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What is meant by 'stability'?
I'm currently learning about substitution and elimination reactions and to do so I need to learn about bases and nucleophiles.
Apparently a good base is an unstable species.
What does stability even mean in chemistry?
I just need to conceptually understand and wrap my head around this concept of a compound or atom or species being 'stable' or 'unstable'. Please don't explain it in complicated terms.

Thinking about something as being 'stable' or 'unstable' is often misleading as stability can depend on many things. Esters, for instance, may be perfectly 'stable' if left out in a vial, but incredibly unstable if stirred in a flask with some aqueous base. Stability should, therefore, always be qualified, for instance thermodynamic stability and kinetic stability are both terms often considered by chemists.

And to take @NotNicolaou’s comment a step further, kinetic ‘stability’ should not be called ‘stability’ at all but better inertness.

Stability has several meanings. A compound can be stable on its own, under the given conditions such as temperature and pressure. This is the case for a lot of compounds. The other definition of stable could be a compound's inability to react with other molecules (like the inability of carboxylic acids to be oxidized). This is probably what you meant. A good base, like $R-O^-$ is unstable because of the +I effect of the alkyl group, which shifts the electron density towards oxygen, which in turn makes the ion a very good and strong nucleophile. This is the reason why such "good bases" tend to cause eliminations  and substitutions in organic molecules.
P.S.: This is the simplest explanation I can provide. If you have any doubts, ask them in the comments.
P.S.S.: Stability of a compound can also be explained on the basis of what @NotNicolaou said.

Stability is when a situation is at a relative free energy minimum OR when there is no possible or reasonable mechanism for something to happen. An historical example: nitroglycerine was a favored explosive in mining etc. Nitro, undisturbed, was stable; but if you shook it or jarred it too hard it exploded, severely damaging the person who shook it; Along comes Alfred Nobel; he figures a way to stabilize nitro and invents dynamite; dynamite sticks can be dropped They are stable! or are they? The profits started the Nobel prizes.
Most compounds are stable if in isolation including strong bases or acids; some compounds are inherently unstable, an example being Pure H2O molecules. They cannot exist, for very long at least, without self-ionizing to H3O+ and OH- ions. So the combination of a strong base mixed with a very weak acid is a mix that might be unstable ie not at equilibrium. It might react if a mechanism is possible and if the base and or acid are strong enough for a reaction.
Stability is an unfortunate word to describe reactivity. A base such as OH- is more reactive than H2O when reacting with acids. Stability depends on the conditions and displacement from equilibrium.


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How do I model atoms with Moebius strips?
This question came to me while I was trying to model the list of elements (Mendeleïev), especially their masses and number of electrons.
It came to my mind that it might be easier to model each element as Moebius Strip with $n$ twists.
If this is possible, is there any literature about this subject? A methodology to do this?
----EDIT----
Ok,
it took me a while to go deep into the subject in order to present "concrete thoughts" but I finally could summarize them into the following spreadsheet:
https://onedrive.live.com/redir?resid=b20ec7aa903ebb07!701&authkey=!AE5lfNOaYgtJWNg&ithint=file%2cxlsx
The model has a R² = 0.9976 (what I think is pretty good as a first try)
Please check the spreadsheet in the following order:
A) Observation VS calculation
B) Racionale ( that presents the reasoning used to make the calculations used by the model)
C) Consequences (the rational consequences of the model --> I'm NOT sure that the reasoning expressed in this table is valid
D) Data ( the data and its source used to make the ratio between 'calculations' and 'observations')

Of interest to you in your quest: The Internet Periodic Table Database.

Topology has also been a weak mathematical area for me, but I don't think that sort of representation would help. The periodic table and the Aufbau diagram are the two keys used to introduce inorganic chemistry. I just don't comprehend how a Mobius strip would simplify either or combine either in some novel way. Science is always looking for new ideas,  so if you have more concrete thoughts on this we'd be happy to critique them. https://en.wikipedia.org/wiki/Aufbau_principle

Hi @MaxW it took me a while to go deep into the subject in order to present you "concrete thoughts" but I finally could summarize them into the following spreadsheet:
https://onedrive.live.com/redir?resid=b20ec7aa903ebb07!701&authkey=!AE5lfNOaYgtJWNg&ithint=file%2cxlsx
The model has a R² = 0.9976 (what I think is pretty good as a first try) and, if the racionale (reasoning) is correct, the consequences seems pretty good to me.

(continuing)
please check the spreadsheet in the following order:
A) Observation VS calculation
B) Racionale ( that are the reasoning used to make the calculations used by the model)
C) Consequences (the rational consequences of the model --> I'm NOT sure that the reasoning expressed in this table is valid
D) Data ( the data and its source used to make the ratio between 'calculations' and 'observations')

I don't have any idea of what you're trying to calculate in the spreadsheet.

I'm trying to calculate the atomic weight of each element in relation to the number of electrons (or protons --> atomic number)

Atomic weight and atomic number could easily be shown in a list ordered by atomic number. The importance of the periodic table is the relationships between the chemical reactivity of the elements.

hmm, well I understand that this spreadsheet may not have any practical use then, but would this make the model incorrect? Actually I'm just trying to make an simpler model to find the atomic  weight of an element.

source: Stack Exchange

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sort: votes


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Single Point vs. 5 Point Calibration in HPLC
Single-Point calibrations aren't that useful for quantitative analysis and there is more room for error in result readings as compared to a 5-Point calibration. Is this why a 5-Point calibration is done instead?
If so, is a more accurate reading the only pro to doing a 5-Point calibration over single point? Anyone with any knowledge on the matter would be greatly appreciated.

You can't regress with a single point, for starters. A calibration curve that encompasses a range of concentrations that are expected for a given analyte will be much more useful that one which is not defined in the response range. There's also the issue of linear and non-linear responses for different concentration ranges (as well as detector response, such as conductivity vs. UV-Vis, for instance).

Thank you very much. Everything makes sense, but can you elaborate on what you mean by "regress"?

Here seems to be a good paper on the topic. https://www.webdepot.umontreal.ca/Usagers/sauves/MonDepotPublic/CHM%203103/LCGC%20Eur%20Burke%202001%20-%202%20de%204.pdf // The point about regression is this. How many lines can you draw through a single point? Especially when doing some sort of instrumental analysis you don't want to extrapolate outside the bounds for which you have created a calibration curve. So if I calibrate between 3 and 50 units of something it would be huge mistake to use that calibration to measure something at 230 units. Also most detectors have some sort of background.

So my English is not perfect, however the goal of having a linear regression curve is to estimate the relationship between concentration response and determine the degree of agreement between the increases in concentration and the response, if for example, a certain wavelength has been experimentally demonstrated that the response of the detectors is directly proportional to the concentration This is that it has a slope of 1 and that the correlation coefficient is 1, this means that throughout this concentration interval the answer and concentration will always have the same increase, so that it will not matter that the sample response is higher or low that the one of the so-called calibration point.
The behavior will always be the same, so the uncertainty of a curve with more than one point or use the factor-response technique will always be despicable, so itself depends on the behavior of the analyte for that exists there are so many methods AOAC that do not need a calibration curve with more than one point and are valid internationally, example: Vitamin A

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Replicating an illustration for the crystal structure of ferrihydrite
I came across this image of crystal structure in an article The Structure of Ferrihydrite, a Nanocrystalline Material [1]:

Does someone know the software in which it was produced? I have been looking for a way to visualize crystal structures of some minerals but I am stuck looking for a piece of software capable of doing it. I am aware of Jmol, but it does not show crystal structures too well.
References

Michel, F. M.; Ehm, L.; Antao, S. M.; Lee, P. L.; Chupas, P. J.; Liu, G.; Strongin, D. R.; Schoonen, M. A. A.; Phillips, B. L.; Parise, J. B. The Structure of Ferrihydrite, a Nanocrystalline Material. Science 2007, 316 (5832), 1726–1729. https://doi.org/10.1126/science.1142525.

I think the software used could be Diamond (http://www.crystalimpact.com/diamond/)

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour]. For remaining questions about the site, visit the [help]. It would be nice, if you could elaborate a little why the software you suggested is a good choice for OP’s problem. We generally prefer elaborate answers where possible. Other than that, I hope you enjoy your stay =)

@Jan Thank you for your comment. I was just answering OP's question about the software used to produce that image, so I thought no more details were needed

Which is why I didn’t downvote (it looks like it’s good) but merely mentioned ;)

Free (as in beer) alternative to Diamond, Olex2 (works on Windows, Linux and MacOS), can also be used to replicate this structure (I used COD-9011573 CIF). Various bonds for cubane $\ce{Fe4O4}$ core, transparent polyhedra and custom coloring schemes are available out of the box:

There is also an extension DrawPlus available on a subscription basis which adds some bells and whistles alongside with extra convenient tools for visualization (though I never used it myself).

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

authors: จตุรงค์ สุขสังข์


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Understanding how potentiometry works
I want to make sure I understand this technique. I am reading information from this website.
From what I can understand, you have it set up like a galvanic cell, but in the middle of the wire connecting the electrodes you have a voltmeter which measures the current (converting it to potential? using V=IR?) but doesn't let any through to the other electrode so no reaction can happen. So the electrons at one electrode can still sense the potential difference between them but the voltmeter blocks them from travelling all the way through the wire? The potential you get is proportional to the concentration of analyte you are measuring according to the Nernst equation.
Please let me know if what I have said is correct or if I have misunderstood.

Yes that is the gist. The half cell potential depends on the concentration of the ion as given by the the Nernst equation. A high impedance voltmeter is used to measure the voltage. "High impedance" means that virtually no current flows. Virtually in turn refers to the concentration of the ion at the surface of the electrode. If a "lot" of current is flowing then the concentration of the ion is depleted at the electrode compared to the bulk of the solution, so the EMF drops as the current increases.


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Why does Iodine form Cations sometimes?
It is reported that iodine forms compounds like $\ce{(CH3COO)3I}$, $\ce{I(ClO4)3}$ and $\ce{IPO4}$ which contains $\ce{I+}$ and $\ce{I^3+}$ and are not bonded covalently as in interhalogen compounds.
Generally, Halogens form anions of the type $\ce{X-}$ to achieve stability by attaining noble gas configuration. However iodine loses electrons when it forms a cation and does not attain noble gas configuration. So why does iodine form cations while no other halogen does?

