text:
Why do only chiral molecules rotate the plane of polarized light and how do they rotate it?
I have learnt that only chiral or unsymmetrical molecules can rotate the plane of linearly polarized light. But, why is it so?
And how can molecules rotate the plane or what does it actually mean by 'rotating'?
Does it mean that the electric field vectors are tilted or turned left or right depending on the type of isomer?

See also: How do Optically Active Compounds Rotate Plane Polarized Light?
This is because optical rotation is a chiral phenomenon.
Take a molecule, and draw arrows depicting the polarization of incoming and outgoing light on it. Even if the molecule is achiral, the molecule with the arrows is chiral.
Chirality can't spring out of nowhere, it can only arise from a chiral process.
Also, see my answer here. Since molecules will exist in all rotations in a given fluid, for an achiral molecule the mirror image molecules cancel each other out.

ManishEarth's second link provides an absolutely excellent explanation for why chiral molecules in solution cannot simply be reoriented.
I would like to add, however, that technically, molecules of the D2d symmetry can rotate light **only if ** they are in a crystal. These are known as birefringent crystals, and they only show optical activity in the crystal form: light going in the x-axis is rotated one way, light going in the y-axis is rotated the other way, and the z-axis is averaged to zero rotation.
There is a paper I've read on this subject, but it's nowhere near basic (I don't even pretend to understand half of it).


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What exactly is a spontaneous reaction?
What I understand by a spontaneous reaction is that - A reaction whose G is negative and which occurs without the input of energy.
Gasoline reacts with oxygen only if a spark is provided. So how can it be a spontaneous reaction?
Do nonspontaneous reactions occur in nature? If so, are they always slow? If they are slow then what is the difference between a slow spontaneous reaction and a nonspontaneous reaction?
Are nonspontaneous reactions active energies more than that of spontaneous reactions?

Spontaneous reactions simply mean that the products will be lower in free energy than the reactants ($\Delta G$ is negative). If $\Delta G$ is negative, then it is energetically favourable for the reaction to occur - in other words, there will be a release of energy due to the reaction.
The magnitude of $\Delta G$ (how large it is) does not say anything about the reaction speed. For example, the reaction between gasoline and oxygen in the atmosphere at room temperature will be extremely slow, despite a large, negative $\Delta G$. This is because this reaction has a large activation energy, $E_A$.
Hold on. When two reactants collide, they interact to form an extremely unstable structure called the transition state. The transition state is high in energy and is NOT energetically favourable, and thus the transition state quickly collapses again and becomes either the reactants or the products. You can see this in the following picture.

Depending on the kinetic energy of the reactants, they might not reach the top of the curve, in which case they will fall back to being reactants. But if they collide fast enough and are positioned appropriately with respect to each other, the top of the curve is reached, and the products will form. The energy required to form the transition state is the activation energy.
With gasoline and oxygen, the activation energy is high. So even though a tremendous amount of energy is released during the reaction, it needs a spark to overcome its activation energy. From here on, the heat generated from the reaction supplies the activation energy.
If this reaction was not spontaneous, we might be able to force the reaction along, but the reaction would not continue on its own.
As you know, not all reactions run to completion. This is because Gibbs free energy is dependent on concentrations of reactants and products, so as the products accumulate, and the reactants are used, $\Delta G$ becomes closer to 0, eventually reaching equilibrium, where $\Delta G = 0$. Visualise this scenario on the above picture. If there is no difference in Gibbs free energy, the reactants will still get to the transition state and turn into products. But the energy required for products to reach the transition state and becomes reactants is equivalent, and thus the rate of the forward reaction is equal to the rate of the reverse reaction.
The difference between reactions that will reach an equilibrium and the ones that run to completion is the magnitude of the $\Delta G$ at standard conditions (denoted $\Delta G°$). This is a measure of how energetically favourable the reaction "inherently" is. If $\Delta G°$ is large and negative, the reaction will run to completion. If it is smaller, the reaction will establish an equilibrium at some point.
That's at least generally true for reactions, where all of the species stay in the same phase I think. If you envision a reaction where one of the products is a gas, that escapes the solution of reactants, then the products will be unable to reform into reactants.

The word 'spontaneous' has different meanings in everyday life and this is unhelpful. I prefer to think of a spontaneous reaction as one that 'is allowed to occur', without any prediction of how fast the reaction occurs. If a reaction is not allowed to occur - i.e. it is non-spontaneous, it cannot occur whatever kinetic tactics (catalyst, higher reactant concentration) we try, except that if we heat up the system, the reaction might be allowed to occur at the higher temp because delta G changes.
My preference would be not to use the word 'spontaneous' at all for chemical reactions but to define a reaction that is allowed as one with a large equilibrium constant. The link between free energy change and K means that a big K means a large negative delta G. The reason for this approach is that equilibrium constants (K) are direct experimental evidence of reaction - a big K means that the reaction has virtually gone to completion.

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How do I write the complete and net ionic equations for Insoluble compounds?
So, I was doing my homework and had to write out the complete equation and net ionic equation for a combination of Lead (II) Iodate and Sodium Hydroxide.
So my equation is given as:
$\ce{Pb(IO3)2_{(aq)} + NaOH_{(aq)} -> ?}$
However my dilemma is that both of these compounds seem to be insolvable and I'm not really sure that I can either write the complete equation or even the net ionic equation, since I have no idea what to expect. If I do it on Wolfram Alpha I get the resulting products as:
$\ce{H2 + I2 + 2Na + 4O2 + Pb}$
However I have no idea if this is right/and or what's even going on here?

I'm not sure what Wolfram Alpha is doing; its answer is wrong!

It seems to me that WolframAlpha is merely breaking the reaction down into its balanced "standard standard" compounds.
Also, I'm not quite sure how sodium hydroxide is insoluble. I just dissolved a whole boatload of it less than 12 hours ago.

That might be a redox reaction Wolfram Alpha is giving you.

While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (111 g/100 mL at 20 °C). The reaction that occurs is a precipitation:
$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -> Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$
Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.

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How can I go about obtaining the activation enthalpy of a reaction using the Arrhenius Equation experimentally?
I plan to do a string of experiments on the kinetics of the decomposition of hydrogen peroxide. I’m going to see the effect of catalyst concentration on rate and the effect of varying hydrogen peroxide concentration on rate, which will then allow me to formulate a rate equation for the reaction.
I would also like to calculate the activation enthalpy of the reaction specifically using the Arrhenius Equation $k=A\operatorname{e}^\tfrac{-E_\mathrm{a}}{RT}$. How would I do this? As far as I’m aware I should use varying temperatures in the reaction to get different values of the rate, but I’m not sure.
So what kind of experiment should I do (varying temperature to see the change in rate?)?
What should I do with my results?
Is the value of $E_\mathrm{a}$ obtained from analysing a specific graph that will be drawn up, and if so what graph would this be?

Experiments at various temperatures will allow you to calculate $E_a$. This is clear from a putting the Arrhenius equation in a different form:
$$\ln{k}=\ln{A} + \frac{-E_a}{RT}$$
A plot (or linear regression) of $\ln{k}$ vs $\frac{1}{T}$ yields $\frac{-E_a}{R}$ as the slope. The same experiment with a catalyst will show the activation energy is decreased.


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Why does ClO₄⁻ only have 3 double bonds?
It seems that most people draw $\ce{ClO4^{-}}$ with three double bonds and the negative charge on the oxygen:
http://www.chemspider.com/Chemical-Structure.109953.html
However, it would seem that it is more stable for the negative charge to be on the chlorine as it is more electronegative? (thus giving 4 double bonds to each oxygen)

Electronegativity of chlorine is $3.16$ and that of oxygen is $3.44$. Oxygen is the second most electronegative element (after fluorine $4.0$). [values are on Pauling scale.]

Very firstly, oxygen is more electronegative than chlorine.!!

I think the best argument comes from resonance.
Drawing perchlorate with three $\ce{Cl-O}$ double bonds and the negative charge on the oxygen allows for three resonance structures to be drawn (four total). This resonance stabilization  dominates the Lewis structure over simply drawing four double bonds to chlorine.
In other words, energetic benefit of charge spread out over four oxygens > charge localized on one chlorine.

You have to be careful to distinguish between the formal structures we draw to count electrons and the actual structure of the molecule. In reality the 4 bonds are the same in the perchlorate ion. But when we are counting electrons we find it bothersome to put fractional charges on atoms because that makes adding up harder. And we want to have single or double bonds for the same reason.
So we draw complete bonds which have two electrons each and balance out the total number of electrons with charge. Moreover the charge should be on the more electronegative atom, hence it goes on oxygen. However, this formalism gives an incomplete picture of the true molecular orbitals that make up the bonding (crudely, nature isn't as fussy as us about doing calculations with fractions and is happy distributing that negative charge over all the oxygens).
A slightly more sophisticated version of our whole-number accounting formalism recognises this via an extension of the simple formalism to include "resonance". Here we pretend that a number of structures "resonate" to give the true structure. In this case we can draw the single bond and negative charge on each of the four oxygens and pretend that each of those structures contributes equally to the true molecule structure.
The idea of resonance is often used to bridge the simple accounting model where bonds have two electrons, with real world structures where the electrons and bonds are spread across several atoms.


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Why is fluoroantimonic acid so strong?
I learned that all strong acids have a 100% degree of ionization and that there are no degrees of strength in acids; however, I also read that fluoroantimonic acid is 20 quintillion times stronger than 100% sulfuric acid !!
Can you please explain the chemistry behind this?

There are degrees of strength in acids, else there would not be strong acids and weak acids. There would just be acids.

I think the way I phrased that is confusing. I meant that there are degrees of weakness but no degrees of strength because all strong acids 100% ionise. That it just what I learnt in high school so it must be wrong :)

The acid strength dichotomy of strong/weak is relative to $\ce{H3O+}$, the conjugate acid of water (see Brian's answer). We can make the divide relative to other conjugate acids, and then some of our "strong" acids may become weak acids.

Dissociation as in $\ce{HB + H2O<=>H3+O + B-}$ is not ionisation.

For many common purposes we say there's weak acids and strong acids, and define strong acids to be ones that dissociate completely in water. This is because if the acid dissociates completely in water, then every proton dissociated will just protonate water, and turn it into $\ce{H_3O^+}$ (hydronium ion). The hydronium ion is the strongest acid that can exist in water as far as I understand, and hence why strong acids are simply strong acids in some contexts.
What you're describing is a super acid, that is not used in aqueous solution Take HCl for example. This is a classic strong acid, that will protonate a wide range of structures. But if we take something that is a VERY, VERY poor base (something that really does not want a proton), such as an alkane, HCl can't protonate it. The proton will rather stay with chlorine than protonate the alkane. That is one of the things you can do with the acid you mentioned.
The strength of an acid depends on many different factors, but it all comes down to how strongly the $\ce{H^+}$ is bound to the acid structure. If it is very loosely bound, the structure is high in energy, and the proton would like to find a more stable structure to partake in. In this case protonating an alkane to form a carbocation and molecular hydrogen is low enough in energy, that the acid can protonate it.
But remember that protonation is an equilibrium. If we achieve a vastly lower energy state by having the acid protonate another structure, then the protonation will be complete. If it is the same energy, half of the acid will have protonated (assuming equal concentrations of acid and base).

