text:
Asymmetry in trigonal bipyramidal geometry
I teach an MCAT course in chemistry. I like to explain VSEPR by saying.. first, imagine arranging electron pairs around the central atom so they are maximally distant from each other, and uniformly arranged. If there are two pairs, they'll just be across from each other. Three? You get a triangle, and so on.
This rhetorical device fails with five pairs. Five is strange because the electron pairs are most definitely not equidistant from each other, with the equatorial pairs being 120 degrees from each other, but 90 degrees from the axial electron pairs.
My question: is this a simplifying geometry for high school / college classes, and the geometry is actually more equally arranged in 3D space among the pairs? Alternatively, if 'trigonal bipyramidal' is actually the geometry, why? What induces the asymmetry? And does that imply that molecules with the same formula and connectivity, but with an atom equatorially versus axially located correspond to distinct stereoisomers?

Welcome to chemistry.se! Let me just quickly point out the obvious: VSEPR-theory is based on solely based on an observation, that many molecules behave like this. There is absolutely no theoretical fundament behind it whatsoever. In this sense it is a oversimplification for chemistry beginners. But since your question is quite broad I am currently not able to provide a complete answer.

As a simple example, take PF5.
The axial and equatorial positions are different, but exchange rapidly through Berry pseudorotation. So in short, there is asymmetry (stemming from orbital theory), and the molecule does not like it.

Yes, 5-coordinate compounds are always a bit confusing.
The best way I've answered this question about VSEPR for students is to do a little demo with balloons. Blow up balloons (i.e., electron pairs) and tie them around a central knot. It will definitely give a trigonal bipyramidal shape - it has nothing to do with chemistry, and everything to do with "fitting 5 things around a central point."
I disagree with @Martin that there's no theoretical basis to the shapes. The theory is purely geometric.
There are multiple possible shapes, including square pyramidal, but these are higher in energy because you have more ~90-degree angles. As discussed in the comments, if all 5 atoms are in a plane, the angles are 72 degrees, which is clearly bad. You could in principal have the top of an octahedron, in which case all angles are 90 degrees, but that's also clearly worse. (In practice, square pyramidal structures have the central atom slightly above the plane, increasing some angles slightly beyond 90.)
So I do the demo with the balloons, talk about square pyramidal and how this alternative is clearly slightly higher in energy.
The Berry pseudorotation, as mentioned above, does convert between the shapes, and it's known that at room temperature the axial and equatorial positions scramble quickly.
Update:
I decided to do some quick calculations using Avogadro and MOPAC on $\ce{PF5}$ in square planar versus trigonal bipyramidal using the PM7 semiempirical method. This took longer to write than to run the calculations (seconds).
Here's the optimized square pyramidal geometry:

The phosphorous isn't in the plane, it's a bit above, to increase four of the bond angles and minimize electron-pair repulsion. Note that the largest angle is ~103 degrees, but the F-P-F angles in the basal plane are all ~87 degrees.
Here's the trigonal bipyramidal geometry:

Now the angles are exactly what we expect from VSEPR. We see 90 degree angles between the equatorial plane and the axial F atoms. And the F-P-F in the plane are all 120 degrees.
Just from the angles we can guess that trigonal bipyramidal should be lower in energy.
For reference, the difference in energy between these two geometries using PM7 is ~2.76 kcal/mol. So there's not a huge difference in energy, but it's there.

Hmm.. "There is simply no way geometrically to create 5 completely equal angles." A planar pentagon satisfies this!

On the contrary, if the five electron pairs are arranged in a planar pentagon around the central atom, the angles between the non-adjacent pairs would be 144 degrees while the angle between adjacent pairs would be 72.
The initial statement was poorly worded, but a planar pentagon doesn't solve the problem.

@JulieThomas Ahhhh right, good point. Here's my thought experiment, and why I have trouble with the inability to get 5 equal angles. Suppose you arrange the 5 in 3D randomly. There are clearly some arrangements which are less evenly spaced than others. But if some are less, and others more evenly spaced, then there must exist a maximally distant arrangement (or set of such arrangements). So, is 'trigonal bipyramidal' that arrangement? Can it be shown mathematically? It seems if that's the case, there's something more than just spatial arrangement to it.

@vector07 The mathematical proof is, in fact, Geoff's example. The the most stable (i.e., lowest energy) configuration is the trigonal bipyramidal with the square pyramidal just a wee bit less stable (higher in energy). Those calculations are the result of minimizing the repulsion among all the electrons arranged around the central atom. It is, in effect, a spatial arrangement, because the repulsion between electrons makes it conceptually appear as though they take up space. That's why the balloon analogy works as a visual aid.

@vector07 Yes, trigonal bipyramidal is exactly the structure that provides a maximally distant arrangement of 5 things around a central point. The reason it's not intuitive is because we don't frequently encounter 5 things around a central point except in chemistry (or the balloon demo).

@GeoffHutchison An optimized PF5 takes on a square planar geometry??  What sort of magical procedure does that?

@GeoffHutchison Also I disagree with the theory being "purely geometric".  The theory really just stems from electrostatics.  The 'geometrics' are a result of that.

@LordStryker While VSEPR may rely on electrostatics, as discussed above it can be purely geometric. There's no electrostatics involved in the balloon demo.

@LordStryker The square planar can be a local minima depending on the method. With some methods, it's a first-order transition state saddle point. With MOPAC it's usually a very shallow local minima.

@GeoffHutchison Not to be semantical but balloons are made up of atoms.  Atoms have eletrons. Electrons repel each other.  Balloons are not ghosts and therefore they don't 'move into' each other.  Its electrostatics.

Let us continue this discussion in chat.

When I studied this in school this year, I too found it confusing as trigonal bipyramidal geometry is asymmetric.
I did some research and found out that in n dimensions maximum of n+1 vectors can be equidistant from each other(like trigonal planar and tetrahedral).(see https://math.stackexchange.com/questions/714711/how-to-find-n1-equidistant-vectors-on-an-n-sphere )
So for 5 vectors, it would require 4 dimensions, which would make a 5 cell( https://en.wikipedia.org/wiki/5-cell#Projections ). Its projection on 3 dimensions gives us a trigonal bipyramidal. I think that it can be proved mathematically too using the fact that it is the projection of 5 cell on 3 dimensions.


text:
Formation of halohydrin vs alkyl dihalide
For the electrophilic addition of $\ce{X2}$ to alkene in $\ce{CCl4}$ solvent, a dihaloalkane is formed. This is a halogen addition reaction, with the intermediate being a halonium ion.

However, if instead of $\ce{CCl4}$, $\ce{H2O}$ is used instead, a halohydrin would be formed.

However, in the second case (halohydrin formation reaction), the $\ce{X-}$ ion is still present in the reaction. Furthermore, $\ce{X-}$ is a stronger nucleophile than $\ce{H2O}$. Why is it then that it is $\ce{H2O}$ that reacts with the halonium rather than $\ce{X-}$, the stronger nucleophile?

The halohydrin is usually formed by means of a solvolysis reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by "encasing" it in a large hydration shell.

Is this reaction regiospecific or regioselective? (i.e. Will the product be 100% halohydrin or perhaps 99% halohydrin and 1% dibromoalkane)? Does the hydration shell exclude the X- completely from reaction? Thanks!

@t.c Unfortunately, I lack practical experience with this reaction so I can't really say how good the yield will be or whether the dihaloalkane is a noticable by-product. But in theory I'd say that yes, there would be some small quantity of dihaloalkane formed but that might even be less than 1% because water is present in an overwhelming excess. You will have a water concentration of about 55.55 mol/L and if you have, say, a bromine concentration of 0.1 mol/L then the difference is already a factor of about 500. But from the bromine an even much smaller amount of bromide is formed.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Chemdoodle to mestranova
Is it possible to copy my chemdoodle displayed structure into mestranovea? I have tried simple dragging (mac) and all the various copy/paste options... any ideas?

ChemDoodle contains many options for communicating chemical data to other applications. The most common way is to use the Edit>Copy as... function and choose a chemical type to be put on the system clipboard as a string.
If you use the copy function in ChemDoodle, a wide range of common data types, from chemical to image, are placed on the system clipboard for a 3rd party application to paste. You can control what is placed on the clipboard in ChemDoodle's advanced preferences.
ChemDoodle also supports round trip editing with applications that will allow it. However, if none of this works with the application you are pasting into, you will have to contact that specific vendor.
More information on working with ChemDoodle data in 3rd party applications can be found in Chapter 11 of the ChemDoodle User Guide: http://www.chemdoodle.com/downloads/UserGuide.pdf
For further help, please open a ticket here: https://support.ichemlabs.com

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Heat Absorption of Liquids
Does same amount of heat absorbed cause the same changes of temperature in the same amount of different liquids? Explain.

Welcome to Chemistry.SE! For formatting help visit the [help] center and for more information about this page, take the [tour]. Your question is phrased a lot like a homework question. Homework questions should show the work or thinking that you have already done in an initial attempt to answer the question. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?

No.
To answer your question we will assume certain things here.

Assume  $C_p = C_v$ for most liquids, where $C_p$ is the heat capacity at constant pressure, and $C_v$ is the heat capacity at constant volume.
Assume that $C$ is constant with respect to temperature, which is not the case in most real fluids. Here, in order to leave out the integration, we assume that it is a constant.

Then,
$Q = C\Delta{}T$
As different liquids have different heat capacities, the same amount of heat supplied to the liquid would result in a different $\Delta{}T$, the temperature change.
The higher the heat capacity, the more heat is required to raise a unit mass(or mole) of substance  by a unit temperature. Likewise, more heat has to be removed from the unit mass of substance in order to decrease the temperature by a unit temperature.
In other words, the higher the heat capacity, the more "reluctant" it is to change its temperature, for the same amount of heat supplied.