Why should iodine not form cations? Fluorine cations can be supplied (albeit not stable for extended periods of time). Thus, any atom can likely form cations, although it is unproven for helium and neon as far as I know.

You should check your data - all three compounds are covalent. Doubt if you can find any solid data on any +3 cation which isn't coordinated to ligands.

@Mithoron maybe except some lanthanides and actinides. 3+ "ions" in the d and p blocks are more likely to be covalently bound.

Iodine is the biggest halogen and by extension the most polarizable.
Diatomic iodine can be oxidized with nitric acid to facilitate electrophilic aromatic substitution.
Iodide ion, as found in sodium iodide, can also be oxidized by concentrated household bleach to facilitate EAS as well.
In the examples you give, there are strong electron withdrawing groups (such as oxygen, which is found in each of your examples). It is plausible that the oxygen has stripped the iodine of its electron(s) in each of your examples.


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Simple Explanation of orbitals
Imagine I was your little brother, how would you explain an orbital to him? (assuming he knows what atoms electrons and neutrons are, and the rest of the basics till that point.)
I have been reading up on this and can't make heads out tails of it.
As a side how do the different energy levels relate to them.
I hope I'm clear, I'm doing my best. I read several books but can't get a clear bearing on this.

Well, the first step is to stop thinking electrons as being very small balls that orbit around the nucleus in a circular path. This is known as the Bohr model. Despite this model being an excellent model at an introductory level, it doesn't tell the entire story.

There are several issues with the notion that electrons travel in a circular orbit around the nucleus, the main one being that they should eventually undergo orbital decay and hence eventually slow down and crash into the nucleus. Of course, this doesn't happen. So how do electrons move around the nucleus?
This is where quantum theory came to the rescue in the early 20th century. It stated that electrons could not be treated as a classical particle and didn't not have a definite position and momentum. So this basically means that we can't exactly know an electron's position and its momentum at the same time. This relationship is given by the Heisenberg Uncertainty Principle.
I won't go into detail about it since to keep this explanation basic, but if you are interested check out this video which provides a really good explanation.
So basically what quantum theory tells us that we actually can't specify where and what a path an electron is going to take. However the Bohr model assumes that you can precisely plot out where an electron is going to be, but we now know that due to the Heisenberg Uncertainty Principle you can’t precisely know the exact position and momentum of an electron. So we actually can't construct an orbit and say that the electron is always going to follow that path.
Therefore, it makes it impossible to plot an orbit for an electron as how can you? If you don’t know where the electron is going to be next, there is no way you can predict its path.
So how can we get around this problem? Well, if we can’t draw the orbit for an electron the next best thing we can do is to plot an electron density probability orbital for it. Also known as atomic orbitals.
An orbital is a well defined region of space. Therefore an atomic orbital is a region of space that shows where the electron will be 95% of the time (usually we take 95% but it could be any number really such as 90% or 75%).
A simple way of thinking of orbitals is that imagine that you have a magic camera that can take a sequence of photos of an electron in a hydrogen atom. The electron appears as a dot. Now if we superimpose all of these images, you will see something like this:

Since we generally take atomic orbitals to be the region of space that shows where the electron will be 95% of the time, the very outer sphere of the above image is said to be the orbital of the electron.
Hopes this helps you understand orbitals. If you have any question, feel free to ask.

I don't think of the Bohr model as wrong, but merely as one of the starting points for the explanation.  First you start off with the planetary model, Then you add quantization via the Bohr model which also adds shells, then probability distributions, and you're there!

I don't think the mechanism that causes accelerating charged particles to lose energy can be called friction.

Thank you so much. What I still don't understand is just because quantum physics states that WE cannot know the definite position of an electron, why does that explain why the electron doesn't undergo orbital decay and crash into the nucleus?

@user20554 Electrons actually do collide with the nucleus! The maximum probability density of the electron in the s orbital is actually at the nucleus position. However at such low energies they don't 'notice' as the wavelength of the electron is much larger than the nucleus (remember electrons aren't small balls, but a mixture of a particle and a wave). .

Another way of looking at it is that the Uncertainty  Principle forbids the electron to be localized on the nucleus, because its momentum would be so uncertain, that such electron would not be bound to the nucleus at all

Hi, could you tell me where did you get that image of the electron positions?

@Typo Sorry, that was a long time ago so I cant remember

Why don't you try with analogies: at least, as a vary basic introduction, I very much appreciate the attempt done here by Goh et al.

An orbital is defined as a region in space in which there
  is a high probability of finding an electron. An analogy for
  an orbital is the possible location of a student based on his
  or her timetable. For example, according to the timetable,
  on Monday at 9.00 am, a student has to be in Lecture Room
  1 for a chemistry lesson. One can say that, over a period of
  weeks, there is a high probability that this student will be
  in Lecture Room 1 at that time on that day. However, one
  cannot be 100% sure that this student will be there because he or she may be absent from school on that day. One
  also usually cannot predict exactly where the student will
  sit in the lecture room, but one can say, with a high probability that this student will be within the confine of the
  lecture room.
  This analogy describes an orbital fairly well, because it
  takes into consideration the probability and region of
  space in the definition of an orbital.

I will definitely start with such a analogy which seems to give a quite simple, although thorough, description of how orbitals should be regarded.

Ok. Forget everything about the orbitals for a second.
You have nuclei. You have electrons around them.
First step: electrons are not balls. Imagine them as being a smeared charge in space. Which is precisely what they are: smeared charge in space, with fuzzy borders, like an unfocused blob. The shape of this blob depends from many factors, but the biggest one is the position of the nuclei.
Now suppose you want to describe the shape of this blob, like if you want to 3d print it. Well an easy way would be to divide space in little cubes and say "charge" if you have some smear or "no charge" if you don't have that smear. In practice, since it's fuzzy, you would say "3.0 charge" in some cubes, "1.3 charge" in other cubes, "0.1 charge" in other cubes and so on. This is a perfectly legitimate way of describing what is called "charge density" in space.
The problem with this approach is that it's rather inconvenient. It has poor accuracy and it scales badly. If you make the little cubes smaller, because you don't like living in a minecraft world, you need a lot more cubes to do that.
So now there's a smarter method to do so, and to explain it, I need you to understand Fourier decomposition. It's not as hard as you think it sounds.
The problem is the following. You have a complex sound wave, like the one produced by blowing a whistle or singing a song, and it turns out that you can create any complex shape of this wave by summing simpler waves together: a careful choice of sines and cosines of certain frequencies (pitches), and with certain intensities (volumes). Play them all together, you get the initial wave back.

This is exactly what you see in your spectrum display in your stereo

Where each column is a different frequency, and the height of the column is the intensity of that particular frequency. It changes all the time because you are playing complex music, but try to play a uniform sound (e.g. a violin playing a single note) and you will see it stays the same throughout.
Now back to the orbitals.
The orbitals are the "sines and cosines" of our problem of describing that blob. We have a complex entity (our smeared blob) and we want to describe it by summing together "something". It does not matter what "something" we use, but it turns out that 3 dimensional functions of that shape has a lot of nice properties which makes the problem much more compact.
Let's make you a simple example. Suppose you have a spherical blob of charge. That's probably described well by a single orbital of spherical shape (an s orbital), exactly like a sine wave from a tuning fork is described well by a single sine function.
Now add an electric field so that the electrons are pulled, and the smeared sphere is now more like an elongated egg. That one is not really described well by a sphere, is it? so you need to describe the lobe, meaning that you need an additional orbital (a p orbital) to add to the mix so that the result is egg-shaped, exactly like you need more than one sine wave to describe the sound of a violin.
That's it, really. orbitals are just convenient 3d equivalents of the sine and cosine. We could well use anything else (and in fact we do, in some cases) and it would work as well, but with some potential disadvantages.

Elucidating the atomic orbital structure of an atom is a fascinating story. It has just been solved within about the last hundred years. It wasn't a linear journey, but a journey filled with all sorts of dead ends and side trips. The Greeks thought that the elements were air, fire, water and earth. Not understanding the atom alchemists spent zillions of hours and a countless amount of money trying to turn other substances into gold.
So the first significant step in solving atomic structure starts with Dmitri Mendeleev who is credited with the notion of a periodic table. This allowed for the discovery of additional elements which were needed to complete the table. However why the arrangement of the table worked was unknown.
About that same time other work was happening. Maxwell created his famous equations linking electric current and magnetism.
The discovery of the electron occurred around about the same time. But it wasn't until about 1900 that the understanding of an electron was fairly complete.
Physicists still didn't understand how to put an atom together. At this point many atomic models were floated to explain various phenomena. One such model was the plumb pudding model. The idea was that an atom was like a blob of pudding in which the electrons were suspended like plums in the pudding.
One big breakthrough came from Rutherford scattering about 1910. These experiments showed that the nucleus (positive charge) was located at the center of an atom and that it had a very small volume compared to the whole atom. This lead to the planetary model of the atom. The negatively charged electrons orbited the positively charged nucleus like the planets orbiting the sun.

This didn't explain other phenomena that were known like line spectra of hydrogen that Balmer had discovered before 1900.

Such line spectra were related via Rydberg formula which related the principle quantum number $n$ to atomic structure.
$\frac{1}{\lambda} = R(\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2})$
in 1913 the Bohr model was developed. This explained that the electrons were arranged in shells, and the filling of the shell structure could be linked back to the periodic table!

The shell structure is basically tacked onto the planetary model by Arnold Sommerfeld and the notion was that for whatever reason the electrons couldn't revolve around the atom's nucleus in "any" orbit but that they had to occupy certain orbits which became known as shells.  Subshells ($l$, $m$ and $s$) were added to Bohr's model to fudge additional orbits to explain the fine spectroscopic structure of some elements. Like the Fraunhofer lines observed in the spectra of the sun.

