Is aqueous medium necessary for complex formation? if so, why? If dry ammonia gas is passed through anhydrous copper sulfate, will it turn blue (due to the formation of tetraamminecopper(II) complex)? Or will silver chloride form diamminesilver(I) complex in liquid ammonia?

If the above complexes do not get formed in non-aqueous mediums, why is it so?

It should be noted that ammonia is a mildly polar solvent and that (for instance) you can prepare $\ce{[Cu(NH3)6]^2+}$ by reaction with liquid ammonia1. The description of the geometry (two pages earlier) of $\ce{CuSO4.5H2O}$ (namely containing a square planar $\ce{[Cu(H2O)4]^2+}$ unit with two additional $\ce{[SO4]^2-}$ above and below the plane) suggests that ammine ligands could fulfill a similar role to water.

Bear in mind that passing ammonia gas over a solid will probably mean that if the reaction happens it will only happen near grain surfaces.

To answer your second suggestion, Housecroft and Sharpe1 identifies a reaction producing $\ce{[Ag(NH3)4]+}$ from $\ce{Ag2O}$ in liquid $\ce{NH3}$.

Hope this is informative.

[1] Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 2nd Edition; Pearson Prentice Hall, 2005; pp. 635–637, 693.


What is the role of the slash ("/") in chemical reactions? I have always been confused about the role of slash in chemical reactions.

In some reactions, e.g. $\ce{NaBH4}/\ce{OH-}$ it means we use both compounds, right?

And in reactions where we write $\ce{Pt/Pd/Ni}$ do we mean platinum or palladium or nickel?

The ambiguity seems to come not so much from Chemistry, but from English punctuation. Slashes are often used in place of "and", "or", or the latin "cum" (meaning combined with). It seems that it unfortunately is just context dependent, though I would argue its use as a replacement for "or" is most consistent grammatically and defining this to be the meaning of the slash would eliminate ambiguity in these situations.


Si and Re facial descriptors I am considering the hydrogenation of 2-butanone with respect to its Re and Si faces.

So if we look at 2-butanone as I've drawn below from above we are looking at its rectus face.

And from below we are looking at its sinister face.

Now, time to consider hydrogenation.

We'll keep the wedge and dash groups as they are and the carbon and oxygen both on lines. In other words we keep two points as is in the originally trigonal planar 2-butanone and when the hydrogen inserts itself, the other two points (a hydrogen and an alcohol group) pop up out the plane to form the classic tetrahedral shape of an $\ce{sp3}$ carbon atom.

So it appears that hydrogenation from the sinister face adds an hydrogen to the central carbon and forms an alcohol, and the alcohol is above the added hydrogen spatially, and this gives us the R isomer. Hydrogenation from the rectal face gives the opposite S isomer.

How's the reasoning? Am I right? I am fairly sure. But not completely sure.

When looking at the corresponding Wikipedia article for prochirality you will find out, that your reasoning is correct for the hydrogenation. Of course if you have a different nucleophile, then the situation would change, as the Cahn-Ingold-Prelog identifiers would change.

See also prochirality in the IUPAC goldbook.


Looking for information about possible ingredients of a fertilizer I have found it in a local store, "20.20.0 - Kompoze" is written on the pack. In the pack, there are yellowish greenish irregular formed little balls. What is this, do you know ? Please help and excuse my poor english.

well , maybe you should have had googled it or spoke to the retailer about it . i dont know why you have no but its constituents are 20% nitrogen (N) and 20% phosphorus (P2O5). Afterall there must be NP written above 20-20-0 and these both must must be written above the KOMPOZE and this all will be enclosed in a red disc.and below all this will be a trademark of toros.


Deciphering stereochemical notation in electrocylization reaction I not agree with this solution, for me the red bond formed must be a dotted line in the drawing, and the hydrogen atom musn't be represent with a dot. Am I right or not?

You are both correct and incorrect. (And this entirely without superimposition of quantum states!)

You are taking an achiral molecule and performing a reaction in achiral media to form a chiral product. Symmetry dictates that the reaction must be racemic, so there must be equal amounts of upward-pointing and downward-pointing carboxylates.

However, the drawing is also incomplete. It does not make sense to discuss the reaction without labelling the entire stereochemistry at the newly formed bond. It is important to realise that it generates only two of the four possible stereoisomers: the two syn ones ((S,R) and (R,S)). You should add an upwards-pointing hydrogen to the right-hand carbon of the new bond.

Judging by your comments, you are only interested in the arrow direction and resulting movement. Well, the arrows are not clear. They are 2D on a 2D plane and trying to signify a 3D movement. You could grab the curved bit of the arrow like a handle and:

either pull it upwards — then you have a rotation which would move the carboxylate below your screen’s plane and create the downward-pointing product ((R) on the α-carboxylate carbon),
or push it downwards — then you have the rotation which would move the carboxylate above the screen’s plane and create the upward-pointing product ((S) on the α-carboxylate carbon).

So the arrows’ directions are arbitrary, we do not know if the are shaped below or above the plane.


Potential wells and symmetry in absorption spectroscopy In a course I am following, we are analyzing the potential wells in molecular absorption spectroscopy.

For molecules with more than one normal mode, we can study the potential wells involved in transitions for each mode and apply the Franck-Condon principle to determine the intensity of each possible electronic and vibrational transition. In these transitions, some of them will not change the geometry of the molecular equilibrium geometry, others will. This leads to think about how the transition change the point group symmetry of that molecule.

Here comes my question: my teacher said that

  modes that are non totally symmetric wont maintain the symmetry of the molecule and therefore the potential wells associated with the electronic ground state and excited state will be one just above the other, with the same equilibrium position and just an increase in energy, while totally symmetric modes won't have this property.

I can't understand why this is true, if there are approximations underlying it and also I can't find any reference for it.

This has been used, later, to distinguish between totally symmetric and non totally symmetric modes in the expression of the transition dipole moment, because for wells not one above the other (totally symmetric) we need to calculate a superposition integral between the two vibrational states, while for totally symmetric modes we know it can be written as a Kronecker delta.

This is a bit tricky to explain, the way I imagine it is this: In a totally symmetric mode where, say, each bond extends in phase the potential minimum need not be the same in the excited state and in the ground state but it might be so 'accidentally'. All the forces are balanced. In a non-totally symmetric mode, one set of bonds stretches as the others compress so the potential has to be the same in both states otherwise there would seem to be an internal force being generated that would move the molecule, which is not possible.

edit:

Other points are

(a) The spectrum reflects the change in geometry between states and thus is that of totally symmetric vibrations only, but see (d) below.

(b) the excited and ground state have to have the same symmetry,

(c) the asymmetric stretch is a symmetric function of normal coordinates, which is what your lower picture shows.

(d) the non-totally symmetric excited state mode need not have the same frequency  as in the ground state, i.e. potential can be wider or narrower. In this case there are selection rules based on the wavefunction's symmetry; both must be the of the same symmetry in the FC factor. The symmetry of vibrational wavefunctions for non-totally sym. modes change with vibrational quantum numbers and it works out that these type of transitions can have $0 \pm2,\pm4..$ transitions only.


Are ther known compounds of Argon? What is their molecular geometry and hybridisation? Can Argon hybridize orbitals and/or form covalent-like/ionic-like compounds?
Is there any study of that? I would be happy to read concrete references.

What kind of molecular geometries for argon compounds are known or hypothesized at the current time?

Is there some peculiarity to apply something like VSEPR or molecular orbital theory to theoretical or experimentally known argon compounds?

Argon is a quite unreactive element, however very few compounds of it are known.

The most popular, and currently the only neutral compound, is argon hydroflouride, $\ce{HArF}$, which was reported by L. Khriachtchev et.al..[1]  In the same issue there is also a comment on the matter by my former supervisor G. Frenking.[2]
Another interesting read on the subject is the prediction of the metastable $\ce{HHeF}$, by M. W. Wong.[3] It generally treats all compounds of the $\ce{HNgF}$ family.

All calculations done so far indicate, that the molecule is linear and only kinetically stable. The highly accurate CCSD(T)/aug-cc-pV5Z level predicts, that the $\ce{HArF}$ molecule is $5.87~\mathrm{eV}$ higher in energy than its fragments $\ce{HF}$ and $\ce{Ar}$. The minimum barrier on this level of theory is a mere $0.35~\mathrm{eV}$ along the stretching coordinate of the hydrogen argon bond. The bending is predicted to be higher in energy. The reported bond distances for $\ce{H-Ar}$ and $\ce{Ar-F}$ are $133~\mathrm{pm}$ and $197~\mathrm{pm}$, respectively.
An NBO analysis on the MP2/aug-cc-pVTZ level of theory shows a strong ionic character of the molecule, it is reported, that there is a strong $\ce{(HAr)+F- }$ charge transfer in the electronic structure of the molecule.
Wong performed accurate calculations on the QCISD/6-311+G** level of theory and combined it with a Quantum Theory of Atoms in Molecules approach. He calculated a positive value of the Laplacian of the electron density $\nabla^2\rho_\mathrm{b}$ for the bond critical point of the argon fluoride bond, which indicates a mainly electrostatic interaction. The contrary is the case for the argon hydrogen bond, where the Laplacian is found to be negative, i.e. charge concentration, and therefore indicating a dominantly covalent interaction.
The following picture is a snapshot of this behaviour at the BP86/cc-pVTZ level of theory.[4]

Solid lines are regions of charge depletion, $\nabla^2\rho_\mathrm{b}>0$, while dashed lines are regions of charge concentration $\nabla^2\rho_\mathrm{b}<0$. Red lines are zero flux surfaces, where $\nabla\rho_\mathrm{b}=0$. And red spheres indicate bond critical points.

This molecule is quite interesting for many reasons, however, since it is only kinetically stable it has to be treated with fullest force of quantum chemical calculations. The calculations I did on a low level DFT method are not sufficient at all on its own, as has been investigated by J. Panek et.al..[5]]

Hybridisation is in a wider sense the mixing of s and p orbitals;[6] and upon further examination of the electron density this will be observed in this molecule. Most likely for the hydrogen argon covalent bond. As hybridisation is just a mathematical transformation of the original atomic orbitals there is no real interpretative value in this case. For example, the NBO analysis at the BP86/cc-pVTZ level of theory give only a negligible 3% s contribution of argon to the $\ce{H-Ar}$ bond.

The VSEPR theory is only a observational concept for well-behaved molecules. It basically lacks a solid scientific basis and should not be used any more, but especially in cases like this, it must not be used.