No water!  $\ce{SbF5}$ is a tremendously strong Lewis acid: high oxidation state, fluorine electron withdrawal, and terrible mismatch of pi-orbital size inhibiting electron back donation.  Solvo-cation, solvo-anion.  $\ce{F^-}$ is a strong Lewis base.  $\ce{SbF6^-}$, given its large size, symmetry, charge delocalization, and lack of polarizabilty (those hard fluorines) is an exceptionally weak base and weak nucleophile.  The proton in $\ce{HSbF6}$ is naked and exceptionally acidic.
The system is anhydrous, so there is no leveling effect making hydronium the most acidic  solvo-cation.

But It is said that it is 10^16 times more powerful than our good old friend H2SO4, You can't do much research on it.. It's definitely most corrosive thing ever


text:
What product will be produced when a sucrose and potassium nitrate mix burns?
What product will be obtained get when a sucrose and potassium nitrate mix, which are the ingredients for smoke bomb, are burned in a combustion container?
$$\ce{KNO3 + C12H22O11 -> ?}$$

You will get potassium carbonate ($\ce{K2CO3}$), carbon dioxide ($\ce{CO2}$), water ($\ce{H2O}$), and nitrogen gas ($\ce{N2}$):
$$\ce{48KNO3 + 5C12H22O11 -> 24K2CO3 + 36CO2 + 55H2O + 24N2}$$
See this yahoo question and this science forums thread.

The reaction equation is as follows:
$$\ce{9.6 KNO3 + C12H22O11 -> 4.8 K2CO3 + 7.2 CO2 + 11 H2O + 4.8 N2}.$$
Therefore $\pu{970 g}$ $\ce{KNO3}$ and $\pu{342 g}$ sucrose, which is $74\%$ $\ce{KNO3}$ and $26\%$ sucrose. This produces $\pu{662 g}$ of potassium carbonate.

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text:
Balancing chemical systems - searching and sorting preliminary information

I was hoping I could get help finding ways to sort information in the problem. What kind of information does "the conversion of toluene to benzene..." give us that is relevant towards solving the problem for values $R_G$ and $R_L$?

What is a lb-mole? Both pounds and moles are units representing amounts.

From your diagram, $R_G$ is the recycled gas - the hydrogen that has not been consumed plus the methane that has been produced. $R_L$ is the recycled liquid - the unreacted toluene. Both go back into the mixer with fresh toluene added at the rate given and fresh hydrogen added to bring the mole ratio of hydrogen to toluene (and hydrogen to methane) to what is given.
$R_L$ is easier. The problem tells you that 80% of the toluene is converted into benzene and 8% of the toluene is converted into biphenyl. What percentage of toluene is left over? The original feed flow of toluene is 3450 lb/hr, but the amount of toluene entering the gross feed to the reactor is higher (includes $R_L$).
$$G_{\text{tol}}=F+R_L$$
$$R_L = 0.12\times G_{\text{tol}}$$
For $R_G$, you need to know how much hydrogen you have going into the mixer. You are told that the mole ratio of hydrogen to toluene in the gross feed is 4:1. Now that you have $R_L$, you can figure out the total toluene in the gross feed, and then do the stoichiometry calculation to get to total hydrogen in the gross feed. Since you have the balanced equations, you can determine how much hydrogen is consumed. If 80% of the toluene is consumed in the generation of benzene, how much hydrogen is consumed? How much methane is produced? Do the same with the conversion of toluene to biphenyl.
Once you are able to obtain the molar flow rates of all molecules out from the reactor, you should be able to calculate the flow rate of the gas exiting from the separator. This gas stream (Let's call it $K$) contains only $\ce{CH4}$ and $\ce{H2}$ of mol fractions $y$ and $1-y$ respectively. The mol fractions can be calculated considering that all $\ce{CH4}$ and $\ce{H2}$ that exists the reactor leaves the separator in $K$.
The molar composition of the $R_G$ and $P$ streams that leave $K$ is the same. This means that you are now able to use a molar flow rate balance on the mixing point between $M$, $R_G$ and the stream to the mixer (Lets's call this $Y$).
$$Y = M + R_G$$
By doing a balance on methane, you can find $R_G$.
$$(0.5)*(Y) = (1-y)(R_G) $$
Since you know $Y$ and $y$, you will then be able to find $R_G$.

Thank you greatly for your help! I was able to calculate $R_L$ and the hydrogen in the gross feed, but I'm having trouble calculating $R_G$ without knowing the purge amount. I know how much H2 is leaving the separator (amount of H2 in gross feed * 0.12) and much CH4 is leaving the separator (amount of B + 2*amount of D), but I don't know how to get information about $R_G$ from this.


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Heating copper wire and sulfur powder
In an experiment in school, we heated copper wire (Cu) and yellow sulfur powder together in a crucible. I was wondering why this reaction cannot produce Copper Sulfate. I know it produces Copper Sulfide but I want to know why don't oxygen from the air, sulfur, and copper combine to form copper sulfate.
Can anybody explain this in simple high school terminology?

By burying the copper wire with plenty of sulphur and then heating the mix, the wire becomes completely submerged in liquid sulphur. The liquid sulphur then gets to react with the copper, and any oxygen that could possibly oxidise the copper would have to diffuse into the liquid sulphur before reaching the copper. However, oxygen quickly reacts with heated sulphur forming sulphur oxides, so it never reaches the copper wire. This is all under ideal conditions, of course.

By burying the copper wire with plenty of sulfur and then heating the mix, the wire becomes completely submerged in liquid sulfur. The liquid sulfur then gets to react with the copper, and any oxygen that could possibly oxidize the copper would have to diffuse into the liquid sulfur before reaching the copper. However, oxygen quickly reacts with heated sulfur forming sulfur oxides, so it never reaches the copper wire. This is all under ideal conditions, of course.
(Originally given as a comment by @Nicolau Saker Neto)


text:
Do ligands with a positive charge exist?
Are there positively charged ligands which can bind to a central metal atom to form coordination compounds?
My thoughts:
I know that ligands are Lewis bases which donate a pair of electrons, and the central metal atom is usually a Lewis acid.
However, a positive charge on a species usually means it is electron-deficient. In that case, there would be no suitable electrons for it to bond to the central metal atom. So, even if positively charged ligands do exist, I think there would be very few. In such cases, it might bond with central atom which is electron rich forming the coordinate bond in the opposite way. Is that possible?

The $\ce{NO+}$ is a very good example of a positive ligand.

Aromatic cations like the tropylium cation ($\ce{C7H7+}$)[1] or the cyclopropenyl cation ($\ce{C3H3+}$)[2] can coordinate to metals, creating sandwich or half-sandwich compounds. The positive charge is shared by all carbon atoms of the rings.
Edit: I noticed OP's additional question about ligand accepting electrons from central atom. This is a good point! Such ligands do exist and are called Z-ligands, as opposed to "normal" ones classified as L-ligands, or X-ligands. They are Lewis acids, but usually still neutral.
Another issue is pi backbonding, which also shows bonding from central atom to ligand.

Anderson, J. E.; Maher, E. T.; Kool, L. B. Electrochemical and spectroelectrochemical properties of the titanium sandwich complexes (Cp)Ti(L) and (Cp*)Ti(L), where L is either cyclooctatetraene (C8H8) or the tropylium ion (C7H7+). Organometallics 1991, 10 (5), 1248–1254. DOI: 10.1021/om00051a009.

Lichtenberger, D. L.; Hoppe, M. L.; Subramanian, L.; Kober, E. M.; Hughes, R. P.; Hubbard, J. L.; Tucker, D. S. Electron distribution and bonding in η3-cyclopropenyl-metal complexes. Organometallics 1993, 12 (6), 2025–2031. DOI: 10.1021/om00030a011.

The nitroso ligand ($\ce{NO^+}$) is also one of the most commonly occurring positively charged ligands.

I would answer as a tentative yes.  Your reasoning is correct that a ligand, which behaves as a Lewis base, is unlikely to have a positive charge (and therefore be electron deficient).  The earliest reference to a positively-charged ion behaving as a ligand (that I've found) is from the journal Inorganic Chemistry (1969, volume 8 page 2331) which describes the synthesis and characterization of Cu(II) and Mn(II) complexes containing a N-methyldabconium cation:

In this case the ligand is positively charged, but there is also a lone pair on the non-methylated nitrogen that is most likely forming the coordinative covalent bond with the metal center.  It is this property (a sufficiently large ligand that has a localized positive charge that is remote from a region with Lewis base character) that would result in a positively charged ligand, and we don't need to invoke a switching of the metal roles (where the metal becomes the Lewis base) as you suggest.  That said, one of the exciting aspects of Inorganic Chemistry is the constant effort to "break the rules".

In addition to already mentioned cationic onium ligands, there is hydrazinium $\ce{H2N-NH3+}$ and its derivatives [1 -- 3].
Crystal structure of hydrazinium-trichloro-copper(II) $\ce{Cu(N2H5)Cl3}$  [3] shows $\ce{[CuN2Cl3]}$ plane with nearly $120^\circ$ angle $\ce{Cu-N-N}$:
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#C88033}{\Large\bullet}~\ce{Cu}$.

Bibliography

Goedken, V. L.; Vallarino, L. M.; Quagliano, J. V. Inorg. Chem. 1971, 10 (12), 2682–2685. DOI 10.1021/ic50106a011.
Govindarajan, S.; Patil, K. C.; Manohar, H.; Werner, P.-E. Dalton Trans. 1986, 1, 119–123. DOI 10.1039/DT9860000119.
Bushuyev, O. S.; Arguelles, F. A.; Brown, P.; Weeks, B. L.; Hope-Weeks, L. J. Eur. J. Inorg. Chem. 2011 (29), 4622–4625. DOI 10.1002/ejic.201100465.

Generally, ligands are neutral or anionic like a Lewis base. Ligands having positive charge are less common. These new category of ligand species are called cationic ligands. These ligands can accept pi-electrons - Z-type ligand and donate lone pair – L-type ligand synergistically making these ligands another class in itself.
The main reason for ligand being non-cationic is the coulombic repulsion between the similarly charged positive metal centre and ligand. Even then, ligands like NO+(nitrosonium), H2N−NH3+ (hydrazinium) and NO2+(nitronium) are commonly found with a positive charge. C7H7+(tropylium) and C3H3+ (cyclopropenyl cation) are well-known aromatic carbocations that can coordinate with metal ions forming a half-sandwich or a sandwich complex.However, cationic ligands which can coordinate through the atom which has a positive charge localized on it are even less reported. These ligands show different modes of bonding like L→M σ donation, M→π back-donation and electrostatic interaction mimicking transition metal to some extent. Among this subcategory of cationic ligands, polydentate cationic ligands have only a few examples
Group 15 cationic ligand
The valence states of these elements are generally +3 and +5. Manuel Alcarazo and his group have generated several cationic phosphines. They have managed to synthesize dicationic and tricationic phosphine ligand species which were able to coordinate with gold(I) and platinum(II) cation respectively. The donating ability of these cationic phosphines was reported to be poorer than that of phosphites and only comparable to extremely toxic or pyrophoric compounds such as PF3 or P(CF3)3. Similar synthetic strategies were applied by Alcarazo to cationic arsines containing imidazolium, cyclopropenium, formamidinium and pyridinium substituents.10 It was seen that cyclopropenium substituent had considerably lesser steric requirement than the other aromatic substituent. Different coordination complexes were explored using different transition metals like Rhodium, Gold and Platinum.

