the reverse, and in order to return to equilibrium, the AcOH concentration must get even lower and the ion concentrations must come back up.

Conceptually, this comes about because the chance of an acetate ion and a hydronium ion encountering each other diminishes as the solution gets more dilute. The dissociation of the acid in bulk water, on the other hand, is just a function of how many undissociated molecules are present.

With regard to the quotes you included, some brief responses are

  as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

The concentration of water is not the same as H3O+, which is decreased by dliution. Water self-ionization will partially compensate, but not very much.

  since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

The dielectric constant of water does have an influence on the $K_a$, but it doesn't change when you add more water.

  since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

If we start with dilute acetic acid in water, we assume that the acetate ions are far enough apart from each other that they do not interfere with each other's solvation, so additional dilution doesn't factor in.

  What motivates or pushes [the acid] to dissociate more?

I think it's more appropriate to say that it reassociates less rather than dissociates more, but that's mostly semantics.


Why is the following reaction not considered a redox reaction? The following question appeared in the Cambridge International AS and A level Chemistry Coursebook:

  Choose from redox, substitution, elimination, addition and hydrolysis to give the type of reaction for the following:
  $$\ce{C2H6(g) + 2Cl2(g) -> C2H4Cl2(l) + 2HCl(g)}$$

The answer given by the answer key was substitution since $\ce{Cl2}$ replaces $\ce{H2}$.

However, why is redox not a suitable word to describe the reaction? After all, the oxidation number of $\ce{C}$ in the reactants is $-3$, while it is oxidised to become $-1$ in the products. Furthermore, the oxidation number of $\ce{Cl}$ in the reactants is $0$ while it is reduced to $-1$ in the products.

Hence, by that logic, isn't the reaction also a redox reaction since $\ce{Cl}$ is reduced and $\ce{C}$ is oxidised?

Thanks in advance for any help.

You are right. The question is definitely misleading. If it would ask for the mechanism, (radical) substitution would be the correct term, as a reduction does not say anything about the mechanism. But "type of reaction" can basically mean anything, and in this case both answers are correct.


Specific Heat of Water Solution If we add an ionic solute to water, will the specific heat of the water be increased or decreased?

I understand that the specific heat is a measure of how much energy we must add to water for the average kinetic energy of the water molecules to go up by a certain unit.

I also understand that hydrogen bonding is what limits the kinetic energy of the water molecules. This accounts for water's high specific heat.

However, when adding an ionic solute to water, such as $\ce{KI}$, we disrupt the hydrogen bonding. So we may expect the specific heat of a water solution to be lower than that of pure water.

On the other hand, there are ion-dipole interactions created, correct? These will act to at least partially counteract the specific heat depression, correct?

From this website:

  Every substance has its own specific heat and each phase has its own distinct value. In fact, the specific heat value of a substance changes from degree to degree, ...

It would mean that a solute in water would change the specific heat from plain water.

My guess is that if the density is greater than water, then the specific heat would be greater (because you are heating the water and the solute). To be certain, you would need to determine it for yourself.


Do chaotropic and kosmotropic affects occur in polar non-aqueous media? For example can I introduce salts that have a chaotropic or kosmotropic affect on ethanol? Or is this phenomenon restricted to water? If this phenomenon does occur in other polar solvents, does it follow the Hofmeister series as well? Are there any papers out there on this phenomenon in polar non-aqueous solvents?

Chaotropic and kosmotropic effects of solutes can occur in any solvent in which interactions between solvent molecules lead to a more ordered structure, like a hydrogen bonding network in water. Such a network will also be present in alcohols like ethanol, although to a lesser degree, because the alcohol molecules have only one proton which can act as a hydrogen bond donor. The Hofmeister (or lyotropic) series for alcohols might still be similar to that for water (not identical), but it will be more different with increasing structural difference between the solvent molecule and water. In fact, the Hofmeister series for water is still not fully understood on a molecular basis, as it is described in this post, and this makes it difficult to predict Hofmeister series for other polar, non-aqueous solvents.


