How to determine the relative contribution of resonance structures when different rules give contradictory outcomes? In order to determine the relative contributions of resonance structures, my textbook gives the following rules (in order):

The more covalent bonds a structure has, the higher it scores.
Structures in which all of the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are especially relevant and make large contributions to the hybrid.
Charge separation decreases the score.
Resonance contributors with negative charge on highly electronegative atoms score higher than ones with negative charge on less or non electronegative atoms. Conversely, resonance contributors with positive charge on highly electronegative atoms score lower than ones with positive charge on non electronegative atoms.

How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?

For example, let's consider the following resonance structures:

$$\ce{\overset{+}{C}H2-\underset{1}{O}-CH3}\longleftrightarrow \ce{CH2=\underset{2}{\overset{+}{O}}-CH3}$$

The following are my conclusions with respect to the above rules:

Structure 2 scores higher than structure 1 as it has more number of covalent bonds.
Structure 2 scores higher than structure 1 as all atoms have noble gas electronic configuration. In structure 1, the leftmost carbon atom has only six valence electrons.
There is no charge separation in either of the two structures and so this rule cannot be used to determine the relative contribution.
Structure 1 scores higher than structure 2 because in the first structure a carbon atom bears a positive charge whereas in the second an oxygen atom carries a positive charge. Since oxygen is more electronegative than carbon or carbon is more electropositive than oxygen, structure 1 scores higher.

Based on conclusions 1 and 2, we could say that structure 2 contributes more. But conclusion 4 contradicts the result given by points 1 and 2. So, how could we determine the relative contributions when the outcomes of the rules contradict each other?

Are some rules (out of the given 4) superior over the other? Or do we determine the overall score by the number of votes on either sides (similar to our community's voting system based on upvotes and downvotes :) ) i.e., structure 2 scores higher since it has two votes (points 1 and 2) supporting it and 1 vote (point 4) against it? If this is the case, what if there are equal number of votes? For example what if points 1 and 2 suggest a structure A scores higher than B whereas points 3 and 4 suggest structure B scores higher than A?

Kindly do not limit your answers to the above two resonating structures. I considered it just to explain the question.

I read the question - How to determine the least stable resonance structures out of a given set? and the answers to it. In fact, that question is based on a problem from the same book I follow. But still, I don't understand how to determine the relative contributions when some rules give contradictory outcomes.

Note: I understood resonance structures are hypothetical, i.e., they don't have an existence in the real world. The molecule is best represented by the hybrid of all the structures. As commented in my previous questions on resonance, I've already read the question/answer - What is resonance, and are resonance structures real? and I feel I don't have any misconceptions on the subject.

The Rules

Actually the rules for determining the relative stabilities (more accurately, contributions) of resonance structures do have their own priorities. Here is one version of them, with descending order of importance listed below (as is indicated by the bold words). The rules can be found in common organic chemistry textbooks (like [1], also supported by [3]).

  In any case, the more the octet rule is fulfilled (for appropriate atoms), the better.
  Resonance structures with charge separation are usually higher in energy than those in which charges can be neutralized, unless rule 1 is not obeyed.
  If charge is separated, then the best structure is consistent with the electronegativity of the atoms (e.g., a negative charge is best on the most electronegative atom).
  The more and stronger the bonds, the better. Adjacent atoms should not have the same sign of charges.

Examples

The first rule is undoubtfully the most important. The example of $\ce{CH2^+-O-CH3 <-> CH2=O^+-CH3}$ is already mentioned. Here I would like to provide an example to justify the importance of rule 1.

  Consider the Lewis acids $\ce{BX3}$ where $\ce{X}=\ce{F},\ce{Cl},\ce{Br}$. We know that $\ce{BF3}$ is the least Lewis acidic despite the fact that $\ce{F}$ is the most electronegative. This fact can be explained by considering the resonance structures $\ce{X2B-X <-> X2B^-=X+}$. According to rule 1, the second resonance structure is more important, so the electron deficiency of $\ce{B}$ is determined by the orbital overlapping between $\ce{B}$ and $\ce{X}$. Therefore, since $\ce{F}$'s $2\mathrm p$ orbitals interact with $\ce{B}$'s best, $\ce{BF3}$ is the least Lewis acidic.

Another interesting example illustrating the relative importance of rule 3 and 4 comes from Grützmacher's work[2].

  The phosphaethynolate anion $\ce{OCP-}$ may attack $\ce{TMSOTf}$ using $\ce{O}$ and $\ce{P}$. At 298K the product is $\ce{TMS-O-C\bond{3}P}$, and at 318K the product is $\ce{TMS-P=C=O}$. To account for this, let's analyze $\ce{OCP-}$'s resonance structures:
  $$\ce{P\bond{3}C-O- <-> O=C=P-}.$$
  The latter has stronger bonds and the former has its negative charge on a more electronegative atom. According to rule 3, the former dominates (actually a ratio of former vs latter 40:52), hence at ambient temperature $\ce{O-}$ attacks more quickly to obtain the kinetic product, while $\ce{TMSPCO}$ is the thermodynamic product.

Theoretical Justifications

You can find theoretical justifications using Hückel Theory as always. I will not show this here because it may be to tedious to do the calculations. However, you can read this article for more information J. Chem. Educ. 2007, 84, 6, 1056.

Ref.

[1]  Grossman, The Art of Writing Reasonable Organic Reaction Mechanisms, Springer, 2002.

[2] Grützmacher et al., Dalton Trans., 2014,43, 5920-5928.

[3] Humbel, J. Chem. Educ., 2007, 84, 6, 1056-1061.


