If a bond dissociation energy is listed, does it refer to homolytic or heterolytic fission? I am perfectly sure than heterolytic fission of a bond needs more energy than homolytic ones, since it requires opposite charge separation. When the bond dissociation energy is listed, which one are they talking about?

Ex: Suppose bond energy of A-B is 500kJ/mol. Is the energy to make A radical + B radical, or A cation + B anion, or A anion + B cation?

This Wikipedia article summarizes the definition of bond-dissociation well:

  Bond-dissociation energy (BDE or D0) is one measure of the strength of a chemical bond. It can be defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at $\pu{0 K}$ (absolute zero). For instance, the bond-dissociation energy for one of the $\ce{C–H}$ bonds in ethane ($\ce{C2H6}$) is defined by the process: $\ce{CH3CH2–H → CH3CH2• + H•}$

Thus the definition of bond-dissociation refers to homolytic fission, and in your example the bond energy of $\ce{A-B}$ is the energy required to make "A radical + B radical".


Egg-Detection in Pastries: An Analysis in Heat-Alterated Molecule Identification I would like to know if anybody here knows of a method to detect the presence of ovalbumin--or any unique, egg-related molecule, in a baked good. Here I am anticipating that the "unique egg-related molecule" might alter under high temperatures, and so the suspect molecule for detection may or may not be any molecule which is present in a fresh, uncooked egg. I'm thinking of something that can be brushed on to a pastry to evolve a color change. Any help is appreciated.

Egg white consists of mostly Ovalbumin. It is a protein, and so will be hard to distinguish from other proteins in a baked product. Egg white changes structure to a solid polymer on cooking. You could dissolve it in a mixture of detergent and hair perm, returning the Ovalbumin to its uncooked state, and then run it through a column to identify it.

There might also be biological assays you could run using enzymes/antibodies which would only interact with Ovalbumin.

I don't think there are any chemical means to identify a particular cooked protein easily.

Ask on Biology exchange?


Ionic compounds dissolve in nonpolar compounds? I think ionic compounds dissolve in just about any compound including nonpolar. My reasoning is this:

Neutral electric charge attracts both positive and negative (This I know from doing things with static electricity and putting it on a neutral surface).

Ionic compounds like strong acids and bases disassociate into individual ions, or if it is not dissolved but melted ionic dimers (like an individual Na+Cl- dimer for example).

Do ionic compounds such as NaCl dissolve in just about every solvent from polar to nonpolar to liquid ionic?

First, I think you have some misconceptions about electrical charge. There isn't neutral charge, it's either positive or negative, if it doesn't have either, it has no charge. Why are you seeing static charge interaction with neutral surfaces? It might be polarization, where a charged particle warps the electron cloud around a neutral atom generating partial charges around the neutral atom. Without knowing more about your experiment I can't say more.

Meanwhile I did an experiment of my own. I took 5 20mL scintillation vials, and added about 0.5g of NaCl to each. I then added 5mL of deionized water, methanol, 200 proof denatured ethanol, isopropanol, or acetone to each vial and tried to dissolve the salt. Only water dissolved all of the salt. I can't say there isn't ANY salt dissolved in the other solvents, but clearly the solubility is solvent dependent, and only water is polar enough to dissolve all the salt.


Determine formula of hydrocarbon from combustion data Source: Oct/Nov 2015 Paper 33 IGCSE Chemistry exam (discounted during marking)
Can someone explain how to find the answer for the last part of the question?
Note: I got the answer pentane with 5 carbon atoms. Is it the correct answer?

Hydrocarbons burn in excess oxygen to form carbon dioxide and water. $\pu{20cm^3}$ of a gaseous hydrocarbon burned in an excess of oxygen ($\pu{200cm^3}$). After cooling, the volume of the residual gas at r.t.p. was $\pu{150{cm}^3}$, $\pu{50{cm}^3}$ of which was x=oxygen.
(i) Determine the volume of the oxygen used. [1]
(ii) Determine the volume of the carbon dioxide formed. [1]
(iii) If the hydrocarbon was an alkane, deduce a formula of the hydrocarbon.

In order to "solve" the problem the following assumptions are necessary:

The initial 20 $\mathrm{cm}^3$ volume of the gaseous hydrocarbon and 200   $\mathrm{cm}^3$ volume of oxygen are at rtp also.
The burning of the alkane (from part iii) also creates water, but at rtp the volume of water is negligible since it condenses into the liquid phase.
All the gases follow the ideal gas law so that volumes are directly     proportional to moles.

(i) Determine the volume of oxygen used.
The starting volume of oxygen was 200 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ of the final 150 $\mathrm{cm}^3$ was oxygen, so $200 - 50 = 150$ $\mathrm{cm}^3$ of oxygen was used.
(ii) Determine the volume of the carbon dioxide formed.
With assumption (2), the final volume was 150 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ was oxygen
so $150 - 50 = 100$ $\mathrm{cm}^3$ of carbon dioxide are formed.
(iii) The hydrocarbon was an alkane. Determine the formula of the hydrocarbon.
Since each carbon atom in an alkane will burn to produce one carbon dioxide molecule, then 100/20 = 5 carbon atoms must be in the alkane. So the alkane might be $\ce{C5H12}$. (Thus the hydrocarbon could be isopentane, neopentane, or n-pentane.)
$$\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}$$

Problems with $\ce{C5H12}$
(A) Given that the volumes were supposed to follow the ideal gas law, 20 $\mathrm{cm}^3$ of hydrocarbon reacts with 150 $\mathrm{cm}^3$ of oxygen. That gives a molar ration of 2 to 15. But the molar ratio of $\ce{C5H12}$ burning is 1 molecule of hydrocarbon to 8 molecules of oxygen, or 2 to 16.
(B) The initial total volume of reactants was 150+20 = 170 $\mathrm{cm}^3$. The final volume of product was 100 $\mathrm{cm}^3$. So the ration of reactants to products, neglecting water, is 17 to 10. From the chemical equation the ration is 9 to 5, or 18 to 10.

So the answer must be a cycloalkane $\ce{C5H10}$ which would react with right ratio (2 to 15) and have the right ratio of products to reactants (17 to 10).
$$\ce{2C5H10 + 15O2 -> 10CO2 + 10H2O}$$
CHECK:
Assuming 22.4 liters for a mole of the hydrocarbon, then each mole of hydrocarbon would produce 10 moles of water.
$$10 \times \dfrac{0.020}{22.4} = 0.0089 \text{ moles of water}$$
Since each mole has a mass of 18 grams and water has a density of $1 ~\mathrm{g}/\mathrm{cm}^3$, then there is:
$$ 0.0089\times 18 = 0.16 \mathrm{cm}^3$$ of water which is indeed negligible.
Also at rtp water will have a vapor pressure of about 18 torr out of 760 torr so the partial pressure of water is pretty small also in the gaseous phase.
PS: Thanks to DGS for pointing out that the alkane must be a cycloalkane.


How to detect harmful gas occurence in carboy bottle To fill with drinkable water, I have lots of empty 19lt carboy bottles. Since they came from different people and companies, I have to check their healthiness to decide wheter they can be sterilise for reuse or not.

I need to detect if a 19lt carboy bottle was used previously for gas, lpg, diesel etc. or is there any other harmful chemical (such as urine) in it.

What could be the cheapest and trustable solution for it? Could I use gas sensors or infrared imaging or something else?

Thanks in advance.

Gas sensors are probably too specific. Infrared imaging is going to have to take account of the carboy material itself. However, sampling the air and performing a gc/ms analysis could work - basically its headspace analysis.


Isolating KI3 from solution Is it possible to isolate potassium triiodide from KI solution? Sources claim that this is possible because potassium is a sufficiently large cation and triiodide is a sufficiently large anion, therefore it is capable of forming a stable solid. This is an explanation based on HSAB, where triiodide is the soft base and potassium cation is the soft acid. Has anyone tried to do so?

Well, @NilayGhosh already commented on stability of $\ce{KI3}$ in aqueous solution, also $\ce{K+}$ is not bulky enough. But as long as OP didn't mention explicitly that it must be water we isolate potassium triiodide from, there is a workaround, if one desperately want to get its crystals (in an altered form though).

From isopropanol one relatively easy obtains red diamond-shaped crystals of (18-Crown-6)-potassium tri-iodide (1) as the result of reaction between potassium isopropylxanthate and iodine:

There are also many other examples in the literature showing how one can man make potassium "bulkier", but I found this particular example the most straightforward.

Of course, there is always a fallback option to crystallize ammonium triiodide $\ce{NH4I3}$ (2) as ammonium is probably the closest cation resembling potassium for this case.

(1) Sievert, M.; Krenzel, V.; Bock, H. Zeitschrift für Kristallographie - Crystalline Materials 2010, 211 (11), 794–797. DOI: 10.1524/zkri.1996.211.11.794
(2) Cheesman, G. H.; Finney, A. J. T. Acta Cryst B, 1970, 26 (7), 904–906. DOI: 10.1107/S0567740870003357


Why enthalpy changes in phase transition when there is no change of temperature? Why there is a change in enthalpy in phase transition?

I'm asking because enthalpy depends only on temperature:

$$\mathrm dH = C_p\,\mathrm dT,$$

and the change in temperature is $\mathrm dT = 0.$

For a single phase, enthalpy is a function of temperature and pressure, and, at constant pressure, $\mathrm dH = C_p\,\mathrm dT.$

However, at a change of phase, enthalpy (per unit mass) undergoes a jump change, even at constant temperature and pressure.  If the phase change is from liquid to vapor, for example, this jump change in enthalpy is called the heat of vaporization.


Reactions of alkynes with sodium/ammonia vs sodium amide What happens when alkynes react with sodium in ammonia ($\ce{Na/NH3}$), or when they react with sodium amide ($\ce{NaNH2}$)?

Are these two sets of conditions different?

Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar!

Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, $\mathbf{Na/NH_3}$ is a way of reducing some organic compounds (including alkynes).

Internal alkynes $\ce{R-C#C-R}$ can be reduced to trans-alkenes using $\ce{Na/NH3}$. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate 4 can interconvert between a cis and trans geometry, and the trans geometry is preferred as this minimises steric repulsion between $\ce{R^1}$ and $\ce{R^2}$. When this anion is protonated, the trans-alkene is formed. This is a nice synthetic procedure for the preparation of trans-alkenes, complementary to Lindlar hydrogenation which produces cis-alkenes from alkynes.