Literature and Notes

Leonid Khriachtchev, Mika Pettersson, Nino Runeberg, Jan Lundell, and Markku Räsänen, Nature, 2000, 406 874-876.
Gernot Frenking, Nature, 2000, 406, 836-837.
Ming Wah Wong, J. Am. Chem. Soc., 2000, 122 (26), 6289-6290.
To avoid copyright issues I performed a quick calculation to illustrate the above raised point.
Jarosław Panek, Zdzisław Latajka, and Jan Lundell, Phys. Chem. Chem. Phys., 2002, 4, 2504-2510.
Hybridisation is a concept that dates back to the beginnings of Valence Bond (VB) Theory. It was originally implemented in the strict fashion of creating $s^{1/(n+1)}p^{n/(n+1)}$ or $sp^n$ hybrid orbitals for $n = \{1,2,3\}$. H. Bent later went on suggesting fractional hybridisation schemes, see Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311. Nowadays the term s-p-mixing in Molecular Orbital theory is almost synonymous with the VB theory term hybridisation.


Why is the energy of an electron always negative? I just wrote my chemistry exam a few hours ago. However, I am still confused about the following question:

  Why is the energy of an electron in an orbital always negative?

I wrote that from the following equation

$$E_n = -R_h\left(1-\frac{1}{n^2}\right),$$

we can see that the energy will always be negative.

Is my answer correct?

First of all, regarding your comment, Rydberg's constant is not negative. If it were, then the equation you gave would have a positive energy all the time.

To be precise, Rydberg's constant is $$R_{\infty}=\frac{m_ee^4}{8\epsilon_0^2h^3c}$$ where all those constants have their usual meanings. Importantly, all of those constants are manifestly positive.

Whether the energy is positive or negative is entirely arbitrary. The energy of an electron can also be formulated as, $$E_n=R_h(1-\frac{1}{n^2}).$$

The only difference between the equation you wrote down and the one I just wrote down is that the above equation sets the zero of energy at the ground state of the hydrogen atom. You'll see that when $n=1$ the energy is zero and as $n\to\infty$ the energy approaches the Rydberg constant which is the ionization energy of the hydrogen atom.

When you say you wrote your exam I assume you mean you took your exam, not that you were writing the exam? Haven't heard that expressions before, but I suppose it makes sense. Maybe you were supposed to write that the negative energy represents the fact that the electron in a hydrogen atom is more stable (lower in energy) when associated with a proton than when isolated by itself.


Any home made tricks to fill up a baloon that floats? These days there are a number of fancy balloons available, some of them are very beautiful. As always, they loose their gas overtime and don't rise anymore.

Some of those new type balloons appear to have been made of plastic and are quite easily re-usable but then you can't be going to the balloon guy every time you want a refill.

Are there any home made techniques that can let you fill those up (so they float again) and which generate a gas which is safe to handle at home without any protective gear?

P.S: This question is not motivated by any urge to save extra money, it's just that I find so many of them lying at home useless when they can be filled up again. Looking for something that's simple to make if possible.

There are some options, not all easily available or safe: Helium is the most used and it most certainly makes them float, being safe at the safe time. However, it's not an home-made trick. Methane is more easily available (in many cities, methane is piped into homes for domestic heating and cooking purposes) but it's very unsafe for such an use, as it could explode. Hot air is also used (think of hot-air balloons) but I wouldn't know how to make it last for a long time as the air would cool down to room temperature very quickly. Lastly, there's hydrogen which could also be produced with some reagents available at home, but it's as dangerous as methane.


How many products does this reaction yield? How many products can we get from the monochlorination of 2,4-dimethylpentane including stereoisomerism? I have seen someone say that it yields 3 structural isomers one of which has two enantiomers making them 4 products in total, but I couldn't find the chiral carbon he was talking about.

The expected products of mono-chlorination of 2,4-dimethylpentane are depicted in following scheme:

The substrate has 12 primary hydrogens (on 1-,1'-, 5-, and 5'-methyl groups),  2 secondary hydrogens (on 3-methylene group),  and 2 tertiary hydrogens (on 2-, and 4-methine groups), as numbered in the scheme. Altogether, total of 3 different hydrogens are present, therefore, 3 different mono-chlorinated products are possible as sketched in the bottom boxes of the scheme. Although 1- and 1'-methyl groups (and 5- and 5'-methyl groups as well) are equivalent in the structure, mono-chlorination of each would give different stereoisomer (strictly speaking, enentiomers). Therefore, counting stereoisomers, you should get total of 4 different products.


Which method can be used to determine the pKa of an acid-base indicator like bromothymol blue? I have a question about how one would determine the pKa of an acid-base indicator such as bromothymol blue.

I know that spectrophotometric analysis is one option, but is it possible to perform an acid-base titration to find the pKa (7.1)? How would one set up this experiment?

I was thinking I would use a pH meter/titrant curve and treat the indicator like a weak base, but what titrant would I use? Would I use a strong acid like HCl?

Yes, HCl would work with a pH meter. If you know the mass of base (indicator) you use, you can as 0.5 equivalents of a strong acid such as HCl and determine the pH. At that point, since the weak base would be half protonated, the pH would be equal to the pKa, based on the Henderson-Hasselbalch equation. Or you could just so the whole titration and find the pKa halfway to the equivalence point.

In this case, you have to make sure you start with the basic form of the indicator, however. Not sure what form it is supplied in.


Why does 2+ oxidation state become more stable relative to 3+ oxidation state for first row of transition metals? I viewed an image showing all the possible oxidation states of each element in the first row of transition metals, and the main oxidation states highlighted in a different colour.

I noticed all elements Sc to Cr had in common a   3+ main oxidation state,
and from Mn to  Zn 2+ was the common main oxidation state.

The reason stated for this in my good textbook was 2+ oxidation state become more stable relative to 3+ oxidation state, I can't quite understand why.

So  Q : why does 2+ oxidation state become more stable relative to 3+ oxidation state  for first row of transition metals?

I had drawn the electron - in -boxes configuartion for all elements in the first row, and looked at Hund's rule and partially and fully filled orbtals, but I have not really seen a clear, fundamental explanation of why, for this question.

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.


What is the best way to remove/break carbon carbon bonds? I am working on synthesis homework and I can not figure out how to remove a carbon substituent from a cyclohexane.

What is the best way to remove the three carbons and exchange them for a ketone?

A possible pathway is as follows:

At high temperatures, formation of the more substituted alkene will be favored. Catalytic amounts of base provide a kinetic pathway for this rearrangement. $\ce{OsO4}$ creates a vicinal diol, while $\ce{NaIO4}$ cleaves the $\ce{C-C}$ bond and oxidizes the alcohols to carbonyls. Additionally, $\ce{NaIO4}$ oxidizes $\ce{(HO)2OsO2}$ back to $\ce{OsO4}$, allowing the reaction cycle to continue.


Electrolytic cell chemical composition during/after charging? I have a pot with some (bottled) water and added (sea) salt. It has a pencil (graphite) and tin foil (aluminum). The pencil is connected to the positive outlet of a solar panel and the tinfoil is connected to the negative outlet of the solar panel (effectively 'charging' the electrolytic cell).

To ensure I don't come to harm, I would like to know:

A) What chemical(s) gas(es) are produced at the pencil?
B) What chemical(s) gas(es) are produced at the tin foil?
C) What will the chemical composition of the water generally be (assuming I've been charging it a while)? (Acidic, toxic, etc?).

Voltage determines the kind of reactions, amperage determines the amounts.  The positive graphite electrode is the oxidizing anode.  It will evolve oxygen and chlorine and be eroded to carbon dioxide, carbon monoxide, and chloro-organics.  The negative aluminum electrode is the reducing cathode.  It will reduce Na+ to elemental sodium that immediately reacts with water to make lye and hydrogen.  The lye will erode the aluminum into more hydrogen and aluminum hydroxide gel, plus some black slime for about 1% alloy-reinforcing precipitated intermetallics in the aluminum foil (usually 8111 alloy).  Lye plus chlorine gives hypochlorite bleach that will erode both electrodes, and do a little disproportionation into chloride plus chlorite, then that perhaps into chloride and chlorate.

Sodium carbonate (washing soda) is a more benign electrolyte, but it is an alkali hazard to eyes and skin.  A glass is a more inert container.  Do not short the electrodes, do it outdoors and stay upwind.  Avoid electrolyte or product contact with your eyes and skin.


Water from coffee condensation Why is it that from a cup of hot coffee, only the water condensates on the inside of the lid? Why aren't there any coffee particles? Does it only have to do with density or is it something else involved?

I agree with Jan. The "coffee molecules", probably mostly alkaloids like caffeine, have larger and heavier structures than water molecules. I would expect most have dipole-dipole intermolecular forces, and (minor) hydrogen bonding may be present for some.

When you heat the solution with thermal energy, when we zoom in to the molecular scale we see this energy us actually kinetic. The molecules move faster, and some at the surface break free of their attractions to their neighbors. Despite water being held together by the strong hydrogen bonding intermolecular force, water molecules will be first and most to gain enough energy to pop out of the liquid solution. As Jan said, there are many more water molecules that coffee solute molecules. Also, the coffee solute's greater weights will make them harder to bring up to speed (though as a general rule of thumb IMFs like H-bonds actually play a more significant role than mass).

So those water molecules that pop out are far from their neighbors, so they constitute what we call a gas or vapor. There is much more water-gas above the solution than coffee-molecule-gas. This water vapor exerts a pressure based on how many molecules there are (and depending on kinetic energy / temperature), and we call this pressure from the gas above the coffee the "vapor pressure of the solution". The vapor pressure of water, despite being low relative to many organic liquids, is far higher than the vapor pressure of the molecules from coffee.

Think about the coffee residue in your pot, or the grounds - do they evaporate? Essentially no. There is essentially no vapor pressure, they are too bound by their IMFs to pop out and become gas (at temperatures a human usually encounters anyway). But water has a high vapor pressure, it is easily evaporated.

So when the gaseous water molecules encounter cooler surfaces, they will transfer some kinetic energy to the surface and transition from energetic gas state to lower-energy liquid state. This is the meaning of "condense". No coffee molecules condense, because none were in the gas phase to begin with (essentially).

Jan had a great point that these concepts are the foundation of the incredibly important process of distillation.


Can we oxidize oxygen? I have encountered this equation:
$$\ce{SO4^2- -> S2O8^2-}$$
In this reaction 2 of the oxygen ions change from $-2$ to $-1$ so they are getting oxidized.

It is because two oxygen ions are connected together.

A question rises: why can't we oxidize oxygen ions in $\ce{NO3-}$ or any other?

Oxygen can be oxidized. However, it is uncommon due to its high electronegativity. In fact, it is highest second only to fluorine. This makes it extremely difficult to remove electrons from oxygen, thus making it hard to oxidize.

One definitive example of oxygen being oxidized is oxygen difluoride. This exception is even listed on many lists of oxidation state rules (see: oxidation states).

And of course, if you go beyond stable molecules, it is possible to ionize oxygen with an electrical charge.


Does chemical compound with formula N₄O₁₀ exist? I do not know if a dimer of $\ce{N2O5}$ (dinitrogen pentaoxide) exist, and if its formula is $\ce{N4O10}$. Can you explain me that please? I do not know where to search for this information.