What is degree of dissociation for polyprotic acids? My textbook says that degree of dissociation= [HA]dissociated/[HA]initial * 100%.
However, it also says that citric acid has a greater degree of dissociation than acetic acid because citric is triprotic whereas acetic is monoprotic.

This reasoning doesn't fit with the top definition because, more protons donated doesn't necessarily mean more acid molecules dissociated (triprotic acid vs monoprotic acid for example; for every acid molecule dissociated, triprotic acid donates more protons).

So my question is, how does one define degree of dissociation for polyprotic acids?

I think this resembles the question Confused about strong/weak dibasic and tribasic acids?
Saying that the degree of dissociation $(\alpha)$ of citric acid is greater because its triprotic is not correct. The degree of dissociation should be calculated for each dissociation separately. The degree of dissociation of citric acid's first dissociation $\ce{C6H8O7 <=> C6H7O7- + H+}$ is greater than that of acetic acid, because its $\mathrm{p}K_\mathrm{a}$ is smaller (3.13 vs 4.76)$^1$, meaning it's a stronger acid.
The second value of $\alpha$ for citric acid, however, will be very to close to acetic acid's, because the second dissociation has approximately the same $\mathrm{p}K_\mathrm{a}$ (4.76)$^1$. The third $\alpha$ will be smaller, since the $\mathrm{p}K_\mathrm{a}$ of the third dissociation is greater (6.40)$^1$.
This can be seen in the common polyprotic acids dissociation graphs, like this one (source):

Note that $\alpha$ is represented on the $y$-axis in a 0-1 scale. For each of the four states of the molecule, there is a different $\alpha$, and different values of $\alpha$ happen simultaneously.
$^1$ $\mathrm{p}K_\mathrm{a}$ values taken from Wikipedia.


The detergent 'Positive 170' and what exactly makes it so corrosive? So in season 5, episode 7 of Dirty Jobs, we're shown how a locomotive engine block is degreased using the detergent 'Positive 170' and we're told that it is extremely dangerous.

Just how 'dangerous' are we talking about here and why?

Thanks.

Hint: read the column in the centre.  The potential risks in damage of tissue, described on the left and on the right hand column increase by increase of the solution's concentration, and temperature (and 170 F [about 76 C] is considerable), and obviously by the time of exposure to this cleaner.

As an example of what may happen to skin under attack of strongly basic solutions (sometimes called lye) if you do not wear appropriate protective gear, see this example. -- Be told, such pictures may disturb you.


Heat of Combustion examples What is the criteria to identify the order of heat of combustion among different organic molecules?
(like the main points to be noted while selecting the order)

The answer being Z>Y>X.
The method tried is that for the bonds to break easily , if enough of electrons are there it would break easily so order being X>Z>Y,
Please tell where I am wrong.

You have to consider the $3$D structure and sterics when dealing with cyclic compounds.  If you look at the geometry of cyclohexene below you can that the ring is not symmetrical, thus the placement of the vinical methyl groups in regard to the double bond matters.

In (X) when the vinical methyl groups are both on doubly bonded carbons they are forced into a coplanar arrangement which is unfavorable and maximally strains the ring's double bond and the methyl group bonds.  In (Y) when they are opposite the double bond the methyl groups are optimally spaced apart and steric strain on the ring and methyl bonds is at a minimum.  When they are not located on the double bond they are neither at a maximum or minimum, just in between so (Z) just falls in the middle.  There are of course other conformers of cyclohexene but the relation will still hold.

The question could have also thrown in a fourth isomer where one methyl group was on a double bonded carbon, but it would not be easy to rank it above or below (Z) without computational methods.


How to determine rotational states of a molecule? How can you determine the possible rotational states of a molecule? Atkins Physical Chemistry book says that for linear molecules there are two possible rotational states, but what is the general method for finding the amount of rotational states for other relatively more complex molecules, say for example $\ce{H2O}$ and $\ce{CH4}$?

Molecules have only 3 (or 2 if linear) rotational modes.  A molecule with n atoms has 3n-6 vibrational states, 3n-5 if it is linear.  The "6" or "5" that is subtracted represents the 3 translational modes that all molecules can move in (x, y, and z directions) plus the 3 rotational modes (rotation about the x, y and z axes).  The 3 rotational modes reduce to 2 if the molecule is linear since you can't observe rotation about its molecular axis (the axis that contains the $C_{\infty}$ symmetry element).


Why there exist no void with co-ordination number 5 Here

      oh = octahedral
      td = tetrahedral
      CN = co-ordination number

My problem is that, that why it is not possible to have a void with CN = 5 ?

My reasoning is:

According to the definition of CN i.e. the number of nearest neighbors of a particle (here a void because we are talking about it).

In the following figure if we treat silver sphere as the void then we say that it is a td void.

But if we place a sphere over this td void then we generate a void whose distance from center of void to the center of all the 5 spheres is equal. And so making the CN = 5

So kindly help me in understanding that why a void with CN = 5 not possible.

Trigonal bipyramidal complexes are abundant.  However, if you want a periodic crystal, you must tesselate volume with translation - the unit cell of the crystal lattice.  A five-fold symmetric figure cannot do this, for the angles do not add up without gaps.  2-, 3-, and 6-fold symmetries are allowed, offering 230 unique 3-D space groups overall.

If you allow two kinds of unit cell, you can get approximate periodicity, as in quasicrystals and gas-water clathrates.  Both can have 5- or 10-fold symmetry overall and give sharp x-ray or neutron diffraction spots (the IUC definition of a crystal).


Is it possible to react chlorine with hydrogen under a UV lamp during the electrolysis of salty water? I have set up an experiment where I am generating chlorine by electrolysing water saturated in NaCl. I know chlorine gas is actually being generated because I hang wet litmus paper in the gas which quickly bleaches. I also know hydrogen is being generated because if I hold a match to the gas it reacts with a popping sound.