So at this point physicists and chemists had cataloged much of the behavior of atoms, but still were lacking a complete understanding of how atoms worked.
In the mid 1920's Schrödinger deveoped his famous equation which completed the puzzle of the atoms structure into the more fundamental electrons and nucleus. From the Bohr model it was realized that electrons have to be in certain orbits because the orbits are quantized. That is to say that each orbital has a specific energy. What Schrödinger showed was that the orbits were not like the lanes at a track and field event that constrained the electrons, but that the electron orbit was distributed in a 3D cloud around the atom. Another aspect of Schrödinger's work and others was that the electron had both wave characteristics and particle characteristics.
Using the Schrödinger wave equation the shape of orbitals could be calculated. Now this was in the 1930 time frame computers were still decades away. So these calculations were done by hand! The "problem" was that solving the equations was only possible for one electron. Using calculus it is nice to be able to solve equations so that the integrals can be easily calculated. However the three body problem doesn't allow such a solution. With modern computers it is possible to calculate numerical solutions even though the integrals don't exists.
So back to the question. An orbital is a mathematical function that describes an electron's 3D path around the nucleus. Rather than an planetary like "orbit" the orbital is a probability function. The density of the orbit varies as a function of radius. Depending on the orbital the probability function also shows an orientation in 3D space. So a S orbital is spherical like in that there is no X-Y-Z preference. the P orbitals though not only have a radial aspect but they have spacial orientations. Thus the three P orbitals are like 3D rotations of a two-leafed rose in 2D space. So P orbitals have lobes oriented along +/- x-axis, the +/- y-axis, and the +/- z-axis.
A word of caution. The orbital representations are very useful for predicting chemical behavior, but they are not "real." This obviously gets a bit crazy when four $sp^3$ molecular orbitals are formed each with 25% S character and 75% P character! It's sort of like thinking that a unicorn is a hybrid of a horse and a rhinoceros.

Some nice history there Max. The thing that impresses me is that the phrase spherical harmonics was coined by Thomson and Tait in 1867.  They were into knots, which makes me think of TQFT, and "vortex particles", as was Maxwell.

No, I am not German, and unfortunately I can't read or speak German either. Reading German might not be so necessary now, but the "real" knowledge about organic chemistry when I went to school was in German. Beilstein was entirely in German.

I'm still writing a paper about sp-hybridisation vs. octahedron structure of Neon and Argon electron configuration. I enjoyed your last paragraph. (Only if you address your comments with @name, someone get a notification; not necessary for the one, who made the thread)

@MaxW I've cited your great answer in my paper  https://www.academia.edu/18391675/About_the_distribution_of_the_electrons_magnetic_dipole_moments_in_the_atoms_Part_1

@MaxW In my paper, mentioned above, I had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals. Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is ... the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model.

I don't know enough physics to know whether this is on-target or not, but...
The idea of different vibrational modes is easy to demonstrate with a big bowl of water and your finger.  Fill the bowl, and use your finger to excite different standing-wave patterns on the surface.
Start at the center of the bowl, move your finger slowly up and down, and you should be able to get a radially-symmetric standing wave that looks something like a two dimensional version of a 1S orbital.  Once you find the fundamental frequency, double it, and you'll get something like a 2S orbital.  Move off-center, and you probably can excite a mode that looks like a 1P, and a 2P.
Unfortunately, you'll have to stop there, because the higher-order modes on the two-dimensional surface of the water don't bear much similarity to the higher order, three-dimensional orbitals.
But what you can show, is that standing waves in the confined space of the bowl are limited to only a few discrete possible kinds, and that for each kind, the frequency seems to always be an integer multiple of some fundamental frequency.

Orbital is an one electron wave function. It describes where an electron (or pair, if needed) is to be expected, somehow ignoring all other electrons.

The question starts with "Imagine I was your little brother", would you really start with talking about one electron wave function ? Come on, poor little boy ....

@mannaia I hope that the poor little brother is not taught about elementary particles without highlighting their wave-particle duality and thereby introducing wave function (in some intuitive way).

An electron is both a particle and a wave.
Sound waves travelling in free air can have any frequency. But a vibrating disc can only resonate at certain frequencies. These correspond to the vibration modes of the disc. The animations at the bottom of https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane show these vibration modes, and there is a one-to-one correspondence with orbitals of s,p and d symmetry (more complex orbital symmetries require 3 dimensions.)
Although this analogy is far from perfect, it shows the principles of a wavefunction. Note that in all cases except 1s, there is a part of the membrane that does not move at all. This is called a node. Nodes can be either planar or circular/spherical. In an orbital, a node is a place where the probability of finding an electron is zero.
The total number of zones where the probability of finding an electron is zero (this includes the zone of infinite distance from the atom) is given by the number in the orbital designation. The number of planar nodes is given by the letter.
Thus all orbitals with designation 3 have 3 zones where the probability of finding an electron is zero. 3s has 2 spherical nodes (the 3rd zone is infinite distance from the atom.) 3p has one planar and 1 spherical node. 3d has 2 planar nodes (giving it the familiar 4-lobed shape.) Note also that there is one 3d orbital with a conical node, but we will consider it as belonging to the same group as the planar notes.
Going back to the analogy of the circular disc, we see that it can vibrate in a number of discrete modes, and its motion can be described as a combination of those modes.
In an atom, the pauli exclusion principle dictates that all electrons must have different quantum numbers. Thus each orbital can contain up to 2 electrons, provided they have different spins. In this way the atom differs from the vibrating disc analogy, in that the disc can vibrate with any amplitude, but an orbital can only contain 0,1 or 2 electrons.
Each electron has a certain energy, depending on what orbital it is in. The energy of the electron depends also on its interaction with other electrons in other orbitals. Largely because of this interactions we find that the orbitals with more planar nodes are at higher energy than the ones with less planar nodes.

Imagine I was your little brother, how would you explain an orbital to him? (assuming he knows what atoms electrons and neutrons are, and the rest of the basics till that point).

There's a bit of a problem there, in that he probably won't know what electrons are. There's no electron model in the Standard Model. It's described as a fundamental particle, but there's no real description of how it's created in gamma-gamma pair production or what it really is. You can find information about spinors and Dirac's belt and the Dirac equation, which is a wave equation. But then people talk about point-particles and probability and everything gets confusing.

I have been reading up on this and can't make heads out tails of it.

IMHO it's easy when you do a bit of detective work. Have a look at the Wikipedia atomic orbitals article and note that electrons "exist as standing waves". The moot point is that when you take an electron out of an atomic orbital it still exists as a standing wave. You made it along with a positron out of photons in gamma-gamma pair production. You can diffract electrons. Electron diffraction "refers to the wave nature of electrons".  The electron is a wave. A standing wave. Standing wave, standing field. The electron doesn't orbit the nucleus like a planet, it's more like an electromagnetic Saturn's ring. It isn't flat, it's spherical, but that's hard to imagine, so in describing it to your little brother, stick with a flat ring. To show it to him, give him a sparkler, and get him to rotate it round and round. This makes a ring-shaped image, which relates to the s orbital below:

Image courtesy of UCDavis Chemwiki
Now get him to rotate the sparkler in a figure-of-eight fashion. This relates to the p orbital in either the x y or z orientation. Then get him to rotate the sparkler in a four-leaf-clover fashion. That relates to a d orbital, and so on.

As a side how do the different energy levels relate to them.

Because we're dealing with electromagnetic waves rather than sparklers, and the energy levels are a bit like gears. See this picture by Kenneth Snelson:

There's always an integer number of wavelengths in the standing wave. The atom is a "quantum harmonic oscillator". The electron wave can effectively change up a gear when it absorbs an E=hf photon wave. When it emits a photon, it effectively changes down a gear.
Of course electron orbitals aren't quite as simple as that, but it should get your little brother off first base.

The electron is not a wave. It has particle properties, too. Every microscopic particle is affected by wave-particle dualism. It doesn’t help to highlight certain wave features and pretend the particle part doesn’t exist.

@Jan : it's like Chinatown: She’s my sister. Slap. She’s my daughter. Slap. She’s my sister and my daughter. The electron is a particle, but that particle isn't some billiard ball, it's a wave. This is best appreciated with the photon, see Wikipedia. Look at the first line: A photon is an elementary particle. Then see this: the energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ).  And that wavelength can be tens of metres. It isn't necessarily microscopic.

I'd no more try to explain what "an orbital" "is" to my hypothetical little brother, than I would to a dog. IF there were some reason to explain them, then I'd have to know what it was. I noticed you didn't even mention whether you were talking about atomic orbitals or molecular orbitals...
If pushed, I'd say that orbitals describe the shape of the volume of space in which the bound electron with a given set of quantum numbers is most likely to be, the shape of the space where that electron spends most of its time.


text:
What is an 'acidic' oxide?
My book says that:

Oxides in higher oxidation states of elements of group 14 are generally more acidic than those in lower oxidation states.The dioxides $\ce{CO2}$, $\ce{SiO2}$ and $\ce{GeO2}$ are acidic whereas $\ce{SnO2}$ and $\ce{PbO2}$ are amphoteric in nature. Among monoxides, $\ce{CO}$ is neutral, $\ce{GeO}$ is distinctly acidic whereas $\ce{SnO}$ and $\ce{PbO}$ are amphoteric.

First and foremost:
How can an oxide be acidic? Is it because of their reactions with other compounds? I know about Arrhenius, Lewis and Bronsted acids and bases but still can't seem to connect!

Secondly, why does the book say "Oxides in higher oxidation states of elements of group 14 are generally more acidic than those in lower oxidation states"?  Is there a logical reason behind this?

Is the trend mentioned in the second question followed by all groups of the periodic table and not just for group 14?
By 'all groups', I am referring to s and p block elements. I don't know much about the chemistry of the d and f block elements - I will study them in a few weeks time, but I reckon they should also follow this trend.

The word "acidic" in "acidic oxide" means that the hydrated form of the said oxide is an acid. It does not mean that the oxide itself is an acid either by Arrhenius, by Bronsted, or by Lewis.

related http://chemistry.stackexchange.com/questions/6751/how-does-one-tell-if-a-specific-molecule-is-acidic-or-basic

Book is wrong about GeO unless it/you meant GeO2. https://en.m.wikipedia.org/wiki/Germanium_monoxide is reported as amphoteric.

@Ivan you may be right but usually an acidic oxide is, or as it reacts it generates, a Lewis acid.