PubChem database N4O10 (molecular formula) search shows that there is no known compound with this formula.

While $\ce{NO}$ (“mononitrogen monoxide”) and/or $\ce{NO2}$ (“mononitrogen dioxide”), having odd number of electrons, will dimerize or combine to form

$\ce{N2O2{: } 2NO -> ON-NO}$
$\ce{N2O4{: } 2NO2 -> O2N-NO2}$
$\ce{N2O3{: } NO + NO2 -> ON-NO2}$

(note the $\ce{N-N}$ bond),

$\ce{N2O5}$ is formed by the condensation of nitric acid $\ce{HNO3}$:

$\ce{ 2 { }O2N-OH ->[-H2O] O2N-O-NO2 }$

(note the $\ce{N-O-N}$ bond), thus can be considered nitric acid anhydride. And unlike polyphosphoric acid formation from $\ce{H3PO4}$, there are no remaining hydroxy groups for further condensation.

With non-standard formal approach, $\ce{N2O5}$ already is a dimer of hypothetical franctional-oxygen-atom-count “mononitrogen(V) oxide” $\ce{NO_{2.5}}$.


Can iodine be used to disinfect water? I think it was Nat Geo.. Not sure exactly.. Where I heard, years ago, that a drop of iodine can be used to disinfect water of say, 1 litre.

But isn't chlorine used to kill all bacteria? I mean I don't completely remember that documentary but I'm pretty sure it was  iodine that they used.

Can iodine be used or does my memory fail me?

Short answer: Both chlorine and iodine in specific concentrations are used for water disinfection.

Iodine has been used to disinfect water for a long time. It is used in concentrations between 2.5 ppm to 7 ppm, but there is some controversy regarding the maximum safe dose of iodine because excess of iodine consumption in humans can lead to various diseases(thyroid disorders being very common). I believe this is the reason we have moved towards more safer options like chlorine for water disinfection.


How are these molecules chiral and achiral? I always thought that a molecule couldn't be chiral if it had a plane of symmetry. Doesn't the chiral molecule here have symmetry diagonally across the OHs?
The achiral one doesn't look symmetrical to me, so I also don't know how it is achiral.

Regarding the first molecule: I understand that the simplistic Fischer projection - with all its bond depicted by solid lines - can be very confusing at first glance in determining optical activity. You may believe it to be optically inactive, hoping for a center of symmetry to exist, but it is not so.
You must keep in mind how the molecule actually exists in 3d! Recall the rules of Fischer projection, the horizontal bonds project outward, while the vertical bonds extend inward. Hence, the molecule would actually look like this:

Now, I hope you can tell, there is neither any center of symmetry nor any plane of symmetry here. Hence, the molecule is chiral.
PS: Your second molecule is chiral by similar logic, as also mentioned in the comments, and as you yourself noted.


Equilibrium Constant Explained I'm not interested in a simple telling of what the properties of the equilibrium constant are and how to use them. I'm more hoping for an explanation of where it came from/how it might be possible to derive it.

My current guess is that it is a probability equation:

The numerator in the equation is proportional to the probability reactants meet and form products in the volume and the denominator is proportional to the probability products meet and form reactants in the volume. This ratio clearly must stay constant (unless the proportionality constants are changed via temperature or such).

Would anyone help?

Equilibrium is the state at which the rates of the forward and reverse reactions are equal.

$$\mathrm{rate_{forward}=rate_{reverse}}$$

Let's take a generic reaction:

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

For simplicity's sake let's have both the forward and reverse reactions be concerts fundamental reactions. Then we can write the rate laws for the forward and reverse reactions:

$$\begin{aligned}
\mathrm{rate_{forward}}&=k_1 [\ce{A}][\ce{B}]\\
\mathrm{rate_{reverse}}&=k_{-1} [\ce{C}][\ce{D}]
\end{aligned}$$

If we set them equal to each other, we can rearrange the equation to get the rate constants on one side and the concentrations on the other.

$$\begin{aligned}
k_1 [\ce{A}][\ce{B}]&=k_{-1} [\ce{C}][\ce{D}]\\
\dfrac{k_1}{k_{-1}}&=\dfrac{[\ce{C}][\ce{D}]}{ [\ce{A}][\ce{B}]}
\end{aligned}$$

The ratio of rate constants is the equilibrium constant. Thus is derived the law of mass action.


Why does Pb have a higher electronegativity than Sn? I recently learnt that electronegativity generally decreases as I move down a group and from right to left within a period. However, according to the table below, Pb has an electronegativity of $2.33$, which is higher than $1.96$ of Sn. There are also a lot other elements, such as Au, Hg, and W, that does not follow the general rule of electronegativity.

  Can someone offer an explanation of why some elements have a higher electronegativity than the elements directly above them?

The explanation is actually well-known.

Elements in Period 6 get f orbitals for the first time. 4f subshell has relatively poor shielding effects, thus it's harder for the atoms to lose the outer electrons, increasing electronegativity. (similar reasons also account for some chemical properties of Period 4 elements, like $\ce{H2SeO4}$'s oxidizability)
Relativistic effects also contribute. For a very informal (and actually incorrect) understanding, just imagine that the electrons in these heavy atoms move fast and their masses increase and their energy decreases.


Why is the equilibrium constant being calculated differently here? In my NMR textbook, it says the equilibrium constant for a dynamic process (e.g., A in equilibrium with B) is calculated by the concentration of A divided by the concentration of B.  Is this a typo?  I thought the equilibrium constant was products over reactants.

It's a matter of definition, really.  Equilibrium implies a reversible reaction, thus the same reaction could he rendered as $\ce{A<=>B}$ or$\ce{B<=>A}$.  So we can render the equilibrium constant as $[\ce{B}]/[\ce{A}]$ or as $[\ce{A}]/[\ce{B}]$.  We need to define which is used, of course, when we give a numerical value.


Is it possible for me to derive Avogadro's number? To my understanding the mole is the unit used to translate between mass on the atomic level and mass in the macro level, defined as the number of atoms in $12$ grams of Carbon-$12$, which apparently turns out to be $\pu{6.0221409e23}$ atoms, and this was defined as Avogadro's Number.

How was this number derived? Is it something I can derive myself? If I had $12$ grams of Carbon-$12$, how would I be able to deduce that it had $\pu{6.0221409e23}$ atoms?

To determine Avogadro's number you have to measure the same unit at the atomic and macroscopic scales.

This was first achieved by Millikan who measured the charge of an electron. The charge of one mole of electrons was already known and  is a Faraday. Dividing both, you get Avogadro's number.

Before that, Josef Loschmidt was the first one to calculate the number of particles in a cubic meter of gas, using the kinetic theory of gases.
This is, of course, also related to Avogadro's number.
Avogadro himself did not give any number. He just stated that equal volumes of different gases at the same pressure and temperature have equal amount of particles.

A more modern way of doing it is determining the density of an ultra pure element and then determine the number of atoms and their distances in a unit cell with X-ray diffraction.

So yes, you could, in principle, derive Avogadro's number with pure monocrystalline carbon 12. You will, however, need very good instrumentation.

As per the answer to this question on Chemistry SE it might become the other way round.
The Avogadro number will be absolute (defined, with 0 error) and then the kilogram will be redefined as a function of the number of atoms in a monocrystalline perfect $\ce{^{28}Si}$ sphere.


Calculating the volume of 1 mole of liquid water I am doing some rough calculations to test my understanding of elementary chemistry. In what follows, I calculate the volume of one mole of liquid water at about $\pu{4 ^\circ C}$ and $\pu{1 atm}$:

Method 1: Using the molar mass and density of water

The molar mass of water is $\pu{18 \times 10^-3 kg mol-1}$, and the density is $\pu{1 \times 10^3 kg m-3}$. Hence, the volume should be

$$V = \frac{(\pu{1 mol})(\pu{18 \times 10^-3 kg mol-1})}{\pu{1 \times 10^3 kg m-3}} = \pu{1.8 \times 10^-5 m3}.$$

Method 2: Using the size of an individual water molecule

If we model the mole of water as a cube, since there are $6.022 \times 10^{23}$ water molecules, each side of the cube would have $8.44 \times 10^{7}$ molecules.

Using the water data page on Wikipedia, I used trigonometry to calculate the H–H distance as being approximately $\pu{150 \times 10^-12 m}$. I would think the dimensions do not change too much when there are many molecules.

From this, I find that the volume is

$$V = [(8.44 \times 10^7) \times (\pu{150 \times 10^-12 m})]^3 = \pu{2.05 \times 10^-6 m3}$$

which is an order of magnitude smaller than Method 1.

What is the reason for this discrepancy?

The molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.

This will change your answer by a factor of 8.

See Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. J. Mol. Liq. 2001, 90 (1-3), 303–312. DOI: 10.1016/S0167-7322(01)00134-9.


Nature of transition metals The transition metals have valence and penultimate shell incompletely filled . so does this mean that the transition metals first fill up their valence shell and then penultimate shell?

The short answer to your question is, usually, but not always.  The energy differences between these levels are very small, so sometimes other factors come into play that can cause exceptions.  Here are a few exceptions:

 Predicted configuration   Actual configuration

Cr    [Ar] 3d4 4s2  ...........  [Ar] 3d5 4s1

Cu    [Ar] 3d9 4s2  ...........  [Ar] 3d10 4s1

Ag    [Kr] 4d9 5s2  ...........  [Kr] 4d10 5s1

Au    [Xe] 4f14 5d9 6s2  .. [Xe] 4f14 5d10 6s1

There is a special stability associated with a full shell, in fact that's probably why you suspected that "transition metals first fill up their valence shell and then penultimate shell."  There also appears to be a stability advantage associated with a half-filled shell and this can explain many of the anomalies.  The reason that a half-filled shell may have some extra stability is that there are no paired electrons (Hund's rule) in such a configuration.


E1 and E2 elimination reactions of isotopically labeled bromocyclohexanes Question
  a) What would the product of the reaction be for an E2 mechanism

  b) What would the products be for an E1 mechanism

  The Br coming out of the page is the 79 isotope while the Br going into the page is the 81 isotope.

There is no mention of reaction conditions.

I am not totally sure what the effect of the isotope will have on bromine's ability to leave. I am guessing that the bromine-81 will leave since it is bigger and can stabilise the negative charge better.

I am also not sure how the products for the E1 and E2 mechanism will be different. Won't they be the same?

Under E1 conditions (typically hot, polar protic solvent), one of the bromines will leave as bromide to give a carbocation. Ignoring the isotope effect, there's about an equal propensity for either of the isotopic bromines to leave, giving about a 50/50 ratio of the two carbocations. Deprotonation at the beta-position gives an alkene. I've only shown the most substituted possible alkene since that will be the most stable and major product.

(Note, the carbocation could rearrange, but I'm ignoring that for this level of analysis.)

