text:
Stability of Allenes and Alkynes,
In Allenes, the $\ce{C=C=C}$ structure makes sure that the $p$-orbital of central carbon overlaps with 2 other carbons and the $p$-orbital of central carbon spreads out over the 3 carbons.
In Alkynes, the $\ce{C#C-C}$ structure makes sure that the energy released during formation of alkynes.
I also read that more the number of non-delocalized $\pi$-electrons on an atoms, the lesser its stability, though I do not know why.
When, it comes to comparing Allenes and their isomeric Alkynes,it has always been stated that  "Allenes are more stable than Alkynes".
One reason I found was that $\pi$-electrons in Allenes are less strongly bound than those in Alkynes.

Why are $\pi$-electrons less strongly bound?  If the reason in incorrect, what is the actual reason?
Why is it that as the number of non-delocalized $\pi$-electrons increases on a carbon, the smaller its stability?

The shape of the two bonds (see the first page of this .PDF) shows why pi bond electrons are less tightly bound. Sigma bonds are formed pretty much along the line connecting the two atomic nuclei, whereas the pi bonds are formed above and below that same line. Naturally, then, the distribution of the latter less concentrated (having more space to cover) than that of the former. Loosely stated, the shape of the pi bond distribution is more roundabout whereas the sigma bond directly touches the two nuclei which makes the electrons in it less strongly bound.
When it can, delocalization of electrons provides greater stability for the molecule because the electron density is distributed well over the delocalized regions, rather than, say, confining it in a small high-energy region surrounded by low-energy regions.

Think about resonant structures. Benzene (C6H6) is a good example. So the inverse of that is: less delocalization, less stability. Hope this helps.


text:
How to estimate reaction products of competitive binding?
I have two reactions,
$$\ce{A + C \rightleftharpoons AC}$$
$$\ce{B + C \rightleftharpoons BC}$$
I'd like to estimate the relative steady state proportions
$$\frac{[\ce{BC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$
and
$$\frac{[\ce{AC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$
without numerically solving any differential equations, given $[\ce{A}]_{init} = [\ce{B}]_{init}$ and thermodynamic properties: $\Delta G, \Delta H, $ and $\Delta S$.
I'm hoping that there is a way to generalize this for more than two reactions in a linear way. It is for a DNA microarray optimization problem in which computational speed is important. Any thoughts?

Are the A,B, and C in the same vat? Also, what is the limiting reagent? What are the rates? Lastly, sorry for being nit-picky, but the double headed arrow is for resonance structures, not equilibrium reactions.

A, B, and C are in the same solution. We can assume that the forward rates are equal for both reactions and the reverse rates are determined by $K_{eq}$ from the Van't Hoff equation. We can also assume excess $C$. If the formula turns out to be pretty ugly, I was hoping for an order of magnitude estimate (some function of both $K$'s).

Why would you have to solve differential equations for this problem?  What algebra have you tried so far?
If C is in excess, then there is no competition!  You should be able to model the reactions as just $\ce{A <=> AC}$ and $\ce{B <=> BC}$.  The equilibrium equation would be (for A) $K_a=\frac{ac}{A_{init}-ac}$, and you can apparently already calculate $K_a$ from the thermodynamic parameters, so that is an equation where the only unknown is $ac$, the amount of AC formed.  The equation is linear!
$ac=A_{init}\frac{K_a}{1+K_a}$
A similar result holds for B of course.
Once you can calculate $ac$ and $bc$, you should be able to calculate the parameters you want to know very easily.
If C is not in excess, the equilibrium equations will not be linear and the balance on C will couple the two equilibria (competition happens!).

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Is it safe to mix potassium alum powder with magnesium oil (made from water and magnesium flakes)?
Will mixing these two together cause a reaction of any sort? If so, is it safe and is there an order in which one should be added to the other?

Safe for what ? Not to create a bomb, not to kill an ant or safe for touch ? Always add eventually relevant  circumstances and background to prevent wrong assumptions and requests for clarifications.

If you are not sure and you have to do the reaction, try it with small sample of $\ce{Mg}$ such as one flake. $\ce{Mg}$ flakes should react in water and will react with potassium alum.

Potassium alum is $\ce{KAl(SO4)2*12H2O}$, the compound that you use to stop bleeding after a bad shave.
I'd say that it won't react, but don't put your health in the hands of strangers on the net.

Finely divided magnesium in water corrodes, liberating hydrogen.  "magnesium oil?"  That is magnesium chloride in water.  It is New Age hype.

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What causes the SN1 reaction?
The SN1 is a two-step mechanism which begins with heterolysis of the C-X bond. What triggers this? Are there external causes that prompt the leaving group to simply leave?
This question is more conceptual than anything, but I can't seem to find answers anywhere.

Recall that SN1 heavily favors protic solvents (polar/H+). In such cases, the nucleophile to-be-leaving-group (to be Nu-) is going to be attracted to the positively charged molecules of the protic solvent (...H+) and away from the molecule it is attached to. The power of the attraction comes from the dipole between C-Nu, where Nu has the negative bit of the dipole.
At the end of this step, when the nucleophile attached group leaves, you often see "Nu-" hanging around, but really it's with the solvent, e.g. H3O+Br- , and it's fairly common to omit the solvent here and in other similar ionic situations, if I'm not mistaken.
Examples that invoke hydrogen bonding make this more apparent.
F-C...(tertiary) + H2O
The dipole here looks like (δ-)F-C...(δ+) because F is more electronegative
The F is strongly attracted to the closer H: (δ+)H-(δ-)O-H(δ+)(δ-)F-C...(δ+) (pretend the HOH is bent :)

Greater nucleophilicity of the attacking group than the nucleophilicity of the leaving group causes SN1 reactions.
The cleavage of the C-X bond keeps occuring always, both the reactant and the product remaining in the solution. The equilibrium shifts towards the final product.

I don't see how this could be remotely correct. The SN1 mechanism is an elementary reaction, there is absolutely no equilibrium involved. The dissociation is in the ideal form independent of the nucleophile.

@Martin-マーチン Isn't the first step of SN1, that is dissociation into ions, an equilibrium? I don't mean to question the rest of your comment, which I think is correct.

I am referring to the dissociation of the ions being in an equilibrium. I am pointing out the fact that the dissociation of ions takes place not just in the reactant, but in the final product as well.  The equilibrium shifts towards the final product since it is more stable.


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Why is the ionization enthalpy of francium greater than that of cesium?
Why is the ionization enthalpy of francium greater than that of cesium, even though it has a larger size? I found no Google result regarding this.

Francium has a slightly greater ionization enthalpy than Cs. This could due to the Relativistic Effects

Here's a very similar question. The answer is the same, as barium and radium are also in the s-block, merely one element to the right of your pair.

The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.
This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.


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How do I distinguish between internal energy and enthalpy?
I am finding it hard to distinguish between the two. Can someone explain the two terms and the difference between them?
I tried learning from Wikipedia, but it said for both the above terms that it is the total energy contained by a thermodynamic system.

Internal energy, $U$, is the total energy contained in a thermodynamic system. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why:
Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy also (like radiation), but these two are common. Thus,
$$dU = q + w$$
The heat transfer can be determined easily by calorimetry, where $q=cdT$, where $c$ is the heat capacity of the system and $T$ is the temperature of the system. Alternatively, it may be easier to measure the temperature change of surroundings with known heat capacity, since $q_{sys}=-q_{surr}$
Work is a little more challenging to determine, since work is defined as the change in volume at constant pressure. $w=-PdV$ (the minus sign is there to show that when the volume of the system increases, the system does work on the surroundings). However, unless you have a closed system, volume changes are difficult to measure. Additionally, pressure may not be constant.
Thus,
$$dU=q+w=cdT-PdV$$
Internal energy changes can be determined easily at constant volume $dV=0$ so $w=0$ so $dU=cdT$. Internal energy changes can also be determined in closed systems at constant pressure (so $PdV$ is evaluatable).
However, in open systems (like a beaker), $dV$ is hard to determine, making $w$ hard to determine. A new thermodynamic state function was devised to remedy this problem.
Enthalpy is a measure of the total energy of a system. More precisely, it is a thermodynamic potential. Enthalpy is defined as $H=U+PV$. $PV$ is a state function, even though $P$, $V$, $PdV$, and $VdP$ are not state functions. Thus the change in enthalpy is:
$$dH=dU+d(PV)=cdT-PdV+PdV+VdP=cdT+VdP$$
Enthalpy changes are easy to determine at constant pressure (most open systems are constant pressure systems, which is why enthalpy is so useful to most chemists), since $dP=0$ and $dH=cdT$. Enthalpy changes can also be determined in closed systems of constant volume where the pressure change can be determined.
To summarize, internal energy and enthalpy are used to estimate the thermodynamic potential of the system. There are other such estimates, like the Gibbs free energy $G$. Which one you choose is determined by the conditions and how easy it is to determine pressure and volume changes.

At constant volume: $dU=cdT$ and $dH=cdT+VdP$
At constant pressure: $dU = cdT-PdV$ and $dH=cdT$
At constant volume and pressure: $dU=dH=cdT$
At variable volume and pressure, $dU$ and $dH$ are both hard to
determine (but $dH$ may be easier since you may be able to numerically estimate $d(PV)$ from pressure-volume data).

Why do you say P and V are not state functions by themselves? They are.


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Relation between number of equivalents of reactants
Consider the reaction $\ce{Sb2S3 + F2 -> SbF5 + SF6}$. If the milliequivalents of antimony, sulfur and fluorine are x, y and z respectively, then prove that $x+y=z$.
This is what I know: Number of equivalents of each reactant in the reaction is equal, and that number of equivalents is (mass of species) /(equivalent weight). The problem I'm facing is that $\ce{Sb}$ and $\ce{S}$ are in the same compound, so I think $x$ should be equal to $y$ (this is clearly incorrect). Also, the mass of reactants used is not mentioned, so I don't know how to proceed with the question.