1.    Nitrenium cation
A pnictogen cation in its +1 monovalent state is analogous to carbene. Series of NHC analogues of nitrogens were exploited by Tulchinsky et al. as they are isolobal in nature as compared to carbene. N-heterocyclic nitrenium ion is one of the very few cationic ligands using mono-cationic nitrogen center to coordinate with electron-rich transition metal ions like Rh(I), Rh(III), Mo(0), Ru(0), Ru(II), Pd(II), Pt(II), Pt(IV), and Ag(I) .These complexes are significantly favoured in polar solvents. To maintain the transition metal centre in near proximity to L-type nitrenium donor, more L-type donors were attached to the scaffold with between the donor and nitrenium sites to favour Z-type interactions too.  From IR spectroscopy it was established that these nitrenium ligands are weaker σ-donors than both the NHC and pyridine, even weaker than phosphenium! From DFT studies, it was shown that the primary bonding component came from the π-acceptor property as considerable binding energy was observed (47 kcal mol-1).[1-2]
Workers of Ray’s team explored another set of nitrenium cation scaffold where they have replaced phosphine side chains of Tulchinsky’s nitrenium with tetraazacyclotetradecane which they abbreviated as (DMC-nit)+. The five-coordinate complexes of Ni(I) and Ni(II) using (DMC-nit)+ ligand, they were able to oxidize the formate to its corresponding carbon dioxide and a proton due to fact that (DMC-nit)+ is a potential non-innocent ligand alluding to NO+ surrogates. It can reversibly undergo one electron reduction to form a neutral ligand π-radical species.3

2.     Cationic phosphine
Unlike previously discussed example of nitrogen, phosphorus can coexist in different oxidation states, generally +3 and +5. Due to this reason, phosphorus found major role in catalysis. Phosphine, is by far one of the most well-established phosphorus ligands, used in metal-complex catalyst design.
The σ donor ability of the phosphines is generally modulated by replacing the substituent based on electronegativity. To increase the π-acceptance of the phosphine ligand, highly electron-withdrawing groups are attached. Prime examples being PF3, PCl3 and P(CF3)3. These phosphines are not only toxic but also P-X (X is the halogen) is quite labile making it not stable enough for industrial purposes. For the similar reasons, these compounds are not air-stable and thus pyrophoric.5
Key alternatives were to attach cationic organic groups as the substituents enabling to modulate the π-acceptor property along with forming stable P-C bond. The backdonation of phosphoryl lone pair onto cationic C make the P-C bond to have double bond characteristics making it further stabilized.[5-6]
These cationic charged phosphinic ligands are more soluble in polar and ionic liquids, which allows these species to conduct reaction even in aqueous conditions which was previously not possible with polyfluorinated phosphines.
Manuel Alcarazo and his group have ushered in the field of cationic phosphines and cationic phospholes, whose donor ability was even poorer than all the reported phosphites, only being comparable to that of toxic PF3 and P(CF3)3. They have managed to dicationic and tricationic phosphine ligand species which were able to coordinate with gold(I) and platinum(II) cation respectively, that efficiently catalyze the alkyne hydroarylation.5
Applying similar principles as that of α-cationic phosphine, Alcarazo explored α-cationic phospholes with different cationic substituents at phosphorus centre like cyclopropenium, imidazolium and iminium. These gold Au(I) complexes having cationic phospholes as ligands showed heightened electrophilicity which was able to catalyze otherwise difficult cyclomerization of tosylanylines.6
The catalytic activity of cationic phosphole depicts a similar trend for cationic phosphines characterized by similar σ-electron-releasing tendencies and low lying LUMO due to the fact that there is a good overlap of π-orbital and exocyclic σ(P-R). The π- acceptor characteristics get even more stronger with the incorporation of α-cationic substituents favouring further mixing of the σ*(P-R)-π*(ring) thus lowering the energy of the LUMO even more.
Apart from attaching cationic substituent, another strategy was to form Arduengo NHC analogues using phosphenium cations. The N-Heterocyclic phosphenium cations (NHPs) are isolobal an isostructural to that of NHCs. The key difference is NHPs are weak σ-donors and good π-acceptors due to the formal positive charge and isotropic s- (as opposed to directional sp2-) character of the “lone pair” orbital on phosphorus, whereas, NHCs are strong σ-donors and weak π-acceptors. This electronically inverse nature allows them to perform reciprocal reactions with the transition metals. This property was clearly demonstrated by the formation the NHC-NHP adduct. These NHP were able to form planar complex with iron and pyramidal complex with molybdenum. These NHPs also formed complexes with elements of group 9 and 10.7. Another report by the same group showcased the shortest ever Pd-P bond (2.1166(17)A°) through the complexation of palladium (0) and unsaturated NHP.8 Other complexes with first & second-row transition metals and coinage metals include of Cr, W, Pt, Pd and Ru.Alcarazo and his group demonstrated the huge potential of these cationic phosphines in catalysis of π-systems. They developed dicationic and tricationic phosphines with no spacer in between. These polycationic phosphines have cyclopropenium group as a substituent which can be thought as three cyclopropyl carbene stabilizing P(III) centre much alike to NHC stabilized P(III) mono- and dicationic ligands.9 These polycationic ligands were weakest lewis donor second to P(CF3)3. Using gold complex of dicationic phosphine, they were able to catalyze cycloisomerization of ortho-alkynated biaryls to its corresponding substituted phenanthrenes including some naturally occurring polyoxygenated like Bulbophyllantrin and Marylaurencinol A.10 Similarily reactivity was shown when Pt(II) complexes with tri-cationic phosphines. Not only, this π-acidic catalyst was able to synthesize phenanthrene but also able to form several homo- and heteroarenes.11

3.    Cationic arsine
Figure :Different complexes of cationic arsine with transition-metal Rhodium

Cationic ligand using arsenic as the donor centre is significantly less reported than the phosphorus as cationic ligand though arsonium salt are well known.
One successful strategy was to add cationic organic substituents to the arsine. Low-valent arsenic cations were stabilized via different aromatic substituents like imidazolium, cyclopropenium, formamidinium, and pyridinium. These cationic arsines possess poorer electron-donating ability than any typical arsine or phosphine except for the fluorinated ones namely, PF3 and AsF3 as determined from Tolman electronic parameter13.
The only differences of this class of ligand from the previous cationic phosphine are only the longer As-C bond and increased s-character on lone pair of cationic Arsine, stemming from the inert pair effect which leads comparatively lower directionality and lower-energy HOMO of its lone pair accounting for poor σ donation.
Extending the reasoning of lower σ directionality of cationic arsine, the σ*(As−C) orbitals is more diffuse in nature as compared to σ*(P−C) ones thus the π-acceptor character in arsines is also diminished with respect to related phosphites
The following figure shows the list of cationic arsines. From the B3LYP-D3/def2-TZVP level computational study, it was found that all the cationic arsines (ref. figure 3, 6, 9, 10, and 13) have reduced σ-donation as compared to tri-phenyl arsine Ph3As which is evident from the lower energy of the HOMO of substituted cationic arsines. The frontier orbital analysis of HOMO shows that all of these arsines have substantial lone pair character at As.
Interestingly, cationic arsine 13 (having the pyridinium ring with very electron-withdrawing −CF3 group attached to) behaves as the weakest donor among the cationic arsines.
These cationic arsines formed complexes with Nickel, Rhodium, Platinum, Palladium and Gold. It is noteworthy that these cationic arsines when used as π-acid ligand for Pt(II) centre, it gave upto 75% yield in this Pt-catalyzed cycloisomerization, while the best ligand combination previously reported for this case  - PtCl2 + GaCl3 gave a mere 12% yield of product.
Another scaffold for cationic arsine involves N-heterocyclic Arsenium ion. It was able to form a dative complex with cobalt cation.14
Authored by Dube from Prof. Alcarazo’s group, low-valent arsenic cations were stabilized via different aromatic substituents like imidazolium, cyclopropenium, formamidinium, and pyridinium. The only differences of this class of ligand from the previous one are only the longer As-C bond and comparatively lower directionality of its lone pair accounting for poor sigma-donation.
4.    Cationic Stibine

Till date, there has been no report of cationic stibine though there has been computational as well synthesis reports of NHC stibinium cation extending from the spectroscopic and X-ray crystallographic results of metallo-arsines and metallo-phosphines where the metal used was Cobalt. 14
Interestingly, N-heterocyclic stibinium cation showed metal to ligand electron donation much reminiscent to Z-type ligand, one of the key features of cationic ligands.
In the respect of antimony being used as a Z-type ligand has been extensively explored by Prof. Gabbai. He has established the Antimony(V) for their Z-type bonding capabilities and thus their subsequent roles in anion sensing like fluorides. One of the prime examples is cationic complex of palladium with triarylstibine for halide sensing. This complex could be thought of coordination complex using Pd(0) and cationic stibine. 15
Group 14 cationic ligands
Cationic Carbene ligands

Another novel cationic ligand is (dppm)2CH which is reported by Langer et. al to coordinate with [ReBr(CO)5] where the former acting as a σ- and π-donating pincer-type ligand
 - Reference: Facial vs. Meridional Coordination Modes in ReI Tricarbonyl Complexes with a Carbodiphosphorane-based Tridentate Ligand
Cationic silylene ligand