If I have this set up in a closed container, how come I cannot get the hydrogen and chlorine to react under a black light? What am I missing?

I advise you not to attempt a reaction of hydrogen and chlorine gas by light!

Do not attempt it unless you know exactly what you are doing and have seen the experiment be performed by others!

You’re actually rather lucky that one of the following happened:

the black light gave the wrong wavelength
there were too small concentrations of hydrogen gas and chlorine gas to react with each other.

The reaction of hydrogen and chlorine is a radical chain reaction and rather exothermic. There is no change in the amount of substance involved, so entropic effects do not play a role. However, it is an explosive reaction and was often used in the introductory inorganic chemistry course at my university to give off a loud bang (and send the tin can in which the gases were mixed flying).

The reaction is triggered by light energy interacting with chlorine; the first (initiation) step is the generation of chlorine radicals by the following equation:

$$\ce{Cl2 ->[$h\nu$] 2 Cl^.}\tag{1}$$

This chain then propagates according to reactions $(2)$ and $(3)$.

$$\ce{Cl^. + H2 -> HCl + H^.}\tag{2}$$

$$\ce{H^. + Cl2 -> HCl + Cl^.}\tag{3}$$

The light needed for the first step must supply chlorine’s bond dissociation energy; blue (visible) light, not UV light is actually required. But just for your safety please let me restate:

The reaction is explosive! Do not attempt it at home under any circumstance! Do not attempt it unless you know exactly what you are doing!


How are molecular orbitals obtained in quantum chemistry? When trying to solve for a molecule's molecular orbitals, we start off with the total molecular hamiltonian. However, given the difficulty of solving the Schrödinger Equation for this complicated hamiltonian we can introduce the Born-Oppenheimer approximation. By assuming that the nuclei of molecules are relatively fixed, our molecular hamiltonian can be reduced to the electronic hamiltonian. Then, we can use the electronic hamiltonian and solve for the electronic wavefunctions of our molecule.

This calculation for the wavefunctions is significantly easier. What is confusing to me is what the solution to the electronic Schrödinger equation represents.

Are these solutions considered the molecular orbitals?

I think you are close to the right idea, but let me try to clarify some points.
The full Schrodinger equation for a molecule should depend on the $3N$ nuclear coordinates, as well as the $3n$ electronic coordinates, where $N$ and $n$ are the number of nuclei and electrons respectively. This is challenging both due to the number of coordinates and, in particular, due to the coupling between electronic and nuclear coordinates. However, with the Born-Oppenheimer approximation, we assume they are decoupled and solve for the electronic part of the wavefunction at a fixed nuclear geometry.
The issue here is that we still can't solve the Schrodinger equation for an $n$ electron system with $n>1$. To simplify further, we can make the Hartree-Fock approximation, which (in one way of phrasing it) assumes that the $n$ electron wavefunction can be formed from the antisymmetrized product of $n$ one electron wavefunctions/orbitals. These orbitals can be determined by solving a nonlinear system of equations and then stitched together in the form of a Slater Determinant to make an approximate $n$ electron wavefunction.
While chemists like to frame discussions in terms of orbitals, strictly speaking orbitals are always an approximation (except for hydrogen) and the electronic structure of a molecule is determined by a $n$ electron wavefunction. Even though orbitals aren't quite real, they still can give a pretty good qualitative description of a molecule's behavior.
For a bit more about the meaning of orbitals, you might want to check out this previous question: Is it correct to talk about an empty orbital?


Calculating standard combustion enthalpy of dodecane from that of heptane & octane These are the problem from UKChO-2015.

  Given the standard enthalpy changes of combustion ($\Delta_\mathrm{c}H$) for heptane ($n = 7$) and octane ($n = 8$) are $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$ respectively, calculate $\Delta_\mathrm{c}H$ of dodecane ($n = 12$).

Any hint on regarding this problem? My efforts are not worth mentionable here

Heptane is $\ce{CH3-CH2(5)-CH3}$

Octane is $\ce{CH3-CH2(6)-CH3}$

Given combustion enthalpies for each of $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$  respectively, we now have a simple set of algeraic equations:

5A + 2B = $-4816~\mathrm{kJ~mol^{-1}}$

6A + 2B = $-5470~\mathrm{kJ~mol^{-1}}$

First we can solve for A as the combustion enthaply of a CH2 in a linear hydrocarbon and we get the value $-654~\mathrm{kJ~mol^{-1}})$.

We can now solve for B,

In the first equation we get 2B = ($-4816 - (654 * 5)~\mathrm{kJ~mol^{-1}}$, A = $-773~\mathrm{kJ~mol^{-1}}$

In the second equation we get 2B = ($-5470 - (654 * 6)~\mathrm{kJ~mol^{-1}}$ , A = $-773~\mathrm{kJ~mol^{-1}}$

dodecane  has CH3-CH2(10)-CH3 , for 10A + 2B.

10 * A  + 2 * B  = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$


Is this a phase change in a gas under Thermodynamic Equilibrium? The above graph shows the molar density (i.e. n/V from PV=nRT) of a gas plotted against pressure. There are two regions with good fit to a linear relationship which I have coloured green and blue (though these are all from a single set of measurements. The visual distinction is just to highlight where the linear fit applies)

My rusty physical chemistry knowledge suggests that the regions where there is a linear fit are in Thermodynamic Equilibrium as the Ideal Gas law would appear to be holding. Given that we have two linear regions I would also conclude that there is a phase change and that the whole is actually in Thermodynamic Equilibrium when we account for the phase change (i.e. n is different between the two regions)

Is this naïve analysis based on a foggy recollection of Phys Chem lectures from 20 years ago correct or is there another explanation for this data (and if so what is the explanation)

Update

To clarify on the hypothesized phase change. The suggestion is that this phase change is a X + X <=> X2 phase change where X and X2 are both gaseous but the X2 form is only thermodynamically stable at lower temperatures (which correspond to the lower pressures). i.e. not the more common gas-liquid phase change

I don't think it necessarily suggests a phase change.  It just seems to me that, at the higher pressures, we are seeing non-ideal gas behavior setting in.