On the other hand, in sodium amide $\ce{NaNH2}$, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion $\ce{NH2-}$ is a very strong base (the $\mathrm pK_\mathrm a$ of ammonia, $\ce{NH3}$, is 38). So $\mathbf{NaNH_2}$ is a way of deprotonating some organic compounds (including terminal alkynes).

The proton attached to the terminal carbon in a terminal alkyne is acidic enough ($\mathrm pK_\mathrm a \sim 25$) to react with sodium amide, forming the corresponding carbanion 7. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.


How is food-grade calcium sulfate made? I'm doing a chemistry project on ionic compounds found on food labels, and I can't find a clear article on how calcium sulfate is made for edible foods. I know it comes from calcium sulfate dihydrate. I have found articles on how they turn that into calcium sulfate for fertilizer, concrete, drywall, and other materials, but not for food products.

Use of calcium sulfate in food and pharmaceutical applications is widespread
and continues to expand. United States Gypsum Company offers two highly refined calcium sulfate products: USG Terra Alba and SNOW WHITE filler. Both fillers are food- and pharmaceutical-grade forms of calcium sulfate, a mineral that appears on the Food and Drug Administration’s GRAS (Generally Recognized as Safe) list of additives approved for nutritional and functional use in food products.

USG Terra Alba ($\ce{CaSO4 \cdot 2H2O}$), the dihydrate form of calcium sulfate, results from fine-grinding and air-separating a select, high-purity white gypsum that contains about $20$ percent water of crystallization.
SNOW WHITE filler ($\ce{CaSO4}$), the anhydrous form of calcium
sulfate, is produced by calcining and milling high-purity white
gypsum$^{[1]}$.

Gypsum is made suitable for commercial use by a process called calcination, which involves heating the mineral to remove approximately three-quarters of its water. Calcined gypsum can recrystallize into any shape with the simple addition of water.$^{[2]}$.

Presumably through repeated calcination, gypsum is made pure enough to meet FDA standards.

$^{[1]}$ USG, Calcium Sulfate Fillers
$^{[2]}$ Encyclopedia.com, USG Corporation


Why is an atom with excess neutrons unstable? In this chart, I can see that stable nuclides (other than hydrogen) have a neutron count greater than or equal to their proton count, and that the neutron:proton ratio for stable nuclides increases with atomic number.
I can also see that in many cases, radioactive nuclides which have excess neutrons undergo beta minus decay, converting one of the excess neutrons to a proton, and moving toward that stable ratio.
I can see why having excess protons might make an atom unstable: they have positive charge and push each other apart. But it seems to me that excess neutrons would make it more stable, adding to the strong force between the nucleons.
Why is an atom with excess neutrons unstable?

Pulling out Theoretical Nuclear Physics by Blatt and Weisskopf (published in 1952, so a bit dated in areas but still a good introduction), one finds (in Chapter VI.1 The Systematics of Stable Nuclei) that the stability of isobaric nuclei (all the same $A$) is determined by:

(1) The symmetry effect: the charge independence and the exchange character of nuclear forces, together with the Pauli exclusion principle, strongly depresses the energy of nuclei with equal or nearly equal numbers of protons and neutrons.

(2) The charge effect: the effect of the Coulomb repulsion of the protons favors nuclei with fewer protons than neutrons. The neutron-proton mass difference also enters the charge effect, but only as a small correction. The charge effect increases in importance with increasing nuclear charge.

(3) In a few special cases a third effect is important: the spin dependence of nuclear forces favors parallel spin over anti-parallel sin of a pair of (extra) nucleons.

So, yes, you picked up on the Coulomb bit (number 2 above), but it is the symmetry effects of the strong force that are more important, particularly for light nuclei. Until one gets to some nuclear physics, the characteristics of the weak and strong forces just aren't familiar to us humans.


What is the Lewis structure of the azide ion? On a chemistry homework, we're asked to draw the Lewis structure of  $\ce{N3-}$.

My answer:

Nitrogen ordinarily has five electrons in its valence; $\ce{N-}$ has six. This makes a total of 16 electrons overall.
If we place the negative charge on the central atom, we get a structure of $\ce{N=N^-=N}$

This has a formal charge of $(-1)$ on each of the side nitrogen atoms and $(2)$ on the central nitrogen atom once you've accounted for the extra electron, so the formal charges cancel out only when you account for the added electron.

This answer is supported by a bunch of online sources (1, 2).

Homework software response:

The correct answer is $\ce{N#N+-N^2-}$

That's all it says. My understanding of why this is valid is that this gives a formal charge of $(0)$ for every atom, as opposed to averaging out to that.

I suppose the second form makes sense, as it's a lower formal charge on the central atom. If so, why do online resources prefer the first? Is there some resonance going on here, and the actual structure has bond orders of $\frac52$ and $\frac32$?

Your proposed structure is wrong. Nitrogen does not exceed the octet in any of its known compounds (and even if $\ce{NF5}$ will be discovered it will not exceed the octet according to everything we know now). However, if you have a formal negative charge that means an additional electron added to the 5 nitrogen usually has; if four of those six electrons are used to build the double bonds, there is still a lone pair on nitrogen for a total of 10 electrons.

If you are having trouble determining Lewis structures, there are four quick calculations that you can perform to help you:

Add up all valence electrons the atoms are bringing into the compound.
Each nitrogen has five electrons plus there is one negative charge (additional electron) so:

$$3\times5+1=16\tag{1}$$
Add up how many valence electrons would be needed so that each atom has an octet (for hydrogen: dublet) of its own.
Each nitrogen would want eight electrons so:

$$3\times8=24\tag{2}$$
Take $(2)-(1)$. This represents the number of electrons the atoms must share, i.e. the number of bonds.

$$24-16=8\tag{3}$$
Take $(1)-(3)$. This represents the number of electrons that do not have to take part in bonds; these must then be distributed as lone pairs.

$$16-8=8\tag{4}$$

Then, start drawing but make sure you have as many lone pairs and bonding electrons as the equations state. Ignoring the lone pairs, we can get the following possible structures for $\ce{N3-}$:

$$\ce{N#N-N}\qquad\qquad\ce{N=N=N}\qquad\qquad\ce{N-N#N}$$

(The exercise of distributing four lone pairs across the three nitrogens so that each ultimately has eight valence electrons is left to the reader because I am too lazy to open ChemDraw to draw the structures.)

After you have done that, you need to take a look at potential formal charges. For that, split each bond homogenously (i.e. give each atom one of the bonding electrons) and count. Compare that count to what an atom should have; the difference corresponds to the atom’s formal charge. (Because electrons are negative, an additional electron corresponds to a charge of $-1$.) When done for those three structures, we arrive at:

$$\ce{N#\overset{+}{N}-\overset{2-}{N}}\qquad\qquad\ce{\overset{-}{N}=\overset{+}{N}=\overset{-}{N}}\qquad\qquad\ce{\overset{2-}{N}-\overset{+}{N}#N}$$

In each of those cases, the formal charges sum up to the overall charge of the molecular ion ($-1$) which is an indication that we have done it correctly. (Again, I have sheepishly left out the lone pairs; you can use my formal charges to determine where they should have been and how many.)

There is no principle of zero formal charges. However, when debating between different structures, a structure with less formal charges is often (not always!) more ‘favourable’. (The actual term should be ‘contributes more to the overall picture’ but that may confuse too much at this stage.)

But which of the three is correct? They all are! In fact, this is what is known as mesomery: we have a number of (resonance) structures that all explain the actual compound a little bit but neither holds the absolute truth. To show this, resonance arrows are usually drawn between the depictions:

$$\ce{N#\overset{+}{N}-\overset{2-}{N} <-> \overset{-}{N}=\overset{+}{N}=\overset{-}{N} <-> \overset{2-}{N}-\overset{+}{N}#N}$$

The key difference between correct structures and your proposition is that the central nitrogen atom can never carry a negative formal charge as it needs to accomodate four bonds to its neighbours which is only possible for $\ce{N+}$.

As for the answer given in the homework response: It is not strictly correct because it is incomplete. All three structures should be marked as correct – until the concept of resonance has been formally introduced at which point only a combination of the three should be.


Confusion about definition of Bond Enthalpy My Chemistry textbook defines bond enthalpy as "amount of energy required to break one mole of bonds of a particular type between two atoms in gaseous state".
Can someone confirm to me whether

  a particular type

is related to bond order, to the atoms involved, both, or something else?

Also, should not the bond enthalpy of all diatomic molecules be large as one mole of bonds to be broken. But this does not happens. Why?

Bond energies are usually given for each bond-type. You can see this table of values, for example (that's a great site). In order to clasify them we could say that particular type means clasify the bond according to the atoms and to bond-order. So you have C=O, C-O, N=C, N-C, and so on.

If you check the site, look that those values are positive, meaning that the energy -as you said- is the necessary to break the bond. If you need to estimate the difference in enthalpy of formation of:

$$\ce{A<=>B}$$

you will need to change sign of bond-enthalpy values (because usually entalphy is of formation).

The energy is calculated per mol, as you say. And it is a very big value, bigger than enthalpy of vibration, rotation or translation.


Is SbF5 symmetrical? SbF5 forms a trigonal bipyramidal shape. There are polar bonds present and I assumed that it is non polar because of it's molecular geometry. However, The molecule itself is not entirely symmetrical (unless I'm thinking of symmetry incorrectly). If I'm not incorrect, can an asymmetrical molecule be non polar?

As mentioned in the comments, $\ce{SbF5}$ in the gas phase has a rather high symmetry, the point group looks like $D_\mathrm{3h}$ to me, containing a principal three-fold rotational symmetry axis (along the axial bonds) and three two-fold axis (along each of the equatorial bonds), among other operations. As mentioned in this text, a dipole moment can only coincide with a rotational axis. Since there are four such axis in the molecule, it cannot have a dipole moment.

Note however, that each of the bonds is certainly polar. They just cancel out, as could be proven by vectorial addition of the bond dipole moment.


Why did life end up with the big 4 elements, and specifically nitrogen? This question is mostly about the human body, but it's also about life in general.

It's not difficult to notice a big different between the abundance of elements on Earth & that in the body. I've been pondering why. Such a huge question, but here I choose to focus on the big 4. Below is my current understanding, please correct me if I'm wrong or add things that I miss.

An umbrella reason for why $\ce{O}$ & $\ce{H}$ are prevalent is because $\ce{H2O}$ is such a wonderful solvent. Life generally depends on reactions to function, and they happen best when immersed in a liquid. Water is good and copious, hence life use it.