Step 1: Balance the equation. Your equation as written $\ce{ Sb2S3 + F2 -> SbF5 + SF6}$ is not balanced. You do not have the same numbers of sulfur, antimony, or fluorine atoms on each side of the equation. You cannot say anything about equivalents or stoichiometry with an unbalanced equation. Hint: Start with $\ce{Sb}$ and $\ce{S}$ and balance $\ce{F}$ last.
Step 2: Determine the relationship between $x$ and $y$. Since $x$ is antimony and $y$ is sulfur and they are in the same compound initially, they have a relationship (but they are not equal). In $\ce{Sb2S3}$, there are 2 sulfur atoms for every 3 antimony atoms. What does that tell you that the ratio of $\frac{x}{y}$ must be?
Step 3: Use what you now know about $x$ and $y$ and the correct stoichiometric equivalents (which should give you $z$) to show that $x+y=z$.

I think you have confused equivalents with moles, there is no need to balance the reaction while applying equivalent concept. Please read this page http://www.askiitians.com/iit-jee-chemistry/physical-chemistry/stoichiometry-and-redox-reactions/law-of-equivalents.aspx

I'm not sure I have. Equivalents in the definition you link to have to do with solution concentration. You can also use 'equivalents' to refer to ratios of reagents added to a reaction: "bromobutane (1.0 equiv.), Mg (1.2 equiv.), and acetone (4 equiv.) were added...". In this case the equivalents are dependent on the stoichiometric ratio (i.e. if the ratio of bromobutane to Mgwas 1:2, then 1.2 equiv. of Mg would be 2.4 moles of Mg for every mole of bromobutane). This use of equivalents is arbitrary (amounts chosen by the experimenter). Thus I assumed this question was about stoichiometry.

I agree with @AyushPateria, balancing isn't required. Further, you said, "there are 2 sulfur atoms for every 3 antimony atoms" in step 2. It's 3 sulfur atoms for 3 antimony atoms right? I think x/y would be 2/3, but I don't understand your step 3. Could you elaborate on that?

The law of equivalence states that one equivalent of an element combine with one equivalent of the other. In a chemical
reaction, equivalents and milli equivalents of reactants react in equal amount to give same number of equivalents
or milli equivalents of products separately.
Accordingly
in reactions like aA + bB -> mM + nN
the following holds good  : meq of A = meq of B = meq of M = m.eq. of N. In a compound MxNy
meq of MxNy= meq of M = meq of N


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Is there a way to turn cellulose into sugar at home?
I know that once you have sugars, you can ferment, distill, and dehydrate to create 95+% ethanol. I've also read about cellulosic ethanol, with the new step there being the breakdown of the cellulose into simpler sugars.  Is there a relatively cheap way to do that yet, or it still only in expensive research labs?

If you want to do it on a small scale at home, why not just buy sugar: it is likely to be a lot cheaper than trying to make it from cellulose.

I'm thinking specifically about how to convert waste grass clippings into useful fuel.

A method used to break down cellulose into sugars is hydrolysis :
You could use enzymes to hydrolyze the cellulose to produce glucose, but enzymes cost a lot and as you are looking for a cheap way, using enzymes is not an option. A second method is to use an acid (e.g. sulfuric acid) to produce the glucose (just google "cellulose acid hydrolysis"  and you'll find some useful links)

Thanks. Do you know of any prices online for the enzymes? I'm curious exactly how much "a lot" is.  But yes I'll look into the acid hydrolysis.

Sigma-aldrich sells enzyme solutions for around 76.9(USD) for 50ml (and for 250ml it is 211(USD))(however this varies from product to product)

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Plutonium having more oxidation states than samarium?
Why does plutonium have more oxidation states than samarium?
Electron configuration of Pu: $\ce{[Rn] 5f^6 7s^2}$
Electron configuration of Sm: $\ce{[Xe] 4f^6 6s^2}$
I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..$\ce{f^6}$..$\ce{s^2}$?
Elements such as scandium (Sc), Yttrium (Y), and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..$\ce{d^1}$..$\ce{s^2}$, but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.
Does this have to do with the ionization energy, the energy level, or...?

This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.
The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.
In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers and have significant coordination chemistry using $f$ orbitals. Curiously, as you go from curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.
This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals.

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Explaining the oxidation states of palladium
Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr]  4d^{10} 5s^0}$
The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$.
How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$?

Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.

Sodium nitrate, $\ce{NaNO3}$

When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.
$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$
Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.

Palladium nitrate, $\ce{Pd(NO3)2}$

Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.
Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.
$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$

But he didn't say solvated in water- Of course the ion in salt going to have some covalent nature.

The OP did not say solvated in water, but then there will always be a counterion.


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Which of these thermodynamic relationships is true for a system at equilibrium?
Which of the following is true at equilibrium
$$T \Delta{S} = \Delta{H}$$
$$T \Delta{s} = \Delta {G}$$
I think it's the first one but I just wanted to be sure.

At equilibrium, $\Delta {G}$ is zero, so according to the equation $\Delta {G} = \Delta {H} - T \Delta {S}$ the first option is correct.


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What is the mechanism of action of anaesthetics?
A range of very different compounds are used in medicine as anaesthetics. They don't seem to have much in common chemically but they all seem to keep people asleep while medics are doing nasty things to them.
Inhaled anaesthetics include a variety of fluorocarbon gases such as halothane and sevoflurane plus barbiturates like thiopental, oddballs like propofol and even xenon:

What do these heterogeneous compounds have in common and how do they work?

Comments are not for extended discussion; this conversation has been moved to chat.

This answer is a general answer, based on some research.
The mechanisms behind anaesthetics is not entirely understood, according to the article Mechanisms of Anesthesia: Towards Integrating Network, Cellular, and Molecular Level Modeling (Arhem et al. 2003), primarily due to, as you alluded to in your question, the fact that anaesthetics come from a very diverse group of chemicals.
In the fragmentary data presented in the article, the conclusions the authors made were that

general anesthetics modulate the activity of ion channels, the main targets being GABAA and NMDA channels and possibly voltage-gated and background channels, thereby directly or indirectly hyperpolarizing neurons in thalamocortical loops, and thereby disrupting coherent oscillatory activity in the cortex. In our view, it does not seem unreasonable that the ultimate target is NMDA channels.

There is also a suggestion that the unconsciousness that results is in a way like sleep, an adaptive response to the chemical input.
Interesting perspectives are provided in the article Why anesthetic mechanism research
has failed, and what to do about it (Hammeroff), where the author suggests, research is needed in investigating anaesthetic mechanisms by

Look closely at the quantum nature of anesthetic binding by van der Waals
  London forces.

In terms of local anaesthetics, the slides included in the presentation Local Anesthetics: Overview state that the mechanism here is due to

Ionized form of weak base
  blocks $\ce{Na^+}$
  channels
  by binding to an internal
  sequence involved in channel inactivation

One of the perhaps most important theory of anaesthesia is that:

Most anaesthetics enhance the activity of inhibitory GABAa receptors
  and other cys-loop ligand-gated ion channels. Other important effects
  are the activation of a subfamily of potassium channels (the two-pore
  domain K+ channels) and inhibition of excitatory NMDA receptors

Almost all anaesthetics (with the exceptions of cyclopropane, ketamine and xenon) potentiate the action of GABA at GABAa receptors (1)

At the cellular level, the effects of anaesthetics are to enhance
  tonic inhibition (through enhancing the actions of GABA), reduce
  excitation (opening K+ channels) and to inhibit excitatory synaptic
  transmission (by depressing transmitter release and inhibiting
  ligand-gated ion channels).(2)

There is some controversy about whether or not xenon potentiates GABAa responses but at present the weight of evidence suggests it does not.
Ketamine is believed to act by blocking activation of the NMDA receptor. (NMDA receptors are also  an important site of action for anaesthetics such as nitrous oxide, xenon )

$\ce{GABA}$ receptors are ligand-gated Cl− channels made up of five subunits (generally comprising two α, two β and one γ or δ subunit).
Anaesthetics can bind to hydrophobic pockets within different  GABAa receptor subunits.
Specific mutations of the amino acid sequence of the α subunit inhibit the actions of volatile anaesthetics but not those of intravenous anaesthetics, whereas mutations of the β subunit inhibit both volatile and intravenous anaesthetics.
This suggest that volatile anaesthetics may bind at the interface between α and β subunits (analogous to benzodiazepines that bind at the interface between α and γ/δ subunits:

Putative anaesthetic binding sites on GABAa receptor subunits. [A] A model of the α1 subunit of the GABAa receptor with a molecule of isoflurane shown sitting in a putative binding site. The transmembrane α-helices (TM) are numbered 1–4. [B] A model of the β2 subunit of the GABAa receptor with a molecule of propofol shown sitting in the putative binding site. (Adapted from Hemmings HC et al. 2005 Trends Pharmacol Sci 26, 503–510.)
References:

Olsen, R.W., Li, G.D., 2011.GABAa  receptors as molecular targets of general anesthetics: identification of binding sites provides clues to allosteric modulation. Can. J. Anaesth. 58, 206–215.
Schüttler, J., Schwilden, H., 2008. Modern anesthetics. Handb. Exp. Pharmacol. 182.
Rang and Dale’s Pharmacology  8th ed H P Rang et al
Goodman & Gilman's The Pharmacological Basis of Therapeutics, 12th Edition

It's been suggested by Thomas Heimburg's group in Denmark (perhaps others) that anesthetics work by a common depression of the gel/fluid transition temperature in nervous membranes.
Some virtues of this view include its consistency with the action of anesthetics being proportional to their partition in lipid (Meyer-Overton rule... and notice this rule involves the same puzzle you raise, namely why this apparent diversity shares some property). Another virtue of the view is its explanation of the pressure-reversal of anesthesia in a quantitative manner.
My (admittedly superficial) understanding is Heimberg's group (and perhaps others) is using this as one main bit of ammo against the familiar Hodgkin/Huxley neural model. In the paper linked, they suggest it is not necessarily inconsistent with a molecular mechanism (of anesthesia), though of course the problem you raise happens to be the motivation for their thermo' picture. Maybe the two cannot be consistent, but that is not the author's statement (if anything it is due to my ignorance). I would be interested to see more comments. Good question!