Reactivity series of metal I have learned about metal and their reactivity series. However, why metals have different reactivity? What are the factors of different reactivity rate for metal? All I can think of is the number of electrons which causing it.

I suppose that with reactivity you mean a metal's tendency to become oxidized.
There are several reasons and it depends on the metals you are analyzing.

If you are referring to alkaline or alkaline earth metals, reactivity increases as you go down the group. They all have a large oxidation potential since it's relatively easy to remove an electron (or two) of its orbital. As you move down the group, atoms are larger and the valence electrons will be less atracted by the nucleus. Therefore, they are easier to remove and the metal will be more reactive.

For d-block transition metals (most of them are considered "transition" metals) it's a whole different story. There are many factors that contribute to its reactivity and it's harder to specify general trends. Many of them (especially the ones on the second and third transition rows) can be oxidized to different oxidation states depending on the conditions. You need to look at their electron configuration and reduction potentials to start rationalizing its properties. Depending upon the electron configuration, they will be able to lose (or even gain) less or more electrons and with different ease.

I don't know your background with chemistry, so I won't write a more detailed explanation but if you clarify that you have a more concrete interest in any explanation, just ask.


Hydrolysis of cyclic acetal What happens to 1,5-dimethyl-6,8-dioxabicyclo[3.2.1]octane when it is hydrolyzed under acidic conditions?

I know that under acidic conditions the first thing to likely happen is protonation. Does it matter which oxygen is protonated first?

Once that oxygen has been protonated it is going to want to return to neutrality.

Now we have our carbonyl restored (sort of). However, we still want to return to neutrality. Water can attack.

Then we deprotonated the attached water and regenerate the acid-catalyst and we have our original starting compound.

Is this the correct mechanism and is the product correct?

Your product is correct, but the mechanism isn't quite right. In the first step, one of the oxygens will be protonated (either is possible). The other oxygen "kicks out" the one activated in the first step. So far so good. However, in the next step, water will attack the "carbonyl" carbon to give a hemi-acetal and a hydroxyl group. Repeating a similar set of steps reveals the ketone and second hydroxyl.

The key difference here is that by adding to the pi system, water reacts with the C-O pi-antibonding orbital, which is the LUMO of the system


Do only metals form ions? Do nonmetals form ions? I am asking this because I was reading something and I came across the phrase "metal ions," which left me to wonder whether or not there are nonmetal ions as well.

In short, yes. Single-element ions are common for both metals and non-metals. Some simple examples: $\ce{F^{-}, Cl^{-}, I^{-}, S^2-}$ Multi-element non-metal cations are a bit more common: $\ce{NH4+}$ and $\ce{H3O+}$ being the obvious examples and there are plenty of anions like $\ce{CN-}$, $\ce{SO4^{2-}}$, $\ce{NO3-}$, etc. Things like proteins and polypeptides are also frequently ions in solution and may be positively or negatively charged depending on the pH.

Simply due to their electronegativities, all the single element non-metal ions are anions in solution, but in the gas phase, all bets are off and you can knock off an electron to get a cation, (or have negatively charged metal ions) so long as it doesn't bump into anything else.


Why does copper have a high enthalpy of atomisation when it has a half filled d sub shell? If I understood correct, the high energy of atomisation for d-block element comes from the ability to make many metallic bonds due to half electron filled d-orbitals, but it seems here that copper has no free d-orbitals yet has high enthalpy of atomisation? Is there a mistake in my understanding or is there a reason for this anomaly?

When we look up the Wikipedia articles for various elements, we find that the heat of fusion and heat of vaporization of copper are modestly below those quantities from iron, cobalt and nickel, but hugely exceeding zinc.