Another reason for $\ce{O}$ to be the most common element in our body is that it can oxidize things, which can become handy.
For $\ce{H}$ is that it's basically a proton, thus can be a "filler" in most situations.
$\ce{C}$ is the backbone of life because it's the element with the best bond force. In other words, not a goldilock but the GOLDilock. Life loves to handle such flexible tool, thus it uses carbon.
That leaves the question for $\ce{N}$. Why do we even use nitrogen? It's not anywhere close to the top of abundance in the Earth's crust. It does not have attractive characteristics like $\ce{O, C or H}$. One may say that it's dominant in the air, but just because there's a lot of $\ce{N}$ doesn't mean we evolve to use it. We can always eat dirt to get some hypothetical elementary replacement for $\ce{N}$, for example. Besides, life began in the ocean and there's little $\ce{N}$ there. Do fish have different big 4? I haven't looked into it, but I doubt it.

Both $\ce{H2O}$ and $\ce{C}$ have something "best" in their belt. I guess $\ce{N}$ must have some stuff of highest quality, too, but I have no clue what it is. So, even though my knowledge of the other 3 elements is crude, I'm bugged by nitrogen the most.

Because chemistry built on carbon, hydrogen and oxygen would be very dull and building complex structures from modular components would be impossible

The key point to remember is that life depends on complex chemistry. Living things are, by definition, things that can replicate and things than can build structures that can do complex chemistry. Metabolism is the process of taking simple things from the environment and making them into complex things. This requires complex machinery, built from simpler chemicals available in the environment.

While it might, in principle, be possible to build very complex carbon-oxygen compounds that could metabillise and reproduce themselves, there is no easy way to do so from simple components. But throw nitrogen into the mix and simpler routes to complexity are available.

Take the molecules that make up most of the complex machinery of all living things: proteins. Proteins make up most of the complex machines that do the job of metabolism and they also constitute many of the structural components that make up living things. But they are built from a small number (about 20) simpler chemicals: amino acids. Amino acids are fairly simple chemicals that might be thought of as the equivalent of Lego bricks for building an incredible variety of living structures. Most consist of only carbon, hydrogen, oxygen and nitrogen. All have the core structure $\ce{NH2CH(R)CO2H}$ (where R can be a variety of organic chunks) which constitutes the equivalent of the standardised bricks in Lego allowing all pieces to fit together so more complex structures can be built. In the case of proteins, that core functionality allows a variety of different amino acids to be "strung" together end-to-end (the amino end reacts with the carboxylic acid end on each acid).

The variety of different structures than can be created by changing the sequence of amino acids is staggeringly, unimaginably large. Many proteins contain hundreds of amino acids and you only need about 60-70 for the possible number of different sequences to exceed the number of atoms in the universe. This means that, by exploiting combinations of simple chemicals (which are not that hard to create with simple chemistry), life can suddenly explore an unimaginably wide range of possible complex structures. Perhaps such structures could be built from fewer elements, but not in a modular way that uses a small range of available components and a single way of joining them together.

The availability of this level of complexity from things that (mostly) consist of C,H, N and O is why the use of nitrogen is key to life. The constraint isn't the availability of the elements (though simple nitrogen-containing compounds are fairly available), it is the accessibility of complex structures from simple components.


Orbital overlap and bond strength In Peter Sykes' A Guidebook to Mechanism in Organic Chemistry, it says that sp3 orbitals have greater relative overlap compared to sp bonds.
We know that the $\ce{C-H}$  bond is shorter in ethyne compared to ethane. So the intuitive idea that more overlap would lead to stronger bond formation is obviously incorrect.
So what is the relation between relative overlap of two orbitals and their bond strength? Is there any use that we can get from this overlap?

I suppose the author was talking about the internal overlap which takes place while the formation of sp and sp3 hybrid orbitals.
What I mean is when one s and one p orbital overlap, the extent of overlapping is less than the overlap that happens when one s and three p orbitals overlap.
To understand it more easily, suppose a ball represent a s orbital while a pencil represent p orbital. Think in mind the extent overlap of three pencils and one ball will be more than the extent of overlap of one pencil and one ball.
But the situation changes, when you overlap two sp3 orbitals or two sp orbitals. sp having more s character is relatively stronger then sp3, thus c-c bond is stronger in ethyne than ethane.


How can I deduce the structure of my compound from this NMR spectra? I have no idea how to determine the substituents on the aromatic ring from this NMR spectrum. There is (or might be) a triplet on my NMR around $\pu{7.59 - 7.65 ppm}$ which has 4 hydrogens in the same environment. If so, there must be two ortho substituents on two aromatic rings but I think the 4 Hs won’t be in the same environment.
I assume the substance to be the one I added here:

The hydrogens in the structure you drew definitely should not be in the same environment — so while overlapping or close signals would be okay, the way your signals look isn’t. To get two signals to be identical, there needs to be some symmetrical way of corresponding the hydrogens to each other.

I can confirm a trans-configured double bond (I assume you can, too?) and two disubstituted aromatic rings. The question remains which type of substitution they have (ortho, meta or para). If you expect two rings with the substituents you drew, remember they must be pretty different in their chemical shifts!

For ortho, as you drew, one would indeed expect a set of doublet, triplet, triplet, doublet for each ring. meta would expect one set of doublet, triplet, doublet and a singlet somewhere. para, finally, would expect two doublets.

Remember that triplets must have a $1\,:\,2\,:\,1$ ratio, or they aren’t triplets. With this information in hand, go back into the aromatic region and think what you could have gotten. I have dropped enough clues, I hope.


What is the most accepted mechanism for the reaction of ketones with m-dinitrobenzene? This is a commonly used confirmatory technique in organic functional group analysis. The test is performed by addition of meta dinitrobenzene to the given organic sample followed by addition of a base such as  sodium hydroxide. The indication is given by a violet colouration.

Is the probable mechanism based on carbanion formation? Also it would be helpful if exceptions to this reaction and other variants in colour produced by different ketones. My main requirement is to know why aldehydes do not respond to this test.

In the Janovsky reaction, an aldehyde or ketone which contains at least one $\alpha$ methylen group reacts with 1,3-dinitrobenzene and a strong base (like $\ce{NaOH}$) to form a colored Meisenheimer complex (source). The enolate acts as a nucleophile which attacks the aromatic ring that is electron-deficient due to the presence of multiple electron-withdrawing nitro substituents.

image source

As the reaction mechanism relies on the formation of an enolate, non-enolizable aldehydes like, for example, formaldehyde or benzaldehyde, do not resond to this test. Likewise, non-enolizable ketones (like benzophenone) cannot react in this way. The color of the formed complex varies with the number of nitro substituents of the benzene ring, with 1,3-dinitro compounds giving blue complexes, while trinitro compounds yield a red coloration (source).

This paper contains a list of the colors produced by various aromatic nitro compounds when reacted with acetone and $\ce{NaOH}$. It is reported that most mononitro compounds do not produce any color at all, with the exeption of 3-nitro-4-aminotoluene (orange) and 3-nitro-4-aminotoluene benzoate (orange-red). Some examples for the colors obtained from the reaction of dinitro compounds are: m-dinitrobenzene (purplish blue), 2,4-dinitrotoluene (blue), 2,4-dinitrophenol (yellow-orange), 2,4-dinitroresorcinol (brownish-green) and p-dinitrobenzene (green-yellow).


Is the equation dU= TdS - PdV always valid (even for irreversible processes?) I really don't think it should be.

$dU = dQ + dW$ is the statement of the first law of Thermodynamics.

$dW$ can be put as $-pdV$ and $dQ $can be put as $TdS$ only if the process is reversible. (From the definition of entropy, the $dQ$ in $dS$ = $\frac{dQ}{T}$ is for reversible paths between the inital and final states.)

But when I checked on the net, it says $dU= TdS- pdV$ is always true for any type of process. How? Many say because it contains state variables?

We assumed the process to be reversible while deriving it, otherwise the most general equation should be $dU  ≤ TdS - pdV$, right?

Where ever I see on the net, they used that equation $dU  = TdS - PdV$ as something always valid and as a base to derive fancy new equations with partial differentials and stuff.

For example, in page 2 in this pdf: https://ocw.mit.edu/courses/chemistry/5-60-thermodynamics-kinetics-spring-2008/lecture-notes/5_60_lecture11.pdf

The differential statement of the first law, $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ is valid for all states---but not all processes---for the reasons you have described. When we apply it, we are always assuming that a reversible process connects the two states of interest; irreversible processes are outside the range of applicability of this equation. States and processes get muddled up a lot, because we can always find a reversible process that connects any two states.

The generalization of this equation for all processes is given by $$\mathrm{d}U = \delta q + \delta w,$$ which is just the statement of energy conservation. We cannot, however, write $$\mathrm{d}U \leq T\mathrm{d}S - p\mathrm{d}V,$$ for this would imply that $T$ and $p$ are defined for a system at all points during an irreversible process, which is not true.


Equilibrium constant. Can it be reached? Consider the reaction below:

$$\ce{A + B <=> C}$$

Suppose that the equilibrium constant for this reaction is $K = 10$.

I then prepare a reaction vessel with volume of $\pu{1 dm^{-3}}$ which contains $1$ molecule of $\ce{A}$, $500$ molecules of $\ce{B}$ and $1$ molecule of $\ce{C}$. Thus the initial concentration of each species (in $\pu{mol\cdot dm^{-3}}$) will be $1/N_\mathrm{A}$, $500/N_\mathrm{A}$, and $1/N_\mathrm{A}$ respectively, where $N_\mathrm{A}$ is the Avogadro constant.

As a result the initial reaction quotient is $N_\mathrm{A}/500$.

Now, am I right in saying that this system cannot reach equilibrium, since the only possible values of the reaction quotient are $+\infty$, $0$, or $N_\mathrm{A}/500$?

Does this arise because there simply is not enough of each reactant/product in order to reach equilibrium? I feel like this system must reach equilibrium but I'm struggling to see how the value of $K$ can be reached in this closed system.

In a bimolecular reaction step (such as the one shown above), the rate constant for the forward reaction has units $\pu{M-1\cdot s-1}$. I have learnt that if the concentration of $\ce{B}$ is increased significantly, then the Gibbs free energy of the $\ce{A + B}$ mixture will increase because the activation energy has a concentration dependence. Is this true, and is equilibrium only reached once this Gibbs free energy decreases again?

This is confusing as I am used to the Gibbs free energies of the reactants and products remaining constant while the relative amounts of each change throughout the course of the reaction.