What could be a high school experiment that involves changing a condition to get the highest yield of a certain product? I am looking for ideas for a chemistry experiment and I am interested in doing one that has to do with the optimization of percentage yields. What could be a high school experiment that involves changing a certain condition to get the highest yield of a certain product?

You could try the so-called Job's experiment. You prepare in advance 100 mL of two solutions of reactants which are supposed to have the same molar concentration. For example : CaCl2 0.1 M and Na2CO3 0.1 M. You take 9 test tubes in the same rack. Write 1 on the first, 2 on the second tube, etc. up to 9 on the last tube.

Now, with a plastic pipette, you put 1 mL of the CaCl2 solution in test tube 1, 2 mL CaCl2 in test tube 2, 3 mL of CaCl2 in test tube 3, etc. up to 10 mL in the tube 10. Then, you take another plastic pipette, and transfer 9 mL of NaCO3 solution in the tube 1, 8 mL of Na2CO3 in test tube 2, etc. up to 1 mL in tube 9. You mix thoroughly all tubes. You must see a white precipitate that settles in the bottom of  all tubes after a while. If you measure the height of the deposit, you will find that the highest amount is in the tube 5, if the stoichiometry is 1 + 1. If the mot important deposit is in another tube, it means that the stoichiometry is 1:2, oe 2:3, or any other ratio.

You may change and use any other combination of compounds. You will find that in the mixture CuSO4 plus NaOH, the stoichiometry is not 1:2, as you might suspect, as the precipitate is not Cu(OH)2, but 3Cu(OH)2·CuSO4.


Is it possible to electroplate from a gallium anode? I have been looking at some papers on electro-deposition of iron gallium alloys and these all use solutions made up of Fe and Ga sulphates.

I am just curious as to why perhaps an anode made of one of the materials might not have been used as part of the process? I mention gallium specifically because the cost of the Ga sulphates is quite high, while the Ga metal is relatively cheap.

My impression is that it is typically difficult to electro-deposit gallium, but I think this was in specific reference to plating from solutions of the Ga salts, I have not yet come across plating from a Ga anode, and I'm curious about this.

Yes, it is possible to electrodeposit gallium. The difficulties are: efficiency due to the need for low electrodeposition potentials, formation of extremely fine hydrogen bubbles which spoils the surface finish, the requirement to use extremely low temperatures and the melting/alloying of gallium with the metal it's plated on, and poor repeatability of the process. There doesn't seem to be a patent describing using a pure gallium anode.

The difficulties are described in these patents:

US20110226630A1 - "Gallium electroplating methods and electrolytes employing mixed solvents":

  ... Gallium electrodeposition electrolytes and electrodeposition methods for solar cell manufacturing processes have many more stringent and special requirements than the electrodeposition methods and solutions employed for many other commonly plated metals such as Cu, Ni, Co, Pb, Sn, Ag, Au, Pt, and their alloys, etc. This stems from the facts that;

  i) Ga is one of the lowest melting point metals in existence, with a melting point of about 30° C.,

  ii) Ga has a high negative electrodeposition potential and thus Ga electrodeposition efficiency is naturally low since high electrodeposition potentials cause hydrogen generation, in addition to Ga deposition, at the cathode surface in aqueous electrolytes,

  iii) hydrogen bubbles generated on a cathode surface form defects such as un-deposited regions unless such bubbles could immediately be removed from the surface,

  iv) Ga has a tendency to form low temperature melting alloys with many alloy-partner materials such as In, Cu, Ag, Pb, Sn, etc. Furthermore, such alloys may form during electrodeposition of Ga onto surfaces comprising any of such alloy-partner materials.

  Electrodeposition solutions employing glycerol are very viscous and difficult to handle. The viscosity of glycerol at room temperature is 1500 centipoise (cP) compared to the viscosity of water, which is 1 cP. Gas bubbles such as hydrogen bubbles formed on the electroplated (cathode) surface during Ga plating in viscous electrolytes cannot be easily removed from that surface and therefore cause voids and other defects in the electrodeposited films. Such defects may be acceptable for some applications of thick electrodeposited Ga globules. However, they cannot be tolerated in electronic device applications such as solar cell absorber formation applications where they cause compositional non-uniformities, morphological non-uniformities, and pinholes etc., all of which negatively impact the device performance.

  Glycerol based plating solutions become more viscous as their temperature is lowered and therefore the problems cited above may get worse at lower temperatures. One other important point about the electrodeposition process for Ga is its sensitivity to the nature of the substrate surface on which the electrodeposition is performed. For example, to form a Cu/In/Ga precursor stack, the Ga film needs to be electrodeposited on an In surface. To form a Cu/Ga/In precursor stack, on the other hand, Ga plating needs to be performed on a Cu surface. One Ga electrodeposition solution that performs well for plating Ga on a Cu surface may not perform well for electrodepositing Ga on an In surface because the electrodeposition efficiency of Ga on one surface may be very different from its electrodeposition efficiency on another surface.