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Why Fe(CO)5 exists, but Fe(NH3)5 doesn't?
Like in title: Why $\ce{Fe(CO)5}$ exists, but $\ce{Fe(NH3)5}$ doesn't?
Edit:
As Jan Dvorak pointed out, it may exist. If so: why is it so unstable?

How do you know it doesn't exist? We can make predictions it would be extremely unstable, we can predict it would spontaneously decompose in STP, we can grep the literature to verify noone has ever synthesized it, but that's all. Of  course, we can still ask why is it so unstable.

@JanDvorak If you elaborate on that a bit, I think it will make for a good answer.

While we cannot be sure that $\ce{Fe(NH3)5}$ does not exist, the bonding between ligand and metal offers a good explanation for why it is expected to be less stable than $\ce{Fe(CO)5}$.
In metal carbonyls like $\ce{Fe(CO)5}$, the bonding between CO and the metal consists of three single bonds, 1 $\sigma$ and 2 $\pi$ bonds. The $\sigma$ bond is formed by overlap of the carbon $sp$ hybrid orbital (which contains the lone pair) with an empty $d$ orbital of the metal. Electron density is donated from the orbital of CO into the metal orbital, so CO acts as a $\sigma$ donor. The two other $\pi$ bonds are formed by overlap of a filled metal $d$ orbital with a pair of $\pi^*$ molecular orbitals of CO. Electron density is donated from the metal orbital into the molecular orbitals of CO, so CO acts as a $\pi$ acceptor.

For the $\pi$ bonding to occur, the metal is required to have sufficient $d$ electrons, and thus preferably a lower oxidation state. This explains the relatively low oxidation state of iron (0) in $\ce{Fe(CO)5}$. The combination of one $\sigma$ and 2 $\pi$ bonds gives the M-CO bond a partial triple-bond character, while the C-O triple bond is weakened by the transfer of electron density into the $\pi^*$ orbitals (reference).

Because of its ability to serve as a $\sigma$ donor and $\pi$ acceptor simultaneously, CO is a stronger ligand than $\ce{NH3}$ which can only act as a $\sigma$ donor with its lone pair in the nitrogen $sp^3$ hybrid orbital. $\ce{NH3}$ lacks a $\pi$ system and therefore cannot act as a $\pi$ acceptor. Because $\pi$ backbonding is crucial for stabilization of a metal center in low oxidation state, ammonia is unlikely to form stable ammin complexes which are analogous to the corresponding metal carbonyls.


text:
Thermodynamics effect of change of temperature on equilibrium
I'm studying thermodynamics and I don't understand, why, when we are asked which effect has a rise in temperature on the equilibrium constant of a reaction, we can't use the formula: $\Delta G= \Delta H-T \Delta S$, and say that if the reation is entropic, the change in free energy will more negative, therefore the reaction will be more spontaneous and the equilibrium shifts towards the products.
In my exercices books, it is said, that, according to le Chatelier's principle,the equilibrium will try to "fight" the new constraint and if it is exothermic,the equilibrium will shift towards the reactants, and if it is endothermic towards the products.
Le Chatelier's principle seems intuitive to me for the compression of a gas: when compressed, the molecules are now more narrow and they combine faster than they separate. But I don't see why it can be used in the raise of temperature.
Thank you in advance

First of all, spontaneity of a reaction depends on the kinetics, not thermodynamics.
Second, I always imagine LeChatelier's principle to work like this: When you raise the temperature for an endothermic reaction, it has more temperature to work with so the equilibrium shifts to the right. And if you lower the temperature for an exothermic reaction, that means there's more room for the energy created by the reaction.
But if you want to use $\Delta G = \Delta H - T\Delta S$ to check how the equilibrium behaves, you will only see the effect the temperature has on the free energy. It relates to the equilibrium via $$ \Delta G = -RT\ln K \Longleftrightarrow K = \exp\left(-\frac{\Delta G}{RT}\right)$$
This means that if you raise the temperature the expression in the exponent gets closer to 0, which in turn means that the overall $K$ becomes larger. So the equilibrium shifts towards the products.
However, I have neglected the influence of the entropy above. The formula should be, when completely written out:
$$ K = \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) $$
And with this you can always see whereto the equilibrium will shift.

Most of this answer is wrong or misleading.  The questioner is clearly asking about thermodynamic spontaneity, which is determined by where the reaction mixture lies relative to equilibrium, not kinetics. And the temperature dependence of the reaction is determined purely by $\Delta H$, not $\Delta S$.   And your statement "This means that if you raise the temperature the expression in the exponent gets closer to 0, which in turn means that the overall  becomes larger. " is just flat-out wrong, as you can see from the second expression you wrote.


text:
Thermodynamics and equilibrium constant
I just wanted to make sure how we can know whether it is $K_c$ (equilibrium constant of concentrations) or $K_p$ (equilibrium constant of pressures) which "comes out" of the equation, $\Delta G^o= -RT \ln(K)$.
Until now, I have always taken $K_c$ when the reaction was in a solution or $K_p$ if it was a gas. But as there is a difference between their values, I'd like to know exactly which one to take.

What you are doing is right. The value to use depends on the definition of the standard state. The values of $K_c$ and $K_p$ are indeed different, but what you are not accounting for is the fact that $\Delta G^o$ also changes with the equilibrium constant used.
The naught over $G$, representing "standard state" establishes some rules. For homogeneous gaseous reaction, the standard $\Delta G$ corresponds to the standard state of all gases, which is defined in terms of pressure $P=\pu{1atm}$, and hence $K_p$ needs to be used with the quoted value of $\Delta G^o$.
For reactions involving solutes and pure liquids, the standard states are defined differently, in terms of their concentrations, and hence $K_c$ is to be used. Since the definition of standard state affects the value of $\Delta G$, we must choose that form of the equilibrium constant which uses the activity of the reacting species in terms of the quantity used to define the standard state of that species. This maintains uniformity.

Thank you for your answer! So, if I have understood well, if the reaction involves the evaporation, let's say, of water. In the standard state, water is liquid so we are going to obtain Kc, the ratio between water vapor concentration (not pressure) in a volume and liquid water concentration in the same volume ?

@Pao Although it still depends on which quoted value of $\Delta G^o$ you are using, but in many heterogeneous cases involving solutions and gases, what you say (all in terms of concentration) is surely one way to go. A better way is to deal with activities, rather than concentrations and pressures. Then, all gases use pressures and all liquids use concentrations, with appropriate coefficients.


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Metallic properties of iodine
Why does iodine exhibit metallic properties at room temperature although being a non-metal? Are there any similar elements like this?

I don't think Iodine is a typically regarded as a metalloid -- see for example https://en.wikipedia.org/wiki/Metalloid Let us all know if there is a contrary reference as iodine is certainly an interesting element L F Ludwig

Iodine is a metalloid. Metalloid elements have one or more allotropes with properties intermediate between those of a metal and a nonmetal.
Metals are lustrous, ductile, and conductive of heat and electricity.
Nonmetals are matte, brittle, and insulators.
Metalloids may be semiconductors, like silicon, germanium, arsenic, and carbon (graphite). They may be lustrous, but brittle, like iodine and tellurium.

But, according to Wikipedia, iodine is a non-metal in lieu of  a metalloid?

From that same Wikipedia article: There is no standard definition of a metalloid, nor is there agreement as to which elements are appropriately classified as such. I classify iodine as a metalloid. My metalloid list also included carbon and phosphrous, but not aluminum.

Fair enough then.

How is iodine a metalloid? Is it not a halogen?

@user3932000 - "Halogen" is the name of the group - the column on the periodic table. "Metalloid", like "metal" and "nonmetal" is a classification of elements based on similar properties. Iodine may be a metalloid and a halogen. Chlorine is a nonmetal and a halogen. Tellurium is a chalcogen (name of the oxygen group) and a metalloid.

All such trends are attributed to periodicity in properties
Generally, going down a group in periodic table metallic character increases which is attributed to the dominance of addition of new principal quantum number over increasing nuclear charge.This results in increasing size of atoms which in turn results in lowering of ionization enthalpy (a parameter that may be used to judge upon metallic nature:lower the ionisation enthalpy,higher the metallic nature).Since ionistation enthapy is inversely proportional to size,hence it decreases down a Group.
That's why in Halogen family Iodine possess metallic character to some extent.
However these terms become relative if we universalize them.For example there are compounds of Iodine like
Fluorides
Iodine fluoride: IF
Iodine trifluoride: IF3
Iodine pentafluoride: IF5
Iodine heptafluoride: IF7
Chlorides
Iodine chloride: ICl
Diiodine hexachloride: [ICl3]2
where Iodine shows metallic character.At the same time we too have
NaI,KI etc. where it acts as a non metal.Hence it become more logical to say it to be a metalloid.(Since also across a Period metallic character decreases.So the two effects viz General increase in metallic character down a group and decrease in it across a period compensate.