Oxides are acidic or basic based on their reaction with a base or an acid.
Here, in group 14 elements, $\ce{CO2}$ is acidic ie; it reacts with a base. When the oxides are amphoteric, they react with both acids and bases.
Oxides of higher oxidation states are more acidic because of higher positive charge. On going down the group the ability to attract electron decreases and electropositive character increases. Therefore an element becomes less acidic and more basic down the group.
I did not understand your last question, but each period has a trend for their acidic and basic nature of oxides.


text:
Steam Distillation - Condensate not going through condenser
Basically what the title says.  I set my equipment up exactly like pretty much every diagram I could find.  There's condensation through the claisen adapter and into the distillation adapter, but for some reason, the condensation won't fall slide down into the condenser tube.
The boiler flask has been at a roaring boil for several hours, the water going through the condenser tube is hooked up properly and ice cold, and the receiving flask is sitting in ice.  Everything is identical to the diagram.  I've tried it with and without the thermometer - and with and without the separatory funnel but the condensate won't get in the consdenser tube.
Any ideas?

This thing will never work satisfactorily. The path to the cooler is too long, and you need a high power steam source.  Thermometer is superflous, you know at what temperature water boils. http://www.bcp.fu-berlin.de/chemie/chemie/studium/ocpraktikum/_Bilder/_Laborpraxis/_Wasserdampfdestillation/bild_wasserdampf1/wasserdampf1.gif

It was a question that got a good answer, so it must have been a good question.

First of all, make absolutely sure that there is nothing blocking the path between the distillation flask and the final vent, and that all the joints are tight.
When beginning the distillation, you should be able to view the condensation line move up within the apparatus after the mixture begins to boil. If it moves up toward the thermometer, but then stops and never makes it up to the condenser, that might indicate that you need to insulate the apparatus above the distillation flask and up to the thermometer. Try wrapping it with glass wool and foil and see if that helps.

Completely erased the previous comment...I actually took a moment to look up glass wool...Going to give it a try

Well, that was it. Wrapped insulation and aluminum foil around the tops of the claisen adapter and it was all good. Thanks iad22agp.


text:
Why Birch reduction of alkyne is E-selective?
From Organic Chemistry(Clayden, Greeves and Warren, 2nd edition), pp. 682:

The sodium donates an electron to the LUMO of the triple bond (one of the two orthogonal $\pi^\ast$ orbitals). The resulting radical anion can pick up a proton from the ammonia solution to give a vinyl radical. A second electron, supplied again by the sodium, gives an anion that can adopt the more stable trans geometry. A final proton quench by a second molecule of ammonia or by an added proton source (t-butanol is often used, as in the Birch reduction) forms the E alkene.

I have some questions on this reaction:
(1) Double bond on the vinyl anion blocks rotation, but how can anion adopts trans geometry between cis and trans without rotation?
(2) Why anion with trans geometry is more stable? If it is because of steric repulsion between $R^1$ and $R^2$, why electronic repulsion between lone pair and $R^1$ can be ignored?

It's not rotation but inversion (remember amine lone pair inversion) from a linear $sp$ form to trigonal planar $sp^2$ that occurs. Lone pair and radical electron are held close to atom and repulse each other a lot so stable to keep them in anti dihedral positions. If the lone pair next to a radical electron was less repulsive, then it would have a lower energy transition state and be more readily formed according to Hammond's Postulate. But experimental data indicates that that is not the case.
Picture Citation: Kurti (2019)

Could you please expand the reference "Kurti (2019)"? Currently, it's not clear whether it refers to a textbook, a paper or a website.

@andselisk it's a presentation by Laszlo Kurti personally offered at a lecture of his. I apologize it's not publicly available online.


text:
How much precipitation occurs for a given ionic product?
Let us say we have a salt AB in water. Suppose the ionic product exceeds the solubility product at a particular temperature. Now, I know that the ions will precipitate out of the solution. My doubt is, how much of the ions will precipitate? By saying "precipitation starts", do we mean that all the ions will precipitate, or will some ions remain in the solution?

Some ions will remain. The solubility product will tell you how much exactly is that "some".

Solubility product is the product of concentration of ions that can be soluble / in aqueous ionic state under given set of conditions. When ionic product is more means, the remaining is not soluble. Only that amount precipitates out. Still there are ions in water, product concentration of which is exactly equal to solubility product.
For basic understanding of equilibrium constant, you can refer: http://www.adichemistry.com/physical/equilibrium/introduction/chemical-equilibrium.html


text:
Why are wavefunctions with same principal quantum number and different azimuthal quantum number orthogonal
The value  turns out to be zero . This is understandable since they are completely independent of each other and also perpendicular in real space. But also turns out to be zero . How do I justify the orthogonality of the s and the p levels?

The mathematical answer to this specific question lies in the fact that s and p orbitals are simultaneous, non-degenerate eigenfunctions of the quantum mechanical $\hat{L}^2$ operator, which is hermitian:
$$\begin{align}
\hat{L}^2\left|\mathrm{s}\right> = [l(l+1)\hbar]^2 \left|\mathrm{s}\right> &= 0 \\
\hat{L}^2\left|\mathrm{p}\right> = [l(l+1)\hbar]^2 \left|\mathrm{p}\right> &= 2\hbar^2 \left|\mathrm{p}\right>
\end{align}$$
since $l = 0$ for s orbitals and $l = 1$ for p orbitals.
It is a simple proof that the eigenstates of hermitian operators that correspond to different eigenvalues are necessarily orthogonal. Suppose, for a hermitian operator $A$, we have $A\left|m\right> = a_m\left|m\right>$ and $A\left|n\right> = a_n\left|n\right>$ with $a_m \neq a_n$. Then:
$$\begin{align}
\left\langle m \middle| A \middle| n\right\rangle &= \left\langle n \middle| A \middle| m\right\rangle^* & &\text{(definition of hermiticity)} \\
a_n\left\langle m \middle| n\right\rangle &= (a_m\left\langle n \middle| m\right\rangle)^* & &\text{(extracting the eigenvalues)} \\
&= a_m\left\langle n \middle| m\right\rangle^* & &\text{(eigenvalues of hermitian operator are real)} \\
&= a_m\left\langle m \middle| n\right\rangle & &\text{(definition of inner product)} \\
(a_n - a_m)\left\langle m \middle| n\right\rangle &= 0 \\
\left\langle m \middle| n\right\rangle &= 0 & &(a_m \neq a_n)
\end{align}$$
You can likewise show that any pair of atomic orbitals in the hydrogen atom are orthogonal, using whichever of the three operators $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ that you need. All atomic orbitals are simultaneous eigenfunctions of these three operators, and have different eigenvalues which correspond to the quantum numbers $n$, $l$, and $m_l$:
$$\begin{align}
\hat{H}\left|n,l,m_l\right> &= E_n\left|n,l,m_l\right> = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 n^2}\left|n,l,m_l\right> \\
\hat{L}^2\left|n,l,m_l\right> &= L^2\left|n,l,m_l\right> = [l(l+1)\hbar]^2\left|n,l,m_l\right> \\
\hat{L}_z\left|n,l,m_l\right> &= L_z\left|n,l,m_l\right> = m_l\hbar\left|n,l,m_l\right>
\end{align}$$
Since no two orbitals have the same three quantum numbers, no two orbitals can have the same eigenvalues for all three operators. Consequently, there must be at least one operator which can be used to show that those two orbitals are orthogonal.

Thanxx for the explanation! Another thing I would like to ask is that since px and py are perpendicular in real space...can this be used to say that their wave functions will be orthogonal?

Yes, you can. Although do note that the $\mathrm{p_x}$ and $\mathrm{p_y}$ orbitals are no longer eigenfunctions of the $\hat{L}_z$ operator (they are linear combinations of the eigenfunctions). To prove it mathematically, you would need the explicit form of the wavefunctions.

Thanxx @orthocresol


text:
Why are lab coats required in chemical laboratories?
Well at first it seems kind of awkward but let me tell you what I was thinking from past few days. Currently I am studying and go to school. In school, whenever we have to go inside a chemistry lab then it’s compulsory for us to wear a lab coat; without one we were not allowed to enter and do experiments. But then one question arises is: Why it is compulsory?
Then I started finding solution for this question. I thought that we wear lab coat just for safety against all those chemicals we use for experiments but then if accidentally someone drops any concentrated acid then I guess that lab coat won’t be able to protect us. So if we wear something solid or something hard which can protect us from those chemical that might be helpful but we only wear a lab coat.
Then I thought that lab coats have so many pockets on them so it might help us to carry a few things but then again why don’t we use something else which is handy rather than a lab coat (still this example is quite useless).
After hours of thinking I am not able to come up with anything.
And also whenever anyone goes inside chemistry lab he/she wears a lab coat which is quite similar to everyone else’s. I.e. everyone wears same type of lab coat (a long, white colored coat) which is quite fascinating.
So my question is really simple: Why do we have to wear a lab coat when we go inside a chemistry lab and how come everyone uses same type of lab coat?

I did a lot of acid/base work. My lab coat was full of little holes after a couple of years. So it will protect you from drops, but not of course from getting drenched. That is what the lab shower is for. If you have to use the shower all clothes off. No time for modesty. Take nothing off over your head. There should be bandage scissors by the shower.

You should also have some sort of protective eyewear. Think of the eyewaer as horse blinders. A student lab is a crowed place. Also turn your head first and look where you are going before darting off in another direction.

It is protection. It should be worn whenever you work with potentially dangerous chemicals. In German, it is even called (part of the) persönliche Schutzausrüstung, i.e. personal protective gear. Lab coats:

protect you against contamination with chemicals — acids, bases, toxic molecules and salts. They will ‘hit’ your coat and ‘stick’ to it rather than reaching your clothes. Don’t say you are careful. Your labmate may not be or an accident may happen. This is also the reason why labcoats are never allowed where food or drink are consumed.
protect you against spills of liquids. You can quickly take off a lab coat and not be contaminated by the liquid in question. This is basically an extension of the first bullet point.
protect you against fire. This might not be obvious, but lab coats are generally cotton which does not burn well. If you are wearing synthetic fibre below the lab coat, ripping that off could give you valuable seconds for not burning to death.
show other people that you are working with something potentially dangerous. Assuming you are dealing with adults that means they will know better than startle you because they can.

Everything else a lab coat supplies is basically just extension of the protection or meant to help people actually use it. Pockets: If it didn’t have any, people would keep using their trousers’ ones potentially contaminating them. Long sleeves: You arms need protection, too. Being long: so do your legs (at least partially).
And of course, it’s pretty cool if you walk in a lab coat, isn’t it? (That was my feeling at least.)