As examples, water has a heat capacity of 75 J/mol (to raise by 1 degree), while methanol is


text:
How does sugammadex (a modified cyclodextrin) draw rocuronium into its lipophilic cavity?
Sugammadex is a modified cyclodextrin: 8 member sugar ring with carboxyl thioether extensions. Rocuronium is an aminosteroid, 4 steroid rings and one protonated amine group.
Sugammadex binds rocuronium really tightly. The reason for this is straightforward - lipophilic interactions with the inner cavity of the cyclodextrin, which is the perfect size because of the thioether extensions. The negatively charged carboxylate groups lock into place with the protonated amine.
But the question asks how the rocuronium is initially drawn into the cavity, which makes less sense. I've only found one reference to this in the literature, in a figure caption that says the carboxylate groups interact with the four steroid rings.
But how? The rings are not aromatic, so there is no anion-pi interaction. One would think the positively charged amine would interact with the carboxylate groups far more often, leading to the wrong orientation. If anything, the hydrophobic rings should be repulsed by the negative charge?

Figure 6: Encapsulation process: Sugammadex carboxyl groups interact with steroidal rings A, B, C, and D, drawing the aminosteroidal NMBA molecule into the cavity where additional non-covalent attractions hold the molecule securely in place.

Reference:NCBI

Could you provide the reference that describes that interaction?

Sorry, I should have done that in the first place. Edited with the figure and a link.

After consulting with a professor, the answer seems obvious.
The two negative charges repel each other when the free cyclodextrin molecule is in solution, resulting in an "open" conformation that has the carboxylate groups facing "out". The thioether chains interact with the rocuronium as it is drawn into the large open face of the molecule. As the positively charged nitrogen moves into place, the carboxylate groups "close", and lock the drug into place.

I'm not convinced that the rocuronium molecule can be "drawn" only from that side of the sugammadex molecule (which is the substituted lower rim side, not the cavity side, but never mind, that’s pretty large γ-cyclodextrin) (but also I'm not cyclodextrin supramolecular chemistry expert).
There are some multi-level computer simulations done in China recently.[1] They have done some molecular dynamics, but they don't provide their trajectory data. Anyway, that trajectory is only several Å, from some pre-docked position to the optimized docked position.
There's a nice older study,[2] not mentioned in your reference, working with sugammadex-like compound with bulkier substituents (carboxyphenylthio instead of carboxyethylthio), and besides hypothesizing on closed and open conformations because of π-π interactions of the aromatics, they show (or remind) that rocuronium can have different conformations where the morpholine residue (at right at your graphics, opposite to the quaternary ammonium) can be either in-plane, or perpendicular to the plane, so that it can nicely accommodate, or somewhat lock-in once it enters the cavity (besides other interactions), but they seem to also assume that it's the (only) entering side, which I'm not convinced of.
Encapsulation is rather statistical dynamics process with thermodynamic equilibrium, than a sucking in.
They also have a nice X-ray structure data of the encapsulation complex,[3] but again, that's not sugammadex, exactly. Its X-ray crystal structure is in the Angew. Chem. article[4], the substrate is in the same orientation, but I think it's not because it can enter from one side only with the signle orientation only, but rather because of the electrostatic stabilization.
References:

Li, L.; Zhou, Y.; Wang, Z.; Wu, C.; Li, Z.; Sun, C.; Sun, T. Theoretical Studies on the Mechanism of Sugammadex for the Reversal of Aminosteroid-Induced Neuromuscular Blockade. Journal of Molecular Liquids 2018, 265, 450–456.
Cooper, A.; Nutley, M.; MacLean, E. J.; Cameron, K.; Fielding, L.; Mestres, J.; Palin, R. Mutual Induced Fit in Cyclodextrin–Rocuronium Complexes. Org. Biomol. Chem. 2005, 3 (10), 1863–1871.
Crystallography Open Database – Information card for entry 7150103
Bom, A.; Bradley, M.; Cameron, K.; Clark, J. K.; Egmond, J. van; Feilden, H.; MacLean, E. J.; Muir, A. W.; Palin, R.; Rees, D. C.; et al. A Novel Concept of Reversing Neuromuscular Block: Chemical Encapsulation of Rocuronium Bromide by a Cyclodextrin-Based Synthetic Host. Angewandte Chemie International Edition 2002, 41 (2), 265–270

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


text:
Non-wetting fluids for precise volume measurements
I need to precisely measure the internal volume of an open steel fixture.  I was planning to do this via displacement, but I'm wondering what the best fluid would be.  Ideally it should have no surface tension so that it readily penetrates every facet of the fixture.  It would also be convenient if it was non-wetting which (if I understand that term correctly) means that it would not adhere to the steel when removed.
One suggestion I heard was to use water-soluble cutting fluid.  Are there any other readily-available fluids with these characteristics, or well suited to this application?

Use water, distilled if you are afraid, and don't care. It wont damage steel if the exposure is brief. To reduce effects of surface tension, wash the fixture with acetone before several times, while working in washed gloves. Water wets steel, so the problem is to remove traces of fat from hands.

Not worried about brief effects on the steel, and I can completely degrease everything with acetone.  But as you note that doesn't solve the surface-tension problem with water, right?

@permeakra: Actually, is there any reason to not mix a little soap (surfactant) into the water to eliminate the surface tension and its associated problems?

There is not problem with it. However, since water wets steel, the only problem is to get rid of bubbles trapped inside.

For a very precise volume measurement, you might want to consider using mercury because of its high density. (You would actually measure the weight of the fluid.)

But doesn't mercury have high surface tension?  Or is its weight enough to force it into every crevice, so one only has to worry about the meniscus?

I'm not sure I'd want to hang around a large bath of mercury. Not to mention possible corrosion issues after contact between the steel and mercury.  Go with water, wipe down afterwards with methanol...

@JonCuster Mercury is transported in steel containers.

@permeakra - Good catch - unlikely to be dipping hot steel into a mercury bath.  Now just don't go dipping aluminum into gallium...


text:
Explicit differential equation based model of protein folding
I am an applied mathematician interested in the dynamics of potential systems - i.e., systems with multiple unique energy minima. One of the best examples of such systems are protein folding potentials. From sources like This one I know that the potential,
$$
V(x)
$$ exists and is a function of the positions of each residue in space and their interactions with each other. Since such a function exists, there should (mathematically) be a set of ordinary differential equations that capture the dynamics described by the potential.
However, although I have seen definitions of $V(x)$ in the literature, for the life of me I can't figure out exactly what the set of differential equations associated with this system is - i.e., the system is usually written out in a generalized form that applies to all proteins, but the exact information required to build such a potential for one such protein is unclear.
Does anyone have an example of of a particular $V(x)$ (for a particular, short, set of amino acids, say) where all the parameters are known and the system can be solved numerically as a set of ordinary differential equations?

What you are looking for is a force field: http://en.wikipedia.org/wiki/Force_field_(chemistry)
Standard force fields for proteins include CHARRM and AMBER. These have relatively simple, well-defined expressions for defining the various types of interactions between bonded and non-bonded atoms. Parameters are available for all standard atom types, and are fit to experimental and/or electronic structure calculations. These force fields are considered "classical" in that they do not explicitly treat electronic interactions, but rather try to capture these interactions in a coarse-grained way. Typically, there are expressions for describing the following interactions:

bonded interactions (bond stretching)
angle interactions (bond bending)
torsion interactions (dihedral bending)
12-6 Lennard-Jones interactions (non-bonded, dispersion forces)
Short-range charge-charge interactions (Coulombic interactions between charged or partially charged atoms)
Long-range charge-charge interactions (see Ewald summation method for more details)

Note that in these classical models, partial charges are generally statically assigned to each atom. The functional form of these different terms varies depending on the force field and sometimes implementation. These interactions are typically pairwise additive, so to get the total potential energy of the system you simply add together all the various interactions between the appropriate pairs of atoms.
There are certainly more rigorous models that attempt to capture other electronic structure affects, however for large molecules such as proteins, relatively simple force fields are generally applied.
Please note that finding the global minimum in potential energy of a protein is not a trivial problem. People have been working on this for years, it's extremely difficult to develop algorithms that do not get trapped in local minima.

I have read a bit about the really bumpy potentials associated with protein folding. Are there any molecules for which the energy potential smoother, so that maybe there are fewer than 20 local minima? On average, do you know how many variables these models have? How many minima?

This list also doesn't give me a great idea of which potential to choose - there are dozens! Even if I choose one, how do I get the (apparently many) parameters for it?
Thank you for your help, by the way. It is really appreciated.

Even a relatively small protein will have a huge number of local minima (thousands, millions, billions, depending on the size). You could start with something like a single amino acid or even an alkane chain to begin testing.

Check out the AMBER page for specific functional forms: http://en.wikipedia.org/wiki/AMBER And you don't have to actually select which potential to  use, that's determined by the force field you select. The hard work for you will be the programming effort and looking up the exact parameters to use for different atom types. That's why most people use free packages such as LAMMPS, Gromacs, NAMD, DL_POLY, etc.

Ultimately the potential is a function of the positions (actually the relative positions) of all atoms in the molecule.

Let us continue this discussion in chat.


text:
Determining weight-percent methanol in water from specific gravity and temperature
The company I'm working with uses a fuel mixture of methanol and water. We use a hydrometer to measure specific gravity, in order to test the composition of fuel received from our suppliers. To determine the weight percent of methanol we use a chart provided by The Methanol Institute that correlated specific gravity to weight percent at different temperatures.