Iodine is in fact a metalloid and not a non-metal, metalloids exhibit the properties of both metals and nonmetals partially, so that is the reason for iodines both metallic and non-metallic properties.Another example is silicon.


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Silver-promoted reaction of alkyl halide with diphenyl sulfide
What is the major product formed when 1-chloro-3-iodopropane reacts with diphenyl sulfide in the presence of one equivalent of $\ce{AgBF4}$?

I know that sulfur is nucleophilic and that it can displace a halide in a SN2 reaction, but the resulting molecule will have a positive charge. Is that okay? What possible use does the product have?

In the presence of $\ce{AgBF4}$, diphenylsulfide is alkylated to 3-chloropropyl diphenylsulfonium tetrafluoroborate. The sulfonium salt is stable and can be isolated as a crystalline product.

This sufonium salt can be deprotonated (e.g. with $\ce{NaH}$ in $\ce{THF}$) to the corresponding sulfonium ylide.

In the latter, intramolecular displacement of $\ce{Cl-}$ yields cyclopropyl diphenylsulfonium tetrafluoroborate, which again is a stable solid.
This cyclopropyl derivative, once converted to its sulfonium ylide S may serve as a reagent in the Corey–Chaykovsky reaction.


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How do you determine a chemical formula given the molecule's crystal structure?
How do you determine the chemical formula of a compound given its crystal structure?

Presumably we have some other information as well? I mean, if I gave you a sample and told you it's a CsCl crystal structure, you can't really say much, since a good number of compounds assume the CsCl structure.

Presumably, if you know the crystal structure, you know the unit cell? Count all the atoms in the unit cell taking into account fractional atoms (that is account for shared atoms) and adjust so the ratios are integers and you have the answer.

If you mean something more like "given its X-Ray diffraction pattern", you might want to edit the question to reflect that.

Knowledge of the crystal structure implies that you are also able to describe the unit cell of the compound. To determine its formula, count all sorts of atoms inside the unit cell, also considering fractional atoms which belong to more than one unit cell. For example, an atom at the corner of the unit cell counts as $\frac{1}{8}$ atom because it is shared between 8 adjacent unit cells. Likewise, an atom at the edge of the unit cell counts as $\frac{1}{4}$ as it belongs to 4 unit cells etc. When you obtain fractional values for the atom numbers, multiply the values of all atoms with a certain number so that all values are integers.
Let us take sodium chloride ($\ce{NaCl}$) as a simple example. Its unit cell, which is shown below, contains $\ce{Na+}$ (violet) and $\ce{Cl-}$ ions (green).

Counting the number of chloride ions in the unit cell gives
$$12 \cdot \frac{1}{4}+1=4 \ \ce{Cl-}$$
because there are 12 $\ce{Cl-}$ on the edges and 1 $\ce{Cl-}$ in the center of the unit cell. For the sodium ions, which are situated at the corners and on the faces (count as $\frac{1}{2}$ because they are shared between 2 cells) of the unit cell, we obtain
$$8 \cdot \frac{1}{8}+6 \cdot \frac{1}{2}=4 \ \ce{Na+}$$
The composition of the unit cell is therefore $\ce{Na4Cl4}$. Since the number of formula units per unit cell is 4, the formula of the compound is $\ce{NaCl}$.
For compounds in which the building blocks of the crystal structure are molecules, the procedure is essentially the same.


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What molecular features create the sensation of sweetness?
There are many compounds that taste sweet. Many of the artificial ones taste much sweeter than natural sugars (which is why they contribute little to energy intake when use as sweeteners).
A range of sweeteners is shown in the figure below:

Structurally there are common features for some. We can see why sucralose (a non-metabolisable sweetener based on sucrose) might taste sweet. Rebaudioside a (a major ingredient of the sugar substitute Steria) shares some of those features. But the very sweet artificial sweeteners seem to have less in common.
Do we know what makes molecules generate the sensation of sweetness? Which features matter and why are some sweeter than others?

We don't exactly know what makes molecules smell good or bad, which is why companies that look for such molecules still have "the guy with the golden nose". I think it's something similar with our taste buds.

(Please be noted that your bottom left structure ‘has a typo’ and is rather 6′-deoxyrebaudioside A; and another typo is in “Steria”.)

For humans, the "sweet taste buds" are the T1R2+T1R3 & T1R3+T1R3 G-protein coupled receptors.
Chemicals that are sweet are the ones that can bind to these receptors. We know some structures that tend to do this, namely ketones and aldehydes (carbonyl groups).
However, there are chemicals that are sweet that we do not know the mechanism for, such as Curculin, which is up to ~2000 times sweeter than sucrose by weight.

While technically interesting to a biochemist, this doesn't tell me anything useful about the detailed molecular features of molecules that taste sweet. Could you explain furter or give more detail.

In response to: "Do we know what makes molecules generate the sensation of sweetness?"
Yes, their affinity to bind to the aforementioned receptors, and carbonyl groups (ketones and aldehydes) are typical indicators of sweetness.
And the topic isn't completely figured out, e.g. why Curculin is sweet is unknown.

^ Grammar weirdness; I mean
a. their affinity to bind to the receptors, which may not have told you anything.

Pressed enter by accident. Apologies!
b. molecules that do this tend to have aldehyde/ketone groups on them.
My last point was simply that we currently can't enumerate all the groups that can bind to the active sites of these receptors. Sorry if it wasn't too much help.

@rch You may want to incorporate the comments into the answer :)


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Balanced reactions at the anode and cathode
A current of 2.25 A is applied to  $\ce{NiCl2}$ solution
A. Write the balanced half reaction that takes place at the anode
B. Write the balanced half reaction that takes place at the cathode
Can someone check if my answers seem logical? I'm not quite sure if I did the problem correctly
Anode: $\ce{O2 + 4H+ + 4e- -> 2H2O}$
Cathode: $\ce{2Ni^{2+} + 4e- -> 4Ni}$

Can you expand on why you think that answer should be correct i.e. why should the species selected by you get reduced/oxidised in the way you think they should?

sure. when I looked at the standard reduction potential for the Nickel its value was -.25 V. When I looked at the Cl its value was 1.36. I knew that Ni was being reduced because because it's value was more positive than the -.83 of the standard reduction rection 2H20 + 2e- __> H2 + 2OH-. Then I looked at the Cl2 and compared it to the equaiton I had above for the anode. The reason I picked that as my answer is because the it had a standard cell potential of 1.23 and Cl2 was 1.32. Since 1.23 is LESS than Cl2, I chose it to be my answer for the oxidation half reaction

You have written two reduction reactions.  One of them needs to be an oxidation reaction.  In a voltaic cell, where does the oxidation occur?

the oxidation occurs at the anode

At the A-node you O-xidize, at the C-athode you R-educe. (Vowel to vowel, consonant to consonant.)
Following this, you will have to generate electrons at the anode, which doesn't happen with your reaction.
What reaction will happen depends on the normal potential of each reaction, but I assume it will be something along the lines of
$$\ce{2Cl- -> Cl2 ^ + 2e-} $$
Edit: I just noticed that the cathode reaction can be simplified to
$$ \ce{Ni^{2+} + 2e- -> Ni} $$

ok thank you for showing me the proposed anode! I keep getting the reduction oxidation equations with the water and H20 mixed up with the actual cathode/anode equations.

The reaction at cathode seems OK, but at the anode another reaction will occur.
Hint:
At the cathode, 4 e$^-$ are consumed. The electrons have to be "generated" in the system.
Remark:
The value of current applied is irrelevant in this kind of question. More relevant will be the applied potential.

Can you please explain to me what you mean by current? Is that the equation I initially put in for the anode.

I meant the Electric current


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Units in stoichiometry
$$C_x H_y O_z$$
1.What are the units of x,y,z.I've read that they're termed as 'stoichiometric coefficients'.
2.What is the unit of mass proportion of carbon(c) in the fuel?
$$c=\frac{M_{carbon}}{M_{fuel}} .x$$

The stoichiometric coefficients is something else than you suggest.
stoichiometric coefficients:
The number of molecules of a given component that participate in the reaction as written.
The indexes $x, y, z$ represent quantities of atoms in a compound. So they do not have any units.
@Q2 Do you know what are $c, M$ and $x$ in the equation? If you do, you can answer your question. If you do not, than you should find it out.

The book doesn't say what M stands for.But,I'm guessing its the molecular weight(g) or the molar mass(g/mol).Which one of these would be of apt use here?

If we assume that both $M_c$ and $M_f$ have the same units than what you got from the dimensional analysis?

I do not want to give you straight answer as the question seems to be part of a homework.


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Why does ZnO turn yellow when it is hot?
Why does ZnO's color become yellow when it is hot? I have read that it is because of physical properties, but I need a more elaborate explanation.

I read on wikipedia that it is a thermochrome compound (see: http://en.wikipedia.org/wiki/Thermochromism )

If anyone is interested, this video shows the effect very well, and also touches on the explanation.

IT is most likely Thermally induced charge transfer between the inherently unstable oxide ion and the Zinc ion resulting in an electronic transition that absorbs blue light. Observe it under a photographic safe light it will not change color.

See https://en.m.wikipedia.org/wiki/F-center

The thermochromism of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed.
Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.

So heating causes $\ce{2O^{2-}}$ to be oxidized to $\ce{O2}$? That seems unlikely to me. Does something else happen to the $\ce{O^{2-}}$?

This may explain the chemical change but it doesn't explain the reason behind the colour change. What causes the colour?

F centres are the reason for the colour see e.g. https://en.m.wikipedia.org/wiki/F-center

O= is endothermic I propose that  Zn++ O=  becomes Zn+O-  This can be checked by heating in an inert, argon atmosphere.  The color change should still be reversible.

