text:
How to designate the position of an amine substituent in an alkyl-chain?
Why is 1-phenylethanamine named like it is using IUPAC rules? Why isn't it 1-phenylethan-1-amine?
I thought you had to designate where the amine group would be? I.e. the first position on the ethane-chain.

In general usage, the practice of omitting locants when there is no ambiguity is widespread. In particular, the locant ‘1’ is often omitted when there is no ambiguity.
Clearly, the locant ‘1’ for the amine group in unsubstituted ethanamine can be omitted since the isomer ‘ethan-2-amine’ does not exist.
Furthermore, the locant ‘1’ for the amine group can be omitted in substituted ethanamines provided that the amine group is the principal characteristic group (i.e. the characteristic group chosen for citation at the end of the name by means of a suffix or a class name), for example 2-chloroethanamine.
Hence, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine is permissible in general usage. Since locants are placed immediately before that part of the name to which they relate, it is clear that that the remaining locant ‘1’ in 1-phenylethanamine refers to the phenyl group.
However, for absolute clarity in preferred IUPAC names, the current IUPAC recommendations are prescriptive about when omission of locants is permissible. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are omitted in preferred IUPAC names in the cases that are mentioned in Subsection P-14.3.4, in particular:

P-14.3.4.2 The locant ‘1’ is omitted:
(a) in substituted mononuclear parent hydrides;
(b) in monosubstituted homogeneous chains consisting of only two identical atoms;
(…)

Thus, for example, methanamine (a) and ethanamine (b) are preferred IUPAC names.
Generally, however, no locants can be omitted if any locants are required to identify the specific isomer.

P-14.3.3 Citation of locants
In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine, while permissible in general usage, is not allowed in preferred IUPAC names. The preferred IUPAC name is 1-phenylethan-1-amine.


text:
Are there square planar complexes with sp2d hybridization?
We were taught (Under the section 'Valence Bond Theory') seven types of geometries a transition metal complex may assume and its corresponding hybridization states,

Linear - $\ce{sp}$
Trigonal planar - $\ce{sp^2}$
Tetrahedral - $\ce{sp^3}$
Square planar - $\ce{dsp^2}$   (Inner d-orbital involved)
Trigonal Bi-pyramidal - $\ce{dsp^3}$  (Inner d-orbital involved)
Square Pyramidal - $\ce{sp^3d}$   (Outer d-orbital involved)
Octahedral - $\ce{d^2sp^3}$   (Inner d-orbitals involved)

We were told that there are quite a few instances (for Octahedral complexes), where the outer, vacant d-orbital takes part in hybridization, so the hybridization state would thus become: $\ce{sp^3d^2}$
Now my question is:
Are there Square Planar complexes in which the outer, vacant d-orbitals take part in hybridization (i.e- are there square planar complexes with sp2d hybridization) ? If so, could someone provide a few examples.

It has been proven again and again, that the contribution of d-orbitals in all these cases is minimal, almost negligible (usually <3%). The question you are asking is referring to an old, disproven model and therefore cannot really be answered in the way you would like to look at it. Have a look at this question to get an idea what I am talking about.

Except if OP is talking about transition metal coordination compounds, in which case d-orbital participation is very real and important but where I would not speak of hybridisation at all (CC @Mart).

You used [tag:coordination-compounds]. Are you talking about transition metal complexes or main-group non-metal molecules?

@Jan Transition metal complexes, sorry. Since we only deal with those at school, I kinda forgot to specify it....

0.1) Hybridization is to be used with caution in inorganic chemistry above school level. It is proven not working for one-electron properties at the very least.
0.2) Depending on the details, you may or may not be taught about hypervalent compounds using d-orbitals of outer shell. While this concept fell out of favor, it still is taught here and there.
1) ignoring 0.*, $\ce{PnX5}$ family where $\ce{Pn=P,As,Sb}$ and X is a halogen (typically $\ce{F}$ or $\ce{Cl}$) adopts trigonal-bipiramidal shape and was viewed as an example of $sp^3d$ hybridisation. Square planar compounds for p-elements are much rarer, but $\ce{XeF4}$ adopt such structure.
2) A rare anion $\ce{[Ni(CN)5]^{3-}}$ may adopt such structure, specifically in $\ce{ [Cr(NH3)6][Ni(CN)5]\cdot 2 H2O}$ Actually, it is often said that square planar complexes may coordinate weakly an additional ion to form a square pyramid and this is why they are typically much more reactive, than octahedral complexes.

The point about $\ce{XeF4}$ still stands. It is square planar and was considered to have d-orbitals involved in hybridization before this model fell out of favor.

"... fell out of favour ..." is a bit lush I would say. It is proven to be wrong, at least the part about hypercoordination. I still think this pretty much sums up the predicament about the theory, hence thumbs up.

$\ce{[Cu(NH3)4]^2+}$ has $\mathrm{sp^2d}$ hybridisation.
A question might pop why not $\mathrm{dsp^2}$?
So basically if it were $\mathrm{sp^2d}$ then the paired electron which was promoted to $\mathrm{4p}$ orbital, would be less bound to nucleus and hence it would be easier to oxidise, but in contrary, $\ce{[Cu(NH3)4]^3+}$ doesn't exist. Hence finally Huggin suggested that the compound should be present in $\mathrm{sp^2d}$ hybridisation in order to support the experimental data.
Same hybridisation is observed in:

$\ce{[Cu(py)2]^2+}$
$\ce{[Cu(en)2]^2+}$
$\ce{[Cu(CN)4]^2-}$

Welcome to the chemistry part of stack exchange, sit down, put on your seat belt and try to enjoy the inflight movie.

source: Stack Exchange

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

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

sort: votes


text:
How to debug a failed Gaussian IRC calculation?
During IRC calculation I found the following error message. Anybody can help me?
IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC-IRC

Error in corrector energy =          -0.0000002157

 Magnitude of corrector gradient =     0.0003105753

 Magnitude of analytic gradient =      0.0003306121

 Magnitude of difference =             0.0001786220

 Angle between gradients (degrees)=   32.1573

 Pt  9 Step number  20 out of a maximum of  20

 CORRECTOR INTEGRATION CONVERGENCE:

   Recorrection delta-x convergence threshold:    0.010000

   Delta-x Convergence NOT Met
 Maximum number of corrector steps exceded.

 Error termination via Lnk1e in /home/adhikary/g09/l123.exe at Sun Aug 21 05:49:50 2016.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While questions about computational chemistry are on topic, questions about the usage of programs are only to a limited extend on topic. Apart from this, some very important information is missing, e.g. the program you are using, the input, what kind of molecule, etc.

I'm voting to close this question as off-topic because it is asking us for a full debug solution.

According to the output, it seems like OP is using Gaussian 09 software package. And according to the error message, the problem is that at 9th step of IRC procedure, Gaussian could not optimize the geometry in 20 steps (which is the default number). I suggest first to increase the number of geometry optimization iterations for each and every IRC step by using the MaxCycle=N option as follows:
IRC=(...,MaxCycle=100)

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


text:
Why does this protein seem to have multiple chains?
I'm looking at the structure of this protein, which I beleive is supposed to be a single polypeptide chain, since on the PDB page it says:

Unique protein chains: 1

But using the 3D viewer, there seem to be a few "loose ends" i.e. it looks like the protein is composed of multiple polypeptide chains.
This is was I see  - you can see a separate chain there at the bottom.
Where am I going wrong here?

Some parts a the protein are a bit more flexible, also in the protein crystal. These parts don't give clear electron density after x-ray diffraction, and are removed from the resulting model structure. Often the N- and C-terminal ends are removed, this is something you'll find in many .pdb structures, but sometimes also loops are removed, resulting in two extra loose ends.
The authors were nice to note in the comments section of the pdb file exactly which residues were removed from the model (#465).
There might be tools available to model loops into the structure so you'll have one continuous chain in the .pdb file, but I really don't know which one is best (of course they're all completely unreliable).


text:
Do Electrons Really 'Spin'?
With regard to the 'Electron Spin Number', lots of websites mention that electrons don't really spin and that the electron spin number has nothing to do with any physical spinning.
However, my chemistry teacher is quite adamant about electrons spinning about their axes in clockwise and counter-clockwise directions.
So my question stands, "Do electrons really spin?"
Could someone cite a reliable source to read up on this (Apart from Wikipedia, thanks)

They do not. Spin is a phenomenon that mathematically obeys the rules of angular momentum but it is not a physical spinning. Also wikipedia is a perfectly reliable source although it may be a bit complicated.

No, they do not. Electrons are tricky beasts. They pretty much don't do anything "normal", except having mass and electric charge. If you ask about the size or position of an electron, you are in trouble. As for the spinning, ask your chemistry teacher where does the axis of that spinning point to, and what if two electrons with perpendicular axes (rather than parallel or antiparallel) approach each other. On second thought, that may be a bad question to ask.

Giving the book, "A brief history of time" by Stephen Hawking will give you an idea about electron spin states, if you are interested in going deep.

http://physics.stackexchange.com/questions/1/what-is-spin-as-it-relates-to-subatomic-particles

Please read this for your question. Hope it helps. https://iopscience.iop.org/article/10.1088/2399-6528/ac8599

It depends on what you mean by "spin".
If you mean "have intrinsic internal angular momentum, independent of its trajectory through space", then yes, electrons spin, and that's what the quantum number is measuring.
Though if by "spin" you mean "undergoes rotation" ("there's a little billiard ball, and if I were to put a mark on it and watch it, the mark would be rotationally displaced around the center in a periodic fashion"), then no, electrons don't spin. They're point particles(*): there's no billiard ball to mark, and even if you were somehow to mark a "side" of a point particle (you can't), a point particle can't undergo internal rotational displacement - there's no internal to displace! Likewise, the "frequency" of the spin is also meaningless - attempt to find out how many revolutions per minute an electron "spins" and your calculator will spit garbage at you. It's not something that has any physically realizable meaning.
So the spin on an electron is a very real thing and not just a convenient bookkeeping label. You can do experiments where you couple the spin angular momentum of an electron to "macro-scale" (or at least non-sub-atomic) angular momentum, and you find that the total sum of angular momentum - both "macro-scale" and electron spin - is conserved. Electron "spin" is as real and functional as a gyroscope's spin is, but not in the same way. The electron has "spin" (angular momentum) without actually rotating.
Bizarre, but that's Quantum Mechanics for you.
*) I'm assuming the Standard Model of Quantum Physics here. If you start to get into String Theory things get all sorts of complicated, but there's no universally accepted model of String Theory, so I'm just going to ignore it exists. (As most Chemists do.)
Also note that the Standard Model doesn't even attempt to predict where this intrinsic internal angular momentum comes from. It's just a property of electrons, taken as a given. Asking "why do electrons have spin?" is like asking "why are electrons charged?": They just are.