Under more strongly basic conditions an E2 reaction is more likely. There is a strong preference for an anti-periplanar arrangement of the leaving group, carbons where the alkene will form, and the deprotonated hydrogen. This stereochemical requirement is more easily visualized in the chair forms, which are not equivalent. The structure on the left is more stable since the methyl group is placed equatorially. In this conformation, there are two hydrogens that are anti-periplanar with the bromine-81 atom (green). A bulky base would react with the least hindered hydrogen (purple) and eliminate the bromine-81 to give the disubstituted alkene shown. A smaller base would be more likely to produce the most stable product, which results from deprotonating the hydrogen in blue.

Just for completeness I've shown the elimination product of the less stable chair conformation. In this case, there is only one anti-periplanar arrangement. If the orange hydrogen is deprotonated, then the bromine-79 atom can be eliminated to give the cyclohexene shown.


Is heating water endothermic? So I got a problem recently which asked to calculate the amount of energy needed to increase the temperature of 200 g of water from 26 to 51 degrees Celsius. The value I got was 20900 J.

So far I feel like it is endothermic because heat had to be added to the water in order for its temperature to go up, so the water essentially absorbed heat, but I'm really not sure.

So far I feel like it is endothermic because heat had to be added to
  the water in order for its temperature to go up, so the water
  essentially absorbed heat, but I'm really not sure.

This is mostly semantics. Such exam questions are annoying when they make you argue about words. The term endothermic and exothermic are reserved for reactions by IUPAC. An endothermic reaction is one where the overall enthalpy change is positive. However one can extend it to endothermic process, which is any process that acquires energy (from external surroundings). Atkin's physical chemistry clarifies it nicely "it must never be forgotten that heat is a process (transfer of energy as a result of temperature difference), not an entity"


Is this bond cleavage correct? Is this bond cleavage of cyclopentadienone correct?

The ring becomes aromatic and stability "should" increase. But my book and data from experiments suggest that the molecule isn't that polar as it would have been from this cleavage. Why isn't this cleavage favorable?
Doesn't aromaticity counter the appearance of a positive charge on oxygen?

The cleavage that you have shown is NOT favourable. The dipole of the carbon double bond O points towards the oxygen , not the carbon. The bond tends to shift onto the oxygen giving  it a negative charge , while the carbon a positive charge. This, however is anti-aromatic so the molecule tends to do a Diels-Alder reaction with itself.

The bond will not cleave the way as you have mentioned . Just for the sake of aromaticity, a carbon double bond oxygen will not cleave in such a manner where the carbon contains the negative charge and the oxygen a positive charge. An oxygen with positive charge is extremely unstable(as it has a high electronegativity) and will show a strong inductive effect which will pull the electrons from the carbon(which is way less electronegative than oxygen) thus reverting to carbon double bond oxygen structure. Aromaticity can't be obtained as oxygen would just pull those elctrons from the carbon atom.

However let us consider the case of cyclopropanone. Here, the molecule becomes aromatic due to the shifting of the double bond onto the oxygen (oxygen gains a negative charge while carbon , positive).


Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene? Air contains 20 % $\ce{O2}$ by volume. How many cm³ of air will be required
for oxidation of 100 cm³ of acetylene?
(a) 1064 cm³
(b) 212.8 cm³
(c) 500 cm³
(d) 1250 cm³

Answer: (d)
My solution:
Equation: $\ce{2C2H2  +5O2 -> 4CO2 +2H2O}$
From stoichiometric calculations:

2 moles of $\ce{C2H2}$ will react with 5 moles of oxygen
1 $\ce{C2H5} \rightarrow $ 5/2 moles of $\ce{O2}$
$\pu{22.4L}$ $\ce{C2H2} \rightarrow 5/2\times \pu{22.4L}$ of $\ce{O2}$
1 volume of air $\rightarrow$ 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.

Let's begin with your answer:
$2\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5\ \mathrm{mol}$ of oxygen.
$1\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5/2\ \mathrm{mol}$ of $\ce{O2}$.
If we suppose that the gases are under standard conditions, then
$22.4\ \mathrm{L}$ $\ce{C2H2}$ will react with  $\frac{5}{2} \times 22.4\ \mathrm{L  } ~ \ce{O2}$.
This means that:
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\ \mathrm{L}$ $\ce{O2}$.
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$ of air.
So, the volume of air is $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$


van der Waals equation for deviation of gases from ideal behavior $(P + an^2/V^2)(V - nb) = nRT$

I understand that the term $nb$ is subtracted from the volume term as the available volume for the gas molecules to move around decreases in the case of a real gas due to the volume of the molecules themselves.

However, I don't understand why the term $an^2/V^2$ is added to the pressure term. Because of the attractive/repulsive forces, shouldn't the pressure on the walls of the container by the molecules also decrease?
(I thought about this and guessed that due to repulsive forces, the pressure must be increasing, but what about due to attractive forces? Does the sign of $a$ depend on the gas and hence affect the equation? I may be wrong here.)

Please clarify!

The term $-an^2/V^2$ is the reduction of pressure due to attractive or cohesive forces between molecules. This is most easily seen by writing the equation as
$\displaystyle p_\text{obs}= \frac{nRT}{V-b}-\frac{an^2}{V^2}$ .

While the constants $a$ and $b$ are considered to be empirical they do have a physical basis. As you mention $b$ accounts for the finite volume of the molecules and $a$ for the attractive force between molecules.

In the centre of the gas these attractive forces are the same in every direction and so cancel out, however, as close to the boundary as it is possible to go the forces act predominately inwards towards the bulk of the gas, rather as happens in capilliary action or surface tension. If the force between adjacent molecules is resolved into components one will lie along the boundary and the other at right angles pointing inwards.  Thus we can imagine over a long enough period of time a steady force close to the boundary and acting inwards. Thus molecules on reaching the surface are deflected both by the surface and by this inwards force.

The size of the force will be proportional to the product of the number of molecules /unit area and the size of the inward force. Both of these will be be proportional to the density ($Nm/V$) or equivalently molar concentration. The reduction in pressure can thus be written as $-a(n/V)^2$ where $a$ is a positive constant that depends on the gas and is determined only by experiment.


Expressions for constant volume and constant pressure heat capacities My lecture handout says:

$$C_V = \frac{\mathrm{d}q_V}{\mathrm{d}T} = \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

I understand that $C_V = \mathrm{d}q_V/\mathrm{d}T$ and that $\mathrm{d}U = \mathrm{d}q_V$, but what's the need for the partial derivative? Why can't you just substitute and get:

$$C_V = \frac{\mathrm{d}U}{\mathrm{d}T}$$

Likewise for the constant pressure heat capacity:

$$C_p = \frac{\mathrm{d}q_p}{\mathrm{d}T} = \left(\frac{\partial H}{\partial T}\right)_{\!p}$$

The difference in internal energy ($\mathrm{d}U$) is a function of many different variables:

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \!$$

Therefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a total derivative, which is not what the equation requires.


Reactivity of alkaline earth metals towards water In general, it seems to be a trend regarding the reactivity of alkaline metals with water which says that as you go down the group, they become more reactive towards water. But I'm curious about which factors explain this trend.

When Group 2 metals react to form oxides or hydroxides, metal ions are formed.So this must be explained by the atomization energy of the metal and also the first and second ionization energies?(Ionization energies fall down the group, it gets easier to form the ions, the reactions will happen more quickly.)

But apart from these two factors, should we also take into consideration the atomic radius, the hydration energy, the ionic radius or something else in order to fully understand why the reactivity towards water increases as we go down the group?  Also,  do the standard potential value has to do with it?

Do not forget solubility effects involving the hydroxide product.  Magnesium is plenty electropositive enough to decompose water, only to passivate with a hydroxide that has limited solubility (you can detect the presence of some reaction by decanting the liquid and then adding phenolphthalein, or by carefully looking for a few hydrogen bubbles during the reaction).  Put magnesium in methanol, which is less protic than water; the methoxide is less passivating and you can actually bring the alcohol to a boil!  See https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3DNMfs3e9OdZQ&ved=0ahUKEwjlnLOxpPDWAhVDKiYKHVFwB5wQwqsBCCgwAA&usg=AOvVaw2TaAZjVq0m8EHL-JluhxGn.


Why is the composition of a liquified vapor the same as that of the vapor? In Organic Chemistry Lab Techniques (Nichols), I found this diagram.

I understand that when going from Point $a$ to Point $b$, the composition becomes richer in the more volatile component (component A).

However, why does condensation of the vapor at Point $b$ not change the composition? Shouldn't the condensate of a vapor be richer in the less volatile component (the reverse of vaporization), by the principle of microscopic reversibility?

There is a misconception in using this diagram. You cannot go from Point a to Point b This diagram should not be read as you do. It should be done this way.

Suppose you start from a liquid made of 25% A + 75% B. It is a mixture. Its representative point on the diagram is on the edge of the rectangle between "Composition" and "100% B". When heating this mixture, the corresponding point moves upwards, but stays liquid. When reaching the curve "Liquid", the first bubble of gas is formed. We are at the point a (red) But this bubble has another composition. The composition of the gas corresponds to b (in red). So it is practically pure A. This is the beginning of the vaporization phenomena.

If you continue heating the liquid, more gas will be produced. But, as A is more eliminated than B by vaporization, the remaining liquid looses some A and practically no B. So in percentage, the liquid will contain more B. Its composition will soon be 76% B and 24% A, then 77% B and 23 % A, etc. But the temperature has to be elevated to allow this richer mixture to vaporize. The representative point of the liquid will move but stay on the sloping curve. After some time, the point of the liquid will be just above the capital L from the word "Liquid" At this moment the gas is mainly A but it starts containing 1 or 2% B. The composition of the gas could be determined by drawing a horizontal line from the point above the letter "L". This horizontal line will cut the curve "Gas" at an abscissa equal to about 1% or 2%. This defines the composition of the gas.

If you continue heating, the liquid will contain more and more B. Its point will stay on the curve "Liquid". The temperature must be higher and higher. The amount of liquid decreases. Simultaneously the composition of the gas which is eliminated contains more and more B. Its composition is always determined by drawing a horizontal line from the liquid to the curve on the left-side of the diagram.

The last drop of liquid disappears when the liquid reaches the Point "bp B".

Hopefully you have understood that passing from point a (in red) to point b (in red) has no meaning. a and b are always present : one describes the composition of the liquid, and one the composition of the gas.


Which is more reactive, chlorine radical or chloride ion? Ions are very reactive for obvious reasons (i.e. Coulomb force); $\ce{Cl-}$ ions will be very quick to form an ionic bond with positive ions (unless inhibited, e.g. through being in a solution; let's consider both $\ce{Cl}$ and $\ce{Cl^-}$ to be in gaseous phase).

Chlorine atoms without reduction ($\ce{Cl}$), aren't charged, yet tend to combine into $\ce{Cl2}$, utilising a covalent bond.