When you heat $\ce{ZnO}$ then Oxygen leaves as $\ce{O2}$ leaving behind $\ce{Zn^{2+}}$ and 2 electrons. $\ce{Zn^{2+}}$ and the 2 electrons move to the interstitial sites of the crystal . This provides excess electrons in the crystal lattice of $\ce{ZnO}$. When light falls on these crystals then these electrons absorb a part of the light in the visible region and hence impart a yellow colour to the $\ce{ZnO}$.

Most probably because the band gap closes, thus allowing electrons from the valence band to be elevated into the conduction band (by absorbing blue light, which makes the reflected color yellow).

Upon heating ZnO, anionic vacancies are created and when an electron occupies that vacancy it imparts yellow colour (metal excess defect/f-centre).

It would be nice to flesh this answer out a bit, particularly on the color centers and their population vs temperature in thermodynamic equilibrium.

In $\ce{ZnO}$, $\ce{Zn}$ is present in the 2+ oxidation state, and the d subshell has 10 electron, and the s subshell has 0. When $\ce{ZnO}$ is heated there is a transition of electron from d to s subshell, causing it to become yellow, and on cooling it becomes white.


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Cheap and not messy solvent for motor oil
I have an old oil cooler of  an old Volkswagen engine (the type where the oil itself is cooled, and uses no water) stained with decades of usage. If I pass water through it, it comes out clear, but if I pass a clear oil through it, it comes out black.
What is the best method to clean it that uses cheap and easy to find substances? I want to use it to cool clean mineral oil, I want my oil to remain clean.
My father suggested to pass mineral oil through it until what comes out is clean, then install it in the system, but that would be horribly expensive and messy, and I would need to discard all that dirty oil properly afterwards. I myself thought of using gasoline, filling halfways, shakink and discarding, until clean enough, but although that would be less expensive, it would be as messy as the first option, and probably dangerous. My sister suggested using water and kitchen wash-up liquid. That would be much less messier, because it is water soluble, but I don know the proportion of water and kitchen detergent tI must use for it to be effective. My last idea was to use some grese remover like Ajax, that would not be so.much messy, either.
So, what do you suggest?

Car repair shops often have a solvent wash station.  The ones I have seen use kerosene, paraffin, white spirit or diesel to cut through the grime.  You could locate shop in your area and ask if you can spend a little while at their wash station, they may have a similar radiator with just a broken bracket lying around they can offer you that is otherwise unused.

In terms of inexpensive, safe, and a little polar to move oxidation products, I'd try soaking in and flushing with hot soybean oil (replace elastomer hoses and seals).  Then flush with kerosene (odorless white spirits, hardware store  paint section), then blow out with compressed air.
Carburetor cleaner was mixed cresols - dangerousy on skin (heavy gloves), but great against deposits.  If it is really nasty in there, I'd go for gun bore cleaner soaking, specifically Ed's Red without the lanolin and substituting MEK (methyl ethyl ketone) for the acetone to lower flammability. At the end, blow it out, flush with kerosene, blow it out.
Contain all wastes.  Beware of skin contact (gloves) .  Do not breath the fumes (ventilation, and maybe a charcoal latex paint mask, then discard).  Eye protection (goggles).  VAPORS ARE FLAMMABLE AND HEAVIER THAN AIR.
http://www.youtube.com/watch?v=2OfaCCJjBEU
Ed's Red Gun Bore Cleaner  (wear gloves!)
1 part Dexron II, IIe or III ATF, GM Spec. D-20265 or later.
1 part Kerosene - deodorized, K1 (can be Kroil or WD-40)
1 part Aliphatic Mineral Spirits, Fed. Spec. TT-T-2981F, CAS #64741-49-9, or may substitute "Stoddard Solvent", CAS #8052-41-3, or equivalent, (aka "Varsol")
1 part Acetone, CAS #67-64-1.
1 lb anhydrous lanolin or  Lanolin, Modified, Topical Lubricant.
Note that Ed's Red is rapid death to plastics, polymers, varnishes, paints, and coatings.

The lanolin is persistent, to deter gun rusting.  That, of course, is not your problem.  Clean, lubricate, protect is the gun challenge.  Weapon Shield is remarkably good - but you could buy another cooler for the cost of using it in this application.

You can use kerosene instead of gasoline in a well ventilated area, but your best bet will probably be grease remover.
Trying different things will let you know what is effective, and each pass will clean the cooler a bit.
Just don't leave it a long time with water, as that might rust it.


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What is the scaling for % completion time in a Michaelis–Menten scenario as a function of substrate concentration?
Say I have a straightforward Michaelis-Menten reaction where an enzyme $E$ at some fixed concentration $[E]$ forms an interaction complex $ES$ with a substrate $S$ (with initial concentration $[S]_0$), and at a rate $k_{cat}$ can irreversibly convert the substrate $S$ to a product $P$.  The rate of interaction complex formation $ES$ is $k_f$ and its dissociation rate is $k_r$, such that overall, we have the reaction:
$E + S \rightleftharpoons^{k_f}_{k_r} ES \to_{k_{cat}} E + P$
Is there an explicit expression for the time $\tau_{v}$ to convert some percentage $v$ of the substrate $S$ to product $P$?  In lieu of an explicit expression for $\tau_{v}$, what scaling for this time should we expect for $\tau_{v}$ as a function $[S]$ (holding $[E]$ fixed)?
Ultimately, for $\tau_{v} \approx 1$ I suppose the steady-state approximation will begin to fail?

In short, yes.
To obtain the relation between τ and v, we need the actual time evolution of the concentration of the product by solving the differential equations (DEs) and then some way to get steady-state concentration. The DEs are non-linear. So we have to find approximate solutions. This paper develops a solution. It's not nice..
I will probably come back to add more. For the time being, I highly recommend those interested to browse through the paper.
And what do you mean by τ≈1 leading to failure of steady-state approximation please?
Edit: The tittle of the paper is "Analytical solution of coupled nonlinear rate equations : I. Michaelis-menten kinetics" by Phillipson

Can you provide the paper title?  The link you provided blocks people outside of your institution.

Hi, I fixed the link and added the title. Enjoy.


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Quantum Valence Theory Reference
At least to me, the contents of chapters 5 & 6 given here are representative of discussions on chemical bonding given in nearly all intro general & organic chem books, the statements in those chapters are mainly memorization (to me anyway). I basically don't find it rigorous enough, leads to more rather than less confusion and doesn't give any in depth discussion (similar to the way the quantum mechanics sections of those books are rote assertion ignoring all of the math).
I'm in the middle of a physics-based quantum mechanics course, so I'm developing the conceptual & mathematical tools to read a serious explanation of lewis bonding, polar-covalent bonding etc... but my book doesn't mention much in detail on this topic either, and apparently not even Pauling's QM book mentions much on the topics.
I'm really hoping somebody would have a quick look at the contents of chapters 5 & 6 of the first link & offer a recommendation of a rigorous reference or four on the contents of those chapters - I really can't face the headache I'll get reading chapters like those again due to all the confusion they lead to by dumbing it down, I need something more rigorous if such a thing exists, thanks a lot :)

Your Essentials of Physical Chemistry seems to be heavy on the "essentials" and light on the "physical chemistry". A textbook used in a physical chemistry course (and not a "physical chemistry for __" course) is what you need. The second portion of the traditional physical chemistry sequence is devoted entirely to the implications of quantum mechanics in chemical bonding. Pauling's book on QM will not cover much modern bonding theory since that book predates most modern bonding theory. The applications of QM to chemical bonding in all but the simplest molecules require computer programs to slog through the maths, since multielectron wave functions suffer from the three(+) body problem and cannot be solved exactly.
For example, in Physical Chemistry by Peter Atkins and Julio De Paula (W. H. Freeman & Company), Chapters 8 through 20 deal with chemical bonding and phenomena first with quantum mechanics and then with statistical mechanics. There are other physical chemistry texts out there as well. These texts will also cover the computational methods seeking to find 1) exact solutions to approximate multielectron wavefunctions (treating multielectron wavefunctions as linear combinations of single electron wavefunctions) and 2) approximate solutions to exact multielectron wavefunctions.


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Explaining a biochemical reaction (protein labelling, e.g. with fluorescent probes)
I am an immunology master student without prior knowledge of biochemistry.
Could someone explain to me the following 6 reactions?
What I want to understand is why exactly the reactions happen at this site ( in terms of donor/acceptor ).  What is the donor, what is the acceptor?

In all cases the electron donor is the heteroatom of the protein... Maybe you might want to look at nucleophilic substitution.

In my opinion are to many reactions I think nobody could fully explain them all in one answer maybe you should make for every one a question with the reaction name in the title so is more useful for other people who might search for the same problem...

All reactions you gave are typically used in protein labelling, e.g. with fluorescent probes.
In order to allow further measurements on the labeled protein, the following conditions have to be met:
The bonds formed have to be stable like the peptide bonds in your protein. You don't want the probe flying around somewhere.
Proteins tend to be a bit sensitive to degradation; boiling in acetic acid is obviously not an option to run the labeling. As a consequence, the functional group to be attached needs to be activated, e.g. with good a leaving group, to allow the labelling of the protein under mild conditions. The succimidyl ester in the first reaction is a good example.

1) Good leaving group for displacement.
2) Cf: polyurethanes and truck bed liner (polyureas).
3) Nucleophile conjugate addition.
4) Good leaving group for displacement.
5) Good leaving group for displacement.
6) Schiff base formation.