  As mentioned above, gallium is a low melting point material with a melting temperature of around 30° C. As a result, when electrodeposited out of aqueous electrodeposition solutions kept at about room temperature (20-25° C.), it often forms rough films comprising molten surface features, especially at high electrodeposition current densities such as current densities greater than about 5 mA/cm$^2$. This is because even though the electrodeposition solution may be at a temperature lower than the melting point of Ga, the local temperature on the cathode surface may actually exceed this melting point due to the heat generated by the electrodeposition current. As further mentioned above, when Ga is electrodeposited on surfaces of materials that easily form alloys with Ga, molten droplets of Ga alloys with low melting temperatures may be formed on such surfaces. If the Ga film is electrodeposited over In and/or Cu, the local heating and Ga melting may actually promote alloying between the plated Ga film and the underlying In and/or Cu because there are low melting alloy phases between Ga and these materials such as In—Ga alloy phases and CuGa$_2$ alloy phase. As a result, the surface roughness of the deposit may further be increased due to the above mentioned reaction and the formation of molten alloy phases. For example, Mehlin et al. (Z. Naturforsch, vol. 49b, p.1597 (1994)) attributed the rough morphology of their electroplated Ga layers to the alloying of the electrodeposited Ga with the underlying Cu surface of the cathode and the formation of a molten CuGa$_2$ alloy.

  Gallium may be electrodeposited from the electrodeposition solution at temperatures below −10° C., preferably below −20° C., most preferably below −30° C., so that local melting of the deposited Ga and its possible reaction with the materials on the cathode surface are avoided. Furthermore, at these low temperatures, the electrodeposition current densities may be increased to levels above 5 mA/cm$^2$, preferably above 10 mA/cm$^2$ and even above 20 mA/cm$^2$ without causing melting and/or alloying on the cathode surface. As a result, the electrodeposition rate and therefore the process throughput may be increased while, at the same time, the deposited film roughness is reduced. All of these benefits are important for the successful use of electrodeposited Ga layers in thin film solar cell manufacturing. For example, the melting point of methanol is −97° C. and the freezing point of a methanol/water mixture is a function of the ratio of methanol to water in the electrodeposition solution. A mixture of 75% methanol and 25% water, for instance, has a freezing point of −82° C. (−115° F.). This means that a Ga plating electrodeposition solution comprising 75% methanol and 25% water may be operated at a temperature as low as about −70-80° C., thus avoiding the melting, reaction and surface roughness problems described above.

  ...

US7507321B2 - "Efficient gallium thin film electroplating methods and chemistries":

  ... Gallium is a difficult metal to deposit without excessive hydrogen generation on the cathode because Ga plating potential is high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas, rather than the Ga film on the substrate or cathode. Hydrogen generation and evolution also causes poor morphology and micro defects on the depositing films due to the tiny hydrogen bubbles sticking to the surface of the depositing film, masking the micro-area under them, and therefore impeding deposit on that micro-area. This causes micro-regions with less than optimum amount of Ga in the film stack.

  Poor plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation phenomenon itself is a strong function of many factors including impurities in the electrolyte, deposition current densities, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer etc. As at least one of these factors may change from run to run, hydrogen generation rate may also change, changing the deposition efficiency.

  Electrodeposition of Ga out of low pH aqueous electrolytes or solutions may suffer from low cathodic efficiencies arising from the presence of a large concentration of H+ species in such electrolytes. Therefore, hydrogen gas generation may be expected to lessen at higher pH values. However, as the pH is increased in the solution, Ga forms oxides and hydroxides which may precipitate as reported in the literature. Only at extremely alkaline pH values these oxides/hydroxides dissolve as soluble Ga species. Therefore, it becomes possible to electrodeposit Ga in a bath of pH>14 containing Ga salts as was done in prior-art techniques using high concentrations of KOH and NaOH in the bath formulation.

  High concentrations of alkaline species, however, cause corrosion problems for the equipment as well as the cathode material itself. There is also a limit of the Ga amount that can be dissolved in the form of acidic Ga salts (GaCl3, Ga(NO3)3 etc) in such solutions before Ga starts to precipitate. Therefore, the pH needs to be adjusted again by further addition of alkaline species such as NaOH and KOH.

  As pointed out above, solutions comprising a large molar amount of caustics are difficult to handle and they also have high viscosity. High viscosity makes the hydrogen bubbles formed on the cathode stick more to the cathode making it very difficult to remove them by stirring or other means of mass transfer. As explained above, such gas bubbles on the cathode surface increase defectivity of the deposited Ga layer.

  ...

US8545689B2 - "Gallium electrodeposition processes and chemistries" and
WO2012028416A2 - "Gallium and indium electrodeposition processes":

  "... Current processes for electroplating metal containing thin films such as gallium and gallium alloys present numerous problems. These problems include, among others, low cathodic deposition efficiency due to excessive hydrogen generation, poor repeatability of the process, partly due to the poor cathodic efficiency, and the poor quality of the deposited films such as their high surface roughness and poor morphology.

  Gallium is generally considered a difficult metal to deposit without excessive hydrogen generation on the cathode because gallium plating potential is relatively high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas rather than for forming the gallium film on the substrate or cathode. Moreover, hydrogen generation and evolution is a causal factor for introducing porosity into the deposited film, thereby contributing to increased surface roughness and microdefects.

  The plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation itself is a strong function of many factors including impurities in the electrolyte, deposition current density, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer, and the like.

  ...

  Summary:

  The present invention is generally directed to solutions and methods for electrodeposition of a substrate. In one aspect, the solution for electrodeposition of the metal containing film comprises a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; an acid selected from the group consisting of an alkane acid, an alkene acid, an aryl acid, a heterocyclic acid, an aromatic sulfonic acid, an aromatic sulfuric acid, hydrochloric acid, perchloric acid, and nitric acid; optionally a metalloid compound wherein the metalloid compound is selected from the group consisting of arsenic, antimony, bismuth, a combination thereof; and a solvent to dissolve said metal salt, wherein the pH of the solution is in a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14, and wherein the solution is free of a complexing agent.