Even in the Standard Model  electrons aren't point particles. They are excitations of an underlying quantum field. And spin, as well as charge, stems from the fact that physical systems are invariant to rotations (spin) and phase (charge). All SM "particles" are irreducible representations of the symmetry group of the universe (Poincare group). https://en.wikipedia.org/wiki/Noether%27s_theorem

SM says they are spherically symmetrical not really point IIRC, and there's research to find its deviation from spherical shape, even.

Despite the success of the Schrödinger Equation in predicting energy levels of the hydrogen atom, experimental observations suggest that it doesn't tell the whole story of electron behavior in atoms.
Firstly, spectroscopists observed "anomalous" splitting in the previously studied "normal" Zeeman effect, sometimes with unequal spacing.

Secondly, some energy levels show splitting that resemble the Zeeman effect even when there's no external magnetic field!
For example, when hydrogen spectral lines are examined with a high-resolution  spectroscope, some spectral lines are found to consist of sets  of closely spaced lines called multiplets.
Experiments such as the Stern-Gerlach experiment, showed that particles are found to posses angular momentum that can't be accounted for the orbital angular momentum alone...
This led two Dutch scientists, Goudsmit and Uhlenbeck, to propose that electrons may have additional motion.
Using a semi-classical model, they suggested the electrons behave as a sphere of charge that spins about its own axis. If so, it would have an additional "spin-angular momentum" (and hence magnetic moment) which if quantized as well, would explain the observed energy level anomalies:
In the Bohr Model, suppose that an electron is not a point charge (i.e- has a finite size), but is a small spinning sphere. Then the electron will not only have orbital angular momentum, but also spin angular momentum associated with the rotation of its mass about an axis.
The sphere carries electric charge , so the spinning motion leads to current loops and thence, to a magnetic moment.
In an external magnetic field, this spin magnetic moment has an interaction energy in addition to that due to the orbital magnetic moment. It is this spin magnetic moment that is responsible for the "anomalous" Zeeman shifts.
Precise spectroscopic analysis, as well as a variety of other experimental evidence , has conclusively shown that electron do indeed have a a spin angular momentum (and an associated spin magnetic moment) that do not depend on orbital motion.
But there's a caveat,
No size has been found for electrons, i.e- they appear to be point particles, and by virtue of that, they shouldn't be able to physically spin about their axes (since there's nothing that can actually "spin" here).
Yet electrons do have an additional angular momentum that isn't associated with orbital motion!
The way out of this conundrum?
Based on the above arguments, it was concluded that electrons, being point particles do not physically spin. The origin or the "spin" angular momentum is therefore, fundamentally quantum mechanical in nature and it is intrinsic to the electron.

The spin of an electron doesn't make any sense in a non-quantum world
Electron spin cannot be adequately explained by any non-quantum analogy imagining a physical spinning object. Other answers have left out one of the weirdest observations that demonstrates this.
If the electron were a classical spinning object then observations would show certain features like the idea that a full 360° rotation would return the object to its initial state (obviously this is impossible to actually observe in something observationally a point particle). But the quantum description of the electron (or any other spin 1/2 particle) implies that a 360° rotation swaps the direction of spin and that a 720° rotation leaves the state unchanged.
There is no simple classical analogy to this, though it emerges from the math of the quantum description of spin 1/2 particles (as Wikipedia explains here).
The essential point is that attempts to explain the observed properties of electrons using simple classical pictures (like a spinning object) fail badly and do nothing to contribute to understanding them. There is literally nothing useful in the classical analogies that helps explain the nature of the electron (like the Pauli exclusion principle which is literally the principle that enables chemistry to exist).

@ matt black. The question is : Is the spin an angular momentum or not ? If it is an angular momentum, it has a meaning in a non-quantum world

@maurice the question is "do electrons really spin?" not "do electrons have angular momentum".


text:
Why are covalent bonds formed?
As far as I know,
-covalent bonds are formed because atoms are unstable
-unfilled valence electrons
-want to fill up their outer rings and become stable
BUT, why does compounds like ClF3 exist? Why do their valence shells have more than 8 electrons? Isn't it about "filling up the shell"?

http://chemistry.stackexchange.com/a/5242/9961

Atoms form covalent bonds as a result of the Octet Rule. The Octet Rule states that all atoms in a molecule need to have eight electrons in their valence shell. This rule can be satisfied by sharing, losing or gaining electrons. A covalent bond allows atoms to satisfy the Octet Rule via sharing.
The reason atoms in covalent bonds satisfy the Octet Rule through sharing rather than losing or gaining electrons is because covalent bonds form between atoms with similar electronegativities. Electronegativity measures the tendency of an atom to attract electrons. Atoms with similar electronegativities are more willing to share electrons than those with different electronegativities.
Up to three covalent bonds can form at one time in a molecule. In a single bond, one pair of electrons is shared between two atoms. A double bond shares two pairs of electrons; it is stronger than a single bond, but it also creates a less stable molecule because it is more reactive. The triple bond shares three pairs of electrons, making it the least stable covalent bond.

Octet rule doesn't cause anything, and all other statements in your answer are more or less inaccurate, imprecise or wrong.

Atoms in period 3 and above can have more than 8 e- in outer shell because the electrons have access to empty d orbitals. Hence molecules like SF6 can form.


text:
Is metallic sodium a strong base?

Chemical properties: Sodium metal and sodium amide ($\ce{NaNH2}$ ) are strong bases. They react with ethyne to form sodium acetylide with the liberation of hydrogen gas.

How can a metal considered a base (that too strong)? To act as base it should either donate a pair of electrons (in this case it's single electron) or accept proton or give $\ce{OH-}$ ion but here it's not qualifying any one. So is that statement wrong?

Formation of monosodium ethynide:
$$\ce{HC#CH + Na -> HC#C^-Na+ + 1/2H2}\tag{13.59}$$
Formation of disodium ethynide:
$$\ce{HC#C^-Na+ + Na -> Na+C^{-} #C^-Na+ + 1/2H2}\tag{13.60}$$

Source: NCERT 11, Hydrocarbons, Page 386

It should be rather called strong reductor, but according to Usanovich's theory it would qualify as base.

No, stricly speaking metallic sodium is not a strong base. Strictly speaking, metallic sodium is not a base at all. (Sodium amide clearly is one, though: due to the amide part.)
Not everything that reacts with an acid is a base. If an acid reacts to form hydrogen and there are no hydrides present, that is a very strong indicator that the reaction was not an acid-base reaction but a redox reaction. And that is what metallic sodium is: a strong reducing agent.
Just because reducing agents can react with acids (proton donors) to generate hydrogen radicals and thus nascent hydrogen gas does not make them bases. For the reaction partner of an acid to be a base, it must accept (and keep!, at least to a certain extent) the proton donated by the acid — or, in the Lewis theory, it must donate an electron pair.


text:
Calculating the ionic product of water
In the question I've to find out the ionic product of water using the facts that the degree of dissociation of water at $18\ \mathrm{^\circ C}$ is $1.8\times10^{-9}$.
My attempt
Let the concentration of water be $1\ \mathrm M$.
$$\ce{H2O <=> H+ + OH-}$$
Initial:
$$\begin{align}
[\ce{H2O}] &= 1\ \mathrm M\\[6pt]
[\ce{H+}] &= 0\ \mathrm M\\[6pt]
[\ce{OH-}] &= 0\ \mathrm M
\end{align}$$
At equilibrium:
$$\begin{align}
[\ce{H2O}] &= (1 - 1.8×10^{-9})\ \mathrm M\\[6pt]
[\ce{H+}] &= 1.8×10^{-9}\ \mathrm M\\[6pt]
[\ce{OH-}] &= 1.8×10^{-9}\ \mathrm M
\end{align}$$
Therefore,
$$\begin{align}
K_\mathrm w &= \frac{[\ce{H+}] \times [\ce{OH-}]}{[\ce{H2O}]}\\[6pt]
&= \left(1.8\times10^{-9}\right) \times \left(1.8\times10^{-9}\right)\\[6pt]
&= 3.24 \times 10^{-18}
\end{align}$$
According to my book answer is $1.0\times10^{-14}$, which I know is correct. I want to know where am I going wrong?

why concentration of water be assumed 1M?

@JM97 They've asked to find the ionic product, so I thought that assuming the initial concentration of water to be 1M will simplify the problem. But I've done something wrong I guess. I don't know where am I going wrong.

molarity is number of moles per litre so please consider the number of moles of water present per liter . Hint: 1kg of water occupies one litre space.

@JM97 There is no such thing as the "number of moles"; it is called amount of substance, just like mass is not called "number of kilograms".

What it boils down to is, you need to take the fractional dissociation $1.8×10^{-9}$ and multiply it by the total amount of water, not just one molar.  The total concentration of water is $1000$ grams per liter, which you convert to moles per liter by dividing by the molecular weight of the water.  After working this out and doing the multiplication you should get the right hydron and hydroxide ion concentrations and then you get the right $K_w$.
This is just a number crunching exercise.  To find the ion product for real, we can use electrochemistry.  Using a highly reversible cathodic reaction for hydrogen evolution, say on a platinum electrode, find the hydrogen reduction potential for 1 M strong acid and for 1 M strong base, e.g. hydrochloric acud and sodium hydroxide.  There is a difference between them, which you plug into the Nernst Equation to get the hydron concentration in the base solution.  The ion product then follows.

The concentration of water in pure water is $55.5$ M. So that the concentration of $\ce{H+}$ is $\pu{55.5 \times 1.8\times 10^{-9} = 1.00 \times 10^{-7}}$ M.

The problem is that percent dissociation does not give a molar concentration hence Maurice's treatment to convert to moles/L. % dissociation is grams/gram.  Since one mole of water gives one mole of H+ 55.5 moles/L gives 55.5 times as many moles H+/L.  [Formation of H3O+ does not affect the stoichiometry only one water disassociates.] Working out the mass balance between H3O+ and OH- is interesting. Exact molecular weights must be used and the mass of the electron taken into account.

What you did wrong was adding the concentration of the hydron and the hydroxide ion. Also, although it hardly matters, the concentration of $\ce{H_2O}$ is considered a constant and ignored in the product.

That seems to be a typo which I fixed. It should have been apparent on reading the line following it, where they have indeed multiplied the concentrations.