Cationic germylene ligand

Cationic stannylene ligand

A retrospective
Many examples in this long answer are of cationic ligands making a dative bond with transition metal cations which makes us overlook the overwhelmingly large coulombic repulsion at this very short distance between two cations. The coulombic force is inversely related with the square of the distance so one can only wonder how large the repulsive force might be if the distance is of the order of bond length i.e. Ao (~ 10-10 m). To put into the perspective: the Coulomb repulsion between two positive point charges at a separation of 3.1 Å is about 110 kcal/mol. If the charges are delocalized over medium-sized molecular fragments, the repulsion will be smaller but still on the order of about 40 kcal/mol. The attraction between two cations was tackled via London dispersion force along with strong dative bond.[108] The inter cationic-ligand must also be considered to decipher the trans conformation in the case of of Ag(I) stabilized by two Ge(II) cationic ligand. The structure was mainly dominated by Coulombic repulsion between two germanium dication over steric clashes. Similar inter-ligand coulombic repulsion was found in the platinum complex by Gebbink’s group.[109]
Cationic ligands allow a remarkable degree of control over the outcome of transition metal promoted transformations and catalysis. In cationic ligands, poor sigma donation and strong pi acceptance goes hand-in-hand that allows improvement of many transition metal catalyzed transformations. Sinhababu using germylene dication was able to perform hydroboration reaction of different aldehydes and ketones.[110]
References:
4      Y. Tulchinsky, M. A. Iron, M. Botoshansky, M. Gandelman, Nature Chemistry 2011, 3, 525.
12    Y. Tulchinsky, S. Kozuch, P. Saha, A. Mauda, G. Nisnevich, M. Botoshansky, L. J. W. Shimon, M. Gandelman, Chemistry - A European Journal 2015, 21, 7099.
3   F. Heims, F. F. Pfaff, S. L. Abram, E. R. Farquhar, M. Bruschi, C. Greco, K. Ray, J. Am. Chem. Soc. 2014, 136, 582.
5 M. Alcarazo, Chemistry - A European Journal 2014, 20, 7868.
12    M. Alcarazo, Accounts of Chemical Research 2016, 49, 1797.
12    E. Haldón, Á. Kozma, H. Tinnermann, L. Gu, R. Goddard, M. Alcarazo, Dalton Transactions 2016, 45, 1872.
6 T. Johannsen, C. Golz, M. Alcarazo, Angewandte Chemie - International Edition 2020.
7     C. A. Caputo, M. C. Jennings, H. M. Tuononen, N. D. Jones, 2009, 990.
8     C. A. Caputo, A. L. Brazeau, Z. Hynes, J. T. Price, H. M. Tuononen, N. D. Jones, Organometallics 2009, 28, 5261.
9 M. Patil, S. Holle, C. W. Lehmann, W. Thiel, M. Alcarazo, 2011, 12, 20758
10    J. Carreras, G. Gopakumar, L. Gu, A. Gimeno, P. Linowski, J. Petuškova, W. Thiel, M. Alcarazo, J. Am. Chem. Soc. 2013, 135, 18815.
11    J. Carreras, M. Patil, W. Thiel, M. Alcarazo, J. Am. Chem. Soc. 2012, 134, 16753.
13    J. Carreras, G. Gopakumar, L. Gu, A. Gimeno, P. Linowski, J. Petuškova, W. Thiel, M. Alcarazo, J. Am. Chem. Soc. 2013, 135, 18815.
14            S. Burck, J. Daniels, T. Gans-Eichler, D. Gudat, K. Nättinen, M. Nieger, Zeitschrift fur Anorg. und Allg. Chemie 2005, 631, 1403.
15      C. R. Wade, I.-S. Ke, F. P. Gabbaï, Angew. Chemie 2012, 124, 493.
[108]   S. Grimme, J. P. Djukic, Inorganic Chemistry 2011, 50, 2619.
[109]   D. J. M. Snelders, M. A. Siegler, L. S. von Chrzanowski, A. L. Spek, G. van Koten, R. J. M. K. Gebbink, Dalton Transactions 2011, 40, 2588.
[110]   S. Sinhababu, D. Singh, M. K. Sharma, R. K. Siwatch, P. Mahawar, S. Nagendran, Dalton Transactions 2019, 48, 4094
References for introduction:
Hydrazinium Ligand (H2N-NH3+)

Hydrazinium (1+) as a ligand - C. K. Prout and H. M. Powell
(https://pubs.rsc.org/en/content/articlelanding/1961/JR/jr9610004177 )
Hydrazinium as a ligand: structural, thermal, spectroscopic, and magnetic studies of
hydrazinium lanthanide di-sulphate monohydrates; crystal structure of the neodymium
compound - Subbiah Govindarajan, Kashinath C. Patil, Hattikudur Manohar and Per-Erik
Werner (https://pubs.rsc.org/en/content/articlelanding/1986/dt/dt9860000119 )
Thermal decomposition of compounds containing the hydrazinium cation as a ligand -
BanerjeeP.K.BiswasN.RayChaudhuri
(https://www.sciencedirect.com/science/article/abs/pii/0040603181850022?via%3Dihub )
Layered cobalt phosphate grafted with hydrazinium ligands: Co2(PO4)(HPO4)N2H5 -
Ekambaram, Sambandan; Sevov, Slavi C. –
(https://pubs.rsc.org/en/content/articlelanding/2000/JM/b004753h )
Spectroscopic and magnetic properties of compounds containing the hydrazinium cation as
a ligand-A.Nieuwpoort1J.Reedijk (https://doi.org/10.1016/S0020-1693(00)94834-5 )
V. L. Goedken, L. M. Vallarino, J. V. Quagliano, Inorg. Chem. 1971, 10, 2682.
(https://pubs.acs.org/doi/abs/10.1021/ic50106a011 ) - Cationic ligands. Coordination of the
1,1,1-trimethylhydrazinium cation to nickel(II)
Monoquaternized pyrazinium cations as examples of ‘non-basic’ nitrogen-donor ligands-
W.L.Darby1L.M.Vallarino (https://doi.org/10.1016/S0020-1693(00)91191-5)
https://chemistry-europe.onlinelibrary.wiley.com/doi/full/10.1002/ejic.201100465
(copper hydrazinium chloride (CuCl 3 –  N 2 H 5 + ) )

Tropylium Ligand (C7H7+)
Electrochemical and spectroelectrochemical properties of the titanium sandwich complexes (Cp)Ti(L)
and (Cp*)Ti(L), where L is either cyclooctatetraene (C8H8) or the tropylium ion (C7H7+)
James E. Anderson, E. T. Maher, and L. B. Kool
Organometallics 1991, 10, 5, 1248–1254
Publication Date:May 1, 1991 https://doi.org/10.1021/om00051a009 American Chemical Society
Cyclopropenyl cation(C3H3+)
Electron distribution and bonding in .eta.3-cyclopropenyl-metal complexes
Dennis L. Lichtenberger, Martin L. Hoppe, Lalitha Subramanian, Edward M. Kober, Russell P. Hughes,
John L. Hubbard, and David S. Tucker
Organometallics 1993, 12, 6, 2025–2031
Publication Date:June 1, 1993 https://doi.org/10.1021/om00030a011
nitrosonium(NO + )

Electronic Structure of the Hieber Anion [Fe(CO)3(NO)]− Revisited by X-ray Emission and
Absorption Spectroscopy- Lukas Burkhardt, Yannik Vukadinovic, Michał Nowakowski,
Aleksandr Kalinko, Julian Rudolph, Per-Anders Carlsson, Christoph R. Jacob, and Matthias
Bauer* [Inorg. Chem. 2020, 59, 6, 3551–3561]
https://doi.org/10.1021/acs.inorgchem.9b02092
Structure of nitrosyltetracarbonylmanganese(0), Mn(NO)(CO)4
https://pubs.acs.org/doi/abs/10.1021/ic50076a019
Acetonitrile Complexes of Selected Transition Metal Cations- Richard R. Thomas,Ayusman
Sen,Wolfgang Beck,Reich Leidl
https://onlinelibrary.wiley.com/doi/10.1002/9780470132593.ch14
Cyclopentadienyl Rhenium Complexes -Francine Agbossou,Edward J. O'Connor,Charles
https://onlinelibrary.wiley.com/doi/10.1002/9780470132609.ch51
B. F. G. Johnson K. H. Al‐Obadi (1970). "Dihalogenodinitrosylmolybdenum and
Dihalogenodinitrosyltungsten". Inorg. Synth. 12: 264. doi:10.1002/9780470132432.ch47.
James K. Hoyano, Peter Legzdins, John T. Malito (1978). "(η5‐Cyclopentadienydnitrosyl
Complexes of Chromium, Molybdenum, and Tungsten". Inorg. Synth. 13: 126.
doi:10.1002/9780470132494.ch21
https://doi.org/10.1016/S0010-8545(00)80259-3 (Principles Of Structure, Bonding, And
Reativity Metal Nitrosyl Complexes)
Properties of metastable linkage NO isomers in Na2[Fe(CN)5NO]·2H2O incorporated in
mesopores of silica xerogels
https://pubs.rsc.org/en/content/articlelanding/2012/cp/c2cp23607a
Infrared Absorption Spectra of Sodium Pentacyanonitrosylosmate(II) Dihydrate in Two
Excited Electronic Metastable States https://pubs.acs.org/doi/abs/10.1021/ic00120a015

I will edit it time to time ..adding more images and references

Umm, did you paste a whole chapter, or something? ;D

And I think you shouldn't repeat, or list, the examples shown by others in their answers. BTW binding a cation doesn't mean the cation reacts with another cation - it could react for example with anionic complex (this is vide your "A retrospective").

@Mithoron It could've been a chapter if undergrads are allowed to publish by themselves .. you always cross-check it in Turnitin if you find something fishy in my article.
I was trying to make an exhaustive list of cationic ligands, but it's turning out too big for a StackExchange answer.
As you can see definition cationic ligand is little open.. alpha-cationic ligand are those which have +e charge on donor atom (cationic phosphine, germylenes) while others have +ve charge and lone pair on different atom but in same molecule (eg. cationic carbene, pyridine, hydrazines)

Also for classical cationic ligands like NO+ .. it is hard to determine whether it is binding to metal centre as a cation or being non-innocent to reduce itself to neutral/anionic NO..

I think the current wording of the question just asks about things that are a cation before becoming a ligand, Therefore excluding stuff like oxidation state - making Andselisk's and bobthechemist's answers most relevant. This question should been "closed as too broad" at the start, I think, but it's  ancient - over eight years! In the future you could make a "Q&A's" both a question and an answer (but please, shorter!) nobody's gonna read a thing like this one! ;)

$\ce{Na2[Fe(CN)5(NO)]}$ is one example.
This is because it is special compound exception and here the oxidation state of the $\ce{NO}$ ion is +1 rather than 0.
Thus the oxidation state of the ferrate ion turns out to be two.
$\ce{NO}$ with +1 oxidation state is named as nitrosonium or nitrosyl, thus $\ce{NO + NO_x+}$ is the nitrosonium or nitrosyl cation.

This was better suited as a comment on another existing answer - Are there positive ligands?

Is there any proof the NO+ remains as it is and doesn't get reduced (NO is a well known non-innocent ligand)? Also Fe is in +3 state then NO should be in (0) ox. state.
Can you cite me the paper which shows Fe is in +2 ox. state?

In 2022 sulfonium ions, containing trivalent sulfur within the ligand, were complexed with platinum-group metals.
Li et al. 1 synthesized and characterized sulfonium-ligand complexes of rhenium(I) and platinum(II), both relatively low oxidation states for the respective central metals and thus indicating a pi acceptor interaction. This is indeed what happens; the sulfonium ions are isostructural and isoelectronic with phosphines and, like phosphines, accept pi electrons into the formally antibonding orbitals between the ligating atom and the organic groups within the ligand. But with positively charged sulfur instead of neutral phosphorus in the ligating position, the pi-acceptance by the sulfonium ligand is stronger, a fact emphasized in the article.
Reference

Ruiping Li, Nitsan Barel, Vasudevan Subramaniyan, Orit Cohen, Françoise Tibika and Yuri Tulchinsky. "Sulfonium cations as versatile strongly π-acidic ligands". Chem. Sci., 2022,13, 4770-4778. https://doi.org/10.1039/D2SC00588C.

The nitrosonium ($\ce{NO+}$) and nitronium ($\ce{NO2+}$) ions both bear positive charges, but are capable of acting as Lewis bases/ligands. In $\ce{NO+}$, a lone pair on the nitrogen serves as the Lewis basic site for dative bond formation, and in $\ce{NO2+}$ an oxygen lone pair plays the same role.

Can you please provide references (links to original research paper) for both the cations (especially the nitronium one)?  Many thanks and regards

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


text:
How to find the units of the rate constant?