Click image for full-size PDF
I'm trying to build a spreadsheet for operators, suppliers, and customers to measure the specific gravity at an arbitrary temperature, enter the temperature and specific gravity measured, and get the weight percent without having to interpret a chart.
In order to do this, I need some sort of formula to calculate the weight percent for any input of specific gravity and temperature  (within a reasonable range anyways).
Here are the approaches I've tried so far:

Fitting a line to the plot above and creating an empirical model. There isn't sufficient resolution in the plot to do this with any useful level of accuracy.
Finding the source information used to create this plot. I called The Methanol Institute and they have no idea where the plot came from, saying it's been used for years and they don't know a source or have any other related data. I tried calling Methanex, a major supplier of methanol which has published similar tables and charts, but I haven't gotten through to anybody so far.
Finding other sources for water/methanol mixes. One of my coworkers found a (printed) table with weight percent methanol in a water/methanol mixture for a wide range of temperatures and covering the weight percent range we're interested in. I scanned and OCR'd the table and created a plot. The table appears to lack sufficient resolution to develop an empirical model as well; with only 3 significant figures the lines appear wavy when plotted:

Calculate using the specific gravity of pure methanol and pure water at a given temperature, then calculate the specific gravity of the mixture using an adjustment for the $\Delta V_{mix}$. This approach seems to work fairly well, but I could only find data for the change in volume due to mixing for 25 °C (page 12 of this PDF from Methanex). This table is "calculated from density and specific volume data" but no additional information is given. A reference table at the end of the document does have the specific volume of methanol at various temperatures, but nothing about change of volume when mixed with water specifically.
I searched at length for methanol/water excess molar volume data. So far I've found data from various sources, but only at 20 and 25 °C, not over a range of temperatures.
Derivation from thermodynamics. I've pulled out my thermo textbook and am reading about equations of state and thermodynamics of mixing. I think given enough work I could probably derive a model from the thermodynamics approach, but I would prefer something based on actual measurements, since theory doesn't always match reality.

The actual question:

Is there an existing resource that would accomplish what I need?
If not, is there a source of good data on water/methanol mixtures at different temperatures that I might be overlooking?
Otherwise, is there a good source of data for excess molar volume over a range of temperatures (at least 18 – 30 °C, preferably wider) that I might be overlooking?
Is there a different approach I should consider?

You are looking for the excess molar volumes of water and methanol.  A quick Google search shows that these quantities have been determined and re-determined dozens of times.  The art is to find which paper has the best values.
The CRC Handbook of Thermophysical and Thermochemical Data and Landolt-Börnstein may have done the sanity check already.

I've found various sources for the excess molar volume (should've mentioned that), but I haven't found sources that report values over a range of temperatures.

The sources are out there: Journal of Solution Chemistry, 2006, 35, 1315-1328 looks good, at least from the abstract.

This comes back to the same problem... all of the data in the paper you referred to were taken at 25 C.

According to the abstract they offer up indices of refraction at different temperatures.  That allows to infer the density, within limits.

Try this equation I found in a company manual...not sure about the source but I tested it against the monographs you can find at say http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanex-TISH-Guide.aspx, p7
and found it to be quite good...
Percent MeOH by wt. in MeOH-Water Solution
= ((32.04 x (F8 x (0.0004745 x ((G8-32) x 0.555)+0.9894)))/(14.03 + F8 x (0.0147 x ((G8-32) x 0.555) - 9.53))) x 100
where F8 = Sp. Gr. of Soln. and G8 = Temp. of Solution in deg F

Welcome to chemistry.SE! Please visit this page, this page and this one to improve and polish the formatting of your answer a little bit.

I'll point out that (G8-32) x 0.555 changes the temp to Centigrade.

Reading values off the plot (which, of course doesn't lend much precision) I was able to come up with the fit $$ww = K0 + K1T +K2\rho  + K3T^2 + K4T\rho ... K9\rho^3 $$with $T$ being the temperature in °C and $\rho$ the density (IOW if specific gravity is measured multiply by 0.998203) and the coefficients
$$K0 =5870.6 ± 70.1$$
$$K1 =-19839 ± 234$$
$$K2 =7.2685 ± 0.23$$
$$K3 =23091 ± 260$$
$$K4 =-17.035 ± 0.507$$
$$K5 =-0.0152 ± 0.000543$$
$$K6 =-9127.5 ± 96.4$$
$$K7 =9.4966 ± 0.28$$
$$K8 =0.017299 ± 0.000596$$
$$K9 =3.498e-06 ± 1.49e-06$$
The 4th order fit with coefficients
$$K0  =-37002 ± 730$$
    $$K1  =1.7281e+05 ± 3.26e+03$$
    $$K2  =-88.78 ± 2.79$$
    $$K3  =-3.0098e+05 ± 5.44e+03$$
    $$K4  =303.88 ± 9.28$$
    $$K5  =0.14152 ± 0.00672$$
    $$K6  =2.3274e+05 ± 4.04e+03$$
    $$K7  =-347.11 ± 10.3$$
    $$K8  =-0.323 ± 0.0149$$
    $$K9  =1.6286e-06 ± 1.57e-05$$
    $$K10 =-67577 ± 1.12e+03$$
    $$K11 =131.8 ± 3.78$$
    $$K12 =0.18456 ± 0.00821$$
    $$K13 =6.6375e-07 ± 1.73e-05$$
    $$K14 =-1.9561e-08 ± 4.39e-08$$
is even a better fit matching the chart to within half a percent or better except in the corners (20%, - 40 °C) where the error is 1 - 1.5% and (90 - 100% at 30 °C) where it peaks at 1%.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Used MathJax. Frankly don't see how this can be improved. It's the way it comes out of Igor. If it's good enough for Igor it's good enough for me.

I don't know who Igor is (if you are referring to a software package it's also irrelevant as WaveMetrics don't define standards, not to mention it's a closed-sourced code which cannot be audited). For proper typography see BIPM/NIST Brochure and IUPAC Green Book if you find links in my first comment insufficient.

I checked out the NIST and IUPAC documents and didn't really find anything about typography there.  NIST wants us to use $t$ for celcius temperature but IUPAC is fine with $T$ reserving $t$ for time. I actually had $t$ originally but changed it to $T$ for clarity. $\rho$ is fine per IUPAC for density. So then I thought maybe it's expressing uncertainty as, for example $−1.9561e−08±4.39e−08$. NIST is fine with this unless the quibble is with using $e^{-8}$ instead of $x 10^{-8}$. If you can enlighten me as to specifically what the problem is that would help.  On a bulletin board who cares?

TL;DR: I'd suggest to use subscripts for numerical indices (e.g. $K_1$), use $\pu{1.23e-4}$ instead of $1.23e-04$ and align set of equations about "=" sign (align environment and &= to denote the nodes to align about).


text:
How to determine the concentration after a dilution with Beer's law?

The molar extinction coefficient of pure antibody is $1.4$ absorbance units per milligram antibody at $\pu{280 nm}$. You have a concentrated antibody solution, so you make a $5$-fold dilution and measure at $\pu{280 nm}$, getting an absorbance of $0.845$. What is the concentration of that solution?

Using Beer's Law this is my work:
$$0.845 = (1.4)\cdot (1)\cdot (c) \\
\therefore c = \pu{0.604 mol L^-1}$$
Then I think we need to factor in the dilution, therefore $c = (5) \cdot (0.604) = \pu{3.02 mol L^-1} $. Is this correct?

Your process looks correct to me, but notice that the extinction coefficient is given in AU/mg, so there must be some additional conversion done to get the units correct.

Beers law is
$$ E= \epsilon_{\lambda}\ \cdot c\ \cdot d$$
Where $c$ is the concentration of the solution. Then, you've found your answer once you've found $c$.
As Jerepierre points out, if you were to account for units in the answer, the question is actually missing some information. The pathlength $d$ is unaccounted for (but we're assuming its $1$), however if it was say $\mathrm{1\ cm}$, then the units for $\epsilon$ would need to be $\mathrm{AU\cdot mg^{-1} \cdot cm^{-1}}$. Which would mean that the units for your concentration would be in $\mathrm{mg}$. But, for concentration we would need to know antibody per some amount of volume. So taking units into account shows that something is missing.
What you're adding at the end is a step you would take to find the original concentration of the undiluted solution. Given that they diluted the concentration 5 times, but we don't know to what factor so we would have trouble solving for that.


text:
Does someone know the name of this equipment?

I think it is used to measure force or something like that. I broke it and have to buy a new one.

Googling for "spring force meter" led me to this page, is it what you are looking for? http://www.carolina.com/spring-scales/force-meter-spring-balances/FAM_954440.pr

This questions seems to belong to Physics S.E.

There are different equipment used to measure force like pivot balance, strain gauge, spring balance, etc.
By seeing the image, we can easily figure out it is spring balance.


text:
Why are so many different organophosphates able to deactivate acetylcholinesterase?
From sarin to tricresyl phosphate, there are organophosphates that are able to deactivate acetylcholinesterase.
What's so special about the simple organophosphate group that makes them able to do it?

Acetylcholine's normal function is to cleave acetylcholine's ester bond to convert it into acetic acid and choline, but certain compounds can interfere. The short answer is that these nerve agents all share similar structural components allowing them to be bound by acetylcholinesterase, where they then either react to alter the enzyme's binding site, preventing activity, or are just very slow to hydrolyze, temporarily inhibiting the enzyme until it can remove the compound.
Not all of these related acetylcholinesterase inhibitors work in exactly the same way, but if you examine their structures and compare them with acetylcholine, the normal substrate of the enzyme, you'll notice some similarities:
Acetylcholine:
Sarin:
Tricresyl phosphate:
You can see that they all have this central double-bonded oxygen (a few have sulfur instead, but are converted to oxygen metabolically), which is important for fitting into the enzyme. The ester bond holding the choline to the carbonyl in acetylcholine is relatively easily hydrolized, but phosphoesters are much slower to hydrolyze. From what I can tell, this seems to be one mechanism behind trecresyl phosphate's toxicity and probably why it's not as acutely toxic as something like sarin or VX. Sarin, on the other hand, has a pretty good fluoride leaving group. This lets it form a strong covalent bond with a serine's sidechain in ACE's binding pocket. This reaction is effectively irreversible, making sarin very dangerous as it totally disables any ACE it reacts with, meaning it takes very little to cause a significant effect.