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


text:
Why during nuclear fusion does the process only release neutrons?
Say we envision subatomic particles like balloons held together by static electricity (analogous to nuclear force).
Now we smash em' together. They make a new element. My chemistry test answer key has us calculate the number of neutrons that leave the reaction by using Einstein's famous equation. But how do we know they were all neutrons that left? Couldn't some protons have been bounced off into the hinterlands, taking with them the energy their mass represents?

Looks you're talking about fusion not fission. Also it's rather physics then chemistry.

I edited the title but you should respond.

In the chart of nuclides, the most stable nuclides have a certain ratio of proton $(Z)$ to neutron $(N)$ numbers. They form the so-called valley of β-stability. Nuclides on right side of the valley (higher neutron numbers) are unstable to decay by β− emission. Nuclides on the left side are unstable to β+ decay or electron capture.

Image taken from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002)
For the light elements, stability is achieved when the number of neutrons and protons are approximately equal $(N=Z)$, for example $\ce{^2_4He}$ with $Z=2$ and $N=2$, or $\ce{^{12}_6C}$ with $Z=6$ and $N=6$. However, with increasing atomic number of the element, the ratio of neutrons to protons for nuclear stability increases to approximately $N/Z=1.5$ for $\ce{^{209}_{83}Bi}$ with $Z=83$ and $N=126$.
The neutron excess of heavy elements explains the release of neutrons during nuclear fission as well as the radioactivity (mostly β− decay) of the fission products. During the usual nuclear fission, a heavy elements is split into two smaller nuclides with a relatively large excess of neutrons. If no neutrons are released during this process, the large neutron excess of the original heavy nuclide $(N/Z\gt1.5)$ would be too large for the two smaller fission product nuclides to be stable. Hence, some of the excess neutrons are immediately released during the fission process (prompt neutrons). Further neutrons can be released later by the fission products (delayed neutrons) or are mostly converted during the beta decay of the fission products.
For example, during the nuclear fission of $\ce{^{235}_{92}U}$ with thermal neutrons, a typical fission reaction can lead to the creation of the fission products $\ce{^{140}_{54}Xe}$ and $\ce{^{93}_{38}Sr}$ and the release of three neutrons:
$$\ce{^{235}_{92}U + 1n -> ^{236}_{92}U -> ^{140}_{54}Xe + ^{93}_{38}Sr + 3n}$$
The neutron excess of $\ce{^{235}_{92}U}$ with $N=143$ and $Z=92$ is very large $(N/Z=1.54)$. It is further increased by the thermal neutron. Although three neutrons are immediately released during the fission process, the remaining neutron excess is still too large for $\ce{^{140}_{54}Xe}$ with $(N/Z=1.59)$ and $\ce{^{93}_{38}Sr}$ with $(N/Z=1.45)$ to be stable. Therefore, $\ce{^{140}_{54}Xe}$ further decays by successive β− emissions, finally to the stable nuclide $\ce{^{140}_{58}Ce}$ with $(N/Z=1.41)$, and $\ce{^{93}_{38}Sr}$ finally decays to $\ce{^{93}_{41}Nb}$ with $(N/Z=1.27)$.

This answer has become invalid when the question was changed from "fission" to "fusion".


text:
What is the reason for an increase of back-pressure of HPLC column?
Recently i have observed that the back pressure in my HPLC reverse phase column increased almost twice, and sometimes after sample injection the pressure increased so high that removed the connector of column.
What is the reason for that, even though I am using HPLC grade solvents, In my point of view may be some impurities stuck inside. I have washed my column with methanol for long but it's seems useless.
Is there anything which can i do to solve this problem, because the column is expensive and don't have funds in lab to buy another one.

Replace the precolumn filter frit, perhaps? That's the first place I would look and the one with the least expensive fix. If you're not using a precolumn filter, by all means do so from now on.

@ToddMinehardt, thanks for your comment, appreciated

When you say almost twice, what does that actually mean? You can get a pressure spike during injection as the valve switches to put the sample on the column. If your initial solvent system is too viscous or flow rate too high, the back pressure may be running high anyway. Using a KrudKatcher or guard forum is good advice.  If yes don't fix your problem and you have the ability to automate a gradient,  halve your flow rate for the first  5-10 seconds so your initial conditions halve your back pressure.  Once the injection spike passes, the full flow rate can be used.


text:
18 Electron Rule For Determining the Stability of Transition Metal Complexes
I chanced upon this 'rule' while I was messing around on Wikipedia, and since we just finished studying about transition metal complexes at school, it naturally drew my interest.
Though I did read the article on it on Wikipedia, I don't feel I've really grasped the concept well enough (in short … I'm kinda lost).
So I hope it isn't too much to ask but, is there someone here who's willing to spend some explaining the 18 electron rule, lucidly enough for a 17 year old to understand? Or if that isn't possible, could someone recommend a site that explains the rule decently enough? Also could you touch upon some exceptions to the rule (at least for period 4 transition metals)?

18 electron rule hardly applies to typical coordination complexes. It is more for organometallics. http://www.ilpi.com/organomet/electroncount.html

@orthocresol The only kinds of complexes we've studied so far are period 4 transition metal complexes. Would this rule apply (fairly well) to those?

No it does not apply. An example of application of 18e rule would be to show why dimerisation of $\ce{Mn(CO)5}$ occurs, or why $\ce{Cr(C6H6)2}$ is more stable than other benzene complexes. $\ce{NH3, H2O, Cl}$ ligands etc - 18e rule isn't relevant at all.

As already mentioned, it doesn't work for coordination complexes, and is used for organometallics (I have posted a discussion to that effect)

A brief description of the 18 electron rule

A valence shell of a transition metal contains the following: 1 $s$ orbital, 3 $p$ orbitals and 5 $d$ orbitals; 9 orbitals that can collectively accommodate 18 electrons (as either bonding or nonbonding electron pairs). This means that, the combination of these nine atomic orbitals with ligand orbitals creates nine molecular orbitals that are either metal-ligand bonding or non-bonding,  and when metal complex has 18 valence electrons, it is said to have achieved the same electron configuration as the noble gas in the period.

In some respect, it is similar to the octet rule for main group elements, something you might be more familiar with, and thus it may be useful to bear that in mind. So in a sense, there's not much more to it than "electron bookkeeping"
As already mentioned in the comments, 18 electron rule is more useful in the context of organometallics.
Two methods are commonly employed for electron counting:

Neutral atom method: Metal is taken as in zero oxidation state for counting purpose

Oxidation state method: We first arrive at the oxidation state of the metal by considering the number of anionic ligands present and overall charge of the complex

I think this website does a good job of explaining this: http://www.ilpi.com/organomet/electroncount.html (plus, they have some practice exercises towards the end)
Let's just focus on Neutral Atom Method (quote from the link above)

The major premise of this method is that we remove all of the ligands from the metal, but rather than take them to a closed shell state, we do whatever is necessary to make them neutral. Let's consider ammonia once again. When we remove it from the metal, it is a neutral molecule with one lone pair of electrons. Therefore, as with the ionic model, ammonia is a neutral two electron donor.
But we diverge from the ionic model when we consider a ligand such as methyl. When we remove it from the metal and make the methyl fragment neutral, we have a neutral methyl radical. Both the metal and the methyl radical must donate one electron each to form our metal-ligand bond. Therefore, the methyl group is a one electron donor, not a two electron donor as it is under the ionic formalism. Where did the other electron "go"? It remains on the metal and is counted there. In the covalent method, metals retain their full complement of d electrons because we never change the oxidation state from zero; i.e. Fe will always count for 8 electrons regardless of the oxidation state and Ti will always count for four.

Ligand Electron Contribution (for neutral atom method)
a. Neutral Terminal (eg. $\ce{CO}, \ce{PR_3}, \ce{NR_3}$) :  2 electrons
b. Anionic Terminal (eg. $\ce{X^-}, \ce{R_2P^-}, \ce{Ro^-}$) :  1 electron
c. Hapto Ligands (eg. $ \eta^2-\ce{C_2R_4},  \eta^1-\text{allyl}$): Same as hapticity
d. Bridging neutral (eg. $\mu_2-\ce{CO}$) : 2 electrons
e. Bridging anionic (eg. $\mu_2-\ce{CH_3}$) ( no lone pairs): 1 electron
f.Bridging anionic (eg. $\mu_2-\ce{Cl}, \mu_2-\ce{OR}$) (with 1 lone pair)): 3 electrons
(with 1 lone pair)
or, $\mu_2-\ce{Cl}$(with 2 lone pairs): 5 electrons
g. Bridging alkyne 4 electrons
h. NO linear 3 electrons
i. NO bent ( lone pair on nitrogen): 1 electron
j. Carbene (M=C): 2 electron
k.Carbyne (M≡C): 3 electron
Determining # Metal-Metal bonds
Step 1: Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence electrons of the metal plus the number of electrons from each ligand and the charge)-- I'll call this T (T for total, I'm making this up )
Subtract this number from $n × 18$ where $n$ is the number of metals in the complex, i.e $(n × 18) – T$ -- call this R (R for result, nothing fancy)
(a) R divided by 2 gives the total number of M–M bonds in the complex.
(b) T divided by n gives the number of electrons per metal.
If the number of electrons is 18, it indicates that there is no M–M bond; if it is 17 electrons, it indicates that there is 1 M–M bond; if it is 16
electrons, it indicates that there are 2 M–M bonds and so on.
At this point, let's apply this method to a few examples
a)  Tungsten Hexacarbonyl (picture)
Let's use the neutral atom method, W has 6 electrons, the carbonyls donate
12 electrons and we get a total of 18. Of course there  can be no metal metal bonds here.
(b) Tetracobalt dodecacarbonyl (picture), here let's figure out the no. of metal-metal bonds. T is 16, R is 12, Total # M-M bonds is 6, # electrons per metal is 15, so 3 M-M bonds.
A few examples where the "18 electron Rule" works
I. Octahedral Complexes with strong $\pi$ - acceptor ligands
eg.  $\ce{[Cr(CO)_6]}$

Here, $t_{2g}$ is  strongly bonding and is filled and,  $e_g$ is strongly antibonding, and empty. Complexes of this kind  tend to obey the 18-electron rule irrespective of their coordination number. Exceptions exist for $d^8$, $d^{10}$ systems (see below)
II. Tetrahedral Complexes
e.g. $\ce{[Ni(PPh_3)_4]}$ ($\ce{Ni^0}$, $d^{10}$ 18-electron complex