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

When an electron transitions from a higher energy state to a lower energy state, the energy has to go somewhere. Usually it is emitted as light of a given wavelength which corresponds directly to the change in energy of the electron.
The energy of light at a given frequency $\nu$ is
$$ E = h\nu\; , $$
where $h$ is Planck's constant.
The frequency of light is directly related to the wavelength $\lambda$ and the speed of light $c$ via
$$ c = \nu\lambda\; . $$
Combining these two equations we get
$$ E = \frac{hc}{\lambda}\; , $$
with which you should be able to calculate the $\Delta E$ values for the transitions.


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Is it possible to cut yourself on pyrite?
I am planning an RPG adventure where the characters might end up getting a large cubic lump of "gold", which in reality, as you might have guessed, is in fact pyrite.
I also wanted to plan for the contingency that the group wouldn't notice that it's not the real deal, so I thought I could maybe let one of the characters cut himself on the crystal. A quick google search yields images such as this:

(Source)
This looks pretty sharp and possibly blood-drawing, but is there a way to know for sure? Any recommendations as to how to handle sharp minerals? Any first-hand experiences?

Yes it is possible.

+1 for properly assigning the source of the image, which, btw., is really beautiful. No experiences on cutting on crystals, though.

The Wikipedia page for Mohs scale shows that pyrite has a hardness of 6-7 (according to the page on pyrite, it's 6-6.5) while plate glass has a hardness of 5.5 and a pocketknife has 5.1. I've cut myself on both knives and glass before, so I'd wager that pyrite should be able to slice.
The catch is that I've never cut myself on a cube of glass or a cube of steel before--a 90 degree edge just doesn't generate enough pressure for that. So either your pyrite must have very nice serrations on the edge or it must have an edge sharper than 90 degrees.
Is that possible? I don't know. I suppose at this point you could invoke the leeway granted to you by fiction.

Anything that can have an edge will cut you, assuming it is thin enough. This is why some science fiction novels will describe 'super sharp' swords as having a width of a single atom/molecule.
To address pyrite, specifically, we'd have to know the shape of your planned fool's gold.

Is it a lump of it, a bunch of coins or some other 'non-sharp' form?  Those probably wouldn't cut anyone.
Is it a 'golden' sword?  If it was thin (sharp) enough then it probably could.

To address your question about the handling of sharp materials - wear gloves, and don't touch the sharp side.  If you can avoid touching it entirely (with tongs, or by lifting whatever it is resting in/on) even better.  As a general rule of thumb contact with the pointy bits is more likely to make you lose that thumb than no contact.

The shape would be a cube, more or less as depicted by the image. So I reckon, probably you as well, that it's possible...


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What is the pKa of 2-methyl-2,4-pentanediol?
What is the pKa of 2-methyl-2,4-pentanediol? Is there a reference that contains a list of pKa values for various molecules that I can consult or is there a way to determine/estimate the pKa of this molecule if such a reference does not exist?

Do you mean pKa? pH is a property of a solution, and is dependent on both the identity of the solute and the concentration of the solute. pKa is an intensive property of a substance that assesses the relative acidity of that compound.

@BenNorris, I guess I did. I know the relationship between the pKa and pH. I just wondered if there was, for example, a table of known pHs for 1M solutions. I don't particularly need to know a precise pH. I wanted to know if it was acidic or not, and how acidic if so.

In addition to the pH vs. pKa confusion, there's also the determine vs. reference problem.  Do you want to know of methods that can determine the pKa of a material or do you want a link to where these properties can be found?  In any case, please edit your question to reflect what you are interested in learning.

Do you mean 2-methyl-2,4-pentanediol?
The $pK_a$ values of 2-propanol (isopropanol) and 2-methylpropan-2-ol (tert. butanol) are 16.5 and 17, respectively and I'd expect your diol to be in this range range, unless deprotonation is featured by intramolecular hydrogen bonding. In that case, the $pK_a$ might be somewhat smaller, around 15.


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Solutions, solvents and solutes
In all the chemistry books and eBooks it says that "like dissolves like"... but when it comes to some molecules like ether, cellulose, vitamin A and $\ce{CO_2}$, it confuses me.
If we take ether, it is true that it will not form hydrogen bonds with water when it dissolves, but they have polar bonds between $\ce{O}$ and $\ce{C}$ and it's not in a linear form like $\ce{CO_2}$ where the dipoles cancel out.
Vitamin A also has $\ce{-OH}$ group but it still is hydrophobic. In some book, I read that this is because it has more hydrophobic molecules or that they have more $\ce{C-H}$ bonds than $\ce{-OH}$. If it is so, octadecanol and nonadecanol also have way more $\ce{C-H}$ bonds than $\ce{-OH}$ groups, so  why is it still soluble? $\ce{CO_2}$ is dipolar — the dipoles cancel out, so why is it still soluble? Cellulose also has a bunch of $\ce{-OH}$ but still it is soluble. I am very confused by this.

Cellulose is not soluble.. Maybe you should change your question title to something like this "Why solubilities of ether, cellulose, vitamin A and CO2 are these." No one with your same doubts can find your question with a title so generic...

When you make assumption for guessing the solubility of the molecule in some solvent you should consider the whole molecule, his structure and his size.

We can start with the ether, in the below figure (from wikipedia the one at right ) is shown a diethylether molecule.

You can see that is quite a non-polar molecule (charge 0,green, in the most part of the surface) so this explains why has a low solubility in a polar solvent like water. Same thing for Vitamin A:

The most part of the molecule is not polar. Only one $-OH$ can't make the difference. And so is for 1-Octadecanol that has a solubility in water of $1.1  \times 10^{-3} mg/L$ and Nonadecyl alcohol (1-nonadecanol) seems something like $4.8  \times 10^{-3} mg/L$ both up to my references have very low solubility in water (where did you find that were not?). For the $CO2$ is true what you have stated it has not a very strong polarity (but in fact it has a little bit!) because is a linear molecule.

This last example let me introduce to an important factor that influence the solubility: molecule size.
Little molecule are more easily solvatated by solvent molecule, this bring us to the case of cellulose:

Cellulose is very hydrophilic and so should be soluble in water, but this doesn't happen because cellulose is a polymer (in figure is show his monomer) that can have a molecular weight very hight. Polymers may swell but can't be solubilize if their chains are bonded with chemicals bonds until the bonds are break.


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Reaction intermediates of MnO2 catalyzed H2O2 decomposition reaction
Manganese dioxide catalyzes the decomposition of hydrogen peroxide to water and oxygen gas.
But what are the intermediates in this catalyzed reaction?

Reaction conditions
As Watts et al. have shown, the decomposition products of this Fenton-like reaction strongly depend on the $\mathrm{pH}$ of the solution.1
If performed in acidic conditions, the reaction generates mostly hydroxy radicals, but no reductants (which would be the hydroperoxide and superoxide anions). If, conversely, the reaction is held in neutral conditions, Watts et al. have shown that the reaction produces significantly more of the aforementioned reductants.
Do et al. have conducted further research into the reaction mechanism at $\mathrm{pH}=7$ and concluded that the observed reaction order may be approximated as a pseudo 1st order reaction, where the ratio $\ce{[H2O2]}/\ce{[#MnO2]}$ is vital to the description of the decomposition rate.2 They have further shown that reactive intermediates, such as superoxide and hydroperoxide anions are generated by the reaction. After slightly modifying the pH towards alkaline conditions, the production rates for the reactive anions increased drastically.
Reaction mechanism
Do et al. have meticulously assembled a table with a proposed reaction pathway:

And finally, I quote one of their closing statements:2

The existence of hydroperoxide/superoxide anion implies that the suggested reaction mechanism could be explained by hydrogen peroxide being decomposed, not only directly on the surface of manganese oxide, but also through a propagation reaction involving intermediates such as hydroperoxide/superoxide anion in solution.

References

Watts, R., Sarasa, J., Loge, F., and Teel, A. Oxidative and Reductive Pathways in Manganese-Catalyzed Fenton’s Reactions. J. Environ. Eng., 131(1), 2005, 158–164, DOI link.
Si-Hyun Do, Bill Batchelor, Hong-Kyun Lee, Sung-Ho Kong. Hydrogen peroxide decomposition on manganese oxide (pyrolusite): Kinetics, intermediates, and mechanism. Chemosphere, Volume 75, Issue 1, March 2009, Pages 8-12, DOI link.

I am adding a more recent study to complement the answer given by tschoppi. This journal addresses the kinetics and mechanism for the decompostion of $\ce{H2O2}$ on transition metal oxide surfaces. However, the paper only considers $\ce{ZrO2}$, $\ce{TiO2}$, $\ce{Y2O3}$. Although  $\ce{MnO2}$ is also a transition metal oxide, $\ce{MnO2}$ may be different enough to lead to different paths or kinetics. The authors even state:

effects such as
  solution pH, type of oxide, temperature, and oxide particle size
  have profound effects on the kinetics and energetics of this type
  of reactions

Thus, take the results of the study with a grain of salt.
I will focus more on the mechanism part of the journal. According to the journal

The kinetic experiments on the decomposition of $\ce{H2O2}$
  together with the experiments on $\ce{HO•}$
  detection show the existence of an adsorption step prior to decomposition. This type of process is also predicted with the DFT calculations. The
  decomposition of $\ce{H2O2}$ follows a similar mechanism for the
  three metal oxides studied. The obtained transition states are
  largely mediated by hydrogen bonding between $\ce {H2O2}$ and surface $\ce{HO}$ groups. Nevertheless, direct interaction between the oxygen atoms of $\ce{H2O2}$ and the metal atoms present in the
  oxide was also observed in the geometries of the transition
  states. The formation of two $\ce{HO}$ radicals as the primary
  product of the decomposition of $\ce{H2O2}$ is confirmed with both the DFT calculations and the experiments. One of these radicals can further abstract a $\ce{H}$ atom initially bound to a surface $\ce{O}$ and form $\ce{H2O}$. The other $\ce{HO}$ radical can adsorb to the surface by forming bonding states with the metal cation.