And one last thing which I also mentioned is that how come everyone uses same type of coat world wide ?

@Shashank Well, because never change a winning team. Or alternatively: Because it is so optimised, one would arrive at the same optimisation level anyway.

@Shashank they're not all the same, some are cotton, some polypropylene, some tyvek


text:
How can we find the eV value of molecular orbital diagrams for a given atom?
It seems like sometimes the electron volt value of the molecular orbital is given but most of the time it just shows relative values of s p d orbitals. Is there some lookup table for such values?

It can be either experimentally determined (look up photoelectron spectroscopy) or just calculated with a computer (lots of QM if you want to find out more - start with the secular equations perhaps).

Exact values of MO energy are rarely given as they are difficult to obtain in anything other than relatively trivial systems. When they are given, it is often not even clear whether they are experimentally derived (Photoelectron spectroscopy, as mentioned by orthocresol), or computationally derived (by calculations to approximate the wave functions and their properties). For this reason, most MO diagrams presented are qualitative, derived from our knowledge of symmetry, orbital energy etc.
There is a book, The Organic Chemist's Book of Orbitals (Salem and Jorgensen) which gives the MO's of many simple, and slightly more complex  molecules along with computationally derived energies (in eV).
For anything more complicated, you could also try searching the literature, however this would probably just turn up very specific examples that people have calculated to support their research.
If you need specific energies for molecules, you might look into computing them yourself. Simple software such as Spartan does an acceptable job at computing MO energy, with a variety of computational methods available depending on how much computing power you have.


text:
In a simple chemical cell
Why do the electrons, arising from the oxidation of the anode, flow through an external circuit? Why does it not just flow through the electrolyte and reduce the cathode? Why do the electrons give the wire preference?

Free electrons running around in water = extremely, extremely unstable.

@orthocresol what do you mean by unstable?

It will pretty much immediately reduce any species it bumps into.

@orthocresol But in a simple chemical cell, where the half cells aren't separated, will electrons only flow through the external circuit, or will there be some flow of electrons through the electrolyte?

External circuit. The electrolyte conducts electricity via ions, not electrons

@orthocresol if electricity is conducted through ions, what are the steps through which a solution of NaCl could conduct electricity?

I will give you the example of molten NaCl. In a solution, the species involved are not likely to be Cl- or Na+ unless the solution is very concentrated. In molten NaCl, Cl- gives 1 electron to the anode. Electron travels from anode to cathode. Na+ picks up 1 electron from cathode. Rinse and repeat.

The crucial point which explains this is that there are two different forms of conduction happening here.
The first kind of conduction, and the one we are most familiar with is electronic. This is a flow of electrons, and is the conduction which flows in the wire and through the load (a motor, LED etc).
The second kind is the flow which occurs in the electrolyte, and this conduction is ionic. It is a flow of charged particles (ions) within the electrolyte. Even though we often speak of the conductivity of an electrolyte, what is really meant is the electrolyte's ionic conductivity.
This is governed by the mass of the ions, their concentration, temperature and the viscosity of the solution.
Electrolytes do not really conduct electrons at all, which is why there is no 'short circuit' through them. The conductivity arises due to several mechanisms - either the ions themselves move through the solution, or electrons (or protons) hop from one ion to the next, conveying the charge.
This will only happen if there is a means of causing the flow of charge through the electrolyte. When it is open circuit, the electrolyte is overall neutral in charge and no current will flow in it.
If an external circuit is connected, it allows electrons to flow from one electrode to the other, and then the corresponding amount of charge will flow through the electrolyte to balance it out, producing current.

Electrical current will follow the path of least resistance (i.e. it will flow through whichever material is a better conductor); in this case, the external circuit. The conductivity of the electrolyte will depend on what's dissolved in it. I.e. pure water has very poor conductivity, while sea water (dissolved NaCl etc.) is a much better conductor. However, neither are as good as a metal such as copper, and hence electrons flow around the circuit.
If the electrolyte did happen to be a more effective conductor than the external circuit (can think of no examples), then presumably the current would flow through the electrolyte instead.
Hope that helps!


text:
Reaction of Alcohol and Metal Hydroxide
Lets say there is $\ce{CH3-OH + NaH}$. The reaction leads to $\ce{CH3-O- Na+ + H2}$. However $\ce{H-}$ from the metal hydroxide is also a strong nucleophile, so why can't the reaction lead to $\ce{CH4 + OH-}$? I realize that $\ce{OH-}$ is a bad leaving group, but then again, an alkoxide isn't a good leaving group either. An alkoxide's $\mathrm{p}K_\mathrm{a}$ is even higher than $\ce{OH-}$, so its a worse leaving group.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by using MathJax, which does a better job of rendering chemistry than HTML does. For more information on how to do so yourself, check out the [help], this meta-post or this one

Also, hydride is a pretty bad nucleophile. Much better base than nucleophile.

@Jan This answers my question. And thanks for the edit!

In this simple case it is pretty obvious that a hydride ion is a much better base than it is a nucleophile. But that doesn’t answer everything; we could consider cases with much better nucleophiles ($\ce{Br-}$).
Another reason is that hydroxide is an even lousier leaving group than hydride is a nucleophile. It is literally impossible to do $\mathrm{S_N}$ displacements of uncharged oxygen atoms unless you are opening an epoxide (in which case you lose ring strain making the overall process favourable. It does not matter whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$; $\ce{RO-}$ will never leave. To remove an oxygen atom, you always need to use strong acids that can protonate it, giving the far better leaving group $\ce{ROH+-R}$ which gives the free alcohol/water when displaced. Thus, the following reaction works well due to the very strong hydroiodic acid (I tried it):
$$\ce{R-CH2-OH + HI -> R-CH2-OH2+ + I- -> R-CH2-I + H2O}$$


text:
How can I prepare Cu2S powder?
How can I prepare $\ce{Cu2S}$ powder?
I have some idea about preparing the ternary composition containing copper, but copper sulfides are difficult due to the large difference between their melting points (Cu: 1084°C, S: 119°C).

Could you expand the intended use of the compound? Maybe a more readily available alternative exists.

i am working in thin film field , and preparing films using thermal evaporation methode is the best for me . to do this the conpound must be in form of powder or smal pieces . mixing the two elements in  a furnace is the best way for me ,but the large differece in melting and boiling points is big problem

Unless the cost is a hindrance, I strongly suggest to buy it. Copper (I) sulfide is commercially available,  and, unlike home-made compound, you don't have to care about characterisation and impurities. Seriously. Otherwise, mixing the elements and heating in a closed crucible in argon atmosphere to 300-400 Celsium should work.  Also, consider looking through literature - to my knowledge, other methods may be used.

i think i agree with you ,buying the compound is the best way. Thanks

By heating copper and sulfur together, you won't get it anyway; you'd get $\ce{CuS}$ instead.

It is possible to obtain $\ce{Cu2S}$ by heating a mixture of finely-powdered elemental copper and sulfur in a crucible up to temperatures of above 450 °C. The powdered form of the educts ensures that the reaction starts as soon as the sulfur has melted. Alternatively, the mixture can be milled in a planetary ball mill for 2 hours, which yields $\alpha$-$\ce{Cu2S}$. However, $\ce{Cu2S}$ produced by these methods has limited long-term stability, as it slowly converts into $\ce{CuS}$ even when left in an inert atmosphere (reference).

You will end up getting CuS.

$\ce{Cu2S}$ can be prepared from copper pyrite $\left(\ce{CuFeS_2}\right)$. I think this is a good method because the following reaction is part of extraction process of copper metal, so the yield would be quite high.
$$\ce{2CuFeS2 + O_2 \enspace \longrightarrow \enspace Cu_2S + 2FeS + SO2}$$
Of course this would require high temperatures, which is attainable in reverberatory furnace. You just have to find one somewhere near you.

You will end up getting CuO.

Classic methods of $\ce{Cu2S}$ synthesis are listed in Bauer's Handbuch der präparativen anorganischen Chemie [1, p. 981]:
Original text (in German):

I. Ein Gemisch von Cu und S in entsprechenden Mengen wird in ein Quarzröhrchen
  gegeben und dieses an der Hochvakuumapparatur abgeschmolzen. Es wird dann bis zum Schmelzen der Substanz erhitzt.
II. In einem evakuierten, zugeschmolzenen Glasrohr befindet sich auf der einen
  Seite reinstes Cu, auf der anderen Seite reinster S [...]. Beide Elemente werden in stöchiometrischem Verhältnis eingewogen. Man erhitzt 1-2 Tage auf 400 °C; danach ist die Reaktion beendet.
Andere Darstellungsmöglichkeiten:
  Kupfer(II)-sulfid wird im Vakuum bis zum Schmelzpunkt des $\ce{Cu2S}$ erhitzt. Die Substanz befindet sich zweckmäßigerweise in einem Graphittiegel innerhalb eines ev akuierten Porzellanrohres [E. Posnjak, E. T. Allen u. H. E. Merwin, Z. Anorg. Allgem. Chem. 94, 95 (1916)].
  $\ce{CuS}$ (durch Fällen aus einer $\ce{CuSO4}$-Lösung mit $\ce{H2S}$) wird in einem $\ce{H2}$/$\ce{H2S}$-Strom reduziert. Optimale Bedingungen sind eine Temperatur von 700 °C, eine Mischung von 4,6% $\ce{H2S}$ und 95,4% $\ce{H2}$ und eine Versuchsdauer von 1 h. Das Produkt ist kristallisiert und recht rein (N. P. Diev u. E. M. Yakimets, Ural. Filial. 1955, Nr. 3,5; C. A. 50, 13638 a).