Dave's mathematical answer is perfectly valid, but I feel like another, perhaps more chemical point of view is based on the the law of large numbers, i.e., the "thermodynamic" limit.

"Equilibrium" is a statistical concept that only applies to large numbers of molecules.  Another way to say it is that equilibrium is a thermodynamic concept and thus only applies in the thermodynamic limit.  What is "large" depends on the particular system of interest.  In particular it depends on the equilibrium constant.  For your reaction of interest, $K_{eq} = 10$.  Thus the amount of $A$ and/or $B$ at equilibrium will be smaller than the amount of $C$.  For the equilibrium concept to apply to a system of $A$, $B$, and $C$ molecules, the calculated equilibrium amount of $A$ or $B$ molecules must be very large, say tens of thousands or so for precise work, or perhaps as low as hundreds for the equilibrium concept to be "approximately" correct.

If the equilibrium constant were enormously large, say $K_{eq} = 10^{100}$, then by the same logic, the system would need to contain on the order of $10^{102}$ total molecules for the equilibrium to be meaningful.  (And by the way that is larger than the number of particles in the universe!)

Speaking loosely, if $K_{eq}$ is $10^{100}$ and there aren't close to that many molecules, the reaction is irreversible.

A probabilistic view of equilibrium for small numbers of particles, such as that mentioned in DavePhD's answer, is occasionally useful but not always valid.  For example, consider if a second reaction consumed $A$ irreversibly, converting it to $D$ at some rate.  If you take DavePhD's probabilistic view of the $\ce{A + B <=> C}$ equilibrium, you would assume that the probability distribution of persistence times for $A$ is exactly the same as in the thermodynamic limit.  However, that is not necessarily true with small systems.  Statistical fluctuations in the $\ce{A + B <=> C}$ system will mean that some of the time, $A$ persists far longer than in the thermodynamic case because due to a fluctuation, more than the "expected" amount of the molecules were in the $C$ form.  Some other times, $A$ will resist degradation to $D$ for far shorter than the thermodynamic equilibrium calculation would suggest.  To treat these cases, more complex equations such as the chemical master equation are required.


Electrolysis of Baking Soda separated by semi permeable membrane I have dissolved baking soda in battery water (de-ionized), in a glass beaker and used a food handling glove (vinyl) as a semi-permeable membrane to separate a graphite electrode from a copper electrode. This is for an experiment that will use the copper electrode for the reaction and not the graphite electrode. When I apply a voltage $(\pu{12 V})$ and use a multimeter to measure the voltage, I get roughly $\pu{11 V}$ passing through the vinyl glove, however, no bubbles are produced, but if I remove the glove and have both electrodes in the beaker, then loads of bubbles are produced. So, will my experiment still work if the voltage is passing through the glove and there are no bubbles?

The primary problem is too low permeability = too high resistence of the gloves, causing high voltage drop where it should be minimal one.

The gloves are not hydrophilic enough for electrolyte soaking to get low resistance.

The remaining voltage - after subtracting the glove voltage drop - is not high enough to cause electrolysis by significant current.

You may try to use a salt bridge, as a passthru vessel with porous glass ends, or a gel formed from an electrolyte solution.

You may also want to try a compartment created from a glued filtration paper, or other hydrophilic porous material, impregnated by an electrolyte, preferably sharing one of ions with both free solutions.

For the paper, be aware  of the same leveling on both compartments, otherwise. the paper will do its major purpose filtering. Eventually, more compact paper may be better, like the thick charts for paper painting.


Confusions about oxidation and reduction My teacher told me that elements get oxidised when:

1)There is a loss of electron

2) There is an addition of oxygen

3)There is a loss of hydrogen

And that they get reduced when:

1)There is a gain of electron

2) There is a loss of oxygen

3)There is an addition of hydrogen

Now I don't understand these things:

1) Most elements will lose an electron (and get oxidised) on the addition of oxygen due to its high electronegativity. But what about OF2? In this case shouldn't oxygen get oxidised and fluorine get reduced due to the higher electronegativity of fluorine? (I know its a covalent bond but I am talking in terms of on which side the electron will get attracted)

2) How does loss of hydrogen (or addition of hydrogen) oxidise (or reduce) an element? Like there are elements that have higher electronegativity than hydrogen then how will they get oxidised on the loss of hydrogen? They should get reduced because of the gain of an electron from hydrogen, right?

The formal definition of an oxidation/reduction is linked to the loss/gain of electrons. Considering the addition or loss of hydrogen and oxygen is not a global rule, but rather a trick generally used in organic chemistry. If you consider organic molecules, thus mainly composed of carbon, "adding" oxygen atoms suggests an oxidation of the carbon, as oxygen is more electronegative than carbon, while "adding" hydrogen atoms suggests a reduction, as hydrogen is less electronegative than carbon.

An example would be the successive oxidation of ethanol to ethanal, then acetic acid : $$\ce{CH3CH2OH ->[-2 e^-][``-2\text{ H}"] CH3CHO ->[-2e^-][``+1\text{ O}"] CH3COOH}$$


NMR: Most deshielded protons (electron withdrawal) Of these two molecules, which one would be expected to have the most deshielded signal?

First off, I think the wording of this question is wack because neither of these molecules has only one NMR signal.

In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ...

I chose E, because I was looking at the benzene's signals. This one has an alkyl group, which is a slight inductive donor, thereby shielding the benzene's protons (slightly). This one also has a benzylic carbonyl group, which, according to my teacher, is only a resonance withdrawer and not in any form a resonance donator. Based on this, I chose E as the more de-shielded.

The other group similarly has a C=O resonance withdrawing group, but a resonance donating -OH group. Given that the resonance effect usually trumps the inductive effect, I said that E is still the more deshielded of the two.

I realize I might have now erred in limiting my analysis of shielding to only the hydrogens on the benzene. In E, the methyl's hydrogens are fairly well-shielded still, being connected to a carbon (albeit an sp2 hybridized carbon). On the other hand molecule F has a long linear chain; the hydrogens toward the end of that linear chain will have well-shielded signals! F's linear chain is also longer than E's linear chain ...

Frankly I think this is a horrible question but in any case the answer is apparently "F" ... do you agree?

I don't believe there is anything really wrong with this question. Yes, both molecules have multiple signals, but one of them contains a signal which corresponds to it being the most deshielded. In fact, one of these, molecule F, contains the most deshielded signal AND the most shielded signal.

"In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ..." not necessarily so. The correlation between electron withdrawing groups and shielding/deshielding is a strong rule of thumb, but can become complicated with aromatic systems and competing electronegativities of substituents. In any case, you need to be looking at parts of the molecule; shielding is quite a localised effect, and as I said above, molecule F actually has the most shielded AND deshielded protons.

In any case, if you consider the electronegativity of oxygen; a proton attached to this will be strongly deshielded. Stick them on an aromatic ring para to a carbonyl group, and you will generate a resonance system like below:

In this case, the phenolic proton is by far the most deshielded of all protons in the molecules shown; I'd expect a chemical shift of over 10ppm. In response to some of the comments about -OH signals being over the place; it is important not to confuse raw chemical shift and averaged chemical shift resulting from chemical exchange with other labile protons, such as the (often observed) exchange of phenol protons with water. In these cases, the resulting chemical shift depends on the relative concentrations of the exchanging species, the exchange rate, and the original chemical shifts of the non-exchanging species.


Role of Cu in Corey-House synthesis What does $\ce{Cu}$ do? And why do we need it? Can we not do $\mathrm{S_N}2$ without it? And is there any other metal that can help in this? Maybe something like $\ce{Ag}$? I'm guessing this because the special thing about $\ce{Cu}$ is that it is worse reducing agent than $\ce{H}$.

Copper is used to avoid lithium-halogen exchange. Additional information on that reaction can be found in this discussion. Organolithiums will often, instead of acting as a nucleophile, form a new organolithium: $$\ce{RLi + R'X <=> RX + R'Li}$$ Using $\ce{Cu(I)}$ allows the formation of an organocuprate, which acts as a nucleophile, allowing alkylation rather than metallation: $$\ce{2RLi + CuX -> LiR2Cu + LiX}$$ $$\ce{LiR2Cu + R'X -> R'R + RCu + LiX}$$ I'm not sure if silver could be used instead; I'm not seeing any reference for it, but I may just not be searching the right terms. The Wikipedia page on organosilver chemistry isn't terribly extensive, but I'm guessing stoichiometric use of silver, while not terribly expensive as transition metals go, would certainly be more so than analogous use of copper.

Update:

It seems that analogous chemistry cannot be performed with silver or gold. Lithium diorganoargentates ($\ce{LiR2Ag}$) are apparently too unstable. Lithium diorganoaurate ($\ce{LiR2Au}$) complexes are apparently either not good nucleophiles, or the triorganogold(III) ($\ce{R2AuR'}$) intermediate complexes are too stable to perform similar chemistry. This is per this 2005 paper. A fair amount of gold chemistry has come out since then, so I'm not sure how much of this has changed since.

Second update (though I should have answered this to begin with):

Also, I'd like to add that this is not an $\text{S}_\text{N}2$ reaction. The organocuprate undergoes oxidative addition into the $\ce{R'X}$ bond. After that, it loses the $\ce{X^-}$ ligand to form $\ce{LiX}$ and $\ce{R2CuR'}$, then undergoes reductive elimination to form $\ce{R'R + CuR}$.


How do I calculate the change in energy of an electron transition? What are the $\Delta E$'s of the transitions of an electron from $n=5$ to $n=1$ and from $n=5$ to $n=2$ in a Bohr hydrogen atom? The wavelength of the first electron transition is $\lambda_1=409~\mathrm{nm}$ and the other electron transition $\lambda_2=1091~\mathrm{nm}$.

How can I solve this using those two given wavelengths?

When an electron transitions from a higher energy state to a lower energy state, the energy has to go somewhere. Usually it is emitted as light of a given wavelength which corresponds directly to the change in energy of the electron.

The energy of light at a given frequency $\nu$ is
$$ E = h\nu\; , $$
where $h$ is Planck's constant.
The frequency of light is directly related to the wavelength $\lambda$ and the speed of light $c$ via
$$ c = \nu\lambda\; . $$

Combining these two equations we get
$$ E = \frac{hc}{\lambda}\; , $$
with which you should be able to calculate the $\Delta E$ values for the transitions.


Laplacian of the electron density Is it always the sign of the Laplacian of the electron density a good criterion to decide on the covalent nature of a bond?