Acetylcholin’s cleavage proceeds via the attack of a nucleophile (possibly enzyme-activated water or a nucleophilic side-chain of the enzyme) to the carbonyl function resulting in a tetrahedral intermediate with a negative charge on the former carbonyl oxygen.
The phosphate groups in various inhibiting compounds are very similar to this tetrahedric intermediate because:

They have a rather positively charged atom in the middle (phosphorus) like carbon would be in acetylcholin;
they have an oxygen bonded to said central atom with a formal negative charge;
they have at least one further oxygen atom that is able to accept hydrogen bonds;
the overall structure is tetrahedral; and
the bond lengths and thus the shape of the phosphate group are rather similar to those of the tetrahedral intermediate of acetylcholin.

Thus, these phosphates are transition state mimicking competitive inhibitors. Enzymes usually happen to stabilise the transition state well (it is one of the most efficient catalytic mechanisms) so anything that mimicks the transition state will usually bind well in the active pocket. However, since the phosphates’ structure is stable, it cannot react like acetylcholin could; thus the transition state cannot break down and it is hard to liberate the inhibitor. The enzyme is almost permanently blocked.

Someone remind me to add an image soon™.

What 'bout this pic? ;)


text:
How can I compute molar concentration if I am given parts per trillion?
A question in my book asks...
You have a gas X that is found in quantities of $261~\mathrm{PPTV}$ (parts per trillion by volume). Compute the molar concentration at $283.15~\mathrm{K}$ and $101325~\mathrm{Pa}$.
Would you just find the volume using $\frac{n\mathcal{R}T}{P}$ for $1~\mathrm{mol}$ of gas, then multiply this answer by $261/10^{12}$?

@hey It is better to use $\mathrm{...}$ instead of \text{...}$ and also the use of braces is highly recommended to avoid breaking away of the argument. See also here: http://meta.chemistry.stackexchange.com/q/443/4945

@Martin I will take care of it next time.

I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.
First of all the parts-per-notation should be avoided as it is not compliant with SI and highly ambiguous.
Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).
The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$
From here you can probably make your best guesses, determining how many moles of compound do you have:
\begin{align}
pV &=n\mathcal{R}T\\
n  &= \frac{pV}{\mathcal{R}T}\\
n  &= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\
n  &= 1.12\cdot10^{-8}~\mathrm{mol} &\implies c&=  1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}}
\end{align}
If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.
Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.

Thanks! I got the same answer but the question was so vague that I wasn't sure if it's correct.

In truth, the $ppX$ designation is defined on a mass basis, not on a volume basis. Some key assumptions must be applied about the affects of the additive on the solvent to convert the mass basis to a volume basis.

@JeffreyWeimer I really do not understand what you want to say with your comment. Where do you get your definition from? As far as I can see, there is no official definition for this scheme. And could you elaborate more on what you mean with 'mass basis' please.


text:
CMAP layout of data
I would like to use the charmm CMAP correction in another program (PyRosetta). I have looked at parmfile.doc and io.doc and this discussion, but nowhere is it distinctly written how the CMAP is laid out.
For example:
PHI1 PHI2 PHI3 PHI4 PSI1 PSI2 PSI3 PSI4 24
!-180
1 2 3 …

Is this read as:
PHI PSI E
-180 -180 1
-180 -165 2
-180 -150 3

Or as:
PHI PSI E
-180 -180 1
-165 -180 2
-150 -180 3

Also out of curiosity, must the CMAP be defined between two consecutive torsion angles, or can the angles be far apart (topologically)?
PS: This is crossposted from here. And I must say using UBB forums is a pain after getting used to the wonderful SE interface:)
PPS: Would this Q find a better home at https://scicomp.stackexchange.com/?

In the latest version of the CHARMM forcefield in par_all36_prot.prm the comments are a bit more informative:
CMAP
! alanine map
C    NH1  CT1  C    NH1  CT1  C    NH1   24
! phi = -180.0
     0.126790      0.768700      0.971260      1.250970      2.121010
     2.720430      2.089440      1.789790

So from that it can be inferred that my example is read as
PHI PSI E
-180 -180 1
-180 -165 2
-180 -150 3


text:
Does hydrogen bonding only occur between water molecules?
Does hydrogen bonding only occur between water molecules?
Is the dipole-dipole intermolecular interaction know as hydrogen bonding because it is between water molecules, or is called that because of the hydrogen molecules present in the bond?

A quick check of Wikipedia reveals: "A hydrogen bond is the electrostatic attraction between polar molecules that occurs when a hydrogen (H) atom bound to a highly electronegative atom such as nitrogen (N), oxygen (O) or fluorine (F) experiences attraction to some other nearby highly electronegative atom. The name hydrogen bond is something of a misnomer, as it is not a true bond but a particularly strong dipole-dipole attraction, and should not be confused with a covalent bond."  We're okay with elementary questions here, but it's best to take the available information and ask

about a conceptual issue related to what you don't understand about the existing material.

That wikipedia explanation above is somewhat imprecise/outdated/slightly obsolete https://chemistry.stackexchange.com/a/88866/17952

NO, Water is not only the example for hydrogen bonding. It's most common example for basics.
Hydrogen bonds occur in DNA, proteins, polymers, etc. But for introductory level it is easy to explain with water.
DNA: Hydrogen bonding between guanine and cytosine, one of two types of base pairs in DNA.

Polymer: Para-aramid structure

Images: Wikipedia


text:
What is the liquid in Mathmos lava lamps?
How can I identify the liquid in Mathmos lava lamps?
I have two of them with broken caps, I need to get rid of them. Can I flush the stuff down the toilet?
It smells funny.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions, especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

If you still have some of the original packaging it might contain some information on disposal.

The company is not very forthcoming with the ingredients in their FAQ; however, they do state in regards for disposal:

Q. How should I dispose of my lava lamp bottle when it reaches the end of its life?
A. Please make a hole in the metal cap with a sharp point. Then pour
  away the liquid. The bottle may then be placed in the glass
  recycling.

However, personally, I go by a general rule that if I do not know what substance is, I do not pour it down the sink.  The authors of the HowStuffWorks webpage How to Dispose of Lava Lamps state the actual ingredients are a trade secret, but the first patent for lava lamps states the ingredients as being:

"solidified globule of mineral oil, paraffin and a dye as well as paraffin wax or petroleum jell, preferably Ondina 17 with a light paraffin, carbontetrachoride, a dye and the paraffin wax or petroleum jelly."

as for disposal, the author of that website made contact with a company, who advised that

They told me that I could wrap the lava lamp in newspaper and throw it in the trash. I was told that it was non-toxic. They warned me not to pour the liquid down the sink because it contained wax.

The wax could make a mess of the pipes from your sink if it were to accumulate and solidify, and there is more than enough rubbish and pollutants in the water ways as it is.
Best to consult your local area's rules and regulations as to disposal.


text:
Ordinary chemical means
What is the meaning of ordinary chemical means in the following:

An element may be defined as a pure substance that can not be
  decomposed by ordinary chemical means.

Essentially it means that you exclude the possibility of splitting an element even further via nuclear means.


text:
Why does salt dissolved in water taste like salt? If it's just Na+ and Cl- ions
When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$. It stays in this form until the water evaporates and then the ions go back to normal $\ce{NaCl}$. So why does water with salt in it still taste like salt? I am asking because if the molecule $\ce{NaCl}$ is broken down into $\ce{Na+}$ and $\ce{Cl-}$ then how does it have the characteristics of $\ce{NaCl}$?

One interesting thing to note is that NaCl actually forms an ionic crystal, which means that even in solid form, you're still dealing with ions.  The charges balance and stabilize one-another, resulting in the characteristic brittle crystalline structure of table salt.

Answers below. Just one question. Why are you assuming that Na+ and Cl- should taste differently when they're apart than when they're together?

@MrLister Neither Cl- nor intact crystals can pass through the ion channel in the taste receptors as it is a very narrow, positively-charged path. Even if it were possible for NaCl to remain as a crystal up to the threshold of the taste receptor, it wouldn't be able to enter it.

When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different).
There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See here

Yes it should be noted that there are many salts apart from sodium NaCl and many taste salty as well, like NH4Cl (ammonium chloride / salmiac) used in Norwegian salty licorice. Also many receptors designed to respond to specific molecules respond to slightly different molecules, like vanilloid receptors responding to piperine and capsaicine and serotonin (5-HT) receptors responding to various drugs. This is actually one of the working principles of drugs and poisons in general.

Have you ever tasted salt that was not dissolved in water?  Think about this.  Have you ever tasted ANYTHING that was not dissolved in, or in contact with, water?  The answer is "No", unless you have the worst case of chronic dry mouth ever to exist in the history of humanity, and if you do have that terrible case of dry mouth, you no longer taste anything.
This is because everything you taste is sampled by your taste buds after having samples of its molecules transported through pores on your tongue by water.  No water, no taste.  And saliva, of course, is mostly water.  If you place a dry grain of salt on your tongue, saliva immediately dissolves a bit of it, and transports that salt through taste pores, where receptors in your taste buds report its flavor.

I'm not a biology expert, so I'm not sure how the $\ce{Na+}$ and $\ce{Cl-}$ ions actually activate certain taste receptors in your tongue such that you can  taste the $\ce{Na+}$ and $\ce{Cl-}$ ions.
However, for the chemistry part, do remember that even when you are putting solid $\ce{NaCl}$ into your mouth, the saliva in your mouth will dissolve it, resulting in the breaking of the ionic bond between $\ce{Na}$ and $\ce{Cl}$. Sodium chloride is so soluble that even a tiny drop of water could dissolve a significant amount of salt - probably more than  what you would put into your food.
Therefore, it doesn't really matter if you taste solid sodium chloride or sodium chloride solution - they will all taste the same, but perhaps varying in intensity.
Interestingly, if you are able to find a sodium chloride crystal large enough, and ensure that your tongue is dry enough, I believe(again I'm not a biology expert) that you would not be able to taste the sodium chloride, as solid crystals cannot diffuse as easily into your tongue and activate the sodium channels.