Translated (in English):

I. A mixture of Cu and S in appropriate amounts is placed in a quartz tube
  and is fused in the high-vacuum apparatus. It is then heated until the substance is melted.
II. An evacuated sealed glass tube contains purest Cu on the one side, and on the other side there is the purified S [...]. Both elements are taken in a stoichiometric ratio. The mixture is heated at 400 °C for 1-2 days till the end of reaction.
Other preparation methods:
  Heating of copper(II) sulfide in vacuum up to the melting point of the $\ce{Cu2S}$. The substance is preferably located in a graphite crucible inside of vacuumed porcelain tube [E. Posnjak, E. T. Allen u. H. E. Merwin, Z. Anorg. Allgem. Chem. 94, 95 (1916)].
  $\ce{CuS}$ (by precipitation from a $\ce{CuSO4}$ solution with $\ce{H2S}$) is reduced in a $\ce{H2}$/$\ce{H2S}$ gas flow. Optimum conditions are the temperature of 700 °C, a mixture of 4.6% $\ce{H2S}$ and 95.4% $\ce{H2}$ for 1 h. The product is crystallized and quite pure (N. P. Diev u. E. M. Yakimets, Ural. Filial. 1955, Nr. 3,5; C. A. 50, 13638 a).

Reactions:
\begin{align}
\ce{2 Cu + S & ->[\pu{400^\circ C}, vac] Cu2S} \tag{1}\\
\ce{2 CuS & ->[\pu{1100^\circ C}, vac] Cu2S + S} \tag{2}\\
\ce{2 CuS + H2 & ->[\pu{600 .. 700^\circ C}][\pu{1 h}] Cu2S + H2S} \tag{3}
\end{align}
These methods are all quite time- and energy-consuming; if you are using CVD the chances are pretty high that you can synthesize pure $\ce{Cu2S}$ in situ, depending on which support you are using. The bottomline is, you need to work in reducing medium or in vacuum to avoid byproducts such as $\ce{CuO}$ or $\ce{CuS}$.
Reference

G. Brauer, Handbuch der Präparativen Anorganischen Chemie, Band II, Ferdinand Enke Verlag 1978, ISBN 3-432-87813-3.


text:
How is the Activation Energy of the Arrhenius equation related to the exponents in rate laws?
I know that activation energy for a particular reaction does not depend on the stoichiometry of the reaction.But how can that be justified from Arrhenius Equation ?Doesnt 'k' change with stoichiometry?Or am I going wrong?
Because we know that in all the kinetics equations, when  the reaction stoichiometry changes the rate constant in the equations have to be altered accordingly.

Rate Law
For the reaction:
$\ce{aA + bB -> cP}$
The rate law is as follows:
$R = k[A]^x[B]^y$
where:

$a$,$b$,and $c$ are stoichiometric constants which have integer values
   $[A]$ is the concentration of species A
   $x$ is the order with respect to species A
   $[B]$ is the concentration of species B
   $y$ is the order with respect to species B
   $k$ is the rate constant.

(1) Neither $x$ nor $y$ depend on $a$, $b$, or $c$.
(2) Neither $x$ nor $y$ is required to be an integer value.
(3) k is measured for a specific temperature. The Arrhenius Equation would predict how k would change with temperature. So k does not depend on any of $x$, $y$, $a$, $b$, or $c$.
Say the reaction was for some sort of gas reaction. I now add a catalyst. I get a new rate:
$R_2 = k_2[A]^{x_2}[B]^{y_2}$
There isn't any way to use the first rate equation to predict what the second rate equation will be. (You could of course try to analyze the function of the catalyst to make predictions on what the best catalyst would be...)

There is a way to think about the problem that will give you an intuition about the key relationships and why they are the way they are. Think about the individual molecular interactions that have to occur for a reaction to happen.
For example, consider some reaction where two molecules have to bang together to cause some chemical change. In many reactions they have to bang together with enough energy for the reaction to actually occur. This is the reason there is an activation energy. Too little energy in the collision and there isn't enough to initiate the chemical change (perhaps a bond needs to be stretched or a repulsive barrier has to be overcome). From this perspective it is easy to see that the activation energy doesn't depend on the concentration.
But the concentration of the molecules does influence the probability that a collision will occur. If there are a lot more of the relevant molecules around, collisions are far more likely and the rate of the reaction will be faster. But concentration doesn't influence the barrier to the reaction for individual collisions.
The high level equations for rates of reaction are, essentially, an embodiment of these ideas. But thinking one molecular collision at a time will give you the correct intuition about the factors that matter in the equation and how they interact.


text:
If you add sodium salt to carbonated water (sparkling mineral water), could you produce sodium bicarbonate?
I know that $\ce{CO2}$ can dissociate into bicarbonate ions within $\ce{H2O}$. If Na is added, then sodium bicarbonate is produced.
Could you then create sodium bicarbonate using carbonated mineral water and sodium in the form of table salt or does that have too many minerals aside from $\ce{H2O}$ in it?

Assuming $\ce{Na}$ is used to denote sodium metal and $\ce{NaCl}$ is used to denote table salt, the following reactions in water occur:
\begin{align}
\ce{Na + 2H2O &-> 2Na+ + 2OH- + H2(g)} \tag {1}\\
\ce{NaCl(s) &-> Na+ + Cl-} \tag{2}\\
\ce{CO2_(g) <=> CO2(aq) &<=> H2CO3(aq) <=> H+ + HCO3-} \tag{3}\\
\ce{H+ + OH- &-> H2O} \tag{4}
\end{align}
So (1), (3), and (4) would give you sodium bicarboante if you dried the solution.
(2) and (3) would give you just sodium chloride back if you dried the solution.


text:
How to determine the strongest oxyacid of chlorine without being given the acid dissociation constant data?
I was told to find the strongest and weakest acids out of $\ce{HCl, HClO, HClO2, HClO3}$, and $\ce{HClO4}$. I knew that $\ce{HClO}$ was the weakest but was stuck between $\ce{HCl}$ and $\ce{HClO4}$ for the strongest (as these are both strong acids). Is there a way to find the relative strength of strong acids without being given the $K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{a}$? Or would I just have to have memorized that $\ce{HClO4}$ is generally stronger than $\ce{HCl}$?

As for your question, I don’t think there is any way to compare the two since there are no real analogies you can use. But I’m not sure enough to actually call this an answer.

Truthfully a lot of "facts" like this are asked in class tests/problems. In the real world if you use those chemicals then you just know, if you don't use them a lot you then you just look the data up. It is easy to rationalize HClO, HClO2, HClO3 and HClO4. You recognized HCl would be an odd-ball. sigh for now you'll just have to memorize it...

Actually, some people would say that $\ce{HClO3}$ is a strong acid too. I don't think there's a real way to compare $\ce{HCl}$ and $\ce{HClO4}$. There are some empirical rules for estimating the $\mathrm{p}K_\mathrm{a}$ of an oxoacid: https://en.wikipedia.org/wiki/Acid_dissociation_constant#Factors_that_affect_pKa_values but clearly they do not apply to $\ce{HCl}$.

Related: How to determine which substance is the strongest base/acid?


text:
How much CO2 gets free when combusting 6 kg n-heptane (C7H16)
I am very new to chemistry and have a question to an exercise.
It's about combustion (complete oxidation) n-heptane ($\ce{C7H16}$).
There were several questions I could answer, for example the chemical equation:
$$\ce{C7H16 + 11O2 -> 7CO2 + 8H2O}$$
But what I can't answer is, how I calculate how much $\ce{CO2}$ gets free when burning 6 kg of the n-heptane.
Can someone help me with this?
Thanks a lot!
Edit:
I am not sure if this information is needed:
molar mass (g/mol): $\ce{C}=12$ $\ce{O}=16$ $\ce{H}=1$
density of n-heptane: $0.684\ \mathrm{kg/L}$

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

For these types of problem, concept of mole is very handy. Here, you know the reaction stoichiometry, that means you have at least three information from that reaction equation. How?
** For every mole heptane, you need 11 mole oxygen for complete combustion
** For every mole of heptane combustion, you will get 7 mol carbon di oxide and 8 mol water.
So, now if you know the mole corresponding to 6 Kg heptane you can easily find out the amount of carbon di oxide produced in mole unit.
The relation is
$$n_{CO_2}=7 \times n_{Heptane}$$
$$n_{Heptane}=\frac{m_{Heptane}}{molecular~weight~of~heptane}$$
Now if you want to also calculate amount of carbon di oxide produced in mass unit or volume unit, you need to use the second equation or ideal gas law equation.

from where do I get the molecular weight of heptane?

It has 7 carbons and 16 hydrogens. add'em up from the atomic weights.


text:
What is the purpose of calculating other orbitals than HOMO and LUMO?
In this video of benzene orbitals, a lot of different orbitals are presented. Even orbitals that, I assume, could never be occupied. What is the purpose of calculating all these orbitals? Isn't all the chemical information needed (bonding, reactions) coming from the HOMO and LUMO (FMO theory)?
Here is an image of a few of the orbitals shown in the video:

Energy diagram of the molecular orbitals explored for the D6h
  structure (left) and the D3h distorted geometry (right)
  (from David Casanova  and  Pere Alemany, Phys. Chem. Chem. Phys. 2010, 12, 15523-15529. DOI:10.1039/C0CP01326A)

Short answer: No, not all information is present in the HOMO and LUMO shapes and energies. I’ll leave it to one of our quantum gurus, though ;)

A practical answer: in most of molecular quantum chemistry, the cost of obtaining unoccupied ("virtual") orbitals is next to nothing because of the algorithms involved. In short, one diagonalizes some matrix (i.e. finding eigenvalues and eigenfunctions, which are the orbital energies and orbitals) and the virtual orbitals are usually delivered anyway.
A more theoretical answer: While a lot of chemistry involves only the ground state, there is UV/vis spectroscopy and photochemistry. For those fields, virtual orbitals are immensely useful, because they will be used for the description of the electronic transitions that happen in those fields. These transitions are not necessarily between HOMO and LUMO because of symmetry, energy scales involved etc.

In general, I'd say you are right, but I think your phrasing is not to the point. You don't get virtual orbitals at "next to nothing" cost, that sounds as if you'd make some extra calculation to get them, but because they are part of the solution. They are not "usually delivered anyway" but they are delivered period. For UV-vis, virtual orbitals are not only "immensly useful" but they are absolutely necessary in almost all cases. The HOMO-LUMO transition is usually only one peak but might not be the most interesting one in photophysics/-chemistry.

@pH13-YetanotherPhilipp There are useful algorithms that do not yield all eigenpairs, but merely give you all of them up to a set threshold (the case in solid state QC). I have applied them in molecular QC. For UV/vis, some DFT fanatics will tell you about approaches that are not reliant on virtual orbitals per se.

Even orbitals that, I assume, could never be occupied.