Yes, and no, and maybe.

Oftentimes the sign of the Laplacian at the bond critical point in terms of the quantum theory of atoms in molecules (QTAIM) is taken as a measurement for the covalency of a bond. That it is not that simple is quite logical from your question alone.

Just as in bonding in general, i.e. there is a spectrum from pure covalent (ideally, but not achievable) and pure ionic (ditto),* there cannot be a clear yes or no answer here, too. I think the case of $\ce{BCl3}$ versus $\ce{AlCl3}$ and respective dimers, which I have written about, demonstrates this feature and how QTAIM can help understand bonding. As with many things it certainly is hard to draw a line. In that sense it is probably a good criterion to judge the nature of a bond. Judging only based on the sign of the Laplacian, while being a rigorous and unambiguous way to look at things, might lead to the wrong conclusions though.

What does it matter anyway? In many cases such a strict separation is not helpful. It should be clear, that there are covalent and ionic contributions in every bond, and reactivity often does not care about that. You'll never obtain a complete picture, if you are looking at a single property.

* And then there is charge-shift bonding (What is charge shift bonding?), but I personally don't see any value in that.


How does polarimetry account for the orientation of the molecules? If you have some sample in solution (of a pure R substance), wouldn't the individual molecules all have a random orientation in space? Or do the photons emitted into the solution cause all of the molecules to align? Or is it just that, statistically, a given percentage of the particles will have an orientation which corresponds to the polarized light source.

Short answer: The light beam interacts with all of the randomly aligned molecules in its path.  There are roughly $\ce{6 x 10^{23}}$ molecules in a mole of material. If we were to dissolve a micro mole of this compound in 10 ml (1 deciliter is a typical polarimetry sample size) of solvent we would have $\ce{6 x 10^{17}}$ molecules in our sample cell.  Since the molecules are randomly oriented, there are enough molecules so that the polarized light beam will sample all possible orientations of the molecule with respect to the light beam many, many times over.  This allows for measurements to be made in an accurate and statistically reproducible manner.

Detailed answer: A beam of light has associated with it an electric and magnetic field.  In simple polarimetry experiments it is the electric field (or vector) associated with the light beam that is of interest.  Let's say the beam is propagating in the Z direction, then the propagating electric vector lies in an infinite number of planes that contain the Z axis.  When we plane-polarize the electric vector associated with the light beam, we remove all of the planes that the electric vector resides in except for one.

The next important point is to realize that a plane polarized electric vector associated with a beam of light can also be represented by (is equivalent to) two circularly polarized beams (or vectors), rotating in opposite directions (clockwise and counter-clockwise), each one-half the intensity of the plane polarized electric vector.  What we wind up with is two helices, one right-handed and one left-handed, traveling through the sample in our polarimeter.

Let's say our polarimeter sample contains a significant concentration of chiral molecules in solution.  Although the chiral molecules are randomly aligned with respect to the light beam, each chiral molecule will interact with the right-handed or clockwise rotating helical electric vector differently than it will interact with the left-handed or counter-clockwise rotating helical electric vector.  Our sample must contain a large number of molecules so that, despite all of the different molecular alignments between our molecules and the beam, when averaged over the entire sample, the difference in interaction of the right- and left-handed helices with our chiral sample will be repeatable.  It is not hard to have a significant concentration of molecules because a reasonably long sample path length is used.

The unequal (dissymmetric) interaction between our chiral sample and the right- and left-handed helices results in a rotation of the plane of polarization of the emerging light beam from its original position - and this angle of rotation is what the polarimeter measures.


Is the proton the strongest acid? Having looked at the various definitions of acids and bases and having refined my understanding of it after learning about the inadequacies of pKa and the novel use of the Hammett acidity function, I would like to ask if the bare proton is the strongest acid?

I would like to define "acid strength" not as the extent of dissociation but more simply as the ability to protonate other chemical species.

I have come to this conclusion after reading through a post on the explanation behind the strength of fluorantimonic acid, being that the bare proton is liberated and that the conjugate base is so well coordinated, allowing the charge to be spread out over a large structure, stabilising it to a great extent.

There is no doubt about the proton being the strongest acid in the Brønsted-Lowry sense. Similarly, in the Lewis sense, this should also be logical as what could possibly more electrophilic than a bare proton?

No
Brønsted theory
In Brønsted theory $\ce{H+}$ isn't an acid at all. Acids lose protons, becoming conjugate bases, and $\ce{H+}$ is the proton itself.
Arrhenius theory
$\ce{H+}$ isn't an acid, because in this theory acids dissociate in water to form hydrogen ions.
Lewis theory
$\ce{H+}$ is an incredibly strong acid, but nuclei of other, heavier elements, for example alpha particles, are arguably stronger. I haven't found hard data for this and it may be rather difficult to get, these aren't your friendly neighbourhood Lewis acids ;)
Protonating agent
Bare $\ce{H+}$ might be the ultimate protonating agent. In proton transfer, with any Brønsted acid you could always try to find an acceptor weak enough that reaction constant would be lower then 1, proton would "prefer" to stay with acid then protonate base. That's not the case with bare proton, which is unbound. Therefore it may beat any Brønsted acid.
Why only "might"? Because whether bare proton can bind to a species depends on it's energy, which needs to be lower then proton affinity of a molecule to which it's supposed to bind. Otherwise proton may ionise the molecule instead, and even fuse with one of its nuclei, if energy is high enough.
Another thing is that acids like $\ce{H4O^2+}$, which are endothermic molecules, could beat a bare proton, because of their repulsive nature - they spontaneously lose protons and "throw them away" with positive charge of their conjugate base!


U- or RO-method for Singlet-Triplet Gap? When I want to compare the energy between two different spin states of the same molecule, i.e. singlet-triplet-gap, do I better use unrestricted or restricted open formalism to compare the energies between both?

Further on, do I need to calculate the singlet state also with restricted open/unrestricted method? At least within Gaussian I did not recognize any energy difference between all of them. (Which sounds logic for me, as if the ground state is a singlet, also the open shell version will find that best version would be to have the same orbitals in alpha and beta.)

To expand on user1420303's answer a bit:

When I want to compare the energy between two different spin states of the same molecule, i.e. singlet-triplet-gap, do I better use unrestricted or restricted open formalism to compare the energies between both?

It depends.
The unrestricted formalism will almost always give lower absolute electronic energies (i.e., closer to the "real" value) than restricted-open will, due to the greater variational flexibility introduced by the separate spin-up and spin-down orbital sets.  So, from a purely energetic perspective, yes, unrestricted is likely preferred.
If you're interested in other properties, though, the spin contamination introduced by the unrestricted formalism may be problematic.  Of course, in systems with sufficient static correlation (viz., a degenerate or near-degenerate ground state) to result in spin contamination (radicals, bond-breaking states, transition metals, etc.), single-reference methods like Hartree-Fock and DFT may not give reliable results anyways.

Further on, do I need to calculate the singlet state also with restricted open/unrestricted method?

It depends.
For most 'non-exotic' organic systems, there is no need to calculate the singlet state using an unrestricted method since, as you note, the results will often be indistinguishable.  For systems with static correlation, though, if you still want to try to use a single-reference method, you will need to explore the unrestricted orbital space, as there may be a lower-energy solution where the alpha and beta orbital compositions appreciably differ.  The primary method I'm aware of for such exploration is the "broken-symmetry" method.  Two papers I know of that discuss the method, albeit not in extensive detail, are the following reviews by Frank Neese:

F. Neese. "Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling." Coord Chem Rev 253: 526, 2009. doi:10.1016/j.ccr.2008.05.014

F. Neese. "A critical evaluation of DFT, including time-dependent DFT,
applied to bioinorganic chemistry." J Biol Inorg Chem 11: 702, 2006. doi:10.1007/s00775-006-0138-1

The method is integrated into the current version of ORCA (see Section 5.9.10 of the v3.0.3 ORCA manual), and tips for running such calculations are available at the ORCA Input Library.  Other software packages likely include this capability; search their manuals for more information.


The spontaneity of a redox reaction Suppose we have a redox reaction and we want to see in what direction the reaction is spontaneous. We do this by comparing the standard reduction potentials of the sub-reactions available in the main reaction and say if
$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$
 then the reaction is spontaneous in the direction mentioned otherwise in the reverse direction. The same result can be obtained by comparing the standard oxidation potentials. However, as every spontaneity has its root in the second law of thermodynamics I am eager to see how the spontaneity of a redox equation in a given direction can be checked based on thermodynamics.

So basically how can we derive the condition
$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$
 for spontaneity of a reaction from the fundamental thermodynamics laws?

To clarify the question better let me add that I already know that the second law for a common chemical reaction yields into the condition $$dG \le 0\,,$$ my question is about to find the condition
$$E^\circ =E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$ from that inequality. Also let me add further that I even already know that the electric work equals $-n F E^\circ_{\text{cell}}$ but I cannot justify for myself why $\Delta G$ is equal to this work. (Note. the electric work is an internal work not crossing the boundaries of the system, it is not the work of the system, and if it was then at best we had $\Delta G=W_{\text{max}}$ but the inequality $\Delta G\le 0$ no longer hold.)

The cell potential is related to the change in Gibbs free energy by this equation:

$$
\Delta G^\circ= -nFE^{\circ}_\text{cell}
$$

where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.

The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction quotient (via the Nernst Equation), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of Tro's Chemistry has a derivation of all of these relations, but to summarize the first relation:

Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:

$$
w = -qE^{\circ}_\text{cell}
$$

Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:

$$
q = nF
$$

Where $n$ is the number of moles of electrons.

We can substitute this back into the work equation to get:

$$
w = -nFE^{\circ}_\text{cell}
$$

Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction,

$$
\Delta G^\circ = w = -nFE^{\circ}_\text{cell}
$$

In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that "things move downhill" – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.


Determination of structure with HNMR I have been trying to solve this NMR question for about four hours and I feel like I'm always really close but never manage to find the perfect combination.

Anyway here's the info in the question

Formula is $\ce{C11H14O2}$

1.triplet at 2.4 ppm (2H)

2.singlet at 3.3 ppm (3H)

3.triplet at 3.5 ppm (2H)

4.doublet at 7.4 ppm (2H)

5.doublet at 7.5 ppm (2H)

6.singlet at 2.0 ppm integrates for 3H

I have tried a number of structures and the one I feel i got very close to was this one, each number on the side corresponds to the signal that I think it is. The X is just the area where I know i am missing a bond, I just don't know how to seal it up.