Consider the reaction $$\ce{2A + B -> Products},$$ when the concentration of $\ce{B}$ alone was doubled, the half-life did not change. When the concentration of $\ce{A}$ alone was doubled, the rate increased by two times. The units of the rate constant is:
a) $\mathrm{s^{-1}}$
b) $\mathrm{L\ mol^{-1}\ s^{-1}}$
c) Unitless
d) $\mathrm{mol\ L^{-1}\ s^{-1}}$

I have tried it solving through this method:
According to me, when the concentration of $\ce{A}$ alone was doubled, reaction rate also increased by two time, implies that reaction is first order according to $\ce{A}$. In same way, the reaction should be 0th order according to $\ce{B}$. So the net order would be 1. So this gives me answer $\mathrm{s^{-1}}$ but the answer is $\mathrm{L\ mol^{-1}\ s^{-1}}$. Where am I going wrong?

@Martin  If the reaction were zero order in B and first order in A, it would be first order overall and have a half-life. If the reaction is first order in B and first order in A, it would not have a half-life (or in other words, the half-life would be concentration dependent) unless it is pseudo-first order because A is much more concentrated than B. I don't think the question makes sense.

@Karsten I'm not sure how I can help you here. I've just added some markup and removed the homework tag. Have I introduced an error in the question?

You are doing it wrong in situation when concentration of B alone was doubled but half life did not change.
In this situation, the reaction order should be 1 according to B as Half life in first order reaction doesn't depend upon initial concentration of reactants.
$$t_{1/2} = \ln2 / k $$
So the final order will be $2$ and the answer of the question will be b.

@orthocresol But if it is first order in B and first order in A, the integrated rate law (starting with equal concentrations of A and B) would not be an exponential function with a half-life.

@KarstenTheis, thank you, I do see your point, but my only edit was to change $K$ to $k$. I was too lazy to check for factual correctness. Please feel free to edit as you wish... (or provide your own answer if you prefer...)


text:
How can I determine the boiling point of a compound using simple distillation?
I've graphed a distillation curve (temperature against volume of distillate) from the data obtained from a simple distillation experiment. The substance can be any of methanol, ethanol, 1-propanol or a  1-propanol and water azeotrope).
How can I find the boiling point from this curve to determine the identity of the compound?
I should also mention that there was leftover nonvolatile dye in the round bottom flask.

We cannot answer your question without knowing at least some more information, like : Is your mixture made of only  two constituants ? Or more ? Is your mixture made of about the same proportion of the constituants ? Or is one of these components present at only $1$ or $2$ percent in the mixture ? For example, ethanol makes an azéotrope at about $4$% water.

Do you mean that the substance could be pure methanol, ethanol or 1-propanol?
In that case there is a simple way to distinguish between the four options. A pure substance will have a T-V curve that consists of 2 straight lines, 1 running from $T=T_{start}$ to $T=T_{b}$ at $V=0$ and one running from $V=0$ to $V=V_{total}$ at $T=T_{b}$. Probably you will have some impurities which cause some small deviations from this perfect hook, but that will be the general shape. Determining the boiling temperature is just a matter of checking where the curve goes straight up. Depending on the temperature at which this happens you can identify methanol ($T_{b}\approx65$), ethanol ($T_{b}\approx78$) or 1-propanol ($T_{b}\approx97$).
The azeotropic mixture makes it a bit tricky, because the boiling points of 1-propanol and water are so close together, but the curve should show a distinct difference not going vertically up at a certain temperature.

If it is a straight line, does that mean that Tstart = Tb?

No, it means that $V=0$ at $T_{start}$ and $T_b$


text:
Phenyl vs methyl cis/trans stability

Compound I is obviously the Z isomer. I thought it was obviously the less stable one too, given that there's a phenyl and t-butyl on the same side. The answer says that since the phenyl is planar, it actually exhibits less steric hindrance than a methyl group.
Is this because the planar phenyl ring can basically rotate its bulkiness away from the t-butyl, where as the methyl, while also free to rotate, always ends up with a hydrogen in the t-butyl's face?
Thanks!

The molecule contains conjugated system of double bonds, being essentially a substituted styrene. The phenyl cannot therefore freely rotate and its steric requirements are difficult to estimate just by hand-waving arguments.
Correct answer needs some quantitative argument, either e.g. heats of formation or calculation.
From the second, the energies come out using MMFF94 (using Avogadro):

Z: 177.6 kJ/mol
E: 175.6 kJ/mol

Also, both the isomers rotated the styrene substantially from planarity. The difference in results is quite small and can easily change when using other method. According to the numbers I'd agree with your answer, Z is less stable.
I would not consider this a valid question in test, the answer is not straightforward.


text:
Liquid crystal current flow
In a liquid crystal cell used for LCD the conductive substrates, usually made of ITO, are in contact with the liquid crystal. The question is: when a voltage is applied to the electrodes to reorient the molecules is there any current flowing through the liquid? The answer should be positive because LCs are conductive materials. But I know that the molecules should reorient themselves under an applied electric field for example between two plates like a capacitor even without being in contact with the LC. So why the electrodes are placed in contact with the LC and why LCDs consume electricity?

LCs materials are dielectric. But they are not perfect insulators and have some non-zero conductivity, allowing a leakage current to flow, slowly discharging the capacitor.
This is caused by mobile ions in the liquid crystal and inside the polyimide layer [2].
Infact a liquid crystal cell can be modeled electrically as this circuit:
               Rl
         _____WWW_____
         |           |
         |     Cl    |
     O--------||---------O

References:
http://www.lintech.org/comp-per/03lcd.pdf p.10-11
http://www.emdmillipore.com/showBrochure?id=200908.037 p.9


text:
Stoichiometric defects in silver bromide
Stoichiometric defects (Schottky and Frenkel defects) in ionic crystals are formed due to irregularities in the ionic arrangement which maintain the stoichiometry and electroneutrality of the solid. It is known that $\ce{AgBr}$ shows both Frenkel and Schottky defects while most other compounds  show only one of these defects; for example, $\ce{AgCl}$ shows only Frenkel defects.
The ratio of the sizes of the cation and anion determines which type of defect is shown by the crystal. If the ratio is closer to unity, Schottky defects are more common, and if it deviates significantly from one, Frenkel defects are more common. Intuitively, this is correct since Frenkel defects involve the movement of the cation (or the smaller ion) into interstitial sites, which is indeed easier if the ion is very small as compared to the other ion, making it more mobile. Generally, the activation energy for cationic movement is lower. And if the ionic sizes are comparable, filling the interstitial sites is not feasible, and hence Schottky defects are more likely to be exhibited by the crystal.
The $\ce{Cl-}$ ion is smaller than $\ce{Br-}$ (i.e. the ratio $r_+/r_-$ for $\ce{AgCl}$ is closer to 1 than it is for $\ce{AgBr}$). Yet, $\ce{AgBr}$ displays Schottky defects while $\ce {AgCl}$ does not. Why is that so? (The size and hence the mobility of $\ce{Ag+}$ should be the same in both cases.)

Usually schottky defect is observed in strongly ionic compounds having ions (cations and anions) of almost similar size. You can see that Ag and Br have comparetively similar size than Ag and Cl, so AgBr shows Schottky defect whereas AgCl won't.

Fundamentally, although ionic radii may offer some guidance as to defect formation, the bottom line is that the silver halides have complex bonding energetics that have to be taken into consideration.  I might suggest 'Defect structures in the silver halides', D.J. Wilson et al., Phys. Rev. B77, 064115 (2008) (presuming you have APS journal access). There the authors present DFT calculations on the formation energies of both Frenkel and Schottky defects in AgCl and AgBr.

What happens if the radius ratio of AgBr is intermediate for the two effects?


text:
What is the role of hydrochloric acid in a copper electroplating bath?
Considering water as a given, the third most important ingredient in a copper electroplating solution, as a short survey of related recipe webpages will show you, is hydrochloric acid (HCl). Here's a sample link: http://www.thinktink.com/faqs/cupltfaq/cupltf01.htm
I can guess that the role of sulfuric acid (the second most important ingredient; the first being copper sulfate) is to provide you with sufficient anions to allow copper cations to easily enter the solution at the anode. But in almost all recipes that I have seen, a small amount of hydrochloric acid is added. I wanted to know how useful it is and what is the opportunity cost associated with it.

This is an hypothesis: i think HCl form a complex with $Cu^{2+}$  such as $CuCl_4^{--}$, this remove copper ions from solution and so allows further oxidation of copper to take place.

As already mentioned by @GM it indeed is used to take out $\ce{Cu^{2+}}$, but it is also used to take out $\ce{Cu+}$.
I know this because I once used a mixture of $\ce{HCl}$ and $\ce{H2O2}$ to etch copper boards, in that process both $\ce{Cu^{2+}}$ and $\ce{Cu+}$ were dissolved by the formation of $\ce{CuCl2}$ and $\ce{CuCl}$ with $\ce{Cl-}$ from $\ce{HCl}$.

The role of the HCl in copper plating and nickel plating solutions is to increase solution conductivity and  increase anode corrosion and thus inhibit anode passivation which can deplete the bath of copper sulfate.
Be careful with with adding too much HCl since it is a very difficult component of a bath to selectively remove, dummy plate or filter and high levels tend to cause deposit stress and low anode efficiency. HCl is good at removing the oxides from copper but does not form a ionic species as sulfuric acid does.

Please have a look at How to change between lowercase and uppercase.


text:
Functional variation in derivation of the Hartree-Fock equations
In Szabo and Ostlund's Modern Quantum Chemistry, the procedure of single determinant energy minimization is presented. Omitting the whole procedure, I have a question about functional variation during the derivation of the Hartree-Fock equations.
Given the single determinant $| \Psi_0 \rangle = | \chi_1 \chi_2 \ldots \chi_a \chi_b \ldots \chi_N \rangle$, the energy $E_0 = \langle \Psi_0 | \mathscr{H} | \Psi_0 \rangle$ is a functional of the spin orbitals $\{ \chi_a \}$. $E_0$ is the expectation value of the single determinant $\vert \Psi_0 \rangle$,
$$
E_0[\{ \chi_a \} ] = \sum\limits_{a=1}^N [a\vert h \vert a] + \frac{1}{2} \sum\limits_{a=1}^N \sum\limits_{b=1}^N [aa \vert bb] - [ab \vert ba].
$$
This equation can be varied:
$$
\begin{align*}
\delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] + [\chi_a | h | \delta \chi_a] \\
&+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \delta \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] + [\chi_a \chi_a | \chi_b \delta \chi_b] \\
&-\frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + [ \chi_a \chi_b | \delta \chi_b \chi_a] + [\chi_a \chi_b | \chi_b \delta \chi_a] \label{1}\tag{1}
\end{align*}
$$
Authors suggest to the reader as an exercise to manipulate this equation for $\delta E_0$ to show that
$$
\delta E_0 = \sum \limits_{a=1}^N [\delta \chi_a | h |\chi_a] + \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] - [\delta \chi_a \chi_b | \chi_b \chi_a] + \text{c.c.} \label{2}\tag{2}
$$
It is clear that the first sum in $\eqref{1}$ can be easily converted to that in $\eqref{2}$ because
$$
[\delta \chi_a | h | \chi_a]^* = [\chi_a | h | \delta \chi_a]
$$
Analogously, for the second sum in $\eqref{1}$ one can show that
$$
[\delta \chi_a \chi_a | \chi_b \chi_b]^* = [\chi_a \delta \chi_a | \chi_b \chi_b]
$$
and
$$
[\chi_a \chi_a | \delta \chi_b \chi_b]^* = [\chi_a \chi_a | \chi_b \delta\chi_b].
$$
I did the same manipulations with the third sum in $\eqref{1}$ and obtained
$$
\begin{align*}
\delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] \\
&+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] \\
&- \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + \text{c.c.} \label{3}\tag{3}
\end{align*}
$$
So, comparing $\eqref{3}$ and $\eqref{2}$ I suppose that one should show that both terms in each sum are equal, i.e.
$$
[\delta \chi_a \chi_a | \chi_b \chi_b] = [ \chi_a \chi_a | \delta \chi_b \chi_b]
$$
and
$$
[\delta \chi_a \chi_b | \chi_b \chi_a] = [ \chi_a \delta \chi_b | \chi_b \chi_a].
$$
So, how to do that?