Reference

Claudio  M.  Lousada,
Adam  Johannes  Johansson,  Tore  Brinck,  and  Mats  Jonsson. Mechanism of $\ce {H2O2}$
Decomposition on Transition Metal Oxide
Surfaces J. Phys. Chem. C, 2010, 116 (17), DOI link.

you said that the statement,
effects such as solution pH, type of oxide, temperature, and oxide particle size have profound effects on the kinetics and energetics of this type of reactions. Makes the paper not that credible.
If you could please state any reason why you said so ,that would be really helpful for me to further understand the problem statement.


text:
Carcinogens, how do they work?
The easiest carcinogenic thing for me to grasp is radiation, as it directly messes with DNA. Then it seems there are other compounds that simply mimic hormones, but these shouldn't necessarily cause cancer, right? Then there are things that can accelerate mitosis or mess with other processes, which can inadvertently mess with DNA. So my question is, in general, how many different flavors of carcinogens are there? I know there are a lot, but can someone list a good few? Maybe some examples? I'm most interested in how plastics and things like smoke cause cancer. I think a few examples would be best in aiding my understanding.
Thank you!
Edit: I also just realized this is the Chemistry Stack Exchange, I think I am more interested in the chemistry side of this question, but it might need to be migrated.

There are at least two different radiationless pathways to wreak havoc on the DNA:

intercalaction: compounds like benzo[a]yrene or ethidium molecules slide between DNA strands and cause transcription errors
alkylation: compounds like N-Nitroso-N-methylurea may transfer alkyl groups to nucleobases

But the list is by no means complete.


text:
Are there natural conditions that could enable the formation of noble gas compounds?
Noble gases were considered to be inert until compounds that include them, such as xenon trioxide (as an example) were found.
My question is, what natural conditions allow the formation of noble gas compounds?

Natural? As in compounds you could find on Earth outside of a lab? I find that highly unlikely (other than clathrates), because most noble gas compounds are derived from the fluorides after direct exposure to fluorine, either with heating or light. Molecular fluorine is never found free because it is far too reactive, and it would be promptly consumed by anything else in the vicinity of the noble gas. The noble gas compounds themselves would be highly unstable and prone to decomposition. The only possibility I can think of is in extremely pressurized environments, perhaps in the mantle.

@NicolauSakerNeto yes, natural - I would appreciate any peer reviewed references that detail any research pertaining to the question.

I can give an example to make you know, one of the condition. Let us understand as to why $He_2$ molecule is not formed. This can be easily understood by calculating the attractive and repulsive forces when two helium atoms approach each other. Each helium atom has two electrons in its 1s-orbital. There are two attractive forces between the nucleus and the electrons of each atom. It has been found that when two helium atoms approach each other, new four attractive forces and five new repulsive forces come into play. Therefore, repulsive predominates and the potential energy of the system...

...increases which leads to unstability. Thus, $He_2$ represents unstable state and a chemical bond is not formed between helium atoms.

I bet that electrical discharges in evacuated metal tubes, filled with noble gases and halogenes don't count as natural conditions, although the compounds formed were happily emitting in the uv :D

This is just thinking out loud, but could this type of reaction occur with lightning? @KlausWarzecha

@Amaterasu Honestly, I really don't know whether a situation like in an excimer laser might happen in the atmosphere under the extreme conditions of a lightning. I wouldn't even dare to speculate on that.

Here is a video about xenon in sand (theory) http://www.youtube.com/watch?v=fZ7_CyLpwPc

As noticed in the comments, there are probably no natural conditions that would allow formation of noble gas compounds; although it should be noted that definition of 'natural' conditions should be specified. Lightning bolts surely carry enough energy to force such a reaction, but it is impossible for them to meet right mixture of gases, due to scarcity of both noble gases and super-strong oxidants and abundance of other substrates; as noted in comments (again), substrate able to create a compound with noble gas will create it with everything else first, and noble gas last.


text:
Program that simulates basic reactions in organic chemistry
To explain - a friend and I have been working on a program that would allow you to draw organic molecules, indicate the conditions, and predict the products.  This would be primarily aimed at basic, first year undergrad organic chemistry (SN1/2, E1/2, applications, potentially (retro)synthesis, etc).
We've realized that it is probably worthwhile to figure out if there are similar existing programs out there.  From what we've read there doesn't seem to be something that meets what we're planning - everything we've found seems to be aimed at professional/higher level chemists, not assisting a beginner.
Basically it comes down to a couple main points:

Are you aware of similar software?
Do you think that this sort of software would be used (either by yourself or students or whoever).
If/when we have a working version, would users here be interested in using/testing it?

I apologize if this doesn't meet the guidelines/standards for questions here, I was kind of iffy about asking it.

As you mentioned, there've been quite some attempts on CAOS (Computer-Assisted Organic Synthesis), such as

WODCA from the Gasteiger group
CHIRON from the Hanessian group
LHASA from the Corey group

I might be awfully wrong, but I think they were/are all closed source.
But you might want to have a look at OSET, an open source Java project on which Ivan Tubert-Brohman worked some ten years ago.

Indeed, I am not aware of any open source efforts. Particularly, there are not yet any open databases of reactions.

A few more links to those listed above:

Peter Ertl's synthetic accessibility score (SAscore)
SYLVIA
CAESA
ArChem - also called Route Designer

There was also a lot of hype a few years ago about an effort called Chematica, but I have not seen an actual shipping product (commercial or otherwise).

I have been working on a similar project (VirtualChemist), which is an organic reaction software focusing on regiochemistry and stereochemistry.
Using reactant Lewis structure as the only user input, it can (1) describe detailed reaction mechanisms at the level of electron flows in elementary reaction steps, ensuring that all reaction steps are not only atom and electron balanced but also atom-mapped; and (2) track molecular 3-D structure changes in every elementary reaction steps. In each reaction step, users can start animation and look at 3-D structures from different perspectives for each molecules and isomers, to aid the understanding of stereochemistry.


text:
How to distinguish between benzyl chloride and chloro benzene?
If I add $aq. \ce{KOH}$ to both of the compounds substitution takes place in the case of benzyl chloride and then if I oxidize with $\ce{KMnO4}$ I will get benzene carboxylic acid which will give effervescence on reaction with $\ce{NaHCO3}$. But, does $\ce{KMnO4}$ have any effect on chloro benzene?

When benzyl chloride is treated with $\ce{NaOH}$ and $\ce{AgNO3}$ , a white precipitate is formed.
Chlorobenzene does not form white precipitate reacting with $\ce{NaOH}$ and $\ce{AgNO3}$.

Can you expand on why this is so and the reactions involved?

Nice! However, you can force the transformation of chlorobenzene to phenol at ~ 450 °C when streaming mixtures of water and chlorobenzene (in the vapour phase) over oxidic catalysts. That's the old (and long replaced) phenol synthesis by Raschig.

Your observations are correct and on the short term, i.e. in the time scale of typical reactions in the lab, you'll hardly see an effect of the permanganate on chlorobenzene.
However, there seemingly have been attempts to decontaminate soil, e.g. on production sites, using permanganate for the complete oxidation of pollutants to carbon dioxide. But these processes are supposed to take much longer ( > 1 year) and usually don't play a role in considerations on what will happen under typical lab conditions.
But referring to the title, I'd rather suggest to simply record $^1H$ NMR spectra of the compounds. Even on an ancient 90 MHz spectrometer, you'll be able to spot the signals of the benzylic protons at $\delta$ = 4.5 ppm.


text:
How can different no. of unpaired electrons lead to different colours of co-ordination compounds?
To make the question clear I would like to take an example. $\ce{K_3[Fe(CN)6]}$ and $\ce{K3[FeF6]}$ shows different colors in solutions. It is given in my textbook that it is because of different number of unpaired electrons.
But, how can the no. of unpaired electrons affect the frequency of light emitted?
The colors of co-ordination compounds are due to d-d transition of electrons. $\ce{CN^-}$ is a strong field ligand while $\ce{F^-}$ is a weak field ligand. Thus, there will be difference in the crystal field splitting energy in both of the given compounds. This, results in a different energy gap between the eg and t2g energy levels in them. Thus, when electrons undergo transition different frequencies corresponding to different energy gap is emitted.
I was hoping this as the answer. But, in my textbook the reason is attributed to only the difference in no. of unpaired electrons. Why is it so?

The crystal field splitting does definitely contribute. But it is an unpaired electron which usually shifts between various available orbitals and paired electrons rarely do so. Moreover, there might also be spin inversions (see Hyperfine structure) which might give additional lines in the spectra. Presence of unpaired electrons helps in producing additional energy levels in electric and magnetic fields (zeeman and stark effect) providing additional lines in the spectra. All these provide establish the uniqueness of the spectral distribution radiated, and hence the colour.

@SatwikPasani Thankyou.....now I understood the concept behind it.Can you lease post this as an answer with any further explanations(if you would like to add) so that I can accept it as the answer.