Edit: After the advice given by you all, I came up with this structure which seems to be pretty sound

Calculate your DBEs (double bond equivalents):

\[\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5\]
Note that a benzene has 4 DBEs!

Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?
Largest couplings between benzene protons are typically ortho couplings. How large are they in your case?
Two sets of doublets at 7.5 and 7.4 ppm might indicate a 1,4-disubstituted benzene. You might want to play with some increment rules here.
You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.
R in (5) might be $\ce{CH3}$.


sp2 hybridization Group V elements I'm confused about $\mathrm{sp^2}$ hybridization in the formation of $\ce{NO3-}$ ion. Valence electrons of N are $\mathrm{(2s)^2}$ and $\mathrm{(2p)^3}$.

Is one of $\mathrm{2s}$ electrons kicked up to one of the $\mathrm{2p}$ orbitals, and then the remaining $\mathrm{2s}$ electron hybridized with the two remaining $\mathrm{2p}$ orbitals which have unpaired electrons? If so, that gives $\mathrm{sp^2}$ orbitals to bond with $\mathrm{p}$ orbitals of the oxygen atoms.  However, there is now that remaining unhybridized $\mathrm{p}$ orbital of nitrogen, which has 2 electrons, so how can it be involved in $\pi$ bonding?

I would appreciate someone explaining this. I'm sure I'm missing something simple, but my book doesn't have an explanation of $\mathrm{sp^2}$ hybridization for Group V elements.

However, there is now that remaining unhybridized p orbital of
  nitrogen, which has 2 electrons, so how can it be involved in π
  bonding?

Everything is more-or-less "right" up to these point. The thing you're missing is that there is only one electron in the unhybridized $\mathrm{p}$ orbital of $\ce{N}$ atom since to get the following usual configuration of $\ce{NO3-}$ ion (image courtesy of Wikipedia)

with two negatively charged $\ce{O-}$ ions one electron has to be moved from $\ce{N}$ atom to $\ce{O}$ atom first. Now note that since all $\ce{O}$ atom are equivalent to each other we need to invoke the concept of the resonance, but for simplicity think first about any single of the resonance structure above.

So, formally we start by having:

one $\ce{N}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^3}$ electron configuration;
two $\ce{O}$ atoms with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and one $\ce{O-}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

In the first step we transfer one of the $p$-electrons from $\ce{N}$ atom to an $\ce{O}$ atom, so we end up with:

one $\ce{N+}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^2}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

At the next step we invoke the $\mathrm{sp^2}$ hybridization of $\ce{N+}$ ion, so that we get:

one $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion with $\mathrm{(1s)^2 (sp^2)^3 (2p)^1}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

Finally, we form bonds as follows:

One $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and an $\ce{O-}$ ion in forming which we use 1 of the total 3 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Another identical $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ and another $\ce{O-}$ ion; here we use 1 of the remaining 2 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Yet another $\sigma$ bond, but this time between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and the $\ce{O}$ atom in forming which we use the last remaining $\mathrm{sp^2}$-electron of $\ce{N+}$ ion and 1 out of total 2 unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion;
What remains now is one unhybridized $\mathrm{2p}$-electrons of $\ce{N+}$ ion and one unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion which are used to form the $\pi$ bond.

Pictorially (image courtesy of Professor Peter Bird):


What happens when potassium iodide is added to a copper (II) salt? $$\ce{Cu^2+ + KI -> ??}$$

I thought of this reaction:

$$\ce{Cu^2+ + 2KI -> CuI2 + 2K+}$$

But since $\ce{CuI2}$ doesn't exist, the $\ce{I-}$ ions must reduce $\ce{Cu^2+}$ to $\ce{CuI}$ and simultaneously get oxidised to $\ce{I2}$ and thus it will proceed the following reaction:-

$$\ce{2Cu^2+ + 4KI → 2CuI + I2 + 2K+}$$

I gave this reaction to my teacher and he gave me an another version of the reaction:

$$\ce{2Cu^2+ + 5KI -> 2CuI + KI3}$$

He says that he performed the reaction experimentally and obtained a white salt ($\ce{CuI}$) in a brown solution ($\ce{KI3}$).

Now, I started wondering what will happen if I alter the reaction and add sodium/lithium iodide instead of potassium iodide. Will it form sodium/lithium triiodide. Definitely not/ Or maybe, it will form.

  Two questions:

  What is the correct reaction?
  What will happen if we add sodium/lithium iodide instead of potassium iodide?

Related questions:

Why is NaCl3 possible?
What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex?
Why triiodide ion does not form ionic bond with cation having +2 charge?
Why Cu+ is unstable in aqueous medium?

When writing a reaction, you should take care to keep charges and atoms balanced. So the very first reaction you postulated is good (it has charges and atoms balanced) but then it goes South.

If you are only interested in the redox process, then it makes sense to think only about those reactions that form part of the redox process.

$$\begin{align}\ce{Cu^2+ + e- & <=> Cu+}\tag{red}\\
\ce{2I- & <=> I2 + 2 e-}\tag{ox}\\
\ce{2Cu^2+ + 2 I- & <=> 2 Cu+ + I2}\tag{redox}
\end{align}$$

The next step, however, is that the $\ce{Cu+}$ ions will combine with iodide to form insoluble $\ce{CuI}$. Thus, we need excess iodide.

$$\ce{2 Cu^2+ + 4 I- -> 2 CuI v + I2}\tag{precipitation}$$

At any of these points, you could decide to include counterions. For example, your iodide could be $\ce{KI}$ and your copper salt could be $\ce{CuSO4}$. That would give:

$$\ce{2 CuSO4 + 4 KI -> 2 CuI v + I2 + 2 K2SO4}\tag{counterions}$$

Or maybe you decide that only $\ce{KI}$ is important, then of course this becomes:

$$\ce{2 Cu^2+ + 4 KI -> 2 CuI v + I2 + 4 K+}\tag{only KI}$$

Finally, you will typically have an excess of $\ce{KI}$, so that that can further react with the iodine generated to redissolve it. (Iodine is poorly soluble in water, $\ce{KI3}$ is well soluble, hence Lugol’s iodine.)

$$\begin{align}\ce{I2 + KI & -> KI3}\tag{Lugol, balanced}\\
\ce{I2 + I- & -> I3-}\tag{Lugol, ionic}\end{align}$$

Thus, a full equation will look like any of the following:

$$\begin{align}\ce{2 Cu^2+ + 5 I- & -> 2 CuI v + I3-}\tag{full, ionic}\\
\ce{2 Cu^2+ + 5 KI & -> 2 CuI v + KI3 + 4 K+}\tag{full, semi-ionic}\\
\ce{2 CuSO4 + 5 KI & -> 2 CuI v + KI3 + 2 K2SO4}\tag{full, balanced}\end{align}$$

Since the alkaline halides are generally very similar to each other, I think that the reaction should work pretty much identically if potassium is replaced with sodium or lithium. Personally, I didn’t have a chance to perform it, since my inorganic student labs only had pre-mixed potassium iodide solutions.


On the oxyacids of +III pnictogens and +IV chalcogens Phosphorus in its $\mathrm{+III}$ oxidation state is known to exist as the phosphonic acid tautomer rather than phosphorous acid one. All salts isolated containing $\ce{H2PO3-}$ are (to the best of my knowledge) hydrogen phosphonates rather than dihydrogen phosphites. Similarly, all $\ce{HPO3^2-}$ salts should be phophonates rather than hydrogen phosphate. Only as organic esters (which cannot tautomerise easily as carbocations are less stable) are both forms observed: trimethyl phosphite and dimethyl methylphosphonate both exist.

A similar picture is true for sulfur in the $\mathrm{+IV}$ state. Sulfurous acid and sulphonic acid are both known as far as I know, and both structures have been isolated in salts. Similarly, methyl methylsulfonate and dimethyl sulfite are both known to the best of my knowledge.

The corresponding elements of the second period do not show this tautomerism. For oxygen, this is due to its high negativity preventing the formation of ‘oxyacids’ altogether. For nitrogen, this is because an $\ce{N=O}$ double bond is more stable than two $\ce{N-O}$ single bonds so that nitrite ($\ce{NO2-}$) is already devoid of protons and cannot tautomerise. However, remnants of this behaviour still exists in organic compounds where both the nitro group $\ce{-NO2}$ and organic nitrites $\ce{-ONO}$ are known.

What is the case for the heavier homologues of both phosphorus and sulfur? Do the elements arsenic/antimony/bismuth and selenium/tellurium/polonium exhibit both an -onic acid and an -ous acid or is one form preferred over the other? Is there a general trend going down the groups or are secondary effects (e.g. lanthanide contraction) more important?

Answers backed with references to crystal structures of anions (or the free acids) and/or corresponding quantum-mechanical calculations (both in references or self-performed) are preferred.

Note that the fully oxidised pnictogen(V) or chalcogen(VI) compounds are not part of this question,

Disclaimer: This isn't really the answer Jan is looking for.

Holleman, Wiberg "Lehrbuch der Anorganischen Chemie", de Gruyter, Vol. 101, notes that:

Selenium

Selenium forms a weaker, but more stable acid based on oxidation state $ +4$. No explicit mention of the structure, which suggests to me that $\ce{SeO(OH)2}$ is the structure.

Tellurium

Tellurous acid's structure is described as unknown.

Arsenic

For arsenous acid (the $+3$ oxidation state based acid) the tautomeric equilibrium is entirely on the side of $\ce{As(OH)3}$ (as opposed to $\ce{HAsO(OH)2}$). The reason given is that $\ce{As}$ does not form double bonds with $\ce{O}$, which is outdated: Current thinking is that only the first row forms double bonds, i.e. no double bonds in $\ce{H2SO4}$ or $\ce{HPO(OH)2}$. Organic esters exist that can be described as $\ce{RAsO(OH)2}$.

Antimony

The $+3$ oxidation state based acid is described as forming $\ce{H+ + [Sb(OH)4]-}$ in water. No mention of tautomerism.