I'd try integration-by-parts.

Is it not just because of the fact that $[aa|bb]$ is in chemists notation, meaning that $[aa|bb] = [a(1)a(1)|b(2)b(2)]$ and so you can just switch them around?

Thanks. I've already solved the problem. The idea is to change indices in a double sum. ∑a,b[δχaχa|χbχb]=∑a,b[δχbχb|χaχa]=∑a,b[χaχa|δχbχb]. I asked this question at physics.stackexchange.com http://physics.stackexchange.com/questions/80718/derivation-of-the-hartree-fock-equations-functional-varitation

Just once I'd like to see a full description of HF that kept consistent, fully explicit notation all the way through.

you're not the only one...


text:
Why are allenes chiral?
How is 1-bromo-3-chloropropa-1,2-diene ($\ce{BrHC=C=CHCl}$) chiral? It doesn't have any carbon that is connected to four different groups.

You probably will never realize how useful your questions were for students in the future.

Chirality is a property of objects in which they lack certain symmetry operations, specifically improper rotations, including the mirror plane and inversion operations. For example, 3-dimensional chiral objects lack mirror symmetry. According to Wikipedia:

The feature that is most often the cause of chirality in molecules is the presence of an asymmetric carbon atom.

This is not the only cause of chirality. Any molecule of low enough symmetry to lack a mirror plane is chiral.
Similarly, stereogenic centers need not be carbon atoms (or any atoms), they just need to be a point in the molecule's space which generates chirality for the molecule. The central carbon atom in low symmetry allenes serves this purpose. This is also an example of axial chirality, which is chirality about an axis (the $\ce{C=C=C}$ axis) instead of about a point (which is point chirality — the more familiar kind).

Asked and answered, but I think one thing that's missing is that allenes are not planar like alkenes or alkynes are. You can refer to this question for an explanation.

The dihedral angle between the two halogens is 90 degrees (ideally). Here's an animation hopefully providing a better view of the 3D structure:

This results in the two mirror images being non-superimposable upon each other, which means the molecule satisfies the criterion for chirality. If you rotate the right-hand molecule by $90^\circ$ such that the blue and white atoms at the top coincide, the green and white atoms will not match each other.

In general, any allene with the formula $\ce{R^1R^2C=C=CR^3R^4}$ will be chiral as long as $\ce{R^1}\neq\ce{R^2}$ and $\ce{R^3}\neq\ce{R^4}$. In this case we have $\ce{H} \neq \ce{Br}$ and $\ce{H} \neq \ce{Cl}$, so the compound in the question is chiral.
If $\ce{R^1} = \ce{R^2}$ or $\ce{R^3} = \ce{R^4}$, then the compound will possess a mirror plane and will therefore be achiral.

To expand on the already touched points: The central carbon of an allene is sp-hybridized meaning it has 2 half-empty p orbitals for pi-bonding purposes. Regardless of which 2 we choose to leave half-empty (px, py, or pz) they are going to be perpendicular to each other, giving rise to the non-planar allene.


text:
which are the possible ways to detect impurity in drugs?
I want some information regarding the protocol for determining any type of impurity in pharmaceutical substances or drug product.
Which type of impurities can be identified by HPLC and MS?
what is the difference between related substances and impurity?

I imagine you are talking about HPLC-MS (or LC-MS), and not the single technique (HPLC and MS). If so maybe this article could be helpful

Have you also looked into elemental analysis? This is a technique that 'counts' the number of Carbon, Hydrogen etc. atoms in the sample. If the molecular formula differs from this, there is bound to be an impurity. Look at the wikipedia-page maybe?

Do you want something else than described fully in Pharmacopeia? http://en.wikipedia.org/wiki/European_Pharmacopoeia
That means, for each drug the full analytic procedure?

There are a lot of types of impurity. Basically, much as any plant that's growing where you don't want it is a weed, any material in your sample that isn't what you want to have in it is an impurity.
Since you asked a question in simple terms I will attempt to answer it in equally simple terms:

Byproduct contamination - The theoretical reaction is usually not the only possible one, especially in organic chemistry such as pharmaceutical work. Areas of the reaction which didn't have just the right mixture of reactants, or wasn't quite at the right temperature, could have produced different products. Additionally, devations from the ideal when adding reactants can result in unreacted quantities of those reactants.
Various forms of chromatography, such as HPLC, CC, GC, MS, etc can detect and even separate these various byproducts from the ideal product. How effective any one of these techniques is depends on the type and combination of the products and byproducts. One of the most basic aspects of any drug research chemist's job is to isolate their product from all of the byproducts of their reaction.

Structural isomer contamination - Two compounds are structural isomers if they have the same molecular formula (number and type of atoms), but have different molecular structures. Tetramethylbutane, for instance, is based on a 4-carbon alkane, to which two additional methyl groups each are added to the carbons in the "2" and "3" positions of the alkane. The resulting molecule has all the same carbons and hydrogens as the totally linear octane molecule, but its structure causes very different behaviors (for instance, it has a much narrower range of temperatures at which it is a liquid).
Your average petrochemist doesn't care too much about the difference between any of the structural isomers of these hydrocarbon alkanes; their primary reaction of interest, combustion, only cares about the required stoichiometric ratio of fuel to air, and the resulting rate and heat of combustion. However, such isomers pose problems in pharmacology for the above reasons; a reaction that was theorized to produce a molecule with one arrangement may in fact produce an entirely different isomer, with similar physical properties as the desired molecule, but much different reaction chemistry including biochemical effects. Chromatography methods based on molecular weight or physical properties like boiling point or solubility may or may not be able to tell you whether you have the right isomer, but other methods might show a difference in the physical properties of the chemical allowing separation.

Chirality and Racemization - Even if two isomers have all the same atoms, and each atom is bonded to the "correct" adjoining atoms, the molecules can have different shapes because the atoms are bonded to different relative positions of their adjoining atoms. These "spatial isomers" or "stereoisomers" have many forms, from structures that exhibit the same "mirror-image" symmetry that your left and right hand do (enantiomers), or that have certain functional groups transposed by rotation around a "chiral center" (diasteromers, including cis-trans and most detro-levo isomers). Again, like structural isomers, these stereoisomers can cause problems for pharmacologists because the different isomers can have different reactive properties in the complex biochemical soup that is the human body.
Sometimes these chiral isomers can be separated from each other; enantiomers and other spatial isomers are often "optically active", meaning they rotate polarized light in opposing directions depending on their chiral arrangement. Certain "chiral resolving agents" can also separate these isomers based on a differing affinity for one isomer over another, and these agents are often used as "stationary phases" of column chromatography including HPLC that can separate and thus "purify" the desired isomer from its chiral mirror. However, a few troublesome spatial isomers will "racemize"; that is, they will readily rearrange between the two stereoisomers when given the right environment, and so any attempt to produce a pure sample of one enantiomer is ultimately pointless.


text:
Which among phenol and 1,2-dihydroxybenzene has the higher boiling point?
My attempt
Based on symmetry: I think that looking for symmetry phenol has a higher boiling point than 1,2-dihydroxybenzene because the two -OH groups projecting out from the benzene ring of 1,2-dihydroxybenzene will not allow close packing of them which results in a lower boiling point.
Based on hydrogen bonding: In this sense phenol would have a lower boiling point because 1,2-dihydroxybenzene would have more inter-molecular hydrogen bonding than phenol.
How do I know which factor wins?

Your structures (grabbed from Wikimedia Commons)
Phenol:

Catechol (1,2-dihydroxybenzene)

The three main influences on boiling point are:

Shape - molecules with a more spherical shape have lower boiling points than molecules that are more cylindrical, all else as equal as possible. Spheres have the lowest surface area to volume ratio of any of the solids. This is where symmetry comes in. More symmetric molecules are more spherical. The shape/symmetry argument does not work here since the two compounds are not isomers. The issue of close packing is relevant to the melting points of solids, since in a solid, most particles do not have translational motion - they are locked in a crystal lattice. In a liquid, such random translational motion is common. In the liquid state, molecules of most compounds are so disorderly arranged that it does not matter how well they pack as a solid. In your case, both molecules are rather like flat disks with a little bit of polarness sticking off one side. Their shapes are not so different as far as surface area to volume goes. Boiling point will be dominated by mass and polarity.
Molecular mass - bigger molecules require more energy to vaporize because they have more mass, and $KE=\frac{1}{2}mv^2$, all else as equal as possible. Mass difference applies in your case because the two molecules are similar enough in shape and polarity, except one has an extra oxygen in its formula.
Polarity - polarity controls the types and number of intermolecular forces, including hydrogen bonding. The more and stronger intermolecular forces between two molecules, the higher the boiling point, all else as equal as possible. Since catechol has two hydroxy groups and phenol has one, you have your strong argument for increased intermolecular forces.

The other easy thing to do is use the internet to look up the boiling points, and then explain the difference using the three factors. Once you know what the boiling points are, it is easy to choose the structural features that support the difference.

Phenol - $181.7 \ ^\circ\text{C}$
Catechol - $245.5\ ^\circ\text{C}$

Molecular mass is a very weak argument to make.  A molecule will vaporize when it has sufficient energy, not when it is moving some particular speed.  More massive particles might be moving more slowly, on average, at a given temperature, but they have the same average energy as everything else at the same temperature.  The better explanation for the connection between mass and boiling point has to do with the number of electrons in the molecule and its increased polarizability.  High MW molecules have lots of electrons and thus experience much stronger van der Waals forces.

Why can't we use the argument that catechol will have intramolecular H-Bonding whereas phenol will form intermolecular, thus phenol should have higher boiling point?