In the first-row transition elements when they're condensed as metals, we don't really have separate $3d$ and $4s$ subshells because the metallic bonding spreads these subshells into bands and the bands overlap in energy.  This article in Chem Libretexts shows this overlap between $s$ and $p$ orbitals in sodium; a similar concept involving $s$, $d$ and even$p$ orbitals in transition metals.  So we can't separately consider $3d$ and $4s$ subshells in a first-row transition element such as copper.  With contributing orbitals from different subshells mixed together we have a significant number of filled and empty states mixed together in the valence band, leading to relatively strong metallic bonding as long as we keep that energy overlap.  Among elements in this period, only when we get to zinc do we see the $3d-4s$ band overlap drop off and the nonbonding character of the filled $3d$ subshells emerge as a major factor.  Therefore, metallic bonding and with it, enthalpy of atomization drop off sharply at zinc, not so much at copper.


Variations in protonation in reaction mechanisms English and German Wikipedia give slightly different mechanisms on Fischer indole synthesis. The thing bugging me is the protonation of N and the source of the proton.

Look at the intermediates 8 (Ger) and 5 (Eng). In the German Wikipedia imines are not protonated, in the English they are. Or look at intermediates 10/11 (Ger) and 7 (Eng) - protonation occurs but on different N-atoms of the aminoacetal. Ammonia won't leave the molecule unless it is previously protonated, but how important is protonation in other steps? Aren't those molecules in an equilibrium with all their protonation variations anyway?
And so does the reaction mechanism even specify the source of the proton? Do specific steps take a proton from the molecule, whereas other steps take it from the surroundings? I always thought the molecule simply grabs the most available proton (the closest and "loosest" one).

German Wikipedia:

English Wikipedia:

English Wikipedia is more correct. Under acidic conditions, carbonyl and imine additions should happen via the oxocarbenium and the iminium ions, respectively.

Also, the in German version, the step between 5 and 6 is not a step; that's a resonance structure. There's also sloppiness in 3 and 10. Those attacks should come from the lone pair. Technically, those steps should've happened before the addition anyway, as I've already pointed out. Bottom line is that, again, the English version looks better.


Nearest branch numbering If the longest carbon chain is symmetric numbering-wise (meaning the numbering schemes are identical both ways) and has all alkyl substituents, should one prioritize the largest alkyl group?

If numbering from each end of the parent chain gives the same locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name. This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.

For example, a correct systematic name is 4-ethyl-5-methyloctane (not 5-ethyl-4-methyloctane) since ethyl is cited first as a prefix in the name.

Another correct systematic name is 4-ethyl-5-propyloctane (not 5-ethyl-4-propyloctane). Again, the reason is that ethyl is cited first as a prefix in the name. The size of the substituent is not relevant in this step.


How can we directly add half cell potentials to measure the EMF of a galvanic cell? We know that $E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}$. For example, in a cell $\ce{Zn(s) | Zn^2+(aq) || Cu^2+(aq) | Cu(s)}$, $$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Cu^2+/Cu}}+E^\circ_{\ce{Zn/Zn^2+}}$$

This makes sense to me. I think of $E^\circ_{\text{cell}}$ as the $E^\circ$ value of the net cell reaction i.e. of $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$.

But, consider this cell: $\ce{Zn(s) | Zn^2+(aq) || Ag^+(aq) | Ag(s)}$ According to my textbook, the $E^\circ_{\text{cell}}$ value here is again defined by:

$$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Ag+/Ag}}+E^\circ_{\ce{Zn/Zn^2+}}$$

But, I think this is not correct.
Consider the fact that this cell is made up of two half cell reactions:

$$
\begin{array}{}
\text{Oxidation}&\ce{Zn(s)}&\ce{-> Zn^2+(aq) + 2e-}&E_1^\circ\\
\text{Reduction}&\ce{Ag^+(aq) + e-}&\ce{-> Ag(s)}&E_2^\circ\\
\text{Net cell reaction}&\ce{Zn(s) + 2Ag^+(aq)}&\ce{-> Zn^2+(aq) + 2Ag(s)}&E_3^\circ\\
\end{array}
$$

and that we can't add $E^\circ$ values directly, since they are an intensive property. Instead, we need to say that $\Delta G_3=\Delta G_1+\Delta G_2$ and then write $2E_3^\circ=2E_1^\circ+E_2^\circ$ or $E_3^\circ=\frac{2E_1^\circ+E_2^\circ}2$, which is definitely $\neq E_1^\circ+E_2^\circ$, as my textbook says.