  The method for electrodepositing a substrate comprises contacting

  (i) a solution comprising a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; optionally an inorganic metalloid compound; further optionally an organic additive having at least one sulfur atom; and a solvent to dissolve said metal salt, wherein the solution is free of a complexing agent; and

  (ii) a substrate; adjusting pH of the solution to a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14; and applying a current to electroplate the substrate with a metal containing film.

  ...


Molar Gibbs Energy for an ideal gas I'm given a problem where an ideal gas is expanded isothermally. I'm given the molar volumes from the start and the end and basically asked to find the change in Gibbs Energy.

I started with the equation

$$G=H -TS$$

We can write the general form $dG$

$$\begin{align}
dG & = dH -SdT - TdS \\
    & = (dE + PdV + VdP)- SdT -TdS \\
    & = (TdS - PdV) + PdV + VdP - SdT - TdS \\
    & = VdP -SdT
\end{align}$$

It's an isothermal process $\to$ (const.) $T$. Hence $SdT \to0$

Now we are left with $dG=VdP$. We need the Gibb's energy in terms of moles and molar volumes. Hence entropy (which is zero anyway) and volume need to change to molar volume. Our new equation is in terms of moles

$$dG_m=V_mdP$$

if we assume $V_m$ to be constant which it isn't but for argument sake we can get the following relation

$$\Delta G_m = V_m\Delta P=RT \ln\Big{(}\dfrac{P_f}{P_i}\Big{)}$$

since $V_m = \frac{V}{n}=\frac{RT}{P}$ for an ideal gas.

My problem is - I would like to know if I can assume the pressure to be constant and have

$$\Delta G_m = RT \ln \Big{(}\dfrac{V_f}{V_i}\Big{)} $$

Is this equation valid for finding the molar Gibbs energy, knowing the molar volumes (initial and final) and at constant temperature.

I think it's wrong because $dG_m = V_mdP$ is valid for constant $T$ but the pressure cannot be constant. I've just been struggling with trying to find a relationship with gibbs energy and molar volume and not knowing pressures.

Your final result is almost correct, although your derivation is somewhat questionable, and, in particular, this equation is wrong: $$\Delta G_m = V_m\Delta P = RT\ln\left(\frac{P_f}{P_i}\right).$$ For one, as you've mentioned, $V_m$ isn't constant, so we can't use the finite difference implied by $\Delta$, and I don't understand how you would have gotten the second equality without having integrated.

Instead, we should have $$\Delta G_m = \int_{G_{m,i}}^{G_{m,f}}\mathrm{d}G_m = \int_{P_i}^{P_f}V_m\,\mathrm{d}P = RT\int_{P_i}^{P_f}\frac{\mathrm{d}P}{P} = RT\ln\left(\frac{P_f}{P_i}\right) = RT\ln\left(\frac{V_i}{V_f}\right).$$ The third equality follows from $V_m = RT/P$, as you note, and the final equality follows from substituting $P = nRT/V$, the ideal gas law, and cancelling out terms.


Forcing atoms closer together Is there a method in chemistry to alter the energy of atomic bonds by forcing atoms closer together in a smaller volume?
Similar to how pressure squeezes atoms closer together.

There are many ways in which bond lengths can be changed, and these will be accompanied by changes in bond energy.  I will give examples for four different classes of substances, though they may not all count as "forcing" the bonds to be shorter in the way in which you mean.  You can look these over, and perhaps clarify your question:
Glasses: Toughened glass is made by rapidly cooling glass, which creates internal stresses in which the interior is under tension (increasing bond lengths), while the surface is under compression (reducing bond lengths).  So while the changes in bond length are still a result of pressure, the pressure is internally, rather than externally, applied.
Molecules: You can get the molecule to undergo a chemical change.  For instance, changing ethane to ethene will shorten the carbon-carbon bond by converting it from a single bond to a double bond.
Metals: Different metallic crystal structures will have different densities, and thus likely also different average bond lengths (the change in density is not due purely to a change in bond lengths, but also to changes in atomic arrangements, including the number of nearest neighbors, next-nearest-neighbors, etc.). Thus you can alter bond length by inducing the substance to change from one crystal structure to another.  As an example, I'll quote from https://www.conservation-wiki.com/wiki/Tin_pest :

Solid tin exists in two allotropic forms, white and grey, each of which contains different chemical properties. White tin has a body centered tetragonal crystalline structure and a density of 7.28, while grey tin has a cubic structure and a density of 5.84 (Faulkner 1965). This transformation is generally accepted to occur around 13C (Gilberg 1991:3). Below this temperature, grey tin is more stable, while white tin is more stable at temperatures above 13C.

Covalent network solids:  Essentially the same effect as what was described for metals, above, except that the transformation involves making and breaking covalent bonds.  A good example is the transformation between diamond and graphite; though note that, in this case, the structure changes from 3D (diamond) to 2D (graphite)
Here's a graphic by Wikipedia contributor Nath321 illustrating this: https://en.wikipedia.org/wiki/Atomic_spacing -- though here the comparison is between a 2D structure:


Electrode Potential at Equilibrium At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. At equilibrium, there is a separation of charges and depending on the tendencies of the two opposing reactions, the electrode may be positively or negatively charged with respect to the solution. A potential difference develops between the electrode and the electrolyte which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential.

Above is an excerpt from my book.

It says that at equilibrium the potential difference develops between electrode and eletrolyte and it is electrode potential.