When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$.

No, not quite. Dissolving something in water does not break the chemical bonds of the molecule*. You still have $\ce{NaCl}$ molecules floating about. That would leave you with elemental sodium and elemental chlorine, and those two fellows are a bit like drunk bikers.
Mixing sodium, chlorine and water will produce all sorts of things like sodium hypochorite ($\ce{NaClO}$, best known as bleach), sodium hydroxide ($\ce{NaOH}$ - caustic soda) and some lesser-known chlorine compounds. Many are rather dangerous, and I don't think we would have survived as a lifeform if a bit of salt on our food turned into bleach (which destroys DNA) after a Perrier chaser.

exception: some molecules will react with water rather than simply dissolve in in it. Depends on a number of factors.

Since when NaCl forms molecules?

You are actually joking, right?

Not saying that this answer isn't wrong, but the OP also said "molecule". david may just have taken the OP's nomenclature and worked with that.

@MrLister True, but we expect those answering the questions to know more than those asking them.  David could easily have explained the difference between things that come in crystals and things that come in molecules.  I can only assume that he thought it was April Fools' Day.


text:
Why is the standard enthalpy of formation of diamond not zero?
Why doesn’t diamond have $\Delta H_\mathrm{f}^\circ=0$, when graphite does? Is it something to do with the definition – diamonds can’t really form at STP, even though it is naturally occurring?

Gives new meaning to the advertising "diamonds are forever". (No, not thermodynamically..)

Because it was not defined as zero and graphite is of slightly lower enthalpy and is defined as zero. The difference seems to be calculated from heats of combustion. https://atct.anl.gov/Thermochemical%20Data/version%201.118/species/?species_number=951 I wonder if this has been remeasured using pure synthetic diamond and graphene if such exist.

If you set both to zero, you create a contradiction. The only element that is set to zero in two contexts is hydrogen, with dihydrogen and the aqueous hydrogen ion.

You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees:

and

Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In the reference above, both absolute Gibbs free energies, as well as free energies with respect to the stable phase at STP, are given for most elements.

Carbon naturally exists as two allotropes, graphite and diamond. By definition, the most stable allotrope at STP (the one with the lowest heat of formation at STP) is assigned a heat of formation of zero.  Graphite has the lower heat of formation and is assigned a heat of formation of zero, while diamond being slightly less stable has a heat of formation of 2.4kJ/mol.


text:
Representation of chemical formulas as strings?
Is there a standard manner of representing chemical formulas as linear strings?
For example, $H_2O$ can be represented with the string H2O. Isotopes are usually specified with a number before the element. But this must be preceded by a space, or it can be get mixed with the number right before: O4 13CH, which is equivalent to $O_4{^{13}C}H$.
I'm just wondering if there is any set of accepted conventions when representing chemical formulas as linear strings.

If I understand your question correctly, the simplified molecular-input line-entry system (SMILES) and the IUPAC International Chemical Identifier (InChI) are two examples of "human-readable" string notations for chemical formulas your are looking for.


text:
Do reverse reactions occur in combustion?
I am currently reading a textbook on combustion. They assume that all reactions go to completion, such that post combustion only unburned fuel/air and actual products remain.
Assuming that combustion is lean or perfect (no fuel left unburned) and not neglecting dissociation effects, can reverse reaction actually occur to reform fuel at equilibrium? In other words, is the equilibrium constant infinity in real life?

I am also wondering, does the inclusion/exclusion of dissociation effects matter? For instance, if I assume that only CO2 and H2O is formed as products, does this change the answer to the question I have above?

Complete combustion only forms CO$_2$ and H$_2$O. What other products do you mean by dissociation?

Infinite in the sense that you cannot measure any finite backward reaction. Not infinite in theoretical sense, as there is some finit activation energy for the backward reactions. Combustion is a combination of thousands of steps with radical mechanism, which mostly have huge Ks. Also, you should not forget the entropic/information factor: if you burn wood, there will no be reaction producing lignin and complex natural products. Even if energy would allow, the information and selectivity to form lignin is not there. So a backword reaction can lead different products than the initial was!

@JeffreyWeimer Dissociation products I was referring to species are like H2, OH, CO etc. I'm not sure if N or NOx is considered a species of dissociation of combustion products since (I think) it is a primarily result of excess O2 and existing N2 which is technically independent of the fuel-air combustion. I think that dissociation effects would make the reaction even more irreversible?

An equilibrium constant is never $\infty$ in theory. In practice, large values of $K_{eq}$ (and slow reverse reaction rates) lead to the equivalent of complete combustion.


text:
About electricity and liquid products
This may sound weird but could one 'zap' some liquid like milk with electricity of a certain voltage for a 'few' seconds hoping to kill any bacteria in it? I know some people suggested some type of radiation on meat ( I think) but could a small charge of electricity on liquid products kill any bacteria in them?

Yes, this is a known phenomenon. It's a process called electroporation. Here's a paper talking about using it for sterilization. Basically, when you apply an electric field to a cell membrane, you can render it porous. At low field strength, the pores are small and can be closed by the cell, but if you apply a larger potential across the cell membrane, cell lysis is the result, killing the cell. It's actually used for killing cancerous tumour cells in humans.
Unfortunately the trick with doing it with something like milk is that, to get a large enough potential across the cells in the liquid, an enormous potential (up to 1 kV/mm) needs to be applied to electrodes immersed in the liquid, even though the cell membrane is only seeing a volt or two. For this to be manageable, you would probably need a thin flow cell or something. It could be a fair bit more expensive than pasteurization. Even though only a short pulse is used, electrolysis will occur, which could affect the taste. How well it works will also be affected by the electrical properties of the solution, which you can't necessarily play with too much in foods without changing the taste, so it might not be the best technique for foods, but there are certainly other areas where it's useful.

Could electroporation be used on pop drinks ( or would the carbonation be a problem)? If so could this be done instead of their problematic preservatives?

Most soft drinks don't contain preservatives anyways, though some contain sodium benzoate (which has been in and out of the news for the possibility of producing minute quantities of benzene below harmful levels). Sterilizing soft drinks and packaging them in a sterile environment is probably possible, but would add a fair bit of cost. I bet most soft drinks are conductive enough for electroporation, but it's probably not worth the effort. If one really wanted to sterilize them, other techniques may be better suited for the volumes required.

Sorry to disagree but all soft drinks in convenience stores and many in grocery stores contain preservatives. The only ones that don't are Orangina and some European style pop drinks. Would sterilizing POP drinks be LESS expensive than using preservatives?

I really doubt it would be cheaper. Sodium or potassium benzoate is super cheap. I don't know where you are, but in North America, only a few of the major soft drinks contain it, normal Pepsi/Coke/etc. don't, though some of the diet versions do. I'm not sure whether some have higher risk of fungal or bacterial growth because of their composition. Soft drinks don't need to last for years so there's not really much motivation in making them totally sterile.

I live in North America and many POP drinks contain preservatives. PEPSI AND COKE BOTH have sodium or potassium benzoate. As does Ginger Ale , Sprite, Root beer etc,. COKE owns most of the POP brands. Would the phosphoric acid in POP drinks kill bacteria ?

It'll help, but there are certainly bacteria that can grow at pretty low pH.

"could affect the taste" - if this is as much energy as I think, it could curdle the milk outright.

@rumtscho It's not, it's a short pulse that causes dielectric breakdown of the cell membranes, it shouldn't cause significant heating.


text:
Why are electrons transferred in an ionic bond?
I've been taught at school that in an ionic compound like $\ce{NaCl}$, electrons are transferred from the $\ce{Na}$ to the $\ce{Cl}$ to form $\ce{Na+}$ and $\ce{Cl-}$ which both have full outer shells. But what causes these electrons to be transferred?

Related question: http://chemistry.stackexchange.com/questions/16922/why-do-atoms-want-to-have-a-full-outer-shell/16930#16930

The modern definition of electronegativity is due to Linus Pauling. It is the power of an atom in a molecule to attract electrons to itself.
When the electronegativity difference between the two atoms  is greater than 2.0, the bond is considered ionic; i.e. there is complete transfer (between the two atoms) of the electrons in the bond.
The electronegativity (according to Pauling scale) of chlorine is 3.0, while the electronegativity  of sodium is 0.9.

Electronegativity is ability to attract electrons in a covalent framework.

In fact, there isn't a sharp distinction between covalent and ionic bonds. Electronegativity values are useful in determining if a bond is to be classified as nonpolar covalent, polar covalent or ionic. If the electronegativity difference (usually called ΔEN) is less than 0.5, then the bond is nonpolar covalent.  If the ΔEN is between 0.5 and 1.6, the bond is considered polar covalent.  If the ΔEN is greater than 2.0, then the bond is ionic.

So is an ionic bond essentially a covalent bond and are the electrons really transferred (or are they just extremely close to the non metal)?

On the Wikipedia page for ionic bonding: It is important to recognize that clean ionic bonding – in which one atom "steals" an electron from another – cannot exist: All ionic compounds have some degree of covalent bonding, or electron sharing. Thus, the term "ionic bonding" is given when the ionic character is greater than the covalent character—that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bonding to be more polar (ionic) than in covalent bonding where electrons are shared more equally.

I totally agree with you: the term "ionic bonding" is given when the ionic character is greater than the covalent character—that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bonding to be more polar (ionic) than in covalent bonding where electrons are shared more equally.


text:
Interpreting the MOs of a molecule
I've been looking at the MO:s of different molecules in Avogadro, fascinated by their magnificent geometries. But how is one supposed to interpret them, so as to understand their reactivity? What are the first interesting things one should note when looking at the MO:s of different molecules? Interesting patterns to recognize? There is a lot of technical information on MO theory but not much info on how to interpret them intuitively. Any other sources or books I should read?