That's an incorrect way of thinking. Orbitals are not occupied. You need occupation of those orbitals to describe the electronic density of the transformation you are doing.
This is, in my opinion, the biggest flaw of explaining chemistry transformation at the quantum level. It's explained the other way around.

So what is the purpose of calculating all these orbitals?

Two reasons:

you always get as many molecular orbitals as initial atomic orbitals you put in, and the atomic orbitals you put in depends on the size of the basis set you choose.

and

Isn't all the chemical information needed (bonding, reactions) coming from the HOMO&LUMO?

Not at all, but it generally is a reasonably good approximation for some reactions, or some transformations. But for some transformations you do need to involve more of them.

what do you mean by "transformation"? reaction?

@sparkler Could be a reaction, could be excitation, could be application of an external field, or a geometrical distortion.

why do you say that orbitals are not occupied? isn't every orbital occupied by 2 electrons?

Ok, first go check my answer here. It's exactly the same idea with molecular orbitals. Orbitals are occupied by electrons because it's a mathematical convenient way of presenting a description of reality, but it's just a mathematical description.

That said, it's is a complex topic not because it's complex, but because I would explain it to you so that it's clear in a couple of hours if we had access to a whiteboard. This place is simply not good enough to explain you static and dynamic correlation.

"you always get as many molecular orbitals as initial atomic orbitals you put in" This isn't true if you eliminate MOs due to linear dependence, where you can end up with fewer MOs than AOs. But this is an artificial limitation and not related to the question/answer.


text:
Is strength of an acid proportional to concentration or dilution?
I have a confusion between two seemingly contradictory factors.
Since pH =-log[H+], it suggests that more the concentration of the acid, lesser will be its pH, and therefore the stronger it is.
But, according to Ostwald's dilution law, the degree of dissociation of an electrolyte is directly proportional to the square root of dilution. Though this exact relationship is true only for weak electrolytes, I believe that the degree of dissociation of any electrolyte would increase with dilution, keeping the temperature constant. And, more the degree of dissociation, more would be the H+ions released, and stronger the acid.
Where am I wrong in my reasoning?
Thanks in advance.(Please give me an answer and not just a comment)

Is strength of an acid proportional to concentration or dilution?

This is an interesting question. It is sort of like being asked "What is your size?". If you are buying shoes then size means one thing, yet it is something different if you are buying pants.
So there are two different "strengths."
Strength can mean how much acid is present. So per unit volume of solution, a solution with 50% concentration has half the strength of the 100% concentration.
Strength can also mean how acidic, in terms of pH, would a specific solution be if the acid concentration were kept constant. In this sense hydrochloric acid is a stronger acid that acetic acid.
So the gist of your problem is that you're trying to draw a comparison between two different uses of the word strength when such a comparison is pointless.


text:
What is the difference between galvanization and sacrificial protection?
If we use zinc in both methods, then the same process occurs: zinc corrodes in preference to iron, and eventually zinc has to be replaced. So what's the difference?

They are used in different media. Galvanization is used to protect iron exposed to non-corroding medium (air), while sacrificial electrodes are used to protect iron in electroconductive and corrosive medium (sea water or moist soil).
Protective zinc oxide layer on galvanized surface is desired to stay undisolved,  while the purpose of sacrificial electrode is to keep being oxidized to assure electron flow to other metal surface. In latter case medium electrconductivity is a must for ion transfer process, while corrosivness is needed for oxidized zinc surface layer disolution.

Galvanising is coating the iron with a layer of zinc in order to prevent it from rusting. However, sacrificial protection is attaching a piece of zinc to the iron object.
Both methods use the same principle to work. Even if the iron is exposed to water or oxygen in the air, the water and oxygen react with zinc, as it is more reactive.

In galvanisation, zinc is coated on the surface of iron to prevent it from corroding.
But sacrificial protection is the process where zinc is kept near iron surface so that zinc corrodes instead of iron. Here, zinc is not coated on iron's surface.

No difference ; the zinc corrodes preferentially providing protective current/potential to the iron/steel. Although  anodes can be a distance away from the steel if they are immersed in a conductive media. Also , anodes can be replaced MUCH easier than replacing a zinc coating.


text:
Can the compounds NOF3 and POF3 both exist?Why?
Today I came across a question (in one of my books) like

Can the compounds $\ce{NOF3}$ and $\ce{POF3}$ both exist? Why?

After some googling I found that both of them do indeed exist. I can understand that $\ce{P}$ can extend its octet due to empty d orbitals. But $\ce{N}$ can't do so. So how can it possibly exist?

Think of $\ce{HNO3}$; it should be pretty much the same.

@IvanNeretin is it because of a coordinate covalent bond?

I am not getting.Moreover say NCl5 does not exist.

Well, yes, you may think of it as a coordinate bond $\ce{N-O}$, or a structure with charges like $\ce{N+-O-}$, which is the same. It's been discussed before, maybe a dozen times. http://chemistry.stackexchange.com/questions/7174/the-lewis-structure-of-hno3

$\ce{P}$ cannot extend its octet either; the d-orbitals are too far removed in energy. In fact, the creation process of both from atoms is identical.

Note that $\ce{NCl5}$ does not exist because $\ce{N-Cl}$ bonds are polarised towards nitrogen as can be shown experimentally for $\ce{NCl3}$.

Both molecules can be considered the acid (hydrate) trifluoride of the respective acid (phosphoric acid or nitric acid). In the case of phosphoric acid, this is immediately obvious: Just replace the three hydroxide groups with a fluorine atom, each. For the nitrogen compound, you initially need to (formally) add water before you can do the substitution:
$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$
(Note: this is not a proper chemical reaction, merely an illustration of the thought processes.)
You could also approach this compound from the ammonium cation $\ce{NH4+}$ by replacing three hydrogen atoms with fluorine and the fourth with oxygen. Or you start off with $\ce{NF3}$ (isoelectronic to $\ce{PF3}$ and then oxidise the nitrogen by adding a single oxygen atom much in the same way you would oxidise from $\ce{HNO2}$ to $\ce{HNO3}$.
In fact, for many of the compounds traditionally explained with d-orbital participation for main-group elements, an explanation adhering to the octet rule (no more than eight valence electrons per main-group atom; two for hydrogen/helium) is often closer to the truth. This includes, but is not limited to $\ce{OPF3, SO4^2-}$, sulphonic acids and many more.

Yes, trifluoramine oxide, $\ce{\overset{-}{O}–\overset{+}{N}F3}$, does exist.
The structure written above is the "conventional" structure with sigma bonds indicated between all four directly connected pairs of atoms. It thus appears that there can be only sigma bonds. Yet the length and spectrometric chatacteristics on the nitrogen-oxygen interaction seem to indicate multiple-bond character.
What happens is explained by molecular orbital theory involving only the usual $s$ and $p$ valence orbitals contributing. The $\ce{\overset{-}{O}}$ function described above is a powerful $\pi$ electron donor; its lone pairs with the proper symmetry can actually overlap with the otherwise antibonding components of the $\ce{N–F}$ bonds. This loses some of the $\ce{N–F}$ bonding, but strengthens the $\ce{O–N}$ bonding giving that interaction some $\pi$ bonding character. The latter effect is stronger given the right combination of input atomic orbitals.
We can think of this as a form of resonance hybridization and bond delocalization, in which nitrogen-fluorine sigma bonds are exchanged for nitrogen-oxygen pi bonds:
$\ce{\overset{-}{O}–\overset{+}{N}F3<-> [O=\overset{+}{N}F2]F^-}$
Note the role of the electronegativity of the fluorine, both stabilizing the negative charge imparted by the $\pi$-acceptance and placing more of the antibonding-orbital electron density on the nitrogen where it overlaps the $\pi$-donor oxygen.
This type of bond delocalization may also be rendered with later-period elements, as in the phosphate or sulfate ion where it occurs simultaneously along all four bond axes; the fact that this model also covers $\ce{NOF3}$ is one reason it has displaced the "outer-$d$ orbital" theory.

There is also trimethylamin oxide, present in fish bodies.

True. But without the strong electronegativity of the proposed acceptor lifands noted above, TMAO lacks the nitrogen-oxygen bond character noted here for the fluorine compound. The nitrogen-oxygen bond is much longer and the oxygen itself much more basic in the methyl compound.

I was focused on electronic structure rather than on electronegativities and bond polarities.


text:
Is the P-Cl bond energy same in PCl5 and PCl3?
Is the $\ce{P-Cl}$ bond energy same in $\ce{PCl5}$ and $\ce{PCl3}$? Why?
According to wolfram alpha both have $\pu{326 kJ mol-1}$ energy. But my study material says that they are not same. I'm confused. Can someone clarify?

Wolfram Alpha does not know apples from oranges when it comes to chemistry. Also, we don't use mathjax in the title.

Hehe :-P!Ok! @IvanNeretin .Do tell your answer also...

Of course they are different. I expect the bond in $\ce{PCl3}$ to be stronger.

@IvanNeretin why is it so?And in PCl5 itself are all the bond lengths same or different?

They can’t be the same, because $\ce{PCl5}$ (in gas phase) has two different $\ce{P-Cl}$ bonds while $\ce{PCl3}$ only has one.

Mind that Bond Energy and Bond Dissociation Energy are different.
Check the wikipedia pages for both.
Co-incidentally they might be similar in terms of bond energy.

The $\ce{P-Cl}$ bond in $\ce{PCl3}$ is stronger ($\pu{322 kJ/mol}$) than that in $\ce{PCl5}\ (\pu{260 kJ/mol})$. However the $\pu{326 kJ/mol}$ value comes from the average of all $\ce{P-Cl}$ bonds found in all known molecules containing $\ce{P-Cl}$ bonds (including but not limited to $\ce{PCl3}$ and $\ce{PCl5}$).


text:
Heitler-London hydrogen model
As I am learning about the valence bond theory, the model of hydrogen that Heitler and London created, seems to make sense. But, the theoretical and the experimental results are not the same.
Experimental results:
For the molecule to form, the distance between atoms should be $r=74.1~\mathrm{pm}$
and the energy on the other hand $E= -4.474~\mathrm{eV}$.
The theoretical results are this way:
$r= -\frac{1}{4}~\mathrm{pm}$ and $E= 89.9~\mathrm{eV}$
My question is, is it known why these deviations occurred?
Or is this considered as a correct calculation with approximation?