But at equlibrium, should not the electrode potential should be zero as at equlibirum there should not be any tendency of reaction to occur?

From paragraph of textbook.
At the electrode-electrolyte interface the metal ions in electrolyte leak out, leaving a negative charge on electrode side of interface. While metal ions migrate at the other side of the interface from the electrolyte solution.
This eventually comes to a stop when the equilibrium is reached. That is the rate of negative charge build-up is equal to the positive charge. Like you say at this point there is no more drive for the reaction to go one way or the other.
However, now because of the charge separation at the interface, there is a potential difference between both sides of this interface which leads to the reduction potential we read out which is not zero. This is where the Nernst equation comes in handy.


Why are the axial bonds of PF5 longer than those of the equatorial bonds? (Hybridization) The axial bonds of $\ce{PF5}$ are longer than those of the equatorial positions.

One explanation is that because the axial bonds are experiencing more repulsion
than those of the equatorial and this leads to longer bonds.

Another explanation is that the axial bonds of p/d hybrids, whereas the equatorial bonds are s/p hybrids.

I am able to comprehend and appreciate the first explanation. However, I am unable to understand how there are differences in the hybrids. I had always considered d hybrids a dubious argument at best, and I believed the hybridization would lead to degenerate orbitals.

However,  the explanation provided would suggest otherwise, that there are different hybridized orbitals of different energies at different portions of a single hybridized atom (excluding lone pairs. I understand those being non-hybridized in a hybridized atom).

Thus, my question is how the hybrid argument that was presented works and how would one be able  to realize the different hybridized states of an atom in a molecule (as an approximation).

Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.

I could go stealing pictures and explanations from ron’s great answer, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.

If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in ron’s answer. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.

From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.


Metallic and magnetic nature of transition metal oxides Which of the following oxides is both metallic and ferromagnetic:
  $\ce{VO2}$, $\ce{CrO2}$, $\ce{MnO2}$, $\ce{TiO2}$?

This is how I went about solving it:

Since ferromagnetic materials also show paramagnetism, so the condition for paramagnetism must be checked.

In $\ce{TiO2}$, $\ce{Ti}$ has no unpaired electrons, it cannot be paramagnetic and as a consequence it cannot be ferromagnetic.

$\ce{CrO2}$ is amphoteric. A metallic oxide would show basic properties in most cases. So $\ce{CrO2}$ may not be the right answer.

$\ce{VO2}$ is basic. All three except $\ce{TiO2}$ are paramagnetic.

Then I googled, I didn't get an exact answer. But I read that $\ce{CrO2}$ is termed as a "half metal" which is related to its magnetic properties, most part of which I didn't understand.

Then I also found articles about half metallic ferromagnetism of $\ce{TiO2}$. But that goes against my assumption.

So my questions are:

  Explain where I'm wrong.
  Why $\ce{CrO2}$ is the answer? (Found that when I finally referred to the solutions of the question paper. However the solutions given
  can have mistakes. So it is possible that $\ce{CrO2}$ may not be the
  true answer).
  If you think $\ce{CrO2}$ is not the answer, then why?

This question should be cross-posted to Physics.SE, it is not a simple chemistry question. I'll give short rationalization of the fact, but beware: it is a rationalisation, not a deduction. It is also given in layman's terms.

We know the crystal structure of $\ce{TiO2}$, it is stereotypical and quite common. It contains octahedrons $\ce{TiO6}$ with oxygens shared between three octahedrons. Since the four compounds have similar composition and the cations have similar size, we may expect them to have similar structures. This is, however, a mere guess, but lets' go with it.

For a compound to have metallic conductivity, it must have mobile electrons, i.e. partially occupied group of orbitals. Thus, $\ce{TiO2}$ is obviously out.

It is known, that $d$-orbitals of a metal in octahedral neighbourhood split: three are lower in energy then two other.

In $\ce{MnO2}$ there are three electrons, so all three lower orbitals are occupied. This makes electrons immobile: for an electron to move to a neighboring atom, it has to pair with one of the already present electrons, which requires energy. I do not expect, however, that $\ce{MnO2}$ shall be a great insulator: the pairing energy is quite low, and furthermore, any impurity shall create charge carriers. So, $\ce{MnO2}$ is probably a semiconductor.

To choose from remaining two, we should recall that ferromagnetism is common among compounds with a lot of half-filled d-orbitals. This suggests that $\ce{CrO2}$ has higher chances, than $\ce{VO2}$ Still, the latter should be paramagnetic and conducting under right conditions.

Again, you should cross-post the question on Physics.SE and probably read a good solid state physiscs book.


Do materials exist that absorb Visible light and emit Near InfraRed light? I am thinking about the possibility of making a hat or other clothing item that might show up funny on cameras, as they are more sensitive to infrared light than human eyes.
Therefore, I was wondering if there are any (non hazardous) materials that emit InfraRed light when they are exposed to visible light.

Do these exist, and would they be suited for this kind of application?

Yes, there are some minerals that shows VIS-NIR luminescence here a couple:

Synthetic ruby $\ce{Al2O3:Cr}$
Kyanite ($\ce{Al2SiO5}$)

(Reference here)

A more sophisticated material is $\ce{LiSrPO4:Eu2+, Pr3+}$ you can read the full article in 11 February 2013 / Vol. 21, No. 3 / OPTICS EXPRESS 3161.

There are also some pigments like egyptian blue or cadmium yellow (CI pigment yellow 37, that is cadmium sulfide) that were also used in the past. Egyptian blue (calcium copper silicate ($\ce{CaCuSi4O10}$))  is safe, other pigments with cadmium are more toxic.

One of the most spread pigment shows VIS-NIR luminescence: Chlorophyll!

Have a look to this site for a nice overview.