Thus, both $\ce{Cl}$ and $\ce{Cl-}$ tend to form bonds. My confusion lies in the following:

Noble gases are chemically inert, $\ce{Cl-}$ has the electron configuration of $\ce{Ar}$, yet it's very chemically reactive.
Neutral atoms aren't as attracted to each other as ions, and could be considered more "chemically stable" than ions with higher charge. However, even though $\ce{Cl}$ has no charge, it still seems eager to react and get to the electron configuration of a noble gas.

In both cases, information seems conflicting, and perhaps a more clear explanation will make me understand the situation. Is one more prone to react than the other?

(N.B.: My knowledge on chemistry, maths, physics etc … stretches to (but doesn't include) quantum physics, so feel free to go all-out.)

I think we can also approach this from an energetics perspective. Reactivity can also be seen to be related to thermodynamic stability in the following manner: The more thermodynamically stable a chemical species is, the less reactive it is. With this in mind, we can look at the enthalpies of formation of $\ce {Cl.}$ and $\ce {Cl^-}$.

Your question is slightly vague as you did not state the specified state of the substance. This can critically affect the answer. The reactivity of a heavily solvated $\ce {Cl^-}$ ion is certainly different from that of a free $\ce {Cl^-}$ ion in the gaseous phase.

By definition, the enthalpy change of formation, $\Delta H_f$, of an element in its standard state (e.g. $\ce {Cl2 (g)}$) is $\ce {0}$. $\ce {2 Cl.}$ are produced from the homolytic cleavage of the $\ce {Cl-Cl}$ bond in $\ce {Cl2}$. Knowing the bond dissociation energy of this bond to be $\ce {+242 kJ/mol}$, we can approximate $\Delta H_f$ of $\ce {Cl.}$ to be $\ce {+121 kJ/mol}$.

Similarly, the gaseous chloride ion can be produced from the gain of an electron by $\ce {Cl.}$. The first electron affinity, $\ce {1st E_a}$, of chlorine is the energy change when one mole of electrons is acquired by one mole of these gaseous chlorine atoms to give one mole of singly, negatively-charged gaseous $\ce {Cl-}$ ions. $\Delta H_f$ of $\ce {Cl- (g)}$ can thus be approximated as $\Delta H_f$ of $\ce {Cl.}$ + $\ce {1st E_a}$ of chlorine . Since this $\ce {1st E_a}$ is $\ce {-349 kJ/mol}$, this enthalpy change of formation for $\ce {Cl^- (g)}$ can be approximated as $\ce {-228 kJ/mol}$.

As can be seen from energetics data, $\ce {Cl^-}$ is more stable than $\ce {Cl.}$ in the gaseous phase. Thus, we may say that the $\ce {Cl.}$ is more reactive in the gaseous phase. Furthermore, being charged, $\ce {Cl^-}$ is solvated by water molecules in aqueous solution, further increasing its stability as it has ion-dipole interactions with the neighbouring water molecules. Electrically-neutral chlorine radicals would not experience this solvation effect, thus they would also be more reactive in the aqueous phase. In non-polar solvents, $\ce {Cl^-}$ and $\ce {Cl.}$ would have similar reactivity as in the gaseous phase since they do not experience starkly different solvation effects. Overall, based on energetics, we may say that $\ce {Cl.}$ is generally more reactive than $\ce {Cl.}$.

Additionally, it is important to point out the difference in the type of reactivity of the two different chemical species. Being negatively-charged, chloride ions participate in ionic reactions, involving charged species. These reactions include nucleophilic substitutions, as well as nucleophilic additions. On the other hand, chlorine radicals mostly participate in non-polar reactions, involving mostly radical species, such as free radical substitutions.


What is the difference between an ionic compound and a salt? What is the difference between an ionic compound and a salt? They both have a cation and an anion bonded.

On one of my homework sheets, it says that $\ce{NaCl}$ is an ionic compound (which it obviously is). How would i determine here if $\ce{NaCl}$ is an ionic compound or a salt?

The GCSE answer is.
An ionic compound is one in which the bonding between the atoms is ionic, for example sodium chloride is an ionic compound. It is formed from sodium cations and chloride anions.
There are substances which can be regarded as being ionic which are not normally viewed as salts. For example some metal oxides such as potassium oxide are probably best described as being ionic solids but in this case the compound is formed from oxygen and potassium.
A salt is the product of reacting an acid and a base together. Many salts are ionic. For example if you were to react hydrochloric acid with zinc and then to dehydrate the product (avoiding hydrolysis) you would end up with zinc chloride.
The PhD chemists view of the matter
One great problem in chemistry is that there are very few perfectly ionic compounds. I hold the view that the more chemistry you know the more one becomes aware of the fact that simple explanations do not work well.
The great problem in my mind is that a great spectrum of bonds exist from perfectly ionic to simple unpolarised covalent bonds. All cations have some ability to pull electron density towards themselves. A simple anion such as a chloride should in the idea world be a sphere of electron density.
The electrostatic attraction of the electron density by a cation on an anion pulls electron density towards the cation. This can be regarded as leading to some covalent character of the bond existing. Thus as a result I hold the view that there are very few true perfectly ionic compounds.
One example would be tetraphenylphosphonium tetraphenylborate which has two very large ions, the interaction between these ions is close to a perfect ionic attraction. Here is a diagram of the tetraphenyl phosphonium cation.

We can see here that the ion has the positive charge deeply buried inside it, here is the solvent accessible surface, I have rendered it in colours to represent the charges at points on the solvent accessible surface.

The atom locations can be seen in the following picture to help you understand what is happening.

Even the tetramethyl phosphonium cation is able to isolate the positive charge away from the solvent accessible surface, in contrast the trimethyl sulfonium cation (Me3S+) presents to the outside world (the solvent accessible surface) part of the ion which is quite positive.

We can also take a molecular orbital view of bonding, a lot of coordination chemistry of transition metals is considered with the ligand field model. The ligand field model is based on the crystal field model which treats all bonding as ionic. However the same results can be obtained for d block metals using molecular orbital theory.
It is also interesting to note for the lanthanides that they do not have strong ligand field splitting effects, this is due to the fact that their f orbitals are close to the centre of the atom and do not interact well with ligands. In many compounds often the bonding of lanthanides is viewed as being ionic.
If we consider for a moment palladium chloride then it is clear that the attraction of the chlorides to the palladium cations results in the formation of covalent bonds. If we move back from palladium to a metal which is less able to form bonds to chloride (cobalt).
Then if we have anhydrous cobalt chloride, we have in the anhydrous solid a Co-Cl distance of 2.453 Å. If we dissolve cobalt chloride in water then the water molecules will react with the cobalt(II) centres to form hexaaqua complexes of cobalt [Co(H2O)6]2+ which have a larger radius than cobalt(II) cations. These are less able to electrostatically attract chloride anions and the bonding of the chloride anions to the [Co(H2O)6]2+ cations will become closer to being a simple electrostatic (ionic) attraction / bonding than the bonding between the chlorides and the cobalt in the anhydrous cobalt chloride.
Here is a picture of the crystal structure of anhydrous cobalt chloride. (H. Grime and J.A. Santos, Zeitschrift Fur Kristallographie, 1934, 88, 136-141) You can see how the cobalt has a octahedral coordination environment in which six chloride anions form bonds to the cobalt. Keep in mind that the chloride anion is a ligand with a weaker ligand field than water. As a result the cobalt adsorbs red light in the form of anhydrous cobalt chloride.

By the way the solid is a layered solid, the solid can be thought of as being like a series of slices of bread (chlorines) with the cobalt as the jam.

By the way the distance between cobalt and chloride in the [CoCl4]2- complex is 2.276 Å, which is a bit shorter than the distance in the anhydrous solid.


A steel sample was analysed for nickel A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$. The absorbance of this solution measured by atomic absorption was found to be $0.410$. A standard $\pu{6 ppm}$ solution of $\ce{Ni}$ gave an absorbance of $0.522$ under the same conditions. Assuming linearity between absorbance and concentration, calculate the $\% \ce{Ni}$ in the steel sample. (8 marks)

Seems so simple but is confusing me.
A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$.The absorbance of this solution measured by atomic absorption was found to be $0.410$.
A standard 6 ppm solution of Ni gave an absorbance of $0.522$ under the same conditions.
In the final diluted sample,
$$\frac{\pu{6 ppm}\ \ce{Ni}}{0.410\times0.522}=\pu{28.03476311 ppm}\ \ce{Ni} \tag{1 }$$
In the first solution
$$\frac{\pu{28.03476311 ppm}\ \ce{Ni}}{\pu{500 ml}}=\pu{0.05606952622 mg}\ \ce{Ni} \tag{2}$$
Supposing the density of the first solution to be near that of water:
$$\pu{1000 g}\times\pu{0.05606952622\times10^{-6} g}=\pu{5.60695\times10^{-5} g} \tag{3}$$
In the steel sample:
$$\frac{\pu{5.60695\times10^{-5} g}}{\pu{0.35 g}}=0.000160198=0.016\ \%\ \ce{Ni}\tag{4}$$
$2$nd attempt:
\begin{align}
\frac{\pu{6 ppm}}{0.522}&=\frac x{0.410}\\
\implies \frac{\pu{6 ppm}\times0.410}{0.522}& =x\\
x&=\pu{4.7126ppm}\ \ce{Ni} \tag{6}
\end{align}
This means you have $\pu{4.7126 mg}$ nickel per $\pu{1 kg}$ of steel (ppm is one millionth, and thus, mg per kg). Then one tenth of that amount of steel $(\pu{0.1 kg}/\pu{1 kg})$ contains $1/10$ of the $\ce{Ni}$. that is $\pu{0.47126 mg}$
Since it was marked up to $\pu{500 ml}$
In the steel sample:
\begin{align}
\pu{0.47126 mg}\times\frac{500}{50}&=\pu{4.7126 mg ml-1}\\
\frac{4.7126}{100}&=0.047126\\
\frac{\pu{0.047126 g}}{\pu{0.35 g}}&=0.134645\\
&=13.46\ \%\ \ce{Ni} \tag{7}
\end{align}
I feel that I am missing a step in calculations. I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my second attempt of finding a correct answer.
This question is from a past paper.

Your first step is incorrect. Think about it this way:
$5$ people have $10$ eyes. This group of people has $14$ eyes. How many people in the group?
The correct solution is:
This group has $14/10 = 1.4$ times more eyes than the known group. Thus, they have $1.4$ times more people. $5$ people $\cdot$($14$ eyes/$10$ eyes) $ = 5$ people $ \times 1.4$ (unitless ratio) $= 7$ people. Here, the units make sense. This is correct.
In your calculation you do ($5$ people)/($10$ eyes $\times 14$ eyes)$= 1$ people/($28$ eyes$^2$). This unit people/eyes$^2$ makes no sense. This is incorrect.
First you find the amount of $\ce{Ni}$ in the final sample.
Your sample absorbs $0.410$ which is slightly less than $0.522$ ($\pu{6 ppm }$sample which). So your sample has $(0.410/0.522) \times \pu{6 ppm} = \pu{4.713 ppm}.$
Water solution has density of $\pu{1 kg/L}$, so $\pu{500 mL} \to \pu{500 g}$. $\pu{4.713 ppm}$ in $\frac{\pu{500 g}}{1\,000\,000} = \pu{0.5 mg}$. $\pu{0.5 mg} \cdot 4.713 = \pu{2.356 mg}$.
The total weight of the sample was $\pu{0.35 g} = \pu{350 mg}$
$$100\ \% \cdot \frac{\pu{2.356 mg}}{\pu{350mg}} = 1.003\ \%$$
You should round it down to $2$ digits, because that was the accuracy of other data. The answer is  $1.00\ \%$.