Tetrahedral complexes cannot exceed 18 electrons because there are no low
lying MOs that can be filled to obtain tetrahedral complexes with more than 18 electrons. In addition, a transition metal complex with the maximum of 10 d-electrons, will receive 8 electrons from the ligands and end up with a total of 18 electrons.
Violations of 18 electron rule
I. Bulky ligands : (eg. $\ce{Ti(\text{neopentyl})_4} $ has 8 electrons) Bulky ligands prevent a full complement of ligands to assemble around the metal to satisfy the 18 electron rule.
Additionally, for early transition metals, (e.g in $d^0$ systems),  it is often not possible to fit the number of ligands necessary to reach 18 electrons around the metal. (eg. tungsten hexamethyl, see below)
II. Square Planar $d^8$ complexes (16 electrons) and Linear $d^{10}$ complexes (14 electrons)
For square planar complexes, $d^8$ metals with 4 ligands, gives 16-electron complexes. This is commonly seen with metals and ligands high in the spectrochemical series
For instance, $\ce{Rh^+}, \ce{Ir^+}, \ce{Pd^2+}, \ce{Pt^2+}$) are square planar. Similarly,  $\ce{Ni^2+}$ can be square planar, with strong $\pi$-acceptor ligands.
Similarly, $d^{10}$ metals with 2 ligands, give 14-electron complexes. Commonly seen for for $\ce{Ag^+}, \ce{Au^+}, \ce{Hg^{2+}}$
III. Octahedral Complexes which disobey the 18 electron rule, but still have fewer than 18 electrons (12 to 18)
This is seen with second and third row transition metal complexes, high in the spectrochemical series of metal ions with $\sigma$-donor or $\pi$-donor ligands (low to medium in the spectrochemical series).
$t_{2g}$  is non-bonding or weakly anti-bonding (because the ligands are either $\sigma$-donor or $\pi$-donor), and $t_{2g}$ usually contains 0 to 6 electrons. On the other hand, $e_g$ are strongly antibonding, and thus are empty.

IV. Octahedral Complexes which exceed 18 electrons (12 to 22)
This is observed in first row transition metal complexes that are low in the spectrochemical series of metal ions, with $\sigma$-donor or $\pi$-donor ligands
Here, the $t_{2g}$ is non-bonding or weakly anti-bonding, but the $e_g$ are only weakly antibonding, and thus can contain electrons. Thus, 18 electrons maybe exceeded.
References:
The following weblinks proved useful to me while I was writing this post, (especially handy for things like MO diagrams)
http://www.chem.tamu.edu/rgroup/marcetta/chem462/lectures/Lecture%203%20%20excerpts%20from%20Coord.%20Chem.%20lecture.pdf
http://classes.uleth.ca/201103/chem4000b/18%20electron%20rule%20overheads.pdf
http://web.iitd.ac.in/~sdeep/Elias_Inorg_lec_5.pdf
http://www.ilpi.com/organomet/electroncount.html
http://www.yorku.ca/stynes/Tolman.pdf
and obviously, the wikipedia page is a helpful guide https://en.wikipedia.org/wiki/18-electron_rule

I personally feel that the oxidation state method is much better — but I probably just always get lost in the neutral atom method ;) In any case, this deserves upvotes!

The Wikipedia lead is as good an explanation as I could do (actually, better). Reading it carefully you'll note that it admits that the rule is violated. Of course, so is the octet rule, especially as you go down the table. At some point in learning chemistry, simple "rules" are more trouble than they are worth. I suggest that this is one of those times. IMHO there are TOO MANY VIOLATIONS of this simple rule for it to be part of my tool box. I suggest that you hold it in the same regard, that it's better than nothing, but not by much and there are almost certainly better alternatives to use to predict coordination in any specific case you are given (except on an exam, perhaps). What are the alternatives? Well, similar complexes for the most obvious. By similar I mean isoelectronic or similarly sized or, even better, both!


text:
Equivalence point of titration
Why is $\ce{[HIn] = [In-]}$ at equivalence point of a titration? I know that at equivalence point moles of known solution is equal to the moles of unknown solution, but I'm not able to relate that fact with the above conclusion. Perhaps I'm don't know something.

@Shuvam at equivalence point the reductant and oxidant are equal in their mole stoichiometric.

@Shuvam hope this helps http://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html

What is $\mathrm{HIn}$? The indicator? If so, this statement is false. The equivalence point is by definition when you have added a stoichiometric amount of the titrant to the analyte.

Ideally, the concentration of indicator is much, much lower than the concentration of the material to be titrated and of that used to titrate. Therefore, less than a drop of standard solution is needed to make the indicator entirely cross the equivalent point.
For most titrations, the equivalence point of the indicator is not the end point. It’s just somewhere near. And the only thing you can actually see and record is said indicator’s colour change. So it’s the only thing you can resort to.
The important choice in a titration is choosing the correct indicator. If you are titrating a strong acid with a strong base (so the titration’s equivalence point is approximately $7$), you want an indicator whose equivalence point is just behind that (at ca. $8$), so that the tiniest of drops added additionally will cause the colour change.

As you might already know, $\ce{[In^-]}$ is the concentration of the indicator that was added to the solution, which basically shows whether the solution is basic or acidic or neutral. The equivalence point of a titration is when the titrant has reacted with the unknown chemical solution so that it reaches an endpoint of the indicator. The equalization of the concentration $\ce{[HIn] = [In^-]}$ is when the color is at its neutral state: neither colors of the indicator is dominant over the other.

I have downvoted this answer because it is not true that the equivalence point of a titration is when a solution is neutral; consider the titration of a weak base with a strong acid, for example. The equivalence point of a titration occurs when all the titrant has reacted.

Oh, right. Stupid mistake. I do believe the rest of the answer should be right though. Is it?

Yes, I believe so.


text:
ice tables equilibirium q
I was given initial concentrations for the following reaction:
$$\ce{2SO2 + O2 <=> 2SO3}\ \ \ K_c =7.5\times 10^{-2}\\
[\ce{SO3}]= 0.05\ \mathrm{M}\\
[\ce{SO2}]= 0.125\ \mathrm{M}\\
[\ce{O2}]= 0.02\ \mathrm{M}$$
What are the final concentrations?
First I calculated $Q$, and found that $Q>K_c$ so the reverse reaction is favoured.
I created an ICE table and wrote the expression for $K_c$ to solve for $x$, but the calculator says there is no solution.
Here is the ice table:
$$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{SO2}] & [\ce{O2}] & [\ce{SO3}]\\ \hline
I & 0.125 & 0.02 & 0.05 \\
C & +2x & +x & -2x \\
E & 0.125+2x & 0.02+x & 0.05-2x\\ \hline \end{array}\\$$
$$7.5 \times 10^{-2} = \dfrac{(0.05-2x)^2}{(0.02+x)(0.125+2x)^2}$$

Exact duplicate of post a few down by a 'different?' user...

http://chemistry.stackexchange.com/questions/58043/ice-tables-question Instead of simply editing you reposted it!

Well, does this not count as shown effort? Let's close the other one as a dupe.

You have done everything correctly so far. Your calculator might not be able to factor the expression as it is written. It is a cubic equation with three roots and two discontinuous points. You may be used to ICE-type equilibrium problems where you can simplify parts of the expression because $x$ is small compared to the initial concentrations. This is not the case in this problem. Factoring the cubic polynomial is not always easy.
Your mass action expression can be rearranged to the following cubic equation so long as $x \ne -0.02$ and $x \ne -0.0625$, both of which are nonsense anyway. If your calculator cannot factor this expression, try using Wolfram Alpha.
$$0.3 x^3-3.9565 x^2+0.201922 x-0.00247656 = 0$$
Trying solving this equation. It has three roots, two of which are nonsense since they yield negative final concentrations. The remaining root is the answer.


text:
Polymerization Isomerism In Transition-Metal Complexes Not True Isomerism?
At school we've just finished learning about Coordination Compounds.
Our textbook has defined Isomers as

Compounds having same molecular formula but different arrangement of atoms, due to which they differ in one or more physical properties are called isomers.

It further goes on to discuss the various kinds of structural isomerism seen in (transition-metal) complexes:

Ionization Isomerism
linkage isomerism
ligand isomerism
solvate/hydration isomerism
coordination isomerism
coordination-position isomerism
polymerization isomerism

Polymerization isomers, according to our textbook, are

those complexes that have the same empirical formula, but differ in molecular mass by integral multiples of empirical mass.

e.g., $\ce{[Pt(NH3)2 Cl2]}$ and $\ce{[Pt(NH3)4][PtCl4]}$ are polymerization isomers.
Yet another book I've read, simply states that polymerization isomerism is not isomerism in the true sense of the word, without elaborating on this statement.
But looking at the definition for isomerism, the way I see it, polymerization isomerism fits the bill, so what exactly do the authors of the book imply by that statement? Are they wrong?

Your second book is of course right. NO is not an isomer of N2O2. I've never heard the term "polymerisation isomerism", sounds like a rather stupid category to me. ;-)

@Karl, I forgot to mention the examples given in the book, just did it now. Also I haven't come across 'Polymerization Isomerism' before, so perhaps it only applies to complexes (and not NO and N2O2) ?

"Polymer" implies that the strucure of the monomer is retained. This is not the case for your example, so in this case i would say it's just wrong.

At the moment, that's rather my opinion than an answer. Let's wait what others have to say.

I’m pretty sure that polymerisation isomerism should mainly be considered a historical term nowadays. Going back in the history of science, all people knew at some point was which elements a specific compound is made out of. That worked well for main group metal salts but not too well for transition metal salts. Later, quantitative analysis established the law of multiple proportions, which basically says that if two compounds are made up of the same set of elements, the ratio of element mass ratios will always be an integer ratio. Sounds hard to understand, here’s an example:
$\ce{CuCl}$ has one chlorine per copper, $\ce{CuCl2}$ has two. So for every $x~\mathrm{g}$ of copper, one of the two will have $1~\mathrm{g}$ of chlorine, the other will have $2~\mathrm{g}$, and $1$ and $2$ form a simple integer ratio.
And all of a sudden there were compounds that were clearly distinct but gave the same ratio of elements, a simple example being $\ce{NO2}$ and $\ce{N2O4}$. To understand this, the term polymerisation isomer must have been introduced.
Nowadays, isomers are defined as having the same elemental composition on a per-fragment basis. So examples such as $\ce{N2O4}$ (which is a different molecule than $\ce{NO2}$) or the platinum complexes you give would no longer be considered isomers — neither constitutional ones nor stereoisomers. Hence why so many people said ‘your second book is right.’


text:
To which block of the periodic table does helium belong?
According to quite a few books I have read, helium belongs to the p-block with reasoning along the lines of "it's an inert gas, so let's clump it together with other noble gases which are in the p-block."
However, on Wikipedia, helium is classified as an s-block element which, to me, makes sense because the last (or so-called "differentiating") electron that is added to helium will occupy an s sub-shell.
But my teachers are adamant that helium is a p-block element, and the fact that our textbook covers helium under the chapter "The p-block Elements" just gives them more reinforcement of that idea.
So: is helium an s- or a p-block element? Could someone provide concrete reasons for and against the block selection?