The book gives sources only sparingly and on a per-section basis. Relevant items appear to be:

R. Paetzold: "Neuere Untersuchungen an Selen-Sauerstoff-Verbindungen", Fortschr. Chem. Forsch. 5 (1966) 590-630.
W.A. Dutton, W.Ch. Cooper: "The Oxides and Oxyacids of Tellurium", Chem. Rev. 66 (1966) 657-675.
M.A. Ansari, J.M. McConnachie: "Tellurometalates", Acc. Chem. Res. 26 (1993) 574-578.
J.D. Smith: "Arsenic, Antimony and Bismuth", Comprehensive Inorg. Chem. 2 (1973) 547-683.
C.A. McAuliffe: "Arsenic, Antimony and Bismuth", Comprehensive Coord. Chem. 3 (1987) 237-298.
GMELIN: "Arsenic", Syst.-Nr. 17, up to now 1 book, ULLMANN, Vol. 5: "Arsenic and Arsenic Compounds", A3 (1985) 113-141.
GMELIN: "Antimony", Syst.-Nr. 18, up to now 6 books, ULLMANN, Vol. 5: "Antimony and Antimony Compounds", A3 (1985) 55-76.

My thoughts on $\ce{As, Sb}$ are that in the MO picture, $\ce{H}$ has insufficient overlap to form a bond that would compete with the alternatives.

I have performed DFT calculations (PW6B95-D3/def2-QZVP//PBE-D3/def2-TZVP) on gas-phase $\ce{Pn(OH)3}$ and $\ce{HPnO(OH)2}$. Of course, these are gas-phase  energies on single conformers, not free enthalpies. For $\ce{Pn} = \ce{N}$, a 10 kcal/mol favorable energy difference towards $\ce{HPO(OH)2}$ was found. For $\ce{As}$, about 35 kcal/mol towards $\ce{As(OH)3}$ was found. The Mayer bond order analysis was not conclusive.


What are the methods of fluoride removal from drinking water? I would like to remove fluoride from my water supply.

Are there any methods, any test kits or water distillation units to remove fluoride from drinking water?

See also:

Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?
A simple and easy way to neutralize fluorides in drinking water
Scientists Discover New Technique to Remove Fluoride from Drinking Water

Read More: http://www.trueactivist.com/scientists-discover-new-technique-to-remove-fluoride-from-drinking-water/

There are many filters which do not remove fluoride from potable water. Boiling or freezing does not help either. Reverse osmosis filtration may be a solution. However they are expensive ($100+).

Activated alumina defluoridation filters are used in areas where contamination from fluoride is common. These can reduce fluoride levels from 0.5 ppm to 0.1 ppm, but these require frequent replacement and are not a solution for domestic treatment. Distillation filters can be purchased to remove fluoride.


Why doesn't nitric oxide react with water? Why doesn't $\ce{NO}$ (nitric oxide) react with water? I know it is a neutral oxide, but I'm really curious as to why that is. Why will $\ce{NO}$ not react with $\ce{H2O}$?

Although nitric oxide is thermodynamically unstable, it is kinetically stable as its decomposition rate is very slow[ref] and so it is considered stable and won't normally react to anything.

So, nitric oxide wouldn't normally hydrolyze easily. But in the presence of air, it forms nitrous acid $(\ce{HONO})$:

$$\ce{4NO + O2 + 2 H2O -> 4 HNO2}$$

However, nitrous acid is very unstable and quickly disproportionates:

$$
\begin{align}
\ce{2 HNO2 &-> NO2 + NO + H2O &(cold dil. soln.)}\\
\ce{3 HNO2 &-> HNO3 + 2 NO + H2O &(warm conc. soln.)}
\end{align}$$

Since $\ce{HNO2}$ is required to prepare diazonium salts (a versatile compound in organic chemistry), it is prepared in-situ by reaction of $\ce{NaNO2}$ and $\ce{HCl}$. Due to its instability, it is prepared for one-time use and is not stored for further preparations.

Reference: J. Phys. Chem., 1958, 62 (3), pp 360–361 https://doi.org/10.1021/j150561a030


Does adding solutes to water alter its triple point? Does adding solutes to water alter its triple point? If yes, how do we find the new temperature and pressure of this point?

Triple points are typically defined as points on the phase diagram of a pure substance where three phases coexist. Their existence requires that the phase rule be observed, which is $$f=C+2-P$$ where $$\begin{align}f &: \text{number of degrees of freedom} \\ C &: \text{number of components} \\ P &: \text{number of phases}\end{align}$$ The number of degrees of freedom is the number of intensive variables that can be altered independently while remaining in a given phase.

In general, when three phases coexist, $$f=C-1$$ If the system consists of a pure substance, $f=0$, as expected (the coexistence region corresponds to a point). If you have a mixture of $C$ components (labeled $i$), you will observe a different triple point for each choice of $C-1$ independent mole fractions $\chi_i$. But for each combination of mole fractions, coexistence of three phases will still occur at regions corresponding to single points.


How to convert ppm to g/m3? I'm not very comfortable with chemistry and I need to convert $\ce{CH4}$ concentration values in $\pu{ppm}$ to $\pu{g/m3}$. Is that possible?

I've already researched a bit and realised that for water you can assume $\pu{1 ppm}$ equals $\pu{1 mg/L} = \pu{1 g/m3}$. But since I'm measuring concentrations in the air this might not be correct.

I really appreciate any help. Thanks!

I am trying to understand ppm, too. As far as I have understood there are different kinds of ppm, which is basically a ratio: it can be a ratio of amount of substance, masses, or volumes.
Assuming your ppm is a molar ratio I did this reasoning:
Indicating with $n$ the amount of substance, with $M$ the molar mass and with $V$ the volume, the concentration of your gas is:
$$c = \frac{n_\mathrm{gas} \cdot M}{V},$$
and defining the $\mathrm{ppm}$ as:
$$\mathrm{ppm} =\frac{n_\mathrm{gas}}{n_\mathrm{total}}\cdot 10^6.$$
Using the gas law:
$$n_\mathrm{tot} = \frac{p\cdot V}{R \cdot T}, $$
where $T$ is the temperature in kelvin and $p$ the pressure in pascal, and substituting, you get:
$$c = \frac{\mathrm{ppm} \cdot M \cdot p}{R\cdot T}\left[\frac{\mu \pu{g}}{\pu{m^3}}\right].$$


What is the reaction between nitric acid and fluorine? Nitric acid, $\ce{HNO3}$, is a strong oxidising agent as well as fluorine, $\ce{F2}$. What will be the reaction when pure $\ce{F2}$ gas is bubbled through 100% pure $\ce{HNO3}$. While searching, many products are suggested including $\ce{HF}$, $\ce{O2}$, $\ce{H2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{NF3}$, $\ce{NO2FO}$, etc.. What is correct?

It was first discovered by Cady that when fluorine is bubbled through $\pu{3 N}$ nitric acid, then a remarkable compound, fluorine nitrate ($\ce{FNO3}$) is formed. It boils at $\pu{-42 ^\circ C}$ and is a very strong oxidizing agent and is capable of liberating iodine from iodides and in gaseous states, it explodes around $\pu{200-300 ^\circ C}$.

References

The reaction of Fluorine with Nitric acid and with solid Potassium Nitrate to form NO3F by Don M. Yost and Alan Beerbower,  J. Am. Chem. Soc. April 1935, 57(4) 782. DOI: https://doi.org/10.1021/ja01307a520
Prof. Dr. Otto Ruff, Dr. Walter Kwasnikie,  Die Fluorierung der Salpetersäure. Das Nitroxyfluorid – NO3F, April 1935, DOI: https://doi.org/10.1002/ange.19350481604 [German paper]


Why do the trends in reactivity not apply for francium? Why is francium not included in the reactivity series?

Why is potassium considered more reactive than francium?

I know that reactivity increases down the group, but why does it not apply here?

Francium is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked:

  Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 minutes) would immediately vaporize any viewable quantity of the element.

This extreme radioactivity makes it hard to determine the chemical properties of francium. Some are known. For example, francium's ionization energy is 392.8 kJ/mol (cesium is 375.7 kJ/mol). Since francium has a higher ionization potential than cesium, which it does because of relativistic effects, francium should be less active than cesium. However, since observable quantities are difficult (nearly impossible) to come by. It is hard to say for sure.


How do you melt magnesium without causing its combustion? How do you melt magnesium without causing its combustion?

I own a foundry capable of reaching temperatures around 2000 °F, and I recently bought some magnesium for melting before doing my research. I cannot find any information on how this process works or if it is even possible.

I have seen molten magnesium; it was covered with charcoal and other fluxes and in a furnace that had a gas shroud. (I think methane that burned any oxygen that got into the shroud atmosphere.) When an immersion thermocouple was put into the magnesium, the open spot immediately burst into flames. I recommend that you not try to melt magnesium; you certainly must get more information than you can get on the internet.


How to calculate the electrical potential and mols of electrons transferred from combined half cell equations? Given 2 half-cell rxn's and their potentials, balance the full cell and solve for ECell:

$$\ce{Fe^{3+} + e^{-} ->Fe^{2+}}; E=0.68V$$

$$\ce{Cr_2O_7^{2-}+14H^+ +6e^- -> 2Cr^{3-} + 7H_2O};E=1.33V$$

So I know that both these half-cells are Gaining electrons (Reduction), and in order to solve for the full cell I need to flip one. I flip Fe Cell to oxidation b/c it has a lower potential and adjust for number of electrons moving. To Calculate ECell: $E(Cr)-E(Fe)$ despite adjustment of mols.

So my full Cell is
$$\ce{6Fe^{2+}-> 6Fe^{3+} + 6e^{-}}$$
$$+$$
$$\ce{Cr_2O_7^{2-} +14H^+ +6e^- -> 2Cr^{3-} +7H_2O}$$

My question is, why is the formula to solve ECell:  $E(Cr)-E(Fe)$ and NOT $E(Cr)-6*E(Fe)$?

I'm pretty sure that more electrons generally = more electricity, so why is it when I multiply the amount of electrons, I don't multiply my half cell E?

TL;DR

This basically boils down to the fact that $\mathrm{E_{cell}}$ is an intensive property, not an extensive property.

Long Version

Lets define some terms first:

Intensive Property:
A property that doesn't change when the size of the sample changes. For example, density and concentration are examples of intensive properties

Extensive Property:
A property that changes when the size of the sample changes. For example, mass and volume are examples of extensive properties

$\mathrm{E_{cell}}$ is the energy in joules of 1 coulomb of charge (1 volt = 1 joule per coulomb). Hence it is an intensive property. Therefore the number of coulombs of charge or number of electrons doesn't effect the value of $\mathrm{E_{cell}}$.

Another way of thinking of it is that by doubling the amount of moles, the  amount of coulombs and joules also double, hence $\mathrm{E_{cell}}$ remains the same.


Flavor transference through the air and plastic bags I am not sure if this is a chemistry question; however, I was wondering:

In my desk I have two plastic, zip-lock bags.  One contains cinnamon tea, the other contains nuts.