Can lattice enthalpy only be determined by the Born-Haber cycle? Since we are able to determine the enthapy of formation experimentally, why it is impossible to determine the lattice enthalpy directly by experiment?

Because you can't take an ionic solid and convert it into a gas of ions. That's not a physically realistic process. You have to infer it using a Born-Haber cycle and other quantities that can be determined experimentally.


how is there no centre of symmetry in this case? the question was whether the following compounds have the center of symmetry or not .the question was :

And the answer is that there is no center of symmetry .how? according to me, there should be the centre of symmetry. I have made it like this:

on asking my teacher how there is no center of symmetry he told me that since it is a Fischer projection, therefore, the horizontal lines are above the planes. therefore there will be no center of symmetry. I understand that if the horizontal lines are out of the plane then there will be no COS. but what I don't understand how come the horizontal lines are above the plane/out of the plane. please explain.

One way to quickly get a visualization without building a 3D model (having a model is more instructive, though):

Go to https://chemapps.stolaf.edu/jmol/jsmol/simple.htm
Click on Load by name and type "(2S,3S)-Butane-2,3-diol"
Click on console and type "rotate branch @2 @4 180" to get a conformation that is easy to compare with the Fisher projection.


Why highly unstable cyclic boron compounds should be stabilised by two electron reduction? I am trying to figure out why highly unstable cyclic boron compounds  could be stabilised by two electron reduction...

My thoughts are because each Boron has an empty p-orbital perpendicular to plane of cyclic compound, pi-orbitals can form like diagrams below.

The two electrons would then go into the pi-bonding orbital, thus stabilising the entire compound.

Is this the correct explanation? If so, how can vibrational spectroscopy be useful in examining the bonding in the dianion?

Your energy diagram is that of a linear moiety, like an allyl cation. For a cyclic structure, it would be somewhat different, but the general idea remains the same: there is one $\pi$-bonding orbital, and when two electrons go to it, everyone wins.

As for vibrational spectroscopy, well, it can help you tell the cyclic structure from linear, if that's what you want.


Will there be a temp change in free expansion I did a little thought experiment, where I place an infinitely long beaker filled with some gas, which has a moveable piston at the open part. So if I place this beaker in vaccume, there will obviously be no work done on the system and hence no loss of energy due to the work done by the system (because the int pressure of the energy will obviously push the piston upwards).
So here's my main question:
If in the Beginning, let's say we have a certain amount of gas placed in the beaker such that the piston is just 1 cm above the base. But after infinite time the lid will probably be at infinity. So if a t=0, the gas occupies a limited space, won't the collisions be high and hence kinetic energy be high, which would mean the temperature of the system is high? But at t= infinity, the gas would occupy an infinite amount of volume which would mean that there are less collisions, so wouldn't this reduce the velocity of the molecules, hence reduce the kinetic energy, hence reduce the temperature?
I feel something is wrong over here because mathematically, I know there shouldn't be any change in int energy, therefore the temperature. So how do we justify this experiment?

In a real gas, the internal energy is the sum of the kinetic energy of the molecules and their potential energy of interaction. An ideal gas is the limit of real gas behavior in the limit of large molar volumes in which the potential energy of interaction of the molecules approaches zero. The kinetic energy of the molecules depends only on temperature, but the potential energy of interaction of the molecules depends on both temperature and specific volume. The temperature of an ideal gas does not change when subjected to free expansion, but it does change for a real gas.  So the deviation from ideal gas behavior caused by potential energy of interaction of the molecules is what causes the temperature of a real gas to change in free expansion.


Effect of incorrect volume reading on calculated density value When measuring the density using the displacement method, a student read the initial volume of the water too high, but the final volume reading was correct.
  This error would cause the calculated density to be too high or too low?  Explain why.

Wouldn't it be too low because the initial volume is too high?

It's asking for the calculated density, not the calculated volume. If the volume, as measured by the student, is too small, then the density (= mass / volume) will be too large. Of course, that's assuming that the mass was measured correctly.


Why is hydrogen bonding more significant than any other interaction between dipoles? Solvents can be categorized as polar and protic. I understand that to be protic there had to be a significantly polar bond involving a hydrogen atom. However, how is the a hydrogen bond different in any way to another dipole-dipole interaction and therefore, what's the need for the distinction between polar and protic solvents?

When considering the hydration enthalpy $\triangle_{hyd} G^{\theta}$ it is not sufficient to simply use the Born equation:
$$\triangle_{solv} G^{\theta}=-\frac{N_Ae^2z^2}{8 \pi \epsilon_0}(1-\frac{1}{\epsilon_r})$$
BECAUSE this only considers Coulombic interactions (i.e hydrogen bonds are not solely coulombic interactions). My textbook seems to suggest that hydrogen bonds are somehow different and more significant. My question is why? Surely there should be no distinction between different polar bonds; what makes $\alpha^{\delta-}-H^{\delta +}$ special?

Hydrogen bonds often have a significant covalent component. This can be seen in many aspects of the hydrogen bond: it is directional, unlike Coulombic interactions, and the bond length is generally significantly shorter than predicted by Coulombic models. There is great variation in the covalent character of hydrogen bonds depending on their environment, with e.g. bond lengths between 2.2 and 4.0 Angstroms known.

Other intermolecular interactions have significant covalent components; there is nothing special about hydrogen per se but it is the most important example.


Why do post-Hartree-Fock methods fail to predict the direction of the dipole moment of carbon monoxide? In carbon monoxide the dipole moment (negative to positive) points towards the oxygen, as I explained it in How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?

A calculation using density functional approximation, the level of theory is BP86/def2-QZVPP, yields the correct direction:
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.19~\mathrm{D} &&
\text{Direction:}~\longrightarrow
\end{align}

 ---------------------------------------------------------------------
 Center     Atomic      Atomic             Coordinates (Angstroms)
 Number     Number       Type             X           Y           Z
 ---------------------------------------------------------------------
      1          6           0        0.000000    0.000000   -0.648987
      2          8           0        0.000000    0.000000    0.486740
 ---------------------------------------------------------------------
[...]
 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=              0.1907  Tot=              0.1907

Using Møller-Plesset Perturbation Theory of second order, i.e. MP2/def2-QZVPP, we obtain
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.30~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}

 ---------------------------------------------------------------------
 Center     Atomic      Atomic             Coordinates (Angstroms)
 Number     Number       Type             X           Y           Z
 ---------------------------------------------------------------------
      1          6           0        0.000000    0.000000   -0.648450
      2          8           0        0.000000    0.000000    0.486337
 ---------------------------------------------------------------------
[...]
 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=             -0.3002  Tot=              0.3002

There is no improvement going to other post-Hartree_Fock methods, as the following table indicates. A positive value means that the dipole is directed towards the carbon; the used basis set is in all cases def2-QZVPP.
\begin{array}{lr}
\text{Method} & \mathbf{q}(\overrightarrow{\ce{CO}})\\\hline
\text{Experimental}^1 & 0.11\\\hline
\text{HF} & -0.14\\\hline
\text{MP2}  &   -0.30\\
\text{MP3}  &   -0.22\\
\text{MP4(SDQ)} & -0.27\\
\text{MP4(SDTQ)//MP4(SDQ)} &  -0.27\\
\text{CCSD} &   -0.25\\
\text{CCSD(T)//CCSD} & -0.25\\
\text{CISD} & -0.22\\
\text{QCISD} & -0.26\\
\hline
\text{BP86} &    0.19\\
\text{PBE0} &    0.12\\
\text{M11}  &    0.06\\
\text{B3LYP} &   0.10\\
\hline
\text{B2PLYP} & -0.07\\
\hline
\end{array}

This failure probably already starts with Hartree-Fock. Where does the deficiency lie in the HF description in this particular case? And why is this deficiency not corrected by post-HF methods? Even the double hybrid functional B2PLYP cannot predict the direction of the dipole moment correctly.

According to Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. the value can be found in J. S. Muenter. J. Mol. Spectrosc. 1975, 55, 490. (I don't have access to that publication.)

Sample Input for Gaussian 09

#p MP2/def2QZVPP
scf(tight) opt(verytight,maxcycle=100)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

carbon monoxide

0 1
C   0.0 0.0 1.135
O   0.0 0.0 0.0

The question itself is void, as all methods with the exception of Hartree-Fock predict the direction of the dipole moment correctly. Deathbreath found a Full CI calculation in Jeremy P. Coe, Daniel J. Taylor, and Martin J. Paterson. J. Comp. Chem. 2013, 34 (13), 1083-1093 of which luckily an open version exists (arXiv:1301.4904 [physics.chem-ph]). The following passage is taken from the latter:

  The groundstate dipole moment of carbon monoxide, although fairly small, when calculated using HF strikingly has the incorrect sign compared with experimental results. Previous work has suggested that the accuracy of the dipole calculation
  depends on the amount of correlation accounted for.14  The bond length (2.1316 Bohr) and the experimental dipole value (0.122 Debye) are taken from Ref. 15. The positive value for the dipole here signifies a polarity of $\ce{C^−O^+}$.
  With a cc-pVDZ basis, two frozen core orbitals and a cut-off value of $c_\mathrm{min} = 5 \times 10^{−3}$, we see in Fig. 1 [omitted] that the MCCI method, starting from close to the incorrectly signed result of the HF single SD, quickly reaches a correctly signed value which converges at around half of the FCI value.

  14 G. E. Scuseria, M. D. Miller, F. Jensen, and J. Geertsen. J.
  Chem. Phys. 1991, 94, 6660.
  15 J. S. Muenter. J. Mol. Spectrosc. 1975, 55, 490.