It is a noble gas with no p electrons. Another example of a dumb arbitrary classification for little gain in understanding.

What Jon says. You've got reason behind you, your teachers only have their faith in a stupid text book. Stand up to them!

http://chemistry.stackexchange.com/questions/13958/block-on-the-periodic-table

Technically it's an s-block, but follow what your teacher wants if you wanna pass.

Here(in textbook), helium is classified in p-block for sake of ease in studying its properties, as helium is noble gas, where all noble gas are considered in p-block. On other side, if we classify helium in s-block it is only noble gas in s-block.

Blocks in the periodic table should not be mixed up with groups (like noble gases). The reason why Helium is considered a noble gas is because its outermost (and at the same time only) shell is fully occupied by its 2 electrons. When you look at the electron configurations in the PTE you can see that the first element which happens to have an occupied p-orbital is boron. Due to this fact boron also has to be the first p-block element, since the block definition is as follows:

An element belongs to the block x, if the electron, which is added in comparison to the previous element (by atomic number), is in an orbital of type x in the ground-state electron configuration of that element.

Or the fast explanation: Why should helium be considered a p-block element, if it doesn't even have any electron in a p-orbital or any shell including a p-orbital?

Jup. One electron in the 1s shell also doesn't make hydrogen an alkali metal.

Helium is an s-block element with properties like those of the p-block noble gases.
In a periodic table more concerned with chemical behaviour you're likely to see He in the p-block.
In a table more concerned with showing the filling of electron shells you're likely to see He in the s-block.
The optimal solution is to show He in the s-block but to colour it as a noble gas or a p-block element.


text:
Can a chemical reaction have two balanced equations?
I was recently studying Xenon compounds and read this reaction:
$$\ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 +  24HF + 3O2}$$
However this reaction can also be written as the following:
$$\ce{4XeF4 + 8H2O → 2Xe + 2XeO3 +  16HF + O2}$$
Now both of these are valid but wouldn't that create problems, like in quantitative analysis or in kinetics of the reaction?

I was thinking about my statement in kinetics and now I realised the various species are independently reacting. Moreover an interesting pattern emerges if you keep adding the difference of these reactions to the first one indefinitely. Thanks to all of you for clearing this for me.

You can expand the second equation so it has the same amount of educt on the left side as the first one:
$$\ce{6XeF4 + 12H2O → 3Xe + 3XeO3 +  24HF + 1.5 O2}$$
Difference on the product side, compared to first equation, is
$\ce{-1 Xe + 1 XeO3 - 1.5 O2}$.
This excess of one $\ce{XeO3}$ can go
$$\ce{XeO3 → Xe + 1.5 O2}$$
, exactly the amount of xenon and oxygen that was missing compared to the first equation.
Your equations describe not one reaction, but (at least) two, which don't have to occur in a predefined proportion. It's not directly clear, and would involve an in-depth investigation, what exactly the elementar reactions are.
So you could make an infinite number of chemically correct, but quantiatively different equations for the hydrolysis of xenon flouride. All depends on the reaction conditions, probably.

The amount of oxygen should be $\frac{3}{2}$, which comes from multiplying the OP second equation by ${6}{4}$. It seems to me at least. Apologize if I'm wrong. And I think you have answered his most important point (the post title): it has to be experimentally proven. Then if we take first as The reaction, then reaction two is not valid. (+1)

What amount of oxygen, and what experiment? The two equations predict twodifferent reactions!

@santimirandarp The oxygen checks  out as far as i can see. 12 atoms on the left, 12 on the right. If you think otherwise, please write down the corrected equation.

The first reaction is actually two reactions:
$$
\begin{align}
\ce{4 XeF4 + 8 H2O &→ 2 Xe + 2 XeO3 +  16 HF + O2} \\
\ce{2 XeF4 + 4 H2O &→ 2 Xe + 8 HF + 2 O2} \\
\hline
\ce{6 XeF4 + 12 H2O &-> 4 Xe + 2 XeO3 +  24 HF + 3O2}
\end{align}
$$

That is just one arbitrary possibility.

I couldn't find any other linear combination of reactions that would give the same result (multiplying every stoichiometric factor in the above schema with the same number does not yield a new result).

It's the only way to split that specific equation into two other, specific reactions, of course. But the exact stochiometry of the reaction is not clear.

This reaction can also be balanced with following factors: 5,10,3,2,20,2

This is just another linear combination of the actual two reactions, with both the first and second with factor 1. Above the factors are 1 and 2 respectively.

If the coefficients that balance the chemical reaction are expressed as an ordered 6-tuple (a, b, c, d, e, f) where a is the coefficient of XeF4, b is the coefficient of H2O, etc., then from a purely mathematical point of view, the set of all possible coefficients are of the form:
(x/4, x/2, [x/12]+[2y/3], [x/6]-[2y/3], x, y)
where x+8y is a multiple of 12, x-4y is a multiple of 6 and x > 4y.
The mathematical reason for the existence of distinct ways of balancing is that the matrix that represents the reaction has two free variables when expressed in reduced row-echelon form.


text:
What does activated carbon do to gasoline vapor?
What does activated carbon do to gasoline vapor? After the vapor is absorbed into a charcoal EVAP canister, what is left to be sent to the engine?
My question is in reference to the work of Tom Ogle and the carburetor he made in the 1970's, and I just want to find out if his invention is plausible.

Fuel enters the ignition chamber while still in its liquid form.
http://fuel-efficient-vehicles.org/energy-news/?page_id=787
I want to know if this is possible

The evap canister is only meant to trap vapor in the tank so it does not come out when the gas cap is removed. Vapor is not sent to the engine - liquid is.

the evap system is designed to keep in fuel vapor and then send the fuel vapor to the engine once it reaches atmospheric conditions. The thing is by sending gas in its liquid form to be ignited is wasteful, since it only burns as a vapor.

Vaporizing the fuel is the fuel injector's job, directly on the piston. Or, on old cars, in the carburetor (on top of the engine). The fuel pump is there to transport liquid to the engine (injector or carburetor). If you cut the fuel line from the pump to the engine compartment, you get liquid coming out. If you open up a carburetor while the engine is running you see liquid being sprayed in to it.

Actually, gas vapour from the fuel tank is directed to the engine, as a means of disposal to avoid it escaping to the air (environmental consideration). The engine operates on a supply of fuel delivered in liquid form up to the point it is mixed with air in either the carburetor or at the fuel injector. But fuel evaporates in the tank; rather than venting to the atmosphere, this fuel vapor is directed to the induction system so the engine will burn it rather than release it unburned.


text:
Why is there an upper limit to the photon energy that can be observed in the discrete emission spectrum of the hydrogen atom?
I think it must have to do with the fact that an incident photon with a sufficiently high energy will ionize the hydrogen atom (i.e. the electron will be ejected). Therefore it won't show in the emission spectrum. Is this correct?

Kind of, but backwards. The well is only so deep, so an electron dropping from right at the continuum down to the 1s state can only release 13.6eV.

The closest an e- can get to nucleus is the 1st, it can't get closer and so can't get to lower E. If an e- is beyond the influence of the nucleus and falls to 1s it can't fall any further  (drop in E any more ). This represents the largest E emission.  Once an e- is beyond the influence of the nucleus it doesn't matter how much further it goes it is still beyond its influence and hence the energy of the transition doesn't increase

Can you please at least spell out stuff fully?

The max E emission is the energy an e- releases upon falling to the 1s  from a distance so far away from the nucleus that nucleus has no hold. If an e- is further away makes no difference.  Can't make it simpler than that.  If you don't understand please be more specific about what you don't get


text:
Bonding between amylopectin and amylose
I know that amylose and amylopectin are major constituents of starch. I also know the chemical structures of amylose and amylopectin. However, I want to determine how amylose and amylopectin bond together. I think that the OH from the amylose bonds with the HO of the amylopectin to form an oxygen bond and a water. Am I correct, or does the OH chain of amylose bond with the O chain of amylopectin?

What are you talking about? They're two molecules. When and why would such polymerisation supposed to occur?

Starch is composed of amylopectin and amylose molecules bonded together. My question is how and where are they bonded together. Is it a dehydration synthesis with the OH and HO groups?

Different molecules can't be connected with covalent bonds as they would become one molecule. If you have a sugar or mix of different sugar molecules, they're connected with hydrogen bonds.

Ok. Thank you. So where would be the hydrogen bond connecting these two molecules? Which two parts of the respective molecules would be connected with each other?

I suggest h bonds will form in all different parts between hydroxy groups and ether groups and between hydroxy and hydroxy. Water will form a hydration shell and will therefore function as some sort of lose connector.

Actually you were close, below I have tried to discuss how this occurs:
Glycosidic bonds between monosaccharide units are the basis for the formation of oligosaccharides and polysaccharides such as starch.
It is possible for a sugar hydroxyl group (ROH) bonded to the anomeric carbon (the anomeric carbon is the carbonyl carbon of the open chain form of the sugar and is the one that becomes a chiral centre in the cyclic form) to react with another hydroxyl (R’-OH) to form a glycosidic linkage (R’-OR).

(A glycosidic linkage is not
  an ether)

Glycosidic linkages can take various forms; the anomeric of one sugar can be bonded to any one of the -OH groups on a second sugar to form an alpha- (as in starch) or beta-glycosidic linkage.
An example of a glycosidic bond (showing formation of maltose):

C-1 of one glucose is linked by a glycosidic bond to the C-4 oxygen of the other glucose
when an ⎯OH (alcohol) of one glucose molecule (right) condenses with the intramolecular hemiacetal of the other glucose molecule (left), with elimination of H2O and formation of a glycosidic bond.
The reversal of this reaction is hydrolysis—attack by H2O on the glycosidic bond. In animal digestion (first phase), it is aided by α (1,4)-glucan 4-glucanohydrolase (salivary α-amylase)
Here is another comparison of the α 1,4 glycosidic linkage and α1,6 glycosidic linkage:

Having discussed about the glycosidic bond, we can refer back to starch:
Starch is composed of two components, α-amylose and amylopectin. α-Amylose is composed of linear chains of D-glucose in α (1,4) glycosidic linkages. The linear linkages in amylopectin are α (1,4), whereas the branch linkages are α (1,6) glycosidic bonds:

Edit
It is also possible for polysaccharide units to form hydrogen bond interactions between neighbouring units, although these are minor and serve to stabilise the helical structure of amylose chains (around  up to 6 residues per turn) but the major bonds are glycosidic bonds
I am sure you now have a clue on how they are bonded. Hope this helps.
References:
Biochemistry (Campbell and Farell)
Lehninger Principles of Biochemistry
Biochemistry (Grisham)


text:
Number of geometrical isomers
How do I find out the total no of geometrical and stereo isomers of the compound given as $\ce{C2FClBrI}$? Is there any way I can list out all the possible structures mathematically without having to think about each structure in my mind? Any suggestion how you would do it would be wonderful.