Welcome to Chemistry.SE. Your question is too broad to be considered a useful question. Could you rephrase your question to narrow the answers?


text:
Are there ligands without lone pairs?
The Nitronium ion, which is a positively charged ligand, in which the donor atom is Nitrogen. The question is, that Nitrogen has a total of 5 electrons, it shares 2 with oxygen and donates 2 electrons to another oxygen and also looses one electron to form a cationic  $\ce{NO2+}$. It does not have a lone pair, not even a single electron. How is it a ligand?
@Dissenter: You explain nitronium for the nitration of benzene, in which a covalent bond is formed. I would like to know this for a coordination bond.
I do not have any example, but I see the nitronium ion as a example in a book of PRADEEP publication which is a TRUSTED publisher.

Are you sure it's not nitro/nitrite, anionic NO2?

Where did you see nitronium described as a ligand in a metal complex? I don't believe nitronium can be, but perhaps you are looking at some unusual situation. The community needs more information to answer the question.

U say about hydrogen proton as a ligand..... can u explain me how it is

thanks for your approach.....search. the regarding information as soon as possible.... pradeep publication is of India

I have edited your post and tried to clarify what you are actually asking. (I might have terribly failed at this.) Could you please provide the reference, where you found this obscurity. A full citation and a screenshot of the relevant section would be much appreciated.

http://www.transtutors.com/chemistry-homework-help/co-ordination-chemistry/ligands.aspx

i can't understand what are u not understanding ....its simple nitronium ion in which there is no lone pair on donor atom that is Nitrogen atom then how it form coordinate bond...how it is a ligand

please help me.

You have not provided any example from a reputable source where nitronium is a ligand in a metal complex. I do not believe it is possible, and until I see a case where it exists, I'm not going to speculate.

ok nitronium the example which i give ignore it ...but hydrogen proton example that u give ... so can u  explain that.....

A ligand is ...

In an inorganic coordination entity, the atoms or groups joined to the central atom.
In biochemistry: if it is possible or convenient to regard part of a polyatomic molecular entity as central, then the atoms, groups or molecules bound to that part are called ligands. Biochemical usage is thus wider, in that the central entity can be polyatomic. Thus H+ may be a ligand for proteins ...

http://goldbook.iupac.org/L03518.html
Therefore, yes, there can be ligands which do not possess lone pairs. One being the hydrogen proton. It doesn't have any electrons at all.
Regarding the nitronium ion - yes - the central nitrogen doesn't have a lone pair - but that doesn't mean it can't participate in coordination chemistry. If the N had a lone pair and could donate it to something, then that N would be a Lewis base. An example of a Lewis base is ammonia. Since the N in nitronium ion has no lone pairs, but it can form complexes with other molecules - the N in nitronium ion is instead a Lewis acid. What happens usually is that some Lewis base attacks the N in nitronium ion and one of the N=O bonding pairs becomes an oxygen lone pair. This free the central N up for bonding. Consider the bottom left part in this graphic:


text:
Is KF the most ionic compound?
I saw somewhere (can't recall where) that KF is the most ionic compound.  I expected CsF.  Does the greater polarizability of Cs allow it to more easily form covalent bonds compared to K?  Does this overcome the fact that K being in n = 4 should bond better with F in n = 2?

use hannay- smith equation and fajan rule togther,you will get your answer.

Yes, this is a subtle thing.
Using the Pauling electronegativities, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be "more ionic."
Unfortunately, an ionic bond requires separating charge, so $\ce{Cs+F-}$. The problem is that $\ce{Cs+}$ is much larger than $\ce{K+}$ and so the dipole moment for a fully ionic $\ce{Cs+F-}$ would be much larger.
We find that the charge isn't fully separated, likely because of the need to stabilize the large dipole moment.
So some people plot a curve of "% ionic character" as determined by the actual dipole moment vs. the expected dipole moment (i.e, full charge separation). We find that this curve reaches a limit ~75-85% ionic character.
IIRC, I think on that basis LiF is more Ionic than KF.

For more, I really like Bruce Robinson's lecture notes:
http://courses.washington.edu/bhrchem/c152/Lec23.pdf via the Internet Archive

The bit about having to separate the dipole moment was very helpful.

I could go to hard-soft acid base theory since $\ce{Cs}$ is a softer cation than $\ce{K+}$ or $\ce{Li+}$ but I think the dipole moment is sufficient to explain the trend.

I always used HSAB theory to rationalize it but wanted more so I came here.

@GeoffHutchison This answer is quite good. May I add something that the OP might find useful:
https://en.wikipedia.org/wiki/Fajans%27_rules
Cheers!


text:
Does everything exist as a gas at p = 0?
Attached is a phase diagram for carbon dioxide. At $p = 0$, it appears that the substance will exist as a gas regardless of temperature.
My questions are:

For any liquid at a fixed temperature, is it possible to vaporize it completely by simply reducing its pressure isothermally?

For any solid at a fixed temperature, is it possible to sublime it completely by simply reducing its pressure isothermally?

What if the solid is at $T \approx0\ \mathrm K$?

In outer space you get a very good vacuum (albeit the temperature is also very low) and we still get asteroids in solid state and space shuttles come down without disintegrating in between.

What a terrible graph, the ordinate is not "not to scale", but logscaled!

Regardless of whether a substance is in a vacuum, it will vaporize if some fraction of its particles have enough energy (due to their temperature) to exceed their binding energy.  Binding energy is the energy contained in the intermolecular bonds holding a substance together.  It is definitely possible for the particles of a substance to have a binding energy that is substantially higher than thermal energy will readily provide.

No.  While most liquids have fairly low binding energies (and thus, relatively high vapor pressures) there are liquids that will not evaporate in a vacuum at an appreciable rate, one example being ionic liquids.  As I wrote above, the attraction between the ions of these liquids is strong enough that they do not measurably evaporate over time.
If you know the Periodic Videos YouTube channel, they've had a couple of videos about these fluids over the past couple of years.  (Here's a link to one, if you're interested.)  One of them featured the behavior of liquids in a vacuum, but I can't seem to find the link; it may have been on a sister channel.
No, consider all of the asteroids, space ships, satellites, and such floating around in space.  For all intents and purposes they will float around until the end of time without evaporating.  Again, the reasoning is as I listed above.  We also don't see the vast majority of solid objects here on Earth slowly evaporating away, which we would expect if sublimation were a concern in a vacuum.  The reasoning is the same as above.
If the solid is very, very cold, it is less likely to experience sublimation since its particles' average energies would be very low.  Even dry ice would not vaporize if its temperature were very close to 0 K.

Here is a phase diagram of both water and carbon dioxide that might help illustrate the situation.  You can see that as the temperature of the substance decreases the vapor pressure of the substance also decreases.  At a sufficiently low temperature the vapor pressure is effectively zero (below the x-axis of this chart.)  There is always the possibility of a single atom or molecule gaining enough energy to vaporize or doing something weird like tunneling off of the surface of the substance, but the process is so slow as to be unmeasurable.  Again, asteroids in the solar system have been hanging around for several billion years without subliming.

Picture from Wikipedia.
Note: Be careful if you look up phase diagrams of common substances.  They often involve only a few points that are actually referenced and the rest is drawn with quite a bit of artistic license.  Compare the phase diagrams of water here and here to see what I mean.  They are often drawn without scales (or with minimal scales) and are sloppy at best.

How does one generally predict whether a substance is a solid/liquid/gas at absolute zero or at zero pressure? Do we need a phase diagram for the substance to know for certain? For CO2 and H2O, per the above phase diagram it looks like both are solid at absolute zero?

@Yandle A phase diagram would be easiest, but a physical chemist might calculate the intermolecular binding energy and compare it to the thermal energy in the system. Based on the diagram I posted both $CO_2$ and $H_2 O$ are solid at 0K. If you look at this diagram, you can see how water behaves at significantly lower pressures still. If you're still curious, I'd encourage you to post a new, clear question. Reference this one or so that people know you've been thinking about it but that this doesn't answer your question.

Solids can sublime more readily at low p that what you've described here.  See:  https://www.mtm-inc.com/ac-20120328-vacuum-compatible-metals-and-alloys.html

@theorist That certainly wasn't my intent.  The article you've linked mentions two elements with significant vapor pressures at temperatures well above room temp.  I guess it comes down to some vagueness in the original question and the intent of the phrase "fixed temperature."  My reading of the question was, "Is it true that at any arbitrary temperature we can sublime any solid away completely, given sufficiently low pressure?"  Rereading this answer, I think my far bigger mistake was failing to point out that vacuum doesn't matter in the slightest.

With a liquid (assuming its partial pressure is less than p_vap), I would say vacuum does matter kinetically, since whether p_ambient is above or below p_vap will have a significant effect on the rate of conversion of liquid to gas, because liquids can boil.  [I don't know whether there are exceptions for liquids with very high intermolecular forces where, at low T, they don't boil even when p_ambient < p_vap.]   Where p_ambient doesn't much matter is w/ solids, which can't boil.  [A minor point: p_ambient does affect p_vap, though the effect is v. small.]  ....

....Having said all that, I was mostly reacting to your assertion that solids don't evaporate, period. I would instead say that everything evaporates, but solids do so extremely slowly; though on cosmic time scales (since you're using space as an example), slowly would not matter.   In particular, all solids evaporate in a near-vacuum, even at low T, not because p_ambient is low, but because, with a sufficiently high vacuum, the partial pressure of the gaseous form of the solid will be below the material's vapor pressure.

@theorist Fair enough.  In my 1, 2, 3 answers I was too cut and dry about the extent to which things do or do not evaporate, though after those answers I wrote something very similar to what you've put here, essentially that there will always be some rate of evaporation, even cases where it is limited to extreme phenomena like tunneling.