The problem indeed is just a technical one, resulting from analysing the wrong density. Gaussian 09 by default only analyses the SCF density (see population, density). In order to obtain values for the post-HF methods, the density=current keyword needs to be specified. The updated table therefore becomes:
\begin{array}{lr}
\text{Method} &
\mathbf{q}(\overrightarrow{\ce{CO}})/\mathrm{D}&
\mathbf{d}(\ce{CO})/\mathrm{10^{-10}m}\\\hline
\text{Experimental}^{15} & 0.122 & 1.128\\\hline
\text{HF} & -0.14 & 1.102\\\hline
\text{MP2}  &   0.26 & 1.135\\
\text{MP3}  &   0.08 & 1.118\\
\text{MP4(SDQ)} & 0.11 & 1.118\\
\text{MP4(SDTQ)//MP4(SDQ)} & \text{n.a.} \\
\text{CCSD} &   0.08 & 1.124\\
\text{CCSD(T)//CCSD} & \text{n.a.}\\
\text{CISD} & 0.05 & 1.118\\
\text{QCISD} & 0.08 & 1.127\\
\hline
\text{BP86} &    0.19 & 1.135\\
\text{PBE0} &    0.12 & 1.122\\
\text{M11}  &    0.06 & 1.119\\
\text{B3LYP} &   0.10 & 1.124\\
\hline
\text{B2PLYP} & 0.15 & 1.129\\
\hline
\end{array}

Sample Input for Gaussian 09 for analysis
(requires the addition of %chk=mp2qzvpp.chk to the sample input of the question)

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
Geom=check
Guess=(Read,Only)
Density=(Check, Current)

dens-curr

0 1

I apologise for getting everyone worked up over a stupid beginner mistake.


Why Inorganic chemistry reaction mechanisms are hard to find? My textbook Chemistry Part 1,as in the attached photo states that[after the KClO3 reaction]
“This reaction which apparently seems to be of the tenth order is actually a second order reaction. This shows that it takes place in several steps.” If that is so, then as stated above it, all higher order inorganic chemistry reactions must have some sort of mechanism. But it is really hard to find the mechanisms(many a times one cannot find them). Why is that so??.

Reactions can be of order $0, 1 , 2$ or $3$ , but never more. No collision may happen between more than three particules simultaneously. If an equation is written with more than $4$ substances on the lefthand side, it means that the corresponding reaction is a sum of several elementary reactions.
Mechanisms in inorganic reactions are hard to establish because they are made of a lot of successive reactions, in cascade. And each step is so quick that it cannot be studied. For example your oxidation of $\ce{Fe^{2+}}$ ion by $\ce{KClO3}$ is probably a series of at least $6$ steps which are all nearly instantaneous : To be able to study such a mechanism, the $6$ intermediate states must be observed for at least some nanoseconds. If the intermediate states are too short living, all we can do then is to make hypothesis without experimental arguments.


Liquid water below freezing temperature I got in my car this morning and the temperature read 13 F (that's around -10.5 C.) While driving down the road, I noticed a bottle of water in my cup holder was frozen solid as expected.

Another bottle of water (cheap, thin-walled, plastic disposable water bottle) contained liquid water! I shook the bottle, not a single crystal of ice!

Shortly thereafter, I noticed the water became slushy, so it had begun freezing quite quickly!

The interior of the bottle was hardly perfectly smooth given the quality of the bottle and the water was half drunk so it was probably not super pure.

What caused this? Was this supercooling?

You may have just experienced supercooling.

Ordinary crystallization upon reaching the freezing point of a liquid requires nucleation sites - places for crystals to form. In absence of nucleation sites, water can be cooled down to lower temperatures until crystal homogeneous nucleation occurs. In water, this can occur as low as $224.8\ \mathrm{K\ (-48.3\ ^\circ C,\ -55\ ^\circ F)}$.

Nucleation can also be induced by disrupting the system, which happened when you shook the bottle. Even so, the fact that it only became slush may attest to the presence of impurities. A $\mathrm{19\ ^\circ F,\ 10\ ^\circ C}$ freezing point depression is not unreasonable depending on the amount of saliva that might have ended up back in the bottle.


Does evaporation occur while boiling as well? If evaporation were to occur when a liquid was being boiled, then the amount of liquid changing phase would also depend on the surface area of the liquid exposed to the atmosphere (as evaporation rate increases with surface area). However I have read that the surface area doesn't affect the amount of liquid changing phase while the liquid is boiling.

So does evaporation not occur while the liquid is boiling?

Evaporation occurs while a random molecule from the bulk collides with the surface molecules and though the upper molecules are less tightly bounded, they gain the energy due to collision and escape from the liquid surface as vapours. But at low temperature this collision frequency is very less. In boiling, we provide heat to the liquid, due to which the collision frequency increases and upper surface molecules more fastly escape. So, boiling includes evaporation. Now for boiling, surface area doesn't matter as the heat which we externally supplied plays a dominant role compared to the surface area of liquid exposed to atmosphere in boiling. The surface area plays negligible role in boiling.


Why can mercury(I) exist, but not zinc(I)? Mercury shows variable valency while zinc does not.
Its electronic configuration is $\ce{[Xe]\:4f^14 5d^10 6s^2}$. So it can donate the $\ce{6s^2}$ electrons and should only be able to form $\ce{Hg^2+}$, right?

How can it show $+1$ oxidation state?
Why does not $\ce{Zn}$ show $+1$ oxidation state?

On the contrary, zinc(I) compounds do exist, though they are rare, and relatively unstable. Most zinc(I) compounds contain a $\ce{[Zn2]^{2+}}$ core, which is analogous to the $\ce{[Hg2]^{2+}}$ cation. The $\ce{[Zn2]^{2+}}$ ion does, however, rapidly disproportionate into zinc metal and zinc(II), and has only ever been obtained by cooling a solution of metallic zinc in molten $\ce{ZnCl2}$ and in a few compounds such as decamethyldizincocene.
This should show you that the chemistry of zinc and mercury really aren't all that different. The $\ce{[Hg2]^{2+}}$ ion's much greater stability relative to $\ce{[Zn2]^{2+}}$  is due to the atypically large ionization enthalpy of the $\ce{Hg}$ atom ($100.7\:\mathrm{kJ\:mol^{-1}}$ greater than $\ce{Zn}$), which in turn due to relativistic stabilization of the $\mathrm{6s}$ orbital. Relativistic effects account for a myriad of the properties that are characteristic of heavy metals, including why mercury is the only metal that is a liquid at room temperature, and even gold's color.


Molecular Structure of Caffeine I'm currently looking at a problem that asks about the structural relationship between the 4 nitrogens the caffeine molecule. I am asked to identify whether they are in perpendicular planar, coplanar, or inverted planar formations.

I'm am a bit lost as to how I should approach this problem. Should I look at hybridization (3 Ns are sp3, 1 N is sp2)? Or resonance (seems that all Ns are involved in resonance)? Or should I be concerned with sterics?

Lastly, what are some examples that might assume a perpendicular planar, coplanar, and inverted planar formation? I don't think I've ever seen such shapes before...

Would someone please point me to the right direction?

At some point in studying organic chemistry, you might have been taught that a nitrogen in an amide group is planar sp2, due to delocalization of its electrons into the carbonyl group. You also might have learned that nitrogen atoms in a pyrrole or imidazole ring are planar sp2 since they participate in the 6 electron aromatic pi bonding system. You would then conclude that all the nitrogen atoms in caffeine are coplanar. But without that essential background information, you might not have been expected to figure that out.


Reactions of ketones and aldehydes Our topic was about ketones and aldehydes (carbonyl group) including hemiketals and hemiacetals and he asked us to do a homework. I cannot find an answer to this one.Please explain me the reaction and the product formed.

PS: The figure at bottom is my best effort to do this question.

But.....6 member lactones are super stable so I wouldn't be surprised if only 1 ring opened/ some recyclized


Dipole moment - calculation of percentage ionic character Question: The dipole moment of $\ce{HBr}$ is $2.60 \times 10^{-30}$ and the interatomic spacing is $1.41$. What is the percentage ionic character of $\ce{HBr}$?

What I know is that the percentage ionic character is observed dipole moment divided by calculated dipole moment $\times\ 100$.

The calculated dipole moment is $\textrm{charge on electron} \times \textrm{radius of molecule}$.

However, how would I convert interatomic spacing into radius?

The calculated dipole moment is charge on electron * radius of
  molecule

That's off a bit, the calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius.
\begin{aligned}
\mu_{100\%} &= \mathrm{electron ~charge \times interatomic ~spacing}\\
\mu_{100\%} &= (1.6 \times 10^{-19})\ \mathrm C \times (1.41 \times 10^{-10})\ \mathrm m\\
\mu_{100\%} &= 2.26 \times 10^{-29}\ \mathrm{C\ m}\\
\end{aligned}

As you point out,
$$\mathrm{\% ~ionic~ character} = \frac{[\mu_\mathrm{obs}]}{[\mu_{100\ \%}]} = \frac{[2.6 \times 10^{-30}]}{[2.26 \times 10^{-29}]} = 0.12 = 12\ \%$$


How does adding sodium chloride to a solution of copper sulfate change the equilibrium? In school we had a lesson about chemical equilibrium. We performed an experiment: in two tubes we poured $\pu{5 mL}$ of $\ce{H2O}$ and placed $\ce{CuSO4}$, the concentration of the $\ce{CuSO4}$ should have been $\pu{0.5 mol/L}$. We homogenize the two solutions in the two tubes and both solutions I think became blue (might have been green, not sure). Then we were asked to put a gram of $\ce{NaCl}$ in one of the tubes and homogenize. The color of the solution changed (also we might have added another $\pu{5mL}$ to the first solution, the one without $\ce{NaCl}$). Can someone tell me what do they think was the purpose of this lesson and how does it correlate to chemical equilibrium?

When you mix $\ce{CuSO4}$ and $\ce{NaCl}$ solutions, the following equilibrium forms:
\begin{align}
  \ce{[Cu(H2O)6]^2+ + 4Cl^- &<=> [CuCl4]^2- + 6H2O},&
  \Delta Η &= \text{positive}\\
\end{align}

Upon heating the mixture, the equilibrium shifts forward (following Le Chateliers principle) to produce more of the yellowish coloured $\ce{[CuCl4]^2-}$ ions, also known as tetrachlorocuprate (II) ion. As the amount of yellow colour in the mixture increases the solution becomes greener due to subtractive mixing of colours, i.e. blue and yellow solutions mix to give a green coloured solution.

The experiment can be found on YouTube, for example: Chemical Equilibrium ($\ce{CuSO4 + NaCl}$)


in the electrolysis of H₂SO₄ solution, when the concentration is increased why does the current increase even when OH⁻ ions reduce? During the electrolysis of $\ce{H2SO4}$, $\ce {OH-}$ ions are more reactive than $\ce {SO4^{2-} }$ ions and so they dissociate. as $\ce {OH-}$ ions dissociate, they are the negative ions that carry current. So when the concentration of $\ce {H2SO4}$ is increased, the concentration of $\ce {OH-}$ ions actually reduces, but then the current still increases.

Near room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid.  For pure water or pure sulfuric acid the conductivity is orders of magnitude lower.

In pure water, there are only $10^{-7}$M of $\ce{H+}$ and $\ce{OH-}$ to carry current.

When a relatively small amount of sulfuric acid  is added, it dissociates fully to $\ce{H+}$ and $\ce{HSO4-}$.  For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$.