Well, do you know what geometry this type of compounds has - you need  to know symmetry of compound.

No but id love to know the more general way to deal with having to tell the total number of isomers of any type of compound

related http://chemistry.stackexchange.com/questions/7975/finding-the-number-of-cyclic-isomers-algorithmically?rq=1 http://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers http://chemistry.stackexchange.com/questions/33356/is-there-a-formula-to-determine-the-total-number-of-constitutional-isomers

FC(I)=C(Cl)Br,
FC(Br)=C(Cl)I,
FC(Cl)=C(Br)I


text:
Mass percent composition
How would my sample be affected if the sample was not totally dry? What would happen to the mass percent composition? I believe that it will increase but someone argues that due to ratios, it will decrease.

If sample is not fully dry, obviously this would increase the mass or weight of sample.

The IUPAC Compendium of Chemical Terminology defines the mass fraction $w$ as the

mass of a constituent divided by the total mass of all constituents in the mixture

$$w_i = \frac{m_i}{\sum\limits_{i=1}^{N}m_i} \tag{1}$$
This means that $w_i$ can adopt values between 0 ( = 0 %) when the compound in question is not contained in the sample), and 1 ( = 100 %) when the sample only consists of the compound.
If your sample isn't totally dry, the mass of water adds to the total mass of the sample.
As a result, the denominator in $(1)$ will become larger and the mass fraction $w_i$ for any component (except the water) will decrease.


text:
What happens to the mobility of Group 1 elements going down the periodic table?
Whilst reading the book, So you want to go to Oxbridge (Third Edition), I noticed that it states that a past question given during an interview for the place on the undergraduate chemistry course (M.Chem) was:

What happens to the mobility of Group 1 elements going down the
periodic table?

I do not understand the question. And so, I would both like to understand the question and know and understand its answer.
Preemptively, Thank you

I am guessing they mean ionic mobilities (first, and only thing that comes to mind right now)? Simple explanation: The size of the cations increases down the group, and group elements form singly ionised cations (have a charge of +1), so the charge density decreases down the group. Extent of solvation is proportional to charge density, and consequently lithium ions would be the most solvated, hence least mobile. Whereas, $\ce{Cs^+}$ would the most mobile. If you want some simple math to support this, I can supply that later.

@Karl Sorry, how is my question in comprehensible?

@Karl A simple grammatical mistake, which I have now corrected - although, I certainly would not say that it is incomprehensible, as the point I was trying to put across was and is obvious. Also, if the book gave an answer, I wouldn't be on here asking for the answer.

"Oxbridge, 3d edition" is some kind of collection of exam questions, without answers, context, anything ? They charge money for that?

@Karl I think it is exactly that; and I'm pretty sure they do charge an ungodly amount of money for that garbage, lol

@Karl It's pretty good. It is a book written by Oxbridge graduates and an Oxford Admissions Tutor about tips on getting into Oxbridge. Because it is short, concise, and has information on every offered course, it does not have the answers to all questions in the book. And, by the way, it is not a collection of exam questions - it is a guide on the entirety of Oxbridge. The only questions in the book are those which have been asked in previous admissions interviews. Although I have the 3rd Edit., the 6th Edit. is being published 01/09/2016. I recommend it, especially due to who its written by.

@getafix Also, it's only £9.99. Anyway, thank you very much for your help, getafix. I now fully understand the question and its answer.

"Oxbridge"? I know "Oxford", and i know "Cambridge", but haven't heard of a merger.

@Karl It is a commonly used portmanteau, I believe.

I know, but they're two universities. Nevermind.

@Karl True, but they are very similar to each other, and are both looking for the same things in terms of admissions.

When you have charged particles in a liquid acted upon by a electric field, the experience a force, and said force accelerates them to a terminal, drift speed.
Note: The physical treatment of the problem, and the notion of terminal speed is quite similar to say a ball falling from the top of a sky scraper through air (a fluid medium)
Now, let's go back to our charged particles in fluids. Let's say we have applied a voltage $\Delta\phi$ between two large, planar electrodes, immersed in the liquid and separated by a distance $l$
The  magnitude of the electric field, and the electric force experienced by an ion of charge $ze$ are given by the following relations:
$$ E = \frac{\Delta\phi}{l}$$
$$F = zeE = (ze).\frac{\Delta\phi}{l}$$
A cation responds by accelerating towards the negative electrode, and an anion responds by accelerating towards the positive electrode. This acceleration, however, is short lived because the fluid medium offers a retarding force $F_{\text{friction}}$.
$$F_{\text{friction}}\propto s\text{, where s := speed} $$
Assuming we have spherical particles of radius $a$, let's say our retarding force is given by Stokes' Law
$$F_{\text{friction}} = (6\pi\eta a)s$$ $\eta$) is the viscosity of the medium.
When the electric force is exactly balaced by the viscous drag, the acceleration drops to zero and this final speed of the particle is called the drift speed ($s_d$).
$$F_{\text{friction}} = F_{\text{electric}}$$ and solving for the drift speed we get,
$$s_d = \frac{zeE}{6\pi\eta a}$$
At this point we note that the drift speed is proportional to the applied electric field and introduce the following definition:
$$s_d = \mu E$$ where, $$\mu = \frac{ze}{6\pi\eta a} $$
and we call $\mu$ the mobility.
From the aforementioned equations/definitions one can deduce that $\mu \propto \frac{1}{a}$. Thus, bigger/bulky ions ought to have lower mobilities, and lower drift speeds in solution.
However, $a$ is not the ionic radius, but the Stokes radius (or the hydrodyanmic radius), i.e the "effective radius" of an ion in solution taking into account its solvation shell (i.e the solvent molecules surrounding the ion). I believe the following illustration is instructive,
The size of this solvation shell depends on how "attractive" an ion appears to a solvent molecule, and that in turn depends on the electric field produced by the ions.
For spherical ions, we consider the electric field on the surface of a charged sphere: $$ E _\text{surface of a charged sphere} \propto \frac{ze}{r^2} $$
here $r$ is the ionic radius. So smaller ions have a larger solvation shell than larger ions (assuming they carry the same charge).
This is the reason why mobilities of alkali metal ions increase from $\ce{Li^+}$ to $\ce{Cs^+}$ (and not the other way around); the lithium ions are effectively dragging a large amount of solvent molecules with them and thus, move slower.

Hello. Thank you very much for your answer, it is extremely helpful and informative. I have one question though. In the equation: $F_{\text{friction}} = (6\pi\eta a)$ why is the flow velocity left out of the end of the equation as shown on the Wikipedia page on Stokes' Law you have referenced? I had never heard of Stokes' Law before, so the answer to my question is probably obvious.

A typo, sorry. I'll fix it right away

Also, shouldn't all references to $(6\pi\eta a)$ be changed to $(6\pi\eta a)s$, such as in the drift speed equation and the mobility equation?

@SaulMcShane Oh no! we are solving for the drift speed, (i.e s) by equating $$F_{\text{friction}} = F_{\text{electric}}$$ I guess my notation isn't good; should've made it more clear.


text:
Sugar burning with the assistance of ash
If a plain pile of sugar (or a sugar cube) is heated with a torch, it just melts and gives off smoke. However, if some paper or wood ashes are added, it burns quite easily. What are the ingredients of the wood ash that allow this to happen? If I just used sodium carbonate instead of wood ash, would it work as well?

Yes. It just keeps the sugar from melting into a closed surface.

According to an old book I once had, the catalyst is potassium oxide. Sodium carbonate probably won't work, but potassium carbonate should do the trick.


text:
Temperature dependence of molarity and molality
My textbook says

Molarity is dependent on temperature, whereas molality is not.

I'm aware of the definitions of both molarity and molality, but consider this scenario:
A solution of $\ce{NaCl}$ in $\ce{H2O}$ is heated. Eventually, all the water will escape and I will be left with only $\ce{NaCl}$ crystals. At the end, there is no solvent to even measure the concentration of what was once a solution. In this case, how does the argument "molality is independent of temperature" still hold true?

In your extreme example, obviously it doesn't. There's no solution to speak of, so you can't really define a concentration, either in terms of molarity or molality. Since, molarity is moles per volume of solution, it is sensitive to temperature (since, volume changes with temperature). No such problems with molality since it is defined as moles over mass of solvent (which doesn't change).

Formally, molality does not change if you take in account the vapor. Otherwise you are just cheating by taking away mass of solvent instead of just changing it's temperature.

You cannot change the composition of a phase and expect that concentrations do not change. Also, applying heat to a solution does not necessarily mean that solvent will evaporate. It does not when the system is closed and consists of the solution phase only.

Both are quantities of solutions. If you evaporate all the solvent, molarity and molality do not mean anything anymore.

Come to think of it, all water will eventually escape even if you don't heat the solution at all. What will happen to molarity?

Recall that molarity $c$ is
$$c=\frac{n_\text{solute}}{V_\text{solution}}$$
whereas for molality $b$ is
$$b=\frac{n_\text{solute}}{m_\text{solvent}}$$
While in the extreme case you mentioned, both quantities are not defined because all the solvent is either boiled away (or for heat-sensitive species, the solvent can even decompose under heating), molarity depends on volume, and the volume of the solution depends on temperature due to expansion/contraction.

Example: 5 moles of solute in 1 liter of solution.
The Molarity at the beginning is (5moles ÷ 1Liter) = 5 M.
When the solution is warmed up and expanded by 1% say, to 1.01L, the new Molarity = (5moles ÷ 1.01Liter) = 4.95 M
P.S. Don't over think this question! The amount of content should not change, all you're talking about is expansion of content when the temperature rises. All matter expands when temperature rises so if you're calculating the molarity of a solution, you may get different readings on a warm day compared to a freezing cold day!

Molality doesn't change with temperature...

You are not answering the OP's query about the temperature dependence of molarity vs. molality.