After about a week of these bags being next to each other, the flavor of the cinnamon somehow leached through both bags and now the nuts taste like cinnamon!

Is there a chemical explanation for this or isn't this a question of chemistry?

Cinnamon tea is most likely flavoured with cinnamalaldehyde (1) and eugenol (2), the main components in cinnamon oil. The latter compound also gives cloves the characteristic aroma.

To my knowledge, ziplock bags are made from polyethylene. They are not meant to be aroma-tight, like vacuum-sealed coffee packages.


How to quantize the shape of a molecule? I am looking for numbers which tell you something about the 3D shape of the molecule, and is also easily calculated.

One of these is the acentricity factor. I am aware of this one. My problem is you calculate acentricit, from thermochemical data, which are usually hard to find, and can contain a great degree of experimental errors.

A more suitable quantity would be constructed from the groups of the molecule. Is there such a thing? Or what are other quantifiers of molecular shape other than acentricit, factor?

There are lots of descriptors of molecular size and shape that come to mind:

Moments of inertia (e.g., how different are they)
Radii (e.g., enclosing the molecule in a best-fit box)
Surface area or polar surface area (there are existing group-based methods)
Volume (e.g., of the van der Waals surface)

Some people have also used 3D shape descriptors like "spectrophores".


How to arrange substances in order of increasing vapor pressure? During the lecture in my chemistry class, my teacher posed the following problem:

Arrange in terms of increasing vapor pressure:
a) $\pu{0.5m}$ sugar
b) $\pu{0.5m} ~ \ce{NaCl}$
c) $\pu{2m}$ ethylene glycol
d) $ \pu{2m} ~\ce{ LiBr}$

The answer turned out to be: d, c, b, a. However, I do not understand why this is the case. I know the formula for vapor pressure is:
$$ \Delta P = -XP^\circ $$
which means that the pressure of the solvent is proportional to the mole fraction of the solute. I'm still not grasping why the aforementioned answer is correct though. Can anyone provide me with an explanation?

To solve this problem, use Raoult's law. In Raoult's law, the vapor pressure of a fluid is lowered proportionally with the mole fraction of solute particles in the fluid.
You know the formula, but you only need it if you want to calculate the actual resulting vapor pressure. We can know which is bigger like this:
A. $\pu{0.5M }$ sugar. Sugar does not ionize, thus there are $\pu{0.5 M }$ particles in the solvent.
B. $\pu{0.5M \ce{NaCl}}$. Sodium chloride dissociates to two ions. Thus there are $0.5 \times \pu{2 M}$ particles in the solvent.
C. $\pu{2M}$ ethylene glycol. Again, this doesn't dissociate, so there are $\pu{2M}$ particles.
D. $\pu{2M \ce{LiBr}}$. Dissociates, resulting in $2 \times \pu{2 M}$ particles.
Thus, listing the concentration of particles in the solvent, you see where the answer comes from:
D: $\pu{4 M}$
C: $\pu{2 M}$
B: $\pu{1 M}$
A: $\pu{0.5 M}$


Reusable and safe endothermic reaction I am looking for a reusable endothermic reaction that is safe to both touch and handle, it needs to be able to be stored in a jacket and light enough to do physical activity. Any suggestions?

There are many endothermic physical processes such as melting ice or sublimation of solid carbon dioxide $\ce{CO2}$. One should also mention thermoelectric cooling effect (Peltier device), which is is rarely effective in practice, yet still an option.

A suitable endothermic chemical reaction would probably be an electrolytic dissociation of certain salts, among which dissolving ammonium nitrate $\ce{NH4NO3}$ in water is probably one of the most well-known and effective examples.


How to choose the level of theory for modelling reactions of polymers? My Main purpose is using quantum chemistry for simulation of some reactions. Actually we are working on making some new polymers and I need to simulate some molecular properties of these polymers after adding some molecules to them. These include bond breaking and bond making as well as evaluation of optical properties.
That's why I started studying quantum chemistry models. I started with the Schödinger Mechanical Model and then Hartree Fock Theory, but as everyone knows these models are not that efficient for molecules larger than lithium. Then I started learning DFT and got interested in semi-empirical models, but could not learn them yet. Finally I am absolutely confused about which method should I select.
This is my question: As I have not enough time to learn all those quantum chemistry models/techniques, which method is currently best option if I just want to learn ONE method. (I have access to some super-computers in our university.)

The question is by far not an elementary one. Short answer is: unfortunately, there is no silver bullet.

For a molecule of around 65 atoms I would go with DFT as the methodology which provides probably the best quality/cost ratio for molecules of this size.

Now, you need to choose functional and basis set, and this is a tricky business. But to start from:

Do not blindly use B3LYP.
Try to search literature for the same types of calculations (energetics of ring opening reactions, optical properties) for similar compounds. Maybe someone already did the so-called benchmark calculations, i.e. compared on how good different functionals in combination with different basis sets are on reproducing the properties (say, specific rotation) either measured experimentally or calculated at some high level of theory (say, CCSD(T) with reasonably sized basis set).
Do your own benchmark calculations with different functionals and basis sets. First, benchmark different functionals with the same basis set (say, Dunning correlation-consistent basis sets, cc-pVTZ or better cc-pVQZ), vary the functional, compare results with experimental values, choose the best one. Once method is chosen, try to decrease and also increase (if possible) the basis set to see the effect of basis set size (with the above mentioned bases it is quite simple: if cc-pVTZ was used to benchmark different functionals, try the best functional in combination with cc-pVDZ and cc-pVQZ). Choose the smallest basis set that reproduces experimental results with desired accuracy.

P.S. You could also ask Google, there are some resources of information here and there on how to choose method and basis set, though, as I said, this question is not elementary, so only general advice can be given.


Has anyone actually prepared Zn(HCO3)2 absent the presence of pressurized aqueous CO2? Under normal conditions, I am not certain, as per Atomistry on $\ce{ZnCO3}$, the following quote:

  The hemihydrate, $\ce{2ZnCO3.H2O}$, was obtained by precipitating the solution of a zinc salt with ammonium bicarbonate and digesting the gelatinous precipitate in excess of the precipitant till it became crystalline.

where I would have expected zinc bicarbonate, and further:

  The basic zinc carbonate, $\ce{5ZnO.2CO2.4H2O}$, can be prepared by the direct hydrolysis of the normal carbonate, or by boiling a solution of zinc, zinc hydroxide, or zinc carbonate in an excess of aqueous carbonic acid.

And finally, a reference to a presumably formed zinc bicarbonate, and at times, citing the employment of pressure:

  Basic zinc carbonates can be converted into the normal carbonate by contact with water charged with carbon dioxide under pressure. The normal carbonate is itself somewhat soluble in water containing carbon dioxide - presumably forming the bicarbonate.

The last quote is unclear, in my opinion, as to whether under normal experimental conditions, should one expect the creation of a stable $\ce{Zn(HCO3)2}$?

We have to distinguish between the bicarbonate in solution and the bicarbonate as a solid.  While many metal cations are compatible with bicarbonate anions in dilute aqueous solution, only in a few cases can a solid bicarbonate be expected as a precipitate.  In most cases, such as with lithium or with hard-water forming metals such as calcium and magnesium, the carbonate is formed instead.  The only solid metal bicarbonates I know of are those of the heavier alkali metals than lithium.


Creating a chemical solution with water and chlorine I bought a swimming pool for my kids and also a measurement instrument for measuring the amount of chlorid in the pool in mg/l. The problem is now that you have to calibrate the instrument before using it. This means you have to know the current amount of chlorine in the pool.

Since the test strips are quite inaccurate and hard to read an exact value. My idea was to just create a perfect solution in a one liter water container.

Now I am not really sure how much of the given pool chlorine I have to put into the 1 liter water to get a 1.5 mg/l solution of chlorine. With that I would be able to calibrate the instrument perfectly.

What I am using is chlorine granules.
The instruction leaflet says it contains "CAS 51580-86-0 Troclosenatrium, Dihydrat (100g/100g)"
Registration: BAuA N-23822.

Would I just have to add 1,5 mg to 1 liter of water to get 1,5mg/l?
Is it maybe that easy?

Is your measuring instrument made for measuring chloride or chlorine ? It is not the same. In your first sentence, you speak of chloride. Later on you speak of chlorine. Anyway, the best thing to do is to buy a bleach solution. The amount of chlorine is printed on the label. It is often $14$% chlorine $\ce{Cl_2}$. By diluting a sample (maybe $1$ mL) of such a solution in 1 liter water, you obtain a solution $0.14$ gram chlorine $\ce{Cl_2}$ per liter, and you can dilute it still more.


Electronegativity of zinc vs copper in galvanic cell I am reading up on how galvanic cell works and I realised that the flow of electrons is from Zn to Cu. But Zn is more electronegative compared to Cu according to periodic table trends

I read it somewhere that when comparing transition metals, it may be different, but it was not  further elaborated. Can someone explain this to me how Cu is more electronegative than Zn?

There are two common confusions in your question. I hope I can clear them up.

  Electrons are not flowing from zinc to copper. Not exactly.

In the Zinc-Copper galvanic cell, the net reaction is:

$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}$$

Zinc is being oxidized:

$$\ce{Zn(s)->Zn}^{2+}(aq) \ce{+2e-} $$

Copper is being reduced:

$$\ce{Cu}^{2+}(aq)\ce{+2e- ->Cu(s)}$$

Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.

  Electronegativity has nothing to do with electrochemistry.

Electronegativity is the attraction an atom has for a pair of electrons in a covalent bond.

Electrochemistry is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this NIST publication.

For zinc:
$$\ce{Zn->Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{  kJ/mol}$$
$$\ce{Zn+->Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$

For copper:
$$\ce{Cu->Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{  kJ/mol}$$
$$\ce{Cu+->Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{  kJ/mol}$$

  Why is zinc oxidized and copper reduced?

Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:

$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -11 \text{  kJ/mol}$$

The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data.

In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.

$$\Delta G=-nFE$$

For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.

How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.

Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn2+. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu2+. Zinc reduces the Cu2+ ions and copper metal plates onto the zinc.

Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from Wikipedia, which mines them from a number of legitimate sources.

SHE:
$$\ce{H2(1 atm) + 2e- -> 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$
For zinc:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$
For copper:
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-}  \ \ \ E^o =+0.34 \text{ V}$$
To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$
This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn2+]>>[Cu2+], then the reverse reaction could be spontaneous.