Predicting boiling points based on similar structures can be a little hand-wavy, given the factors that can contribute. I'm not sure that comparing symmetry between phenol and 1,2-dihydroxybenzene (catechol) is strictly a direct comparison. Not only do have have an additional functional group to contend with, but you are also introducing molecular weight changes too. Perhaps comparing catechol to 1,3-dihydroxybenzene (resorcinol) or 1,4-dihydroxybenzene (hydroquinone) might allow for better arguments for symmetry: catechol (bp 245); resorcinol (bp 277); hydroquinone (bp 277).
For me, having more functional groups present which increase the opportunity for intermolecular forces to do their bit is more likely to influence boiling point than symmetry.


text:
How exactly is activation energy defined?
In a common interpretation of the Arrhenius rate equation
$$k = A\exp\left(-\frac{E_\mathrm a}{RT}\right),$$
the activation energy $E_\mathrm a$ is understood to represent the difference in the energy of the reactants and the minimum energy required to form the activated complex or the transition state, as shown in the following diagram

But this must be wrong! As $E_\mathrm a$ is the minimum energy itself which the molecules must possess to reach the activated complex. That is, it is not the difference but the minimum energy level itself. This is the only way in which it will remain constant i.e. independent of temperature (otherwise with increase in $T$, $E_\mathrm a$ will go down) and can be used in the Arrhenius Equation where the exponential term denotes the fraction of molecules having energy higher than or equal to $E_\mathrm a$ according to Maxwell-Boltzmann distribution. Sometimes, threshold energy ($E_\mathrm t$) is defined as the minimum energy required for reaction to proceed and $E_\mathrm a = E_\mathrm t-\mathrm{KE_{average}}$. If this definition is true, then threshold energy and not activation energy should be in the Arrhenius equation. I therefore need to clarify the actual definition of activation energy $E_\mathrm a$.

You could say Ea is the minimum energy the molecules must possess RELATIVE to the initial state. I've only seen it defined in this non-absolute fashion.

@Brian But that makes $E_a$ temperature dependent since the initial state is temperature dependent. And how does that hold up against the intuition behind Arrhenius equation, the exponential factor being the fraction of molecules having energy greater than or equal to that required for reaction(Threshold energy)?

This is simply from wikipedia: Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature. http://en.wikipedia.org/wiki/Arrhenius_equation

Can you clarify a little your question? Others may fully understand, but for me it is not clear why you think one or other definition is wrong (or contradicting). The statement  "it is not the difference but the minimum energy level itself" makes no sense to me. Also, it is not clear if you are talking about a single potential energy surface or about an ensemble of molecules - I suspect it is not clear to you, neither.

You are right that the energy of the initial state is a function of temperature. But in practice for this equation the energy difference $E_\mathrm a$ usually is fixed for an arbitrary temperature (in the middle of the diapason of experimental measurements) and assumed temperature-independent.
From Wikipedia:

While this equation suggests that the activation energy is dependent
on temperature, in regimes in which the Arrhenius equation is valid
this is cancelled by the temperature dependence of $k$. Thus, $E_\mathrm a$ can be
evaluated from the reaction rate coefficient at any temperature
(within the validity of the Arrhenius equation).

As to the physical meaning of activation energy, citing Wikipedia:

Both the Arrhenius activation energy and the rate constant $k$ are
experimentally determined, and represent macroscopic reaction-specific
parameters that are not simply related to threshold energies and the
success of individual collisions at the molecular level.

In other words, activation energy $E_\mathrm a$ is a semi-empirical quantity and is not considered to have a strictly defined physical meaning (at least on molecular level). Note that the Arrhenius equation
$$k=A\exp\left(-\frac{E_\mathrm a}{RT}\right)$$
is itself considered to be a semi-empirical relationship.
In the IUPAC Gold Book (DOI: 10.1351/goldbook.A00102), activation energy is also defined to be an empirically determined quantity, in line with the above:

An empirical parameter characterizing the exponential temperature dependence of the rate coefficient $k$:
$$E_\mathrm a = RT^2 \frac{\mathrm d (\ln k)}{\mathrm dT}$$


text:
How to calculate crystal field stabilisation energy?
How to calculate it? I know that it depends on the coordination compound and the number of electrons present in t2g and eg orbitals and the final answer is expressed in terms of crystal field splitting parameter. But, how to know how many electrons are there in t2g and eg orbitals?
I think rather than a broad answer and example would help. So, consider the compound $\ce{[Ni(CO)4]}$ and how to calculate CFSE for it?

forgot to mention wether the election is in high spin or low spin..if low spin d electrons will first fill up the t2g level then the eg level.rest is same
.

Step 1: Look up Nickel Carbonyl and find out what geometry it has.

We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.

Step 2: Find the appropriate crystal field splitting diagram for this geometry.

Step 3: Figure out how many $d$ electrons there are.

Nickel is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations. For more information on $d$ electron count, go to this Wikipedia page.

Step 4: Fill in the $d$ electrons.

You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?

Step 5: Determine $\Delta E$.

$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This Wikipedia page walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:

$e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)
$t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)

Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.
$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$

We have to see the $d$-orbital of the metal atom how many electrons there are, but in excited state not in ground state; then see the spins whether it is low spin or high spin because $d$-orbital is divided into $t_{2g}$ and $e_g$ with respect to their energies difference then apply the formula. The formula is

$$\mathrm{no.\ of\ electrons\ in\ t_{2g}\cdot(-0.4)}+\mathrm{no.\ of\ electrons\ in\ e_g\cdot(0.6)}$$

Example: I have $\mathrm{[Co(F)_6]^{3-}}$ now see cobalt has $3d^7$, $4s^2$ system in ground state but in excited state it loses three electrons in the formation of ions and two elctrons from $4s$ and one from $3d$ orbital so thus Cobalt gets $3d^6$ configuration. Now it is of high spin so $4$ electrons go to $t_{2g}$ orbital and $2$ electrons go to $e_g$ orbital. By applying formula,

$4(-0.4)+2(0.6)=-1.6+1.2=-0.4$

Thanks for your answer.  Could you format it a bit better, it's a bit confusing as it stands.

The process is quite simple. .. to determine CFSE  we need to know that the diffidence of energy levels of two sets have been arbiter taken as 10 dq. . Where t2g set is more stable then eg set octahedral and vice versa in tetrahedral complex.
-0.4x + 0.6y will help to calculate it.
X indicates electrons in t2g and y in eg.
Sum up all and find the diffidence ...
That gives cfsc
Usually tetrahedral splitting is shown in terms of octahedral as
4/9 OS

authors: Στρατηγός Παναγιώτης


text:
Carbon Free energy vs Temperature Diagram
I've recently read that the Carbon takes a unique position as a reducing agent , and i think it has to do with the very special form of its Ellingham diagram  , but i am still not able to make a connection , So what makes exaclty the carbon a good reducing agent ?

Are you referring to the uniqueness of carbon as reducing agent as a consequence of considerable negative $\Delta_fG$ of its oxide formation, or the negative slope of $\Delta_fG(\ce{CO})$ which makes it theoretically possible to reduce most oxides at a sufficiently high temperature? Let me know so I can elaborate on whichever concern you are presenting in the question.

@SatwikPasani yes , that's exactly what i'm referring to .

I mean the whole statement .

Here is the link to a set of Ellingham Diagrams
At high temperatures, smelting and carbon reductions have several practical challenges some of which are energy efficiency, formation of metal carbides, equipment concerns etc. Therefore, to assess the reducing ability of any reducing agent, we must work within practical limits. Let us select a temperature of $1600\ \mathrm{K}$ as our upper bound beyond (which is still quite high) which, practical concerns outweigh any benefit of having a good reductant. At this temperature $\Delta_\text fG(\ce{CO2})$ is approx $-394.39\ \mathrm{kJ/mol}$ and $\Delta_\text fG(\ce{CO})$ is $-504\ \mathrm{kJ/mol}$. These values are high enough and can already reduce several oxides of metals like $\ce{Co, Fe, Cu, Pb, Ni, Bi}$ and $\ce{MnO2, CrO2}$ etc (subject to practical constraints). This gives it a good reducing character, better than $\ce{H2}$ or ammonia.
Another thing to notice, is that, although the variation of $\Delta_\text fG$ for carbon dioxide is flat, for carbon monoxide it slopes downward and hence with increasing temperature, it can reduce many more oxides further expanding its capacity to reduce metal oxides. $\ce{V2O5, VO2, SnO2, WO3, Mn3O4, MoO3, GeO2}$ can be reduced by carbon. (All this just implies thermodynamic favourability; it can still be unfavourable kinetically, or due to practical restraints.) Hence, the negative sloping $\Delta_\text fG$ for $\ce{CO}$ makes carbon a very good reducing agent, and it can reduce a huge variety of metal oxides, within a considerable temperature limit.


text:
why does FeCl₃ have such a specific smell?
When I use $\ce{FeCl3}$ solution for etching circuits it gives off an acidic smell.
I was wondering what is the cause of the smell.
I don't think it's caused by the $\ce{Cl2}$ from $\ce{2FeCl3->2FeCl2 + Cl2}$ because chlorine has a different smell.
The product used is the one in this picture.

Welcome to Chemistry.SE. Could you tell us a little more about your $\ce{FeCl3}$ solution? Is it just $\ce{FeCl3}$ in water? Are there additional components? Do you make the solution or is it commercial? It if is a commercial product, identifying it would be helpful. This is information that can help us provide a better answer. I also cleaned up the chemistry in your post. We have the MathJax plugin for chemistry and maths. You can find out more on our [help] page.

Thanks for the tips Ben. It is a commercial product but its label isn't very informative. I think there might be impurities as it is used for etching hobby printed circuit boards. http://www.vitacom.ro/images/products/PictureVitacom/media/c/CLORURA%20F38_500.jpg

The only clue I have been able to find is a one-off sentence in the Wikipedia article for $\ce{FeCl3}$, for which no reference is given:

Iron(III) chloride undergoes hydrolysis to give an acidic solution.

If so, then the reactions are probably those along the conversion of $\ce{FeCl3}$ to $\ce{Fe2O3}$, producing hydrochloric acid $(\ce{HCl})$. The hydrogen chloride gas that escapes from the solution has a pungent acidic acrid smell.
Possible reactions include:
$$\ce{FeCl3 + H2O <=> FeOCl + 2HCl}$$
$$\ce{2FeCl3 + H2O <=> Cl2FeOFeCl2 + 2HCl}$$
$$\ce{2FeOCl + 2H2O <=> Fe2O3(s) + 2HCl}$$
Because these reactions are in equilibrium, solutions of $\ce{FeCl3}$ can be stabilized by addition of $\ce{HCl}$. Addition of $\ce{HCl}$ also promotes the formation of the tetrachloroferrate ion $\ce{FeCl4-}$, which is more resistant to hydrolysis.
$$\ce{FeCl3 + HCl -> FeCl4- + H+}$$
Ferric oxide (Iron(III) oxide $\ce{Fe2O3}$), is insoluble. Does your solution develop reddish precipitate as it ages? That would be  evidence in favor of hydrolysis as the source of $\ce{HCl}$.

I observed some murky-brown precipitate when I used the solution for etching copper. I thought it was $\ce{Cu2Cl2}$ which precipitates. Now I've looked it up and $\ce{Cu2Cl2}$ is white not brown.

I learned something useful. Adding $\ce{HCl}$ to the solution will make it last longer by converting $\ce{Fe2O3}$ to $\ce{FeCl3}$. Thanks Ben!