I believe I have correctly presented everything here, and cannot figure out my mistake. Where am I wrong?

You are writing $\Delta G_3=\Delta G_1+\Delta G_2$. But it should be $\Delta G_3=\Delta G_1+2\Delta G_2$ (I am considering  $\Delta G_2$ as free energy change for the conversion of $\ce{Ag+}$ to $\ce{Ag}$, which is considered 2 times in the final reaction) Because $\Delta G$ also depends on the number of moles of reactants and products, if you add two times the second reaction, the $\Delta G$ will also be used as twice its value. There you've made a mistake.

Thus the equation will be $$2E_3^0 = 2E_2^0 + 2E_1^0$$ So you will have $E_3 ^0 = E_1^0 + E_2^0$.


Find the volume of the carbon dioxide product Below you can see the chemical reaction I'm working with:

$$\ce{CH_3CH_2OH(l) + 3O_2(g) -> 2CO_2 (g) + 3H_2O(g)}$$

I've got so far:

How can I calculate the volume of the product of carbon dioxide gas (the pressure is 1 bar, and the temperature is 20 deg Celsius).

You need to determine the limiting reactant. Use the equation and the number of moles of $\ce{CH_3CH_2OH}$ and $\ce{O_2}$.
In the reaction, you need 1 mol of $\ce{CH_3CH_2OH}$ and 3 mol of $\ce{O_2}$ to get 2 mol of $\ce{CO_2}$. If you have less than 1 mol of $\ce{CH_3CH_2OH}$ but 3 mol of $\ce{O_2}$, you cannot get 2 mol of $\ce{CO_2}$. In such a case, $\ce{CH_3CH_2OH}$ is the limiting reactant. The amount of $\ce{CO_2}$ will thus be lower than 2 mol.

Using the number of moles of the limiting reactant, you should be able to calculate the number of moles of $\ce{CO_2}$ produced. Use proportions!

The last step will involve using the ideal gas equation, by assuming that $\ce{CO_2}$ behaves like an ideal gas. $$PV=nRT$$
You have the pressure $P$, the temperature $T$, $n$ (the number of moles of the $\ce{CO_2}$), $R$ is a constant. Find the volume $V$.


How are poisons discovered? Does someone have to die/be poisoned from it first? How are poisons discovered? Does someone have to die/be poisoned from it first?

Or are there other ways of discovering the harmfulness of a substance? Perhaps everything is tested on other animals prior to testing them on humans?

Alle Dinge sind Gift, und nichts ist ohne Gift, allein die Dosis macht dass ein Ding kein Gift ist (The dose makes the poison) - Paracelsus

Poisons (I'm going to use this as an umbrella term for "toxins" and "venom" as well. Bear in mind though, they are not the same thing) have been known since antiquity.

Back in the good old days, you figured out if something was poisonous or not by eating/touching it (or getting someone else to do it), i.e- "discovering" a poison was simply a matter of chance.

These chance encounters alone, led to the discovery of numerous poisons.

With the advent of Chemistry, the gents in white lab-coats figured out that compounds that bear resemblance to the already well-known poisons, in terms of their functional groups and structures, are also toxic (albeit, to vastly different degrees). Poisons could now be identified a priori (you could tell it would end badly if you were exposed to such a substance, but not how badly).

And no, you can't really quantify a poison's effects without testing it on something. Nor can you tell how much of something would be needed to kill or severely maim.