These two words sound similar but they are not synonyms, even though both of them are used for representing solution concentration. By definition, molarity is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION). Molality is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT). Most of the time scientists use either molarity or molality to represent solution concentration, but MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes, and that is why.)

Of course, I know the definitions properly! My only doubt was in the fact that even though molality is supposed to be independent of temperature, if I increase the temperature of the solution by heating it, some(or all)of the solvent evaporates and not only does the volume of the solvent change, but also the mass of the same and hence, molality will change. What point am I completely missing? From the other answers, I gathered that this argument will not hold good for extreme cases like the one that I mentioned in my question and must not try to apply it to such situations. Am I correct?

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". Also important: We use a capital M for the molecular mass, and a miniscule m for the mass. We refer to molarity in more technical terms as amount concentration and use c for it, while molality uses b as symbol.

Molarity is a temperature-dependent parameter. The volume of a solution increases with an increase in temperature diye to increase in intermolecular spaces and vice versa. Hence, molarity is affected by temperature change.
As
Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

Okay so no one answered OP properly. Molality(m) is independent of the temperature as long as the temperature change does not actually change the amount of solute. I understand that most resources would not mention this simple fact and expect you to understand it presumptuously. Again, molality(m) is independent of temperature as long as the amount of solvent is not changing.
Example 1: temperature changes --> solution volume changes --> molarity (M) changes --> molality (m) constant
Example 2: temperature changes --> solution volume changes --> amount of solute changes --> molarity (M) changes --> molality (m) changes
See the difference?


text:
Writing equation of reaction in stoichiometry of redox titrations
While solving some questions on the stoichiometry on redox titrations, I come across this question:

Calculate the volume of $0.05\ \mathrm{M} \ \ce{KMnO4}$ solution required to oxidize completely $2.70\ \mathrm{g}$ of oxalic acid in acidic medium.

I know how to solve this question. The only problem I am facing is that the reaction equation is not given. How do I know what the products of this reaction would be? Can someone help me in how to write the equation of the reaction by the information given in the question.

You can assume that oxalic acid will be oxidized to $\ce{CO2}$ and $\ce{H2O}$ while $\ce{MnO4^-}$ will be reduced to $\ce{Mn^{2+}}$.

@aventurin assume...? Here I can see the solution to find this out but in an exam how am I supposed to know?

Permanganate  in acidic solution is a well-known oxidant (permanganometry, https://en.wikipedia.org/wiki/Permanganometry). So either by intuition or by looking at the standard electrode potentials  https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29 and the Nernst equation.

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

See the following steps:

We know what $\ce{KMnO4}$ will do by reading the question:

Calculate the volume of […] $\ce{KMnO4}$ solution required to oxidize completely […] oxalic acid.

We need the structure of oxalic acid. (This is knowledge required to answer the question, no way around.)

Structure of oxalic acid. Image taken from Wikipedia, where a full list of authors is available.
Determine oxidation numbers of all the atoms involved. Shortcut: hydrogen has $\mathrm{+I}$, oxygen has $\mathrm{-II}$ and carbon has $\mathrm{+III}$.
Determine what can be oxidised. Only carbon really can.
Determine what it can be oxidised to. The final product must have $\ce{C^{IV}}$, which is most common as carbon dioxide or carbonate.
Solve the redox equation using either $\ce{CO2}$ or $\ce{CO3^2-}$. Both will work.

Well this works here as carbon can only be oxidized to 4+ from 3+ but in the reduction of potassium dichromate by FeSO4 how would we know the change in the oxidation state of chromium...? Would we have to memorize it or is there a way to figure it out?

@OsheenSachdev You did ask about oxalic acid, so I gave you that answer. For chromium, it’s basically either remember that chromium(III) is a stable oxidation state or correctly interpret the green colour at the end of the reaction to mean chromium(III).

Firstly, oxalic acid with both carbon atoms in +3 oxidation state, which undergoes oxidation has only to form CO2
Secondly, the formation of products from the reduction of KMnO4 is dependent heavily on pH of the solution. But like in the question, if the pH of the solution is not mentioned, then it's assumed to be acidic, and the product is Mn(II), since most of the volumetric titrations are done in an acidic solution provided by acids which will not participate in the redox reaction whatsoever.


text:
Super atoms and their orbitals
I was reading a science magazine the other day when I came across an article on superatoms. I am now wondering if these structures form their 'superorbitals' in a similar way to the pi rings in benzene molecular orbitals or is something different going on?

So what are superatoms? Can we have an explanatory link?

Fullerenes certainly do form their 'superorbitals' (in fact, normal molecular orbitals, just very large) in a similar way to the $\pi$ rings in benzene. Not sure whether they count as superatoms, though.

explanatory article: https://www.newscientist.com/article/2098038-first-superatom-molecules-pave-way-for-new-breed-of-electronics/


text:
Magnesium Bicarbonate Water Reaction With MSM - Methylsulfonylmethane
The following is a recipe for a mineral water containing magnesium and bicarbonate.
Magnesium hydroxide powder is dissolved in a plain bottle of carbonated mineral water resulting in magnesium bicarbonate. This reaction causes the $\ce{CO2}$ to be consumed and the bottle caves in on itself.
But when I added the supplement MSM (methylsulfonylmethane) and the bottle is shaken, the water inside fizzes and pressure builds up again.
So my question: what causes the bottle to re-pressurize and fizz?
And is it still safe to drink, maintaining the original minerals, magnesium bicarbonate and msm?

On the reactions happening
It helps to have clear view of what actually happens at what point of the experiment. Before you add anything to your mineral water, $\ce{CO2}$ and water are in the following equilibrium (I am explicitly ignoring all other ions present, even if they are not spectator ions because they too can take part in acid-base equilibria):
$$\ce{CO2 (g) + H2O <=> CO2 (aq) + H2O <=> H2CO3 <=>\\ <=> HCO3- + H+ <=> CO3^2- + 2 H+}\tag{1}$$
$$\ce{CO2 (g) + 3 H2O <=> CO2 (aq) + H2O <=> H2CO3 + 2 H2O <=> \\ <=> HCO3- + H3O+ <=> CO3^2- + 2 H3O+}\tag{1'}$$
(Equation (1') is merely an equvalen version of equation (1) with extra water molecules to show that naked $\ce{H+}$ does not exist in solution.)
Under equilibrium conditions, most of the $\ce{CO2}$ is either dissolved ($\ce{(aq)}$) or present as minute gas bubbles ($\ce{(g)}$) and the amount of carbonic acid is rather low. Adding $\ce{Mg(OH)2}$, an Arrhenius and Brønsted base, shifts these equilibria rightwards by consuming protons:
$$\ce{Mg(OH)2 (s) + 2 H+ <=> Mg^2+ (aq) + 2 OH- (aq) + 2 H+ <=>> Mg^2+ (aq) + 2 H2O}\tag{2}$$
Since reaction (2) will actively push equilibrium (1) rightwards, gaseous $\ce{CO2}$ is dissolved and then reacted, lowering the bottle’s overall pressure. Now you are adding methylsulphonylmethane. This compound can act as an Arrhenius and Brønsted acid by the following equation:
$$\ce{H3C-SO2-CH3 <=> H3C-SO2-CH2- + H+}\tag{3}$$
Hence, it is able to shift equilibrium (1) back leftwards, counteracting the force of reaction (2). The acidity of dimethyl sulphone (as it is also called) is due to the strong $-I$ effect of the sulphone group, which has a formal $\ce{S^2+}$ atom in its centre. The resulting anion can be understood in a similar way as phosphorus ylides can.
Since equation (3) pushes the equilibrium (1) leftwards, it again becomes more favourable to release $\ce{CO2}$ into the surroundings, i.e. form gaseous $\ce{CO2}$, i.e. increase the bottle’s pressure.

On the safety aspects
You should never drink anything that has come into contact with laboratory grade chemicals. These are not tested to be safe for consumption, they are tested and controlled only to be safe for use in a lab (and hence in chemical reactions). There is at least one urban legend of a fatal accident happening due to mixed-up chemical bottles.
That taken care of, it seems that dimethyl sulphone be a relatively safe additive. According to Wikipedia, its $\mathrm{LD_{50}}$ value is larger than $17.5~\mathrm{g / kg}\text{ body mass}$, making it a safe chemical. It is sold as a dietary supplement, showing that the risk seems to be very low. As such, it will probably be safe to drink the soda after addition of both compounds, provided the pH value is where it should be (i.e. dimethyl sulphone has been added sufficiently to counteract the pH increase induced by $\ce{Mg(OH)2}$.)

I have to disagree with the second part of the answer. The pKa of MSM is around 31 which is waaay more than for carbonic acid or bicarbonate ion. MSM is an extremely weak acid and there is no way that it will shift the acid/base equilibrium to any visible extent. I guess that the evolution of carbon dioxide after addition of MSM is a physical action, most likely it decreases the solubility of CO2 by disrupting the hydrogen bond network of water.

@vapid I would say the solubility of the bicarbonate salt is decreased. Assuming the MSM reduces the ability of the solvent to solvate ions, the reaction $\ce{Mg^{2+} +2HCO3^- <=> MgCO3(s) + CO2(g) + H2O}$ (alkaline earth bicarbonates always "precipitate" in this manner) would be pushed forwards.


text:
Why do SN2 reactions of alkyl halides proceed differently with KCN and AgCN?
Alkyl halides react with $\ce{KCN}$ to form alkyl cyanides as the main product, whereas the use of $\ce{AgCN}$ leads to isocyanides as the chief product. Why does this happen?

I’m up to answering this as it is a very interesting phenomenon. I probably won’t be able to until in a few days time. In any case let me drop the hint that any argument involving HSAB is wrong.

@Jan, I didn't see HSAB mentioned by the OP, but its certainly a somewhat plausible argument based on the relative 'hardness' of N vs C nucleophiles.

might be useful: http://actachemscand.dk/pdf/acta_vol_25_p2327-2336.pdf

@NotWoodward Well, I studied at the LMU, Professor H. Mayr’s home university, so HSAB generally has a weak stand when explaining reactivity. Especially in this case, HSAB has been proven wrong, I believe by the Mayr group themselves.

The cyanide anion is an example of an ambident nucleophile, these are nucleophiles which, usually due to delocalisation, are able to attack an electrophile from two or more atoms. In the case of the cyanide anion reaction from carbon leads to nitriles, whilst reaction from the nitrogen leads to isocyanides.
To give the full IUPAC definition from the Gold Book:

A description applied to a chemical species whose molecular entities each possess two alternative and strongly interacting distinguishable reactive centres, to either of which a bond may be made in a reaction: the centres must be connected in such a way that reaction at either site stops or greatly retards subsequent attack at the second site. The term is most commonly applied to conjugated nucleophiles. [...]