The crystal field splitting does definitely contribute.But it is not the only thing to contribute to the colour of the coordinate complex solution. The colour is equivalent to the spectra (emission or the absorption) of the solution, and hence by enumerating the factors contributing to the emission spectra of a solution, we can compare the colours of two different solutions.
It usually is an unpaired electron which shifts between the various available orbitals and paired electrons rarely do so, since the energy required to unpair acts as a barrier for its' transitions. Moreover, there might also be spin inversions (see Hyperfine structure) which might give additional lines in the spectra. Presence of unpaired electrons helps in producing additional energy levels in the electric and magnetic fields of the ligands (zeeman and stark effect), providing additional lines in the spectra. All these establish the uniqueness of the spectral distribution which the solution absorbs, and hence the colour.
The number of unpaired electrons also determines the resultant magnetic field of other species in the solution, which alters the energy levels of the solvating molecules.


text:
What is the process of turning ionic silver solution into colloidal silver solution?
I've looked at a number of different colloidal/ionic "silver solution generators."  After my research I'm under a few impressions and if they're wrong please let me know.
I also am wondering what the product is that one generator sells to add to the ionic silver to create colloidal silver and is it safe to ingest?  I've already seen enough information to believe that ionic silver is safe. I'm wondering if the additive that turns ionic silver into colloidal silver is safe.
Impressions:
When you set up a silver generator, either homemade or store bought, you send a current through two pieces of silver in distilled water.  The particles released into the water are ONLY ionic silver.
Will colloidal silver form over time?  Will ionic silver precipitate or remain suspended?
I could easily build a machine to do this rather than buy one from a dealer that seems to withhold info to either prevent comparisons or to prevent laymen from building their own.  I'm looking for any and all comments related to this topic.  Thanks
To reply to GM's answer: I hesitate to put up pictures or links to the commercial generators since I'm not sure if that would violate the TOU and I have no interest in advertising for anyone.  I imagine if the commercial generators, which work on same principal as the home made ones, didn't generate at least ionic silver they'd have some explaining (lawsuits) to do to someone and they've been on the market for years.

Polyphenol in green tea may be a better reducing agent. Maybe, possibly, cook ionic silver then put in a cup of green tea. Using distilled water for both. Or is this just too simple

A common reducing agent would be ascorbic acid(vitamin C). I am not sure though if it can reduce Ag+ to colloidal Ag.

Ingesting colloidal silver is a bad idea. I'd trust the Mayo Clinic more than I would some quack web site. http://www.mayoclinic.org/healthy-lifestyle/consumer-health/expert-answers/colloidal-silver/faq-20058061

@MaxW Thanks for the concern but this isn't a debate about the merits or how to use it. The question is "how does, or can, one colloidal silver to ionic silver.

If you provide a scheme of this mysterious  "colloidal silver generator" would be very interesting, from what I see from googling I see only things like this.

Up to my knowledge this "colloidal silver generator" has no sense. From instructions notes it seems there are two silver bar in a solution of pure water, as you say, this in fact should create a very slow (because there are no electrolytes) electrolysis of water into $O_2$ and $H_2$ but I don't see any reasons for the producing of colloidal silver.
In chemistry laboratory to produce colloidal silver usually you start from a solution of ionic silver $Ag^{+}$ and reduce it to metallic silver $Ag$.
$$\ce{Ag^{+} + e^{-} -> Ag} $$
Caution the following is a reaction not a recipe do not ingest the ultimate product of this reaction!!!
Once time I've made a true colloidal silver with Lee-Meisel method (J.Phys. CHem 1982,86,3391-3395) if you look at the original paper there are many methods however the following (method c) is used most.

(c) $AgNO_3$ ($90 mg$) was dissolved in $500 mL$ of $H_2O$ and brought to boiling. A solution of $1 %$ sodium citrate ($10 mL$) was added. The solution was kept on boiling for cs. 1h.

The solution should be greenish yellow. If you examine the reaction the $AgNO_3$ is the source of $Ag^{+}$, sodium citrate is the reducing agent that reduce $Ag^{+}$ to $Ag$ and stabilize it. I think colloidal silver generator would be for anyone who makes colloidal solution a dream I assure you is a very complex and tedious process.
Moreover both $Ag^{+}$ and colloidal silver are very difficult to store, the first may  undergo photo-reduction to the metal the second is unstable as many colloidal solution because colloidal solution have grater energy due their larger surface (More energy many times means more instability). Without any stabilizer (such as t-BuONa or citric acid) I don't think the colloid will last long.
However I don't see why you should drink something like this. In fact $Ag^{+}$ has antiseptic properties such as  Mercurochrome,HydrogenPeroxide or Lysol(R) but this don't mean drinking Lysol, Hydrogenperoxide or mercurochorme every morning will make you feel better. I strongly suggest you to do not drink such thing ever.

Thanks for the input.  People have been ingesting various silver concoctions for hundreds if not thousands of years.  A small handful of people have turned blue b/c they've made improper concoctions or taken way too much. The silver ends up in their skin and when they expose themselves to the sun they basically develop the silver, similar to photography, and turn a bluish color.  This is way beyond very rare though and so far no other negative side effects have been exposed... ionic/colloidal silver is a known topical and internal antiviral, antibacterial, antifungal and is sold as such.

@hortstu people have ingested various toxic concoctions for thousands, this doesn't mean you have to continue to ingest them. There are many more effective and selective antibiotics. $Ag^{+}$ kills every biotic form and may cause to you argyria.

other than argyria, which is extremely rare amongst people using silver, what other toxic effects are proven? None that I can find. There are extensive studies showing the benefits of silver.  "Effective and selective" antibiotics is a big part of the reason we have antibiotic resistant bacteria are becoming a bigger and bigger problem.  Finally antibiotics don't work on viral or fungal infections.

also this isn't relevant to the question.

@hortstu I'm quoting from your question "and is it safe to ingest?" for me the answer is "no":  "In 1999, the FDA prohibited the sale of over-the-counter drugs containing colloidal silver or silver salts because they had not been shown to be safe and effective.". $Ag$ builds up in the tissues of the body moreover up to my knowledge there is no evidence that $Ag^{+}$ can have effects on viral infections. However you are free to do what you want I only tell you what a know!

If I worded my question poorly I apologize. I meant is the additive that turns ionic silver into colloidal silver safe for ingestion. I already believe silver in moderation to be safe... and we can debate the efficacy and intent of the FDA somewhere else.

@G M   "In 1999, the FDA prohibited the sale of over-the-counter drugs containing colloidal silver or silver salts because they had not been shown to be safe and effective."  The FDA never proved Vioxx to be safe or effective either but they allowed that on the market... Also they may not allow it in drugs but as a dietary supplement it is commercially available.   http://store.peacefulmountain.com/products/ionic-colloidal-silver

@hortstu ok, however keep to investigate!  I hope someone else could give you a better answer. Maybe you should change the title including "How "colloidal silver generators" works?" I hope your question will have more attention from community!

@ GM thank you. I appreciate your help. I think I understand how the generator works to create ionic silver. I'm trying to find the steps from there to colloidal.

@hortstu There are better ways to get rid of money. I suggest pouring lighter fluid on it and throwing a lit match on top.  More seriously though there is no proven benefit from silver ingestion.  Argyria will make you purple for the rest of your life. If you are dead set on ingesting silver, just buy an ounce of silver nitrate and make a dilute stock solution but keep it out of light.  An ounce will cost about $30 at most and last for life a silver generator will rip you off for $300 to give you what you could have bought for $30.

Adding to G M's answer, there are more ways to produce colloidal, or more properly nanoparticles from silver nitrate.
The fastest one is using sodium tetrahydroborate (or borohydride). It takes a couple of seconds and if you don't deliberately put something to the solution nanoparticles will be stable for months (most probably longer). A polymer can be used to prevent coagulation of colloids yet as explained we did not need.
Citrate is also an option or any other similar organic acid but they are not as an effective reducing agent as borohydride so you need to heat the solution almost to boiling to get some activity. On the contrary, we always used ices to at least reduce the activity of borohydride to control the reaction.
Other than these, I have also synthesized silver nanoparticles by brewed, boiled or simply left tea, and also Artemisia absinthium (there are actually tens of articles on making silver nanoparticles using inorganic or organic chemicals, or even direct extracts of tea orange etc.
I emphasized nanoparticle more than colloid here since you will only have colorful silver solutions when particles are in between 10-100 nm in diameters. Yellow for smallest and black for largest, red and purple/violet between.
In short, these are the proper methods to produce silver nanoparticles in chemical or green synthesis.


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How can molecularity be the order of the rate determining step?
In my textbook the molecularity is defined as(along with the standard definition) the order of the rate determining step(RDS). It is also given that the order can be a fraction but molecularity is always a whole number. But, then if the order of RDS is a fraction molecularity will be a fraction which is a against the definition. So, where is the problem? Is it that the order of RDS can only be a whole number?

What your book, and most other elementary books present, is at best a generalization, not applicable to all cases.
Molecularity, as per the collision theory, is defined only for elementary reactions. One of the things that define an elementary reaction is the fact that the entire transformation from reactants to products occurs simultaneously, and follows from a properly oriented, appropriately energetic collision of the reactant species. If the reaction is not a direct consequence of a single collision, it is not an elementary reaction, and hence molecularity is meaningless.
Now, what usually happens is that out of several elementary steps constituting a reaction, one is very slow as compared to others. So whatever the rate of that elementary reaction, becomes practically the rate of the entire reaction. And since it is an elementary reaction, its order is its molecularity, and to an approximation, the molecularity of the slowest step then becomes the rate of the equation.
But, in most cases, the slowest step is not slow enough for it to entirely determine the rate. It is the slowest step, and it is an elementary reaction and therefore its molecularity is its order. But now, its order is not necessarily the order of the entire reaction, since that was an approximation valid only in cases where the slowest step was so slow that it entirely determines the rate. Then, the order of the entire reaction, due to the influence of the other steps (elementary in itself), can be fractional, while the order of the slowest step, is still its integer molecularity. ( here the term slowest step is preferred over the misnomer rate determining step, as it does not entirely determine the rate).
Also keep in mind, that many a times what people refer to as the "rate determining step" or the "slowest step" is not an elementary reaction. then its order need not be an integer, even if it is slow enough for the order of the net reaction to be equal to the order of that step.