I think you have some typo or flaw in your question.  "E = 89.9 pm" doesn't make sense.  You are saying energy equals a length. The edited version "E = 89.9 eV" has energy units, but the theory never predicted such as large value.  And "r = -1/4 pm" doesn't make sense either.  Distance can't be negative. Also, even it were positive 1/4 pm, the theory never predicted such a small magnitude value.
The initial calculations of Heitler and London were:
r = 80pm
E = -2.4 eV
A major source of error was inability to compute integrals.  Additionally, the calculations neglect ionic contributions.
See Computational Imperatives in Quantum Chemistry for a historical overview.

Yes it was a mistake..

The first paragraph should really have been a comment to the question.

@Jan the answer needs to emphasize that the values in the question (even after the edits) have no relationship to Heitler and London.

@DavePhD Yes; but I was explicitly referring to the bit ‘$E = 89.9~\mathrm{pm}$’. That’s not really relevant at all, that was a b** typo of some sort that shouldn’t (imho) be addressed in answers but in comments. The bit after that, stating the original values, is fine by me.


text:
dsp3 and sp3d hybridization — what is the difference and which applies to square pyramidal?
What is the difference between $\mathrm{dsp^3}$ and $\mathrm{sp^3d}$ hybridization?
Are they one in the same?
Also, I have a book that says that in compounds where the central atom is $\mathrm{dsp^3}$ hybridized, it's shape is square pyramidal, like that of $\ce{BrF5},$ but the hybridization of $\ce{Br}$ in $\ce{BrF5}$ is $\mathrm{sp^3d^2}.$ Why is it so? Is my book wrong or am I missing something?

$dsp^3$ hybridization cannot result from square pyramidal geometry because it only produces five hybrid orbitals whereas square pyramidal requires six. $sp^3d^2$ is a possible hybridization for square pyramidal geometry.

So , my book was wrong and I was right , right ? Even if so , then what is the difference between dsp3 and sp3d , if there is one ?

Well, your book is wrong. Both hybridisations don’t really exist. Main group elements only hybridise s and p-orbitals.

So there are no sp3d , sp3d2 or sp3d3 hybridizations ?

Yes,  they exist. XeF2, for example, exhibits sp3d hybridization on the Xe.

@quintopia No it doesn’t. Xenon is essentially unhybridised as is fluorine. Xenon’s $\mathrm{p}_z$ orbital forms a four-electron-three-centre bond with the two fluorine atoms. Xenon’s 4d orbitals are fully populated and belong to the core orbitals; they won’t participate in bonding. Xenon’s 5d orbitals are empty and energetically remove; the 6s orbitals are closer. Thus, xenon has no available d electrons for ‘hybridisation’.

@Jan there are indeed sources, albeit outdated, claiming that the hybridization of S in SF6 is sp3d2. Those sources would be referencing the expanded-octet model of hypervalence. Nowadays, of course we know that this model is flawed and superseded. However, the model is not wrong...

@DHMO There are also sources (albeit even more dated) on the positive effects of smoking on health, so … ;)

@Jan Indeed, the model of sp3d2 is inaccurate, but the current model which classifies every element as sp3/sp2/sp/no hybridization is not more correct. It is still inaccurate. But we still teach our students that our carbon in ethane is sp3-hybridized.

[VOCl4]2-  has dsp3

We can split this into two cases: main group elements and transition elements.
Main group elements do not use their d-orbitals for bonds and thus cannot be hybridised in any d-containing way. Bromine in $\ce{BrF5}$, to cite your example, is unhybridised. The $\mathrm{p}_z$ orbital forms a traditional two-electron-two-centre bond with one of the fluorines. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals each form a four-electron-three-centre bond with two additional fluorines for a total count of five fluorines. The bond order of these four-electron-three-centre bonds is $\frac12$.
Bromine’s 3d orbitals are completely populated and effectively belong to the core orbitals. They do not participate in bonding. Bromine’s 4d orbitals are energetically even further removed than the 5s orbital is — remember the aufbau principle. Thus, neither the lower nor the higher d-orbitals can participate in bonding to any notable extent.
This extends well to all main group elements as hinted above. Practically no main group element compounds will include d-orbitals.

For transition metals the case is slightly different since we are in the midst of the d-block where d-orbitals can play a role. However, invoking hybridisation for transition metals puts a big blanket over the bonding situation declaring things equal which really aren’t. A much better description for transition metals would be to use a molecular orbital scheme as I have done in countless answers of mine. Most importantly, the bonding situation can, in my opinion, not be understood without this depiction.
Additionally, many transition metal geometries provide unintuitive ‘hybridisation’ solutions — octahedral being the notable exception which actually hybridises six orbitals (three d-orbitals are nonbonding). For example, a tetrahedral metal complex should be understood as using all orbitals for bonding. Likewise, the square-planar ligand arrangement only has one nonbonding d-orbital.
My suggestion is never to use hybridisation approaches for transition metal complexes.

Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. That would be the transition metal case above. And placing the d behind s and p would use d-orbitals from the same shell. That would be the main group case. Thus, the difference between the two (tl;dr) is:
The first is very strongly discouraged, the second is wrong.

@Jan While your answer is technically correct and is the method that would most likely be used in an upper division/graduate inorganic or physical chemistry course, I think the OP's question was from the perspective of general chemistry where things are kept a bit more simplified. For example, in the general chemistry textbook I use by Petrucci, et.al., they discuss the hybridization model for both main group and transition metal chemistry. They also point out that experimental data indicates difficulties with $\ce{sp^3d^2}$ hybridization for $\ce{SF6}$ and suggest an alternative which involves 4 covalent bonds to the sulfur and 2 ionic bonds, i.e. $\ce{SF4^2+(F^-)2}$. No general chemistry textbook that I have seen (and I have seen quite a few) discusses 4-electron-3-center bonds because it becomes too complicated at the general chemistry level and not necessary for those students who are not chemistry majors. MO theory is discussed at the general chemistry level but only for diatomic molecules and ions from the second period, again because more atoms or higher periods makes it more complicated and it's just meant to give them a taste of what it is.

While this is possibly true, it is also supposed to be a comment and not an answer.

But the OP's question remains unanswered it would be really helpful if you explain the difference between the $SP^{3}D$ and $DSP^{3}$ hybridization. Because no reference websites seem to touch that point.


text:
How do I call the functional group of this alkane?
Given the following chain of carbon atoms (imagine they are all saturated with the correct number of hydrogen atoms), how do I call it? 4-(dimethyl-ethyl)-octane? I'm having trouble naming the short carbon chain extending from the 4th carbon atom of the octane. Can you help me out please?

Under the 2004 provisional recommendations by IUPAC, one possible systematic name is 4-(2-methylpropan-2-yl)octane, which indicates that the longest chain in the substituent is three carbons long, is connected to the main carbon chain via the second carbon (-2-yl), and has a methyl group on the second carbon (2-methyl).
4-(1,1-dimethylethyl)octane is also possible, and indicates that there is an ethyl substituent on the fourth carbon in octane, which is further substituted with two methyl groups on the first carbon where it branches off (see section P-29.4).
However, the preferred IUPAC name is 4-tert-butyloctane; the tert-butyl group is used to describe a $\ce{-C(CH3)3}$ group and is preferred over either of the two above systematic names (see section P-56.2.2.2).
I don't have access to the 2013 recommendations but I believe it should not be different.

In the 2013 edition, the structure and wording of several rules have changed, which makes them difficult to compare to the 2004 draft. Anyway, the resulting names and the preference for “4-tert-butyloctane” has not changed. However, note that the current IUPAC recommendations do not use parentheses in “4-tert-butyloctane”.

@Loong Okay, edited.

4-tert-butyloctane is allowed according to IUPAC

Could you elaborate a bit? This isn't really helpful to the OP nor to anyone else visiting this post. Is 4-t-butyl octane the preferred IUPAC name? How did you name the compound like that? And so on, are questions you can answer in order to write a decent post.

2,2-dimethyl-3-propylheptane
Choose the 7 carbon chain instead. Not necessary to be the longest chain in this case

Yes it is. Choosing the longest all-carbon chain is always necessary for saturated alkanes.


text:
Isopropyl alcohol to clean germs on countertops
My wife recently found some website recommending mixing 91% isopropyl alcohol, white vinegar, water, and corn starch and using it to wipe down countertops. This would be where we put our raw chicken and pork after washing them.
I'm concerned that this won't kill the germs as well as harsh chemicals would, although I am as eager as she is to replace those hard chemicals. I need to know, from a chemist, whether isopropyl alcohol really will kill all those E. coli and Salmonella and whatever else the raw poultry is leaving behind.
As a bonus, can anyone recommend any better solutions that we can make on our own than the one mentioned above?

Every biology lab I have been in so far that dealt with live bacteria used 70% ethanol solutions for wiping down. Not exactly the definition of a ‘harsh’ chemical ;)

see also http://www.cdc.gov/hicpac/Disinfection_Sterilization/6_0disinfection.html#a1 and http://cmr.asm.org/content/12/1/147.full

The corn starch is probably intended to be a thickener and may supply your most active bacterial culture medium, I would leave that out.  High purity alcohol on its own should be pretty good, water and vinegar will dilute it though the vinegar action may assist with cleaning but may not make much difference to bacteria counts.

Don't place raw meat on the countertop. Use a bowl or other container instead. Also, this is in my opinion really a question for biology stack exchange.

As @BuckThorn states, don't spread everything over a counter. Use a cutting board, that can be more easily disinfected. https://www.eatright.org/homefoodsafety/four-steps/separate/cutting-board-safety

A 70% “rubbing” alcohol solution will kill those pesky bacteria usually in ten seconds or less. It will also kill a host of viruses, but not all of them.
For your common household disinfecting, a 70% solution will do the trick. Especially for E. coli, salmonella, staph, and a few others that would be risk bacterium when preparing poultry, etc.
A final bit of information: 70% will be a more effective carrier than the 90-99% equivalent. Why? Because the water mix (30%) will act as a carrier or transport mechanism to get the alcohol into the bacterium cells in order to terminate them! The 90% mix simply is not very effective at penetrating the cell membranes.
See this for additional information:  https://www.cdc.gov/infectioncontrol/guidelines/disinfection/disinfection-methods/chemical.html