This answer explores the case of gallium at its triple point; the vapor pressure is so low that you have to account for the noncontinuous, atomic/molecular nature of the material.  You find that the required volume per atom is abominably large, millions of cubic meters.  The only way to reckon such a result is to realize that with high probability there is not even one atom of gallium vapor in the vicinity of a sample ostensibly at the triple point.  And that's at $300$ Kelvins, versus open space generally being much colder.
Similarly, many solids in space, if they are rocky or metallic, have very low prbabilities of any atoms evaporating in space at any time, with the result that the solid hunk will have a long lifetime before it evaporates even at zero absolute pressure.

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


text:
How does copper reduce dilute nitric acid to nitric oxide and concentrated nitric acid to nitrogen dioxide?
Here are the 2 equations:
\begin{align}
\ce{3Cu + 2NO3- + 8H+ &<=> 3Cu^2+ + 2NO + 4H2O} & E &= 0.62~\mathrm{V}\\
\ce{Cu + 2NO3- + 4H+ &<=> Cu^2+ + 2NO2 + 2H2O}& E &= 0.45~\mathrm{V}\\
\end{align}
The Nernst equation tells us that as we increase the concentration of nitric acid the first equation becomes less spontaneous.
My question is: how? What is the mechanism of the two reactions that decides between the first and second reactions depending upon how much nitric acid is available?
Also, any qualitative explanation for what the Nernst equation shows would be appreciated. Why does having more protons available end up in fewer protons being consumed by the reaction?

It was pointed out that the second equation above is improperly balanced.  Here's the correct equation:
$$\ce{Cu + 4H+ + 2NO3- <=> Cu^2+ + 2NO2 + 2H2O}$$
So, as pointed out in the comment by Neto, the more concentrated acid does consume more protons per reaction.
Does anyone have any insight on the mechanism of the reaction and how it changes with nitric acid concentration?

Note that the second equation is improperly balanced. After correction, you'll see that it actually consumes more protons per mole of copper than the first equation, which is what one would expect for a reaction in more concentrated acid.

Equation balance corrected.

If we examine the Nernst equation: $E = E^\circ-\frac{RT}{zF}\ln Q$, the logarithmic term is what changes with concentration. The reaction quotient $Q$, gets smaller as the ratio of reactants to products increases, meaning that the log term will decrease (and become negative below $Q = 1$), so both equations will become more favourable as the ratio of reactants to products increases.
The correctly balanced second reaction is $$\ce{Cu + 2NO3- + 4H+ <=> Cu^2+ + 2NO2 + 2H2O}.$$
If we look at just the logarithmic term for each reaction, assuming the products part of the reaction quotient is unity and that the proton activity is equal to the nitrate activity:
\begin{gather}
\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{10}}\tag{cyan}\\
\frac{25.7~\mathrm{mV}}{2}\ln\frac{1}{[\ce{NO3-}]^6}=\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{18}}\tag{red}
\end{gather}
y axis: log term (mV), x axis: $[\ce{NO3-}]$

As expected, when multiplied to normalize the number of electrons transferred, the reaction that consumes more reagents becomes more favourable more quickly as the reagent concentration is increased and less favourable more quickly as the concentration is decreased.
Conceptually, the reason for this is that the reactions travel down a slope towards the equilibrium point (see figure) and the reaction that consumes more of the reagents to produce a given amount of the products involved in the equilibrium reaches equilibrium with fewer reactions. Because ∆G is simply the partial derivative of the Gibbs energy with respect to reaction extent, the reaction that reaches equilibrium with less reaction extent has a steeper slope.

In addition to the thermodynamics you can also consider the mechanism.  $\ce{NO2}$ is a mixed acid anhydride, reacting with water as follows:
$\ce{3 NO2 + H2O -> 2 HNO3 + NO}$
Adding water to dilute the nitric acid promotes this reaction, so you don't see the $\ce{NO2}$ as a product as you would with the concentrated acid.  The regenerated nitric acid reacts again with copper, and the $\ce{NO}$ is the nitrogen-bearing product that comes off given a sufficient water addition.

More on the mechanics of the interaction of nitric acid with transition metals, to quote a related source on the electrolysis introduction of solvated electrons into HNO3 in place of via  oxidation of copper:
$\ce{Cu -> Cu(I) + e-}$
$\ce{Cu(I) -> Cu(II) + e-}$

One major synthetic process for hydroxylamine is the hydrogenation of HNO3 on a catalytic surface,[2] where HNO3 is catalytically hydrogenated to nitrous acid, which is then reduced to hydroxylamine and ammonia. The electrochemical reduction of nitric acid, nitrous acid, and their salts was proposed using amalgamated lead electrodes by Tafel in 1902.[3]

In terms of radical chemistry involving the hydrogen atom radical, which is cited as the major reducing specie by Buxton in solution (see Eq (3) on page 2 here) likely involving the following reactions:
$\ce{H+ + NO3- ⇌ HNO3}$
$\ce{H+ + e- ⇌ •H}$
$\ce{NO3- + •H -> •NO2 (g) + OH-}$
$\ce{•NO2 + •NO2 ⇌ N2O4}$
$\ce{N2O4 + H2O ⇌ HNO2 + HNO3}$
The formed nitrous acid can further engage in reactions including the following:
$\ce{3 HNO2 ⇌ HNO3 + 2 NO + H2O}$
To quote a source in the reversal nature of the above reaction:

These observations indicate that the true oxidant is not nitric acid, but that the effective agent is nitrous acid, which is formed in aqueous solutions of nitric acid or nitrogen dioxide according to the following equilibria:
$\ce{HNO3 + 2 NO + H2O ⇌ 3 HNO2}$


text:
Can we compare the electron shells with the orbits of the planets around the sun?
Can we compare the electron shells with the orbits of the planets around the Sun, is this a good comparison?
Is it true for all the atoms that they can have only 2 electrons on the first shell, but why is it so, why 2?

The model of the atom you use should depend upon what you are trying to figure out.  What do you want to know about the atom?

Electron configuration in the atom.

Don't waste your time with the planet thing.  Just learn the filling order.  To understand more learn quantum numbers.

The comparison with planets is problematic. In fact, it's so problematic that it's one of the reasons that the development of quantum mechanics was necessary and this classical picture of the atom was entirely discarded. If the electrons were like planets, they would act like rapidly spinning charges. That would cause them to radiate out a lot of energy. As they lost energy they would crash into the nucleus - and so destroy the atom in a flash of radiation almost instantly. In short, atoms simply could not exist if they acted like planets.
This is why the shells "exist" in the first. The shells are the only place an electron can go. The first shell is the lowest energy an electron can take, which stops it crashing into the nucleus as it would if it were circling it like a planet.
As for the second half of the question, it is true that the first shell contains, at most, two electrons. You really need to learn a bit more about quantum mechanics and quantum numbers to really get to grips with why, though.
Perhaps the shortest answer is that these "shells" you're talking about are not a complete picture of the structure of electrons in an atom. Each shell is divided into smaller parts called orbitals and each orbital can contain two electrons. It happens that this first shell contains only one orbital - so it takes two electrons. The second shell has four orbitals (one that is similar to the orbital in the first shell and three others) - so that adds up to eight electrons in the second shell.
You can see this implied on some of these "shell" diagrams where electrons are grouped in pairs, representing how each orbital has two electrons:

The reasons behind all of that would make for a far longer answer, though. That would range from "because that's what the quantum numbers say" to "because those are the solutions to Laplace's equation say" and more. Which I think is a bit beyond the scope of what you're asking right now, but I think it is more or less thoroughly covered here if you need it.

We can compare the orbitals of electrons to the orbitals of planets, and although this is not a perfect model, it can be useful(such as looking at electron velocities in a similar fashion to planetary velocities, subject to relativistic effects Why is gold golden?).
One use is to find the escape velocity of the electron, treating it like a planet trying to escape orbit. So yes, we can usefully make such a comparison.
Secondly, the answer to your question is multi-layed. In maths, 5+5+5+5+5 is equivalent to 5x5 could be seen as 5 rows of 5, in total 25 objects. Alternatively, it could be represented as 52, and later we know that when multiplying by the same base, you add the indices: 51 x 51 = 51+1 = 52
In a similar fashion, shallowly its given that there can only be 2n2 electrons per shell, where n is the shell number. For the first shell, 11 = 1 so 2n2 = 2x1 = 2
This also gives the second shell to have capacity for 8(2x2x2) and the third as 18(2x3x3).
On a deeper level, in Quantum Mechanics, electrons have 4 quantum numbers(http://en.wikipedia.org/wiki/Quantum_number) which are the coordinates (n, l, ml, ms). Any electron in an atom can be represented as combinations of these 4 quantum numbers. The first, n, is the shell number, just like before. The middle 2 numbers, L and mL are complex, but eventually cause the n2 part of the equation.
The last number, ms is its spin. An electron can either be in spin 'up' or 'down', only 2 states are possible. This is where the 2 comes from in 2xn2, as for every combination of (n, l, ml) there is two possibilities, one where the fourth is 0(meaning spin is 'down') and another when it's 1(where spin is 'up). Due to the Pauli exclusion principle(http://en.wikipedia.org/wiki/Pauli_exclusion_principle), no two electrons can have identical Quantum Numbers. This means atleast one of the 4 must be different. Due to this, once all of the combinations of the first middle two have been filled in a shell, and once an equal number with the opposite spin(but same middle two quantum numbers) have been filled, the first number, n, must change to accommodate more electrons. This establishes the limit for electrons per shell as 2n2.
Keep in mind the 2n2 limit applies only to shells other than the valence shell(outer most shell). The valence shell is ruled by the octet rule that places the limit at 8 for the outermost shell. This is also ruled by the orbitals and the energy to fill the shells, but that requires a deeper understanding.


text:
Proton NMR signals and rings
Does pyridine or furan have the more shielded proton NMR signals?
I know that both are aromatic, so both have rather deshielded signals due to ring current combining constructively with the applied magnetic field.
However, which would have the more shielded signal? I know that shielded protons have more electron density around them. I guess then the question boils down to whether nitrogen or oxygen is more electron-releasing, as both nitrogen and oxygen have at a lone pair participating in resonance.