This is where we bring in the idea of a median lethal dose $\pu{LD_{50}}$ which is, simply put, the minimum amount of a substance required to kill off half of all the animals in a particular test group.

$\pu{LD_{50}}$ values for a particular substance depend on the animal used.

Of course, the only way to get an accurate $\pu{LD_{50}}$ (which itself, is really a "mean/average" value of sorts) for a human is to actually poison someone, which doesn't sound very nice. So you do the next best thing, you measure it for a rat/chimp, couple it with your knowledge of the poison's mechanism of action, and extrapolate the value to something that would kill a person.

Another way to establish degree of toxicity, which doesn't involve killing animals, would be by exposing a cell/tissue culture (and not the entire animal) to the potential poison .


Why does calculating formal charges help us draw correct Lewis structures? Reading about how to draw Lewis structures, all I know is that formal charge is the charge assigned to an atom in a molecule assuming that electrons are shared equally, regardless of differing electronegativity between atoms. It is just a concept used to arrive at the correct Lewis structure and does not represent actual charges on atoms. But what I don't understand is, why does using this concept lead us to the correct structure?

For example, right now I am looking at the example of the sulfate ion. I get to this point before formal charges need to be calculated:

All the atoms have an octet around them (though I'm aware sulfur can have more than eight electrons around it), but it is not the correct structure. Why does considering the formal charge give the right structure, if as stated above it doesn't reflect any actual charges? It also says on Wikipedia "Formal charge is a test to determine the efficiency of electron distribution of a molecule. This is significant when drawing structures.", but I am not sure what it means by this.

Formal charges are helpful for two reasons:

at lower levels, you use them to show your instructor that you can properly count electrons and ‘keep the books’. You would need to realise that each oxygen has an additional electron and therefore a formal negative charge while sulfur is lacking two electrons and therefore has a formal positive charge. Adding up the formal charges should give the charge of the molecule.
Since all lone pairs are typically drawn at this level, lacking formal charges can easily be rederived.

at higher levels, lone pairs are typically not drawn unless they are important for some reason (e.g. if they take part in a resonance mechanism discussed at this very moment). Therefore, the only information on the number of electrons around an atom we can extract from the formula is the formal charge.
If at higher levels, you draw a tetravalent sulfur atom without formal charges, we must assume that it is actually a formally neutral sulfur and that there is an additional lone pair. This, of course, makes a big difference; compare the electronic states of sulfur in $\ce{SO4^2-}$ and $\ce{SF4}$.


Reasons for negative iodoform test Why does 2',6'-dimethylacetophenone not give iodoform test?

As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate.

But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered.

Hence, the triiodo intermediate is well anticipated.

However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.

Notice that the Burgi-Dunitz trajectory, which we may assume the incoming nucleophile to take, is hindered by the large iodine atoms and the methyl group.

It is quite safe to assume that the attack at the carbonyl carbon is unfavoured, preventing the release of the $\ce{CI3-}$, and ultimately $\ce{CHI3}$ never appears.

EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.


Why is glycol used as an antifreeze? Lowering of the freezing point on adding a solute is a colligative property of solutions, it doesn't matter what the nature of the solute is (of course we have to assume ideal solutions). So when it is required to use an antifreeze - why are glycerol, glycol, methanol, etc. commonly used? For a given decrease in freezing point, a given value of molality is required - in other words a particular number of moles for the liquid's given mass. For the same number of moles, it should be simply more economical to use any substance with lower molar mass. Solubility in the solvent will obviously play a role - but there are many more solutes which are more soluble than alcohols. Why isn't $\ce{NaCl}$ used, for instance? Its molecular mass is not very high, and it is soluble to an appreciable degree. It's also more economical (moreover it will dissociate in water and generate twice the number of moles, so twice the decrease in boiling point).

The main reason ionic substances like salts are not used is because they tend to be corrosive to metals.

Polar organic liquids tend to be the best because many are very soluble in water, and they are fairly inert under the conditions antifreeze is used.