Protonation of ketone and ester If there is a ketone and ester functional group in the same compound and I want to protonate with $\ce{HCl}$, which one will protonate first? Is it the carbonyl group in the ketone or the carbonyl group in the ester and why? Is it related to the $\mathrm{p}K_\mathrm{a}$ of the functional group?

As was mentioned in the comments, with reference to the Myers $\mathrm{p}K_{\mathrm{a}}$ table, the $\mathrm{p}K_{\mathrm{a}}$ values of a protonated ester and a protonated ketone are largely similar (in the region of $-7$ to $-6$).

Therefore, it will likely depend on external factors, which group is protonated first. Are there electron withdrawing or electron donating groups in the vicinity of either? Can the proton on either maybe intramolecularly hydrogen bond to something?

All these effects will also slightly modify the $\mathrm{p}K_{\mathrm{a}}$ of the functional group you’re observing. The one with the higher $\mathrm{p}K_{\mathrm{a}}$ will be protonated first. Thus, it is impossible to give a general answer here, it depends on which molecule you have and which groups are where.

Side fact: $\ce{HCl}$ has almost the same $\mathrm{p}K_{\mathrm{a}}$, so the protonation will be very close to equilibrated between all three components.


Is it realistic to have a natural oxygen source in a cave? Is it realistic to have a natural oxygen source in a cave? I am writing a story, about people who live underground in a land that is freezing outside. Heat is taken care of by geothermally heated groundwater. How do I make it realistic for these people to breath in the caves?

According to this page, the CO2 buildup in the cave might also be a problem.

But, if there was a river that ran on the surface, then oxygen and nitrogen would dissolve into the water. The stream could then pass into the cave and release some of the dissolved oxygen into the oxygen deficient environment in the cave. It would also remove the CO2. Of course, if the surface of the planet is frozen, then the river would have to avoid freezing somehow. But perhaps there is a hot spring on the surface.

My intuition is that the rate of oxygen dissolving out of the water would be very slow and it could be increased by increasing the surface area of the water. In the human lungs, there are many, many tiny branches (Alveoli) that allow the lungs to have a huge surface area so that the slow diffusion of oxygen from the air into our blood can go fast enough for us to survive. Perhaps the cave-people can accomplish something similar if they let their oxygen-rich stream pass over a tall, broad waterfall that would cause the river to break up into tiny droplets, thereby increasing the surface area. This has the added benefit of looking super cool in case your story gets made into a movie :)

Other options:
1) they could generate electricity and use that to make hydrogen and oxygen from water.
2) http://www.mpg.de/621120/pressRelease201003241?filter_order=L


Why does ozone have a dipole moment different than 0 First of all, I know that the reason for this is the bent structure of the molecule.
However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule.
But ozone only has bonds between two oxygens, so why is the dipole moment of each bond not equal to zero?

The electrons are not shared equally among the oxygen atoms.  The central oxygen atom alone donates three electrons to the covalent bonding whereas the other two oxygen atoms combined donate the other three (there is a 3-center, 4-electron pi bond in which two of the rlectrons are shared between just the end atoms).  Since the central oxygen atom is providing an excess of bonding electrons, it becomes positively charged.  The other oxygen atoms, contributing fewer electrons to the bonds and retaining more "lone pairs", take the compensating negative charge.
Trisulfur is a similarly polar elemental species, with a bent structure and unequal electron sharing similar to ozone.  Sulfur dioxide and disulfur monoxide, the latter found along with trisulfur on the Jovian moon Io, have a similar unequally-shared-electron structure with sulfur in the middle.  But we take little notice in the latter two cases because the sulfur, being less electronegative than oxygen, would be positively charged anyway.


Release of sulfur dioxide from odored fuel gases Often, fuel gases such as butane and natural gas (methane) are mixed with some foul-smelling impurities such as thiols (mercaptans) and hydrogen sulfide. Presumably, anything containing sulfur would burn to sulfur dioxide. The odor of sulfur dioxide is not easily noticed while burning such fuels, but I presume that the large amount of fuel gases burnt in the world would cause a significant amount of $\ce{SO2}$ emissions.

Why is this not usually brought up? Are the amounts actually so small that they are completely insignificant?

According to this North American Energy standards board publication , the total concentration of sulfur in natural gas, both natural and from oderant additives, must not exceed $\pu{70 ppbv}$. Concentrations are actually more commonly about a tenth of this. For example, from the same source, the Union Gas system typically contains about $\pu{5 ppbv}$ sulfur.  Furthermore, the sulfur content of coal is many orders of magnitude higher than that of oderant-enriched natural gas at about $\pu{0.5 to 5\%}$ sulfur, and the worldwide combustion of coal is much greater than that of natural gas (although much of the sulfur from coal combustion is now scrubbed out at the source of combustion). Other fossil fuels like diesel also contain significant amounts of sulfur (regulatory sulfur content of diesel fuel in the US and EU was 10 ppm as of 2009).

The global atmospheric background concentration of $\ce{SO2}$, according to
Owen, Lewis A.; Pickering, Kevin T (1997). An Introduction to Global Environmental Issues. Taylor & Francis. pp. 33–. ISBN 978-0-203-97400-1, is around $\pu{1 ppbv}$, meaning the sulfur content of the Union Gas product, for example, is only five times that that found throughout the Earth's atmosphere.

Summary - TL/DR:
Based on the background $\ce{SO2}$ concentration and the overwhelming sulfur emision source from the combustion of coal and other fossil fuels, the sulfur originating from natural gas oderant additives represents a negligible contribution to the global atmospheric $\ce{SO2}$ concentration.


Why is the cyclopropenium ion aromatic? I don't understand how the 'cyclopropenium' ion is aromatic. According to my understanding, the carbons within the central 3 carbon ring must each be sp2 hybridised with a single electron in a p orbital - since each carbon is bonded to 3 other atoms and has 4 valence electrons. How does that lead to aromaticity, since this doesn't fullfil Huckel's rule?

It does meet Huckel's rule.  When you have $m$ carbon atoms in a conjugated ring with a positive charge of $q$, there are $m-q$ pi electrons.  Here $m=3, q=1$ therefore the number of pi electrons is $m-q=2=4×0+2$.

A Huckel model calculation indicates that cyclopropenyl cation has roughly as much stabilization as benzene, but it's distributed over three instead of six carbon atoms.  Thus aromaticity is especially powerful in a cyclopropenyl cation ring.   This explains how a positive (formal) charge can be stabilized on as few as three carbon atoms and the ring holds together despite steric strain.


Is it possible to "dope" glass to change it's refractive index? For many purposes, like in the making of achromatic lens combinations, we need transparent media of different refractive indices. Let's assume I want a transparent medium of refractive index of 1.4, how would I go about getting such a material? Can I "dope" it by dissolve some impurities into molten glass mixture and then solidify it later? How would I choose my impurity to add? How do I calculate how much to add?

Glass doesn't crystallize at all, that's the point. Consequently, unlike crystalline materials, it doesn't have fixed composition. In a way, it can be said that glass consists of impurities. And yes, of course, you may change its composition this way or that, so as to tune the refractive index to your purposes.

Further reading: crown glass, flint glass.


Number of water molecules to contribute one molecule to every milliliter on earth How many millilitres of water would it take so that, when they are thoroughly dispersed around the world, it would yield an average of one molecule of the original sample of water in each millilitre of water in the world?  For the sake of argument, assume that the H and O atoms do not dissociate, and the water molecules remain intact.

I calculated it once and got something around 40 ml, but I forgot how I made that calculation.

A very quick Googling tells me that the volume of the oceans is 1.3 billion $\mathrm{km}^3 = 1.3\times10^9\;\mathrm{km}^3 = 1.3\times10^{18}\;\mathrm{m}^3 = 1.3\times10^{24}\;\mathrm{cm}^3 = 1.3\times10^{24}\;\mathrm{mL}$.

To get one molecule into each of those mL, we'd need $1.3\times10^{24}$ molecules of water, or $\frac{1.3\times10^{24}}{N_A} = \frac{1.3\times10^{24}}{6.02\times 10^{23}}$ moles of water, about 2.2 mols.  Since water is 18 g/mol, that comes out to 39 grams of water or about 39 mL.


What is the difference between ψ, |ψ|², radial probability, and radial distribution of electrons? I am very confused regarding these four terms- ψ,|ψ|², radial probability and radial distribution

I know that ψ is called the wave function but is it the same as radial probability?

I also know that |ψ|² is called the probability density of finding an electron inside an atom but is it the same as radial distribution—which is the probability of finding an electron in a volume $\mathrm dV$ inside a nucleus?

All the four terms are different and they represent different concepts in quantum mechanics.Firstly, the term $\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\Psi(x,y,z,t) = \sqrt{\frac{8}{abc}} \sin\left(\frac{n_x\pi x}{a}\right) \sin\left(\frac{n_y\pi y}{b}\right) \sin\left(\frac{n_z\pi z}{c}\right) e^{-2\pi iEt/h}$$ is an example of a wave-function.

But the $|\Psi|^2$ is mathematically defined as $\Psi\cdot\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\Psi(x,y,z,t)|^2 = \frac{8}{abc} \sin^2\left(\frac{n_x\pi x}{a}\right) \sin^2\left(\frac{n_y\pi y}{b}\right) \sin^2\left(\frac{n_z\pi z}{c}\right)$$ Here the complex part will not appear as in the previous example because $\Psi$ is multiplied with its complex conjugate.

The probability of finding the particle in a unit volume element $\mathrm dV$ is $|\Psi|^2 \mathrm dV$. In spherical polar coordinates, it is $|\Psi|^2 r^2 \, \mathrm dr \, \sin\theta \, \mathrm d\theta \, \mathrm d\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\pi$ as, $\int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta \, \mathrm{d}\theta \, \mathrm{d}\phi =4\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\Pr(r) = |\Psi|^2 4 \pi r^2 \, \mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.


What is the name of the diagram in this question? How do you read it? My teacher calls it an 'orbital diagram' but every time I search for information something completely different pops up. I assumed it was a fancy Bohr diagram, but the electron placement is always weird. I understand electron configuration , but I cannot understand why each electron goes where on this diagram. This example is really simplistic, but other ones that I've seen make no sense in terms of electron placement.

So, Bohr's diagram is a super-simplified version of this, almost to the point where it is arguably incorrect.

This diagram, which shows the orbitals of an electron, shows where an electron has a >90% chance to be. That means that if an electron is in any of the orbitals in your picture there, and you take many snapshots of the atom, it will show a higher concentration inside the orbital. You will find quite a few outside the orbital, but the biggest concentration will be inside those "bubbles".

Basically it shows where you are most likely to find the electrons around an atom.

If you google "different orbitals" you will find picture similar to that of yours. (I tried just now). Hope that clears most of it up, if not I really reccommend you to watch Khan Academy's videos on the subject. That really helped me.