As you noted, the hobby-reagent grade chemicals are quite impure, so what you smell is either HCl, which you can buy separately and check, or any of the impurities, which reacts in this very dirty mixture to anything. Probably not worth research. It also causes coloration of CuCl (should become CuCl2 soon). BTW, I etch PCBs with much cheaper process: https://sites.google.com/site/ssavcoblog/electronics/pcb


text:
Vertical lines in a Pourbaix Diagram
I know that because of the absence of the hydrogen ions in a reaction ($n_{H+} = 0 $) , the correspondent E-ph line have to be horizental , but what exactly the types of reactions (if there is) that characterizes the vertical lines of an E-ph plot ?

A Pourbaix diagram is a special kind of phase diagram. As in a phase diagram, the lines of all kinds represent phase boundaries. Each line represents the conditions in which two phases are in dynamic equilibrium. Without information on what the phases enclosed by lines look like, a phase diagram is pretty meaningless. Below is the phase diagram of water. Imagine how hard it would be to interpret without the labels. We still need to go look up what is meant by Solid IX - is it ice IX or ice 9?

Here is the iron-carbon (steel) phase diagram. Again, the phases are labeled.

In a Pourbaix diagram, the axes are pH and applied electric potential $(E_0)$ in volts instead of temperature and pressure (pure substance) or temperature, pressure (sometimes omitted), and composition for mixtures. The lines represent boundaries, but the "phases" are now the predominant iron-containing ion in solution. These diagrams seem to be primarily for aqueous systems, and so there is never an absence of $\ce{H+}$ ions. They may not be needed in the redox equation, but they are in solution. Otherwise, how would we have different pH along the horizontal axis?
Below is the Pourbaix diagram for aqueous solutions of iron. Notice that not all lines are horizontal or vertical. Each line represents a set of conditions ($E_0$ and $\text{pH}$) where two ionic species are in dynamic equilibrium.

For example, at $E_0=1.0\ \text{V}$ and $\text{pH}=2$, the predominating species is $\ce{Fe^{3+}(aq)}$. If we increase the $\text{pH}$ to $5$, the predominating iron species becomes $\ce{Fe2O3}\cdot\ce{nH2O}(s)$, rust. The reaction might be (we increase the pH by added $\ce{OH-}$):
$$\ce{2Fe^{3+}(aq) +6OH- (aq) -> Fe2O3(s) + 3H2O(l)}$$
Note that this reaction is redox neutral - all elements remain in the same oxidation state. Vertical lines represent redox neutral transitions/reactions.
If we start at the same place $E_0=1.0\ \text{V}$ and $\text{pH}=2$, and change $E_0$ from $1.0 \ \text{V}$ to $0.4 \ \text{V}$, the predominating species becomes $\ce{Fe^{2+}(aq)}$. When we lower $E_0$, it is like adding electrons to the system (it is really encouraging electrons to flow through the system in a specific way), we get this reaction, which is pH-neutral.
$$\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}$$
As you noted, there is no $\ce{H+(aq)}$ in this reaction. Horizontal lines are pH-neutral transitions/reactions.
A diagonal or curved line is neither pH or redox neutral. For example, let's go from $E_0=0\ \text{V}$ and $\text{pH}=6 \ (\ce{Fe^{2+}(aq)}$ to $E_0=0\ \text{V}$ and $\text{pH}=10 \ (\ce{Fe2O3\cdot nH2O(s)}$ crossing a diagonal line. The reaction now involves a change in oxidation state and in pH. Since we are traveling horizontally, the electrons lost by $\ce{Fe^{2+}}$ must go somewhere.
$$\ce{2Fe^{2+}(aq) + 4HO- (aq) -> Fe2O3(s) + 2H2(g) + H2O(l)}$$


text:
How is iron(III) chloride changed from solid to solution?
I'm conducting an experiment using iron(III) chloride in aqueous solution, but am also aware that it's usually a solid.
How is iron(III) chloride experimentally converted to a solution?
And can you please explain how the conversion happens and its mechanism.

You can prepare a solution of $\ce{FeCl3}$ by dissolving the appropriate amount in water.
Let's say you want a $0.1 \text{ M}$ solution of $\ce{FeCl3}$. You would do this by dissolving 0.1 mole of $\ce{FeCl3}$ in 1 liter of water.
$$0.1000 \text{ mol} \times \dfrac{162.2\text{ g}}{1\text{ mol}}=16.22 \text{ g}$$
You would need to dissolve 16.22 g of $\ce{FeCl3}$ in enough water to make 1 liter of solution.
This solution is not stable over time. $\ce{FeCl3}$ hydrolyzes in water. See my answer to this question. The hydrolysis can be slowed/reversed by adding $\ce{HCl}$. Since solutions of $\ce{FeCl3}$ are already acidic, there is no harm usually in adding more acid. For this 0.1 molar solution, you might choose to dissolve 16.22 g of $\ce{FeCl3}$ in enough 0.1 M HCl to make one liter of solution.

How is iron(III) chloride experimentally converted to a solution?

Iron (III) chloride dissolves in water via acid base chemistry. Iron is the Lewis acid in this case (electron pair acceptor) and water is the Lewis base (electron pair donor). Iron (III) becomes solvated with six water molecules, forming an octahedral complex via the water's oxygen atoms. This pulls electron density away from the water's hydrogens, making them available to be pulled off in solution by other water molecules, a process called hydrolization. Thus the solution becomes more acidic; as iron (III) becomes solvated it increases the concentration of H3O+ in solution.


text:
How much water vapour can a gram of Uppsalite absorb at NTP?
This site states:

It turned out that Upsalite had the highest surface area measured for an alkali earth metal carbonate; 800 square meters per gram.

Assuming that a sphere may enclose the maximum volume for a given surface area, the radius for a sphere enclosed by $\pu{800 m^2}$ is $\pu{7.9788 m}$. It's corresponding volume is $\pu{2127.656 m^3}$ - which leads to the incredible figure of something in excess of 2000 litres of water ... per gram of dessicant.
How much water vapour can a gram of Uppsalite absorb at NTP?

One fallacy in your question is assuming that the particles 1) are spherical and 2) only accessible on the outside. The press release linked says that this material is nanoporous - the particles are perforated by pores "all having a diameter smaller than 10 nano meters". The extra surface area is inside these pores. Spheres have the lowest surface-area-to-volume ratio. Other shapes have higher ratios (in other words, more surface area for less volume). These nanoporous materials maximize the surface area to volume ration - minimizing the volume. If these particles are $\ce{MgCO3}$, then their density is $\pu{2.958 g/cm^3}$ (Wikipedia). One gram of this material has a volume of:
$$\pu{1 g}\times \pu{1 cm^3 g^-1}\times \left(\frac{\pu{1 m}}{\pu{10^2 cm}}\right)^2=\pu{3.38E-7 m^3}=\pu{3.38E-3 L}$$
The surface-area-to-volume ratio is then:
$$\frac{A}{V}=\dfrac{\pu{800 m^2}}{\pu{3.38E-7 m^3}}=\pu{2.37E9 m^-1}$$
A second fallacy is that the volume of water that can be absorbed is the same as the volume of the particle. The particle is not absorbing water per se. The water is not taken into the interior volume where the $\ce{MgCO3}$ is. That would require displacement of the magnesium carbonate, and where would it go? The water goes into the pore of the material - it adsorbs to the surface inside those pores.
You can read more about the science at the original research article:  A Template-Free, Ultra-Adsorbing, High Surface Area Carbonate Nanostructure. PLoS ONE 2013, 8 (7). DOI: 10.1371/journal.pone.0068486.

Yes: nanoporous materials have very high surface area, which makes them suitable for catalysis, adsorption, fluid separation, etc.

"in excess of 2000 litres of water... per gram of dessicant [sic]."
1) What is the volume of one gram of desiccant?
2) What is the volume of 2000 liters of water?
3) What is the cognitive dissonance?
y = sin(1/x) in the interval (0,1).  Rotate about the x-axis. The volume of the cylinder is pi.  The surface area of the curve is infinite.  Wet that surface for a great storage reservoir.  If budget is tight, (1,0.0001) will hold the same volume.  Find the fallacy.
Ditto metal-oxide frameworks and gas adsorption to lower pressure of stored STP volume.  It not about the BET area, it's about the closed cycle - including adsorption and desorption enthalpies.


text:
Does all processes that consume energy from outside result in negative entropy change?
In any spontaneous process entropy increases.In biological systems disordered elements are converted into highly ordered tissues etc(-ve entropy change) which make up living organisms. But, it requires energy from the sun(ultimately). So,does that mean all processes that consume energy from outside result in negative entropy change? Endothermic reactions consume energy in the form of heat from outside so, is all endothermic reactions causing the entropy to decrease?

Entropy increases in a spontaneous evolution of an isolated system… other than that, there is no general law on the evolution of entropy.

does that mean all processes that consume energy from outside result in negative entropy change?

No. When you heat a solid, it increases its entropy.

endothermic reactions causing the entropy to decrease?

Endothermic reactions which cause entropy to decrease have both ∆H positive and ∆S negative, so they will have a positive ∆G, i.e. their are not thermodynamically favorable.

So, is there a general answer for-In which types of reactions does entropy increase/decrease?

@RajathKrishnaR I have edited my post yesterday to reply to your question (at the top)


text:
Solvated solids
I have a very basic question concerning solvation. If a solid (solute) was dissolved in a liquid (solvent), is the solute still in the solid state? or does it convert into another state of matter when this happens?

Good question. Theoretically, the physical phase of the matter does not change; crystalline salt dissolved into water does not become a liquid or gas. The dissolution reaction is theoretically a kinetic change, not a thermodynamic one as the traditional solid-liquid-gas phase changes are (at least at 1atm). However, the solute really isn't a "solid" anymore, either; depending on the nature of the solute and solvent, the two materials may interact chemically, and in fact when dissolution occurs, it's because it is thermodynamically favorable for the solute to be in a dissolved state as compared to a solid.
It is often useful to describe dissolved solutes as being in this dissolved state, especially if the solvent is water, in which case it is in an "aqueous" state. You see this in chemical equations where the states of the reactants and products are of interest, such as when a solid or gas precipitates from an aqueous reaction:
$$\ce{2Na(s) + 2H2O(l) \to 2NaOH(aq) + H2(g)\uparrow}$$
Notice that the sodium hydroxide produced by addition of sodium metal into water is virtually never "pure", even if the reactants are mixed in ideal ratio. The thermodynamically-favorable product is an aqueous mixture, which is treated as a liquid even though "pure" sodium hydroxide is an off-white solid powder.

Thanks for your answer. I can understand your explanation for the ionic materials, such as Sodium. But if this was for a glucose or some other organic compound, does this still apply? You would not have a chemical compound like the NaOH in the case of organic materials.

True. Unlike the molecules themselves dissociating into ions, the dissolution of an organic molecule like a sugar is a little different, involving the water getting in between the hydrogen bonds that form the solid. However, the sugar is still sugar, and it hasn't become a liquid or gas in its own right, so the explanation still holds.

Thanks for your reply buddy. I would like to think it is still in the solid form, but couldn't find any evidence for that.I am not sure what you mean "explanation sill holds". Do you mean that there is still now answer to this problem?

What I meant was that my answer - that no thermodynamic "phase change" has occurred, but the substance may be thought of as being in a "dissolved" kinetic state - applies to organic solids and other non-ionic soluble solids.