Here is the proton nmr spectrum for pyridine.

Here is the proton nmr spectrum for furan

For comparison, benzene has a single proton nmr signal at 7.27 ppm.  In both pyridine and furan there are different types of protons, so multiple signals are seen in both cases.
Here are the resonance structures for pyridine

and furan.

Pyridine removes electron density from the alpha and gamma positions by resonance while furan adds electron density to all four ring positions by resonance.
In addition to resonance effects there are also inductive effects and ring currents that will modify the proton chemical shifts.  Inductively, the alpha positions in both molecules will be deshielded.  Ring currents will also deshield the protons as you've noted.
All of the signals are in the general vicinity of benzene.  The alpha-protons in both molecules are the most deshielded due to inductive effects, with furan's alpha signal upfield from pyridine's due to the electron donating resonance effect of the furan oxygen.  The beta proton in furan is the most shielded since it is not inductively deshielded and, again, the oxygen in furan is pushing electron density into the ring via resonance which will cause shielding.

looks like the spectra overlap a bit; so perhaps this question was poorly put. As you pointed out, there are multiple signals for both, some more shielded and some less shielded.

Ron's answer addresses the empircal data, but I feel compelled to comment that your assertion that pyridine has a nitrogen who's "lone pair participating in resonance" is not correct.
In pyridine, nitrogen is sp2 hybridized and forming a pi-bond necessarily puts the lone pair perpendicular to the ring, and so it may NOT donate the lone pair via resonance.  For the sake of completeness, pyrrole (like furan, but with nitrogen), will have a lone pair that is involved in resonance, and hence be an electron rich ring, similar to what was explained by Ron.
The essence of this problem is comparing the geometry of this lone-pair.  The switch between nitrogen and oxygen is a red herring, and questions like this are asked to make sure you are able to form judgments about which factors are more strongly contributing to the observed phenomena.  In this case the resonance dominates the difference in electronegativity.


text:
Why does K⁺ have 0 valence electrons?
If an electron is removed from an atom of $\ce{K}$, why does it then have 0 valence electrons as it states in my General Chemistry textbook? I would think that if this happened, it would have the same exact electron configuration as $\ce{Ar}$, which has 8 valence electrons, correct? What's going on here?

This question hardly makes sense, if one considers finer features of bonding. Potassium doesn't really exists as bare cation in chemical entities. It may coordinate weakly, but still its typical coordination number is six.

What you say is correct.  The  [Ar] configuration we are left with does have 8 valence electrons.  But I think it is just semantics.  Elemental potassium has an [Ar] 4s1 electron configuration.  One would say it has one valence electron.  If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because "1 - 1 = 0".

You are correct in that it has 8 valence electrons, not 0. If we look at the electron configuration of potassium (K), we see that it has one electron: [Ar]4s1. Obviously, removing that electron gives us [Ar] (same configuration as K1+), which is a noble gas and has 8 electrons.

The definition of a valence electron is an electron involved in bonding. As the electrons in the 3rd shell of K are not involved in bonding, it does not make sense to call them valence electrons. K forms metallic bonds with its one 4s electron, and ionic bonds by giving it up to form K+. On the other hand, argon does use the electrons in the 3rd shell to form (covalent/dative) bonds, in the very limited number of compounds it forms. This (3) is the longest list I could find http://www.endmemo.com/chem/common/argon.php

So you're saying that even though it has a full octet, the electrons in the outer shell in this case are not considered to be valence electrons? What about an anion with a full octet?

Yes, in an anion with a full octet, the electrons are involved in bonding, so they are valence electrons. Cl- has a full 8 valence electrons (one of which is perhaps captured from a K atom.) BTW, they're still available to be donated to O atoms to form anything up to ClO4-. On the other hand, a K atom had a valence electron but lost it to become K+. That valence electron (if it could be identified) is now somewhere else, orbiting the nucleus of an anion.

Ar is isoelectronic with Cl-, but due to its lack of charge, it is very difficult to get it to share its valence electrons to form species like ClO4-. K+ is isoelectronic with Cl- and Ar, but due to the positive charge it is literally impossible to get K+ to share or give up its 3s electrons (except in the plasma phase), so they are not "bonding" or "valence" electrons.

Valence electrons are generally regarded as being 'the outermost electrons' for a given atom.  Therefore, with neutral potassium, there is one valence electron.  If we take away the outermost electron, we now have a new set of outermost electrons being the 8 electrons in the 3s and 3p orbitals.  Ron is correct in his assessment of semantics.

I think you need to consider that if you take last electron from the orbital it's there, but it's empty. Orbital itself is a bit abstract thing and if you accept that it can be empty then you can have 0 valence electrons - that's why your book says what it says.
If I'm not right it will be a lesson for me too :)


text:
How can I turn Pine-Sol into an oil?
I'm wanting to make  a Pine-Sol scented candle.  However, in candle making you have to add a fragrance oil.  People make them all the time by adding vanilla, etc and canola oil, vegetable oil, etc.  I'm wondering what proportions I need to do this with Pine-Sol since it has chemicals in it??  I'm using the Original Pine-Sol - it contains 8.7% Pine oil in it if that helps.  That's all I got from the label.
Thanks for your help!
Mandy

With the proviso that nearly all matter you've ever come in contact with is chemicals, I would be hesitant to add Pine-Sol to a volume of hot wax.  If it is sufficiently hot, the wax could spatter, and we have no idea what compounds other than pine oil are in the stuff, so it's probably best to avoid vaporizing and inhaling them either in making or in burning the candle.  If pine oil will do the trick, scent-wise, then that's probably a safer bet.

I can answer the chemistry part of your question... the candle-making part is better suited to a candle making forum.
There is no easy way of converting the other 90% of (older) pine-sol into pine oil and newer pine-sol has 0% pine oil. Best to purchase pure pine oil.
Pine oil is cyclic:
The other pine-sol ingredients are esters and other linear compounds. So using pine-sol to make pine oil would be ridiculous.
I am not certain how much pine oil is needed for a candle to be scented though. It may just be enough. For that you will have to just experiment... or ask on a candle forum.

The answer would be fractional distillation. The idea behind fractional distillation is to evaporate the lighter, low boiling point molecules, while collecting 'fractions' some of which contain the desired aroma chemicals. This requires testing each fraction by smelling to determine if the desired pine-smelling fraction has been distilled. However, this is difficult to perform using standard household materials. Since this requires a closed distillation system, I'd recommended other alternatives to capture that pine scent.
Alternative Ideas:

Using Pure Gum Spirits of Turpentine - This is also known as fractionated essential oil of pine, and is derived from using the process I described above, except with all-natural pine wood (and perhaps leaves, bark, etc.) and is currently available on Amazon for $12 per 4 oz. I'd go with this one, since you would be buying pure pine oil. Not to mention you could probably dilute it back into pine-sol with the proper solvent.
Using Aromachemicals - The major constituents of pine oil include alpha-pinene and beta-pinene (smell of forest air), camphene (woody/herbal), cis-3-hexanol (scent of freshly cut grass), linalool (floral, fresh) as well as others described in my source below. I've listed the ones that are more commonly used in the fragrance industry, and are reasonably easy to obtain.

Source: Pine Oil GC-MS
http://faculty.ksu.edu.sa/badjah/Documents%20Badjah/Pinus%20essential%20oil%20analysis.pdf

Pine sol soap molecules can be desoapified into fatty acids and K+ ions, by adding water and an acid. The acid breaks down tha soap molecules and then adds H+ ions to the fatty acid anions. These anions accept H+ ions because fatty acids, and especially those with long hydrocarbon chains, are very weak acids.
This reverses the process that was done when pine oil was made into soap.
So You now can collect the fatty acids from the top layer.
You can use pine sol, vinegar, water and salt to do this in a container. I have lined out this process in the comments below.

Add equal amounts of vinegar (5%) and pine sol to a container. This should turn the mixture first in to a milky white blend, and then into a grayish blend with white lumps in it.
This happens because the soap in pine sol is partly broken down, and some of the fatty acids have been released from the soap. So the soap molecules now encapsules the fatty acids, and that is what soap is supposed to do.
Add some water if the micture has become to thick.

Now,  add table salt (NaCl), stir and wait for 20 miuntes.
The micture is now clear with the fatty acids from pine sol on top, and water and water soluble ions below. You can now collect the fatty acids from the top layer.

This happens because the salt forces the acetic acid in the vinegar to transfer it's H+ ions to the fatty acid anions. It does this because the fatty acids in the pine sol are even weaker acids than acetic acid.


text:
How does steric hindrance cause sulfur hexafluoride to have almost no reaction chemistry?
Sulfur hexafluoride ($\ce{SF6}$) is, according to the Wikipedia page, a hypervalent substance.  However, according to the page in regards to $\ce{SF6}$ reactions:

There is virtually no reaction chemistry for $\ce{SF6}$. The main contribution to the inertness of $\ce{SF6}$ is the steric hindrance of the sulfur atom, whereas its heavier group 16 counterparts, such as $\ce{SeF6}$ are more reactive than $\ce{SF6}$ as a result of less steric hindrance

The linked steric hindrance provides many examples of how that chemical process works, but specifically, what steric hindrance (or any other) mechanism results in $\ce{SF6}$ possessing virtually no reaction chemistry?

As you can see:

 Sulfur (16) is significantly smaller than Se (34). That size difference
 means that there is more room between the fluorine atoms on Se than on
Sulfur. That is what steric hindrance means. Think of steric hindrance
 as limited access. The smaller the atom the less access there is.
Oxygen is so small that less than 6 fluorine atoms can crowd around it
because of steric hindrance... probably more like four.
So because there is more room between the fluorine atoms on Se there is
more opportunity for other atoms to attack the central atom Se, but
sulfur is completely shielded by inert fluorine.