All the luminesce processes have not a high quantum yield, so to record them, most of the times, you have to avoid reflection. Tungsten, halogen lamps and sun emit a lot of NIR radiation so is possible that reflected light overhang the luminescence. When I performed NIR luminescence I used VIS LEDs and a long pass filter at 850 nm.  If you don't have result, first thing you have to try is to remove the ICF (IR-Cut off-Filter) filter from your camera so NIR light can reach the sensor more easily.


Why does (3E)-3,4-dimethylhex-3-ene yield a meso dihalide when reacted with Br2/CCl4 at room temperature? The answer is B. But why? What makes it so different than the others?
I know that it attacks from the back (anti addition). But can't it do that for all and result in a meso compound?

A meso compound is a molecule with multiple stereocenters that is superimposable on its mirror image. Meso compounds are easily identified by their internal plane of symmetry.

Using this knowledge, and given that $\ce{Br2}$ undergoes anti addition to alkenes, there are only really two answers that could possibly be meso compounds, B and D. Let us consider the products:

Notice how only B has an internal plane of symmetry, and is therefore a meso compound.


Calculating Heat Release from Bomb calorimeter When $0.400\ \mathrm g$ $\ce{CH4}$ is burned in excess oxygen in a bomb calorimeter that has a heat capacity of $3245\ \mathrm{J/^\circ C}$ a temperature increase of $6.795\ \mathrm{^\circ C}$ is observed. What is the value of $Q_V$?

I thought it was the regular $Q=mc\Delta T$ problem but it isn't and not only that I did the equation but it didn't work.

I am not sure what to do right here at the moment

  Using the data determine standard enthalpy change for the combustion of methane

Also I would like the definition of the standard enthalpy change and what it tells you and the purpose of it.

Heat of combustion is the amount of heat (enthalpy) liberated when 1 mole $\ce{CH4}$ reacts (combusts) with oxygen.

In other words, what is the heat of reaction in this case?

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$

Then you must ask yourself, what happens after the heat is produced i.e. where does it go? It quickly turns into an accounting problem...

Take a look at: Constant Volume Calorimetry


Solute and Solvent classification in a Solution When I look up the definition of solute and solvent, the definition by the Oxford English Dictionary is
solute

/ˈsɒljuːt,sɒˈljuːt/

noun

the ***minor*** component in a solution, dissolved in the solvent.

What does minor imply here? Does it classify them as solute and solvent as per mass/volume comparison? Or does it refer to their concentration?

By major/minor they mean whichever component has the greater/lesser presence in mass or moles.


Is there a non-violent, practical way of making hydrochloride with only Hydrogen and Chlorine? A way of making HCl gas is to put Chlorine and Hydrogen into a test tube and expose it to UV light.

This reaction is however fast and strong and I would like to store the Hydrochloride from it.

Another way would be to dissolve Chlorine in water, which would make Hydrochloride, but most of it would go in water and make dilute hydrochloric acid. I could however store the HCl fumes from the dilute acid. I do not know about the practicality of this.

What is the best, and exceptionally practical way of making Hydrochloride when the only resources I have are Hydrogen and Chlorine?

The best and practical way is to sell hydrogen and chlorine, and buy yourself some $\ce{HCl}$ (unless it is prohibited in your jurisdiction). However, I would recommend that only if you have some experience in chemistry and understand quite well the dangers thereof.

The technical answer to your question is: build a galvanic cell with platinum anode and graphite cathode, and feed them with hydrogen and chlorine, respectively. It will perform your reaction in a pretty controllable and non-violent way, and also will give you some electricity for free. Not that I expect you or anyone else to be actually doing this, though.

Why would you need gaseous $\ce{HCl}$, really?


Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'? I was reading Cannabinoid Chemistry - 6.2 Extraction & pondered over the following:

  Cannabis must be dried be(fore) it is extracted, because it is not possible to remove more than 50% of the cannabinoids from fresh material. THC-Acid is difficult to extract. If you plant(sic) to convert the THCA to THC, the plant material should be thoroughly decarboxylated by heating it under nitrogen at 105℃ for 1 hour before performing a solvent extraction.

Why is decarboxylation done 'under nitrogen'?  My first guess would be that at 105℃ the plant begins to oxidize (but that is just my naive guess).

I've also sourced this graph:

Effect of heating time and temperature on the THC content of an n-hexane
marihuana extract after heating on the glass surface in an open reactor.  Journal of Chromatography 520 (1990).

Which suggests that 122℃ for 27-30 minutes will be even more effective for decarboxylation, but if the sample begins to oxidize at 105℃ it seems even more important to get the right mixture of gases around it at 122℃.

[…] it seems even more important to get the right mixture of gases around it at 122℃

No, it's only important to keep the oxygen away while you are heating the sample.

Have a look at the structures: the thermal decarboxylation of the THCA isn't different from the decarboxylation of salicylic acid.

But with the carboxyl group or without, (upon heating,) phenols are easily oxidized by air.

In principle, the sample could be heated under vacuum too. In this case, you might want to use a cooling trap to collect volatile material. But usually, one does that anyway to protect the vacuum pump.


Why is the melting point of TiCl4 lower than TiCl3 I was wondering why the melting point of $\ce{TiCl4}$ was below that of $\ce{TiCl3}$

The melting point of the former is  $\ce{-25 Celsius}$ whereas the latter is $\ce{425 Celsius}$ .

Thanks!

-The high melting point of  $\ce{TiCl3}$
is mainly due to the short bond length $\ce{Ti-Ti}$ , which infer
 a strong metallic character to these bonds between titanium ions in  $\ce{TiCl3}$.

-While in $\ce{TiCl4}$, the main intermolecular forces are of van