To answer your actual question about why $\ce{KCN}$ forms nitriles, whilst $\ce{AgCN}$ forms isocyanides, March's Organic Chemistry (7th ed., p 449) gives a pretty concise answer in which we consider the extent to which the cyanide anion is associated with the counter-cation in solution.

All negatively charged nucleophiles must of course have a positive counterion. If this ion is $\ce{Ag+}$ (or some other ion that specifically helps in removing the leaving group), rather than the more usual $\ce{Na+}$ or $\ce{K+}$, then the transition state is more $\mathrm{S_N1}$ like. Therefore the use of $\ce{Ag+}$ promotes attack at the more electronegative atom. For example, alkyl halides treated with $\ce{NaCN}$ generally give mostly $\ce{RCN}$, but the use of $\ce{AgCN}$ increases the yield of isocyanides ($\ce{RNC}$).

                           [See footnote]

This, of course, is not a complete picture, as solvent effects greatly change the reactivity. In protic solvents (MeOH, to give a possible example) the most electronegative atoms (nitrogen in this case) will be solvated to a greater extent via H-bondng. In aprotic solvents (THF/DMF, to give possible examples) the anion isn't solvated quite so much (on either atom of the nucleophile), but the cation is somewhat more solvated, leaving the nucleophile more able to attack (via the most electronegative atom as this is the most nucleophilic in the absence of other factors).

Footnote: As an aside at this point, it's important to realise that these are not fully covalent bonds (in $\ce{AgCN}$ and $\ce{KCN}$), despite having some level of covalent character. We're somewhat bastardising the Winstein model of nucleophilic substitution in which we define $\mathrm{S_N1}$/$\mathrm{S_N2}$ based upon the separation between the cation and the anion at the point when substitution takes place, with the two mechanisms being extremes of a continuum:

If you haven’t already, the Mayr review that Jan’s answer cited is worth a read. They don’t mince their words - they basically say that HSAB theory (and the related concept of charge vs orbital control) is a myth and directly criticise a passage from March which is almost the same as what you quoted! Further contradictory observations include the reaction of TBAB cyanide with MeOTf and Me3OBF4, both of which lead to C-methylation, even though both are ‘hard’ methylation game agents. The replacement rationalisation they offer is based on Marcus theory.

The different reactivities of $\ce{KCN}$ and $\ce{AgCN}$ with alkyl halide, giving nitriles and isonitriles, respectively, has often been cited as a prime example of Pearson’s hard and soft acids and bases (HSAB) theory. The reasoning was that ‘hard’ electrophiles preferentially attack the ‘hard’ nitrogen centre while ‘soft’ electrophiles prefer the ‘soft’ carbon centre. This was often further explained by a change from an $\mathrm{S_N2}$-type mechanism (with $\ce{KCN}$) to a more $\mathrm{S_N1}$-type mechanism (with $\ce{AgCN}$). This has been challenged multiple times, and a review paper by H. Mayr, M. Breugst and A. Ofial sums up the evidence against it and provides a much more logical explanation.[1]
The first piece of evidence the authors present is the reaction of syn- and anti-2-bromo-3-(methylthio)butane. Thanks to neighbouring group effects, these two reactions all procede via an $\mathrm{S_N2}$-like transition state, yet the products are as expected as shown in scheme 1.

Scheme 1: Reaction of both the syn- and anti-isomers of 2-bromo-3-(methylthio)butane with $\ce{KCN}$ and $\ce{AgCN}$.[2]
Furthermore, Mayr et al. point out that even in cases where an $\mathrm{S_N1}$-like mechanism is impossible, such as the 1-chloroadamantane, isonitrile formation is possible using $\ce{TiCl4}$ and $\ce{TMSCN}$ in dichloromethane.[3]
The authors move on to explain the reactivity of the cyanide anion with their general nucleophilicity scale. In this, two parallel curves are given: one for carbon-attack whose reactivity is generally higher and one for nitrogen attack. They note:

From Figure 5 [sic!] one can extrapolate that the unsubstituted
  benzhydrylium ion ($E=5.9$), a-alkyl benzyl cations ($E$ ca. $3$ to
  $9$), and tertiary alkyl cations ($E$ ca. $8$) will undergo
  barrierless reactions with both termini of the free $\ce{CN}$ ion in
  acetonitrile.[1]

Thus, explaining the ambident reactivity of cyanide can be done on much simpler grounds. In most cases, the attack of the carbon atom to yield a nitrile is orders of magnitude faster than the attack of the nitrogen atom to yield an isonitrile. This changes once the diffusion barrier is reached for very strong electrophiles when both compounds should be obtained. To generate isonitriles, one needs to block the carbon centre. This can be done by using $\ce{TMSCN}$, where carbon is covalently bound to the $\ce{TMS}$ group’ silicon, or by using silver(I) salts. For these, strong silver-pseudohalide complexes are formed:
$$\ce{AgCN + CN- <=> [Ag(CN)2]-}$$
The strong silver–carbon bond prevents attack of the more reactive carbon centre.

References:
[1]: H. Mayr, M. Breugst, A. Ofial, Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.
[2]: J. C. Carretero, J. L. Garcia Ruano, Tetrahedron Lett. 1985, 26, 3381. DOI: 10.1016/S0040-4039(00)98303-5.
[3]: T. Sasaki, A. Nakanishi, M. Ohno, J. Org. Chem. 1981, 46, 5445. DOI: 10.1021/jo00339a050.

I’m reading the Mayr review (your ref 1) now and it’s revelatory. The very first paragraph of the article calls out the exact passage in March quoted in NotEvans’ answer... I don’t want to bounty this as from previous experience it just leads to every answer being upvoted, but I sorta wish there was a way to highlight your answer as being ‘more in line with modern research’.

Because of high charge to size ratio of Ag(I), it forms a polar covalent bond with the carbon atom, whereas K(I) forms a purely ionic bond with the carbon atom, so the attack as a nucleophile in AgCN occurs from the available lone pair of the nitrogen atom whereas in KCN the attack takes place from the negatively charged carbon atom which is more nucleophilic. However the reaction(with AgCN) is rendered speed if carried out in a polar protic solvent, and so mentioning of the solvent is extremely important.

The ionic radius of $\ce{Ag+}$ is between that of $\ce{Na+}$ and $\ce{K+}$ . I hope, there is not doubt that $\ce{NaCN}$ is ionic.

Bonds are never purely ionic, especially when the prospective anionic center is a relatively low electronegativity atom like carbon.  Ionic bonding is an approximation that is reasonably accurate for some interactions.

I know that the bond can never be purely ionic, but the electronic effect of K(I) is such that the bond is much more ionic than you think you would predict comparing their relative electronegativities (or by using the Hanny Smith's formula)


text:
Stability of unsaturated syn/anti 3-ladderane
In an oral exam for physical organic chemistry, one student was asked to explain the differences in the ionization potential ($IP$) of syn and anti tricyclo[$4.2.0.0^{2,5}$]octa-3,7-diene (unsaturated 3-ladderane, see Figure).
Apparently, the $IP$s were as follows:
$IP_{1,syn}=9.04$
$IP_{2,syn}=9.38$
$\Delta IP_{syn}=0.34$
and
$IP_{1,anti}=8.97$
$IP_{2,anti}=9.93$
$\Delta IP_{anti}=0.96$
Koopman's Theorem $IP=-\epsilon_{HOMO}$ might help us here. I think it leads us to an orbital diagram. How do I have to combine the $\sigma$-bonds with the $\pi$-system to be able to explain the difference in the $IP$? Maybe you know a paper for it? I checked out Joseph J. Gajewski's Hydrocarbon Thermal Isomerizations (2004), but he only discusses mechanistic features, no theoretical aspects.

Any through-bond interaction that occurs between the cyclobutane sigma bonds and the two pi bonds will be the same for the syn and anti isomers.  Only the through-space interaction between the two pi bonds will differ in the syn and anti isomers.  In the through-space case the syn isomer will show a greater splitting between the two pi bonds.  Are you sure the IPs you've assigned to the anti isomer aren't really the IPs for the syn isomer?

I copied it from an exam protocol. It might very well be that the examinee mixed it up. Do I follow you correctly? You are saying, that the splitting in the syn-config should be larger because of the antisymmetric through-space overlap (through bond being equal). Consequently, the HOMO of the syn would lie higher, i.e. less stabilized, i.e. smaller IP (which would be less negative according to Koopman). This would indeed be indicative of mixing the two IPs.

Yes,the HOMO would be slightly destabilized and the orbital below the HOMO (HOMO-1) would be slightly stabilized in the syn isomer due to any through space orbital interaction. In the anti isomer the HOMO and HOMO-1 would occur at roughly the same energy.  The difference in IP1 and IP2 in the anti isomer is attributable to the electrostatic repulsion from placing a second charge in the molecule.  In the syn isomer you also have an even greater electrostatic effect plus the through-space effect.  It would be interesting to separate these two effects.  I wouldn't be surprised if the

electrostatic effect alone accounted for most of the difference.

I found a paper, that investigates this very question: 10.1002/cber.19731060105


text:
Causes for the irregular electron configuration of the lanthanides
I know what the electron configurations of the lanthanides are, but I was asking myself, why they are so irregular.
The configutation of Lanthanum is $\mathrm{5d^1\ 6s^2}$, but according to the Klechkowski rule, the $\mathrm{4f}$ orbital has a lower energy level and should be filled up before the $\mathrm{5d}$ orbital. Why isn't that happening?
The irregulatity of Gadolinium is more comprehensive. The electron pairing energy is higher than the energy difference of fitting that electron in the $\mathrm{5d}$ shell.
Then we come to Ytterbium which has every shell filled up. So why does it's most stable oxidation number is still $\mathrm{+III}$? I would understand why it can leave 2 electrons (it's outermost shell is the $\mathrm{6s}$-shell) but after that, we have to remove a single electron of the filled $\mathrm{4f}$ shell. Isn't that extremely unfavorable?
I hope that somebody can explain why all this happens, I would really appreciate it.

related: http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

The relative energy of the orbitals is not static. It depends on the charge of the nuclei and electron configuration. Klechkowski rule is just a rationalization and just as most rationalizations it does fail. In general, the less nodes an orbital has, the less is its energy (which is why $n+1 s/p $ orbitals are filled before $n d$ orbital) However, the precise relative energy is in a very delicate balance between 'node' factor, charge factor and repulsion with core electrons. It cannot be fit into universal simple rules, even if ad-hock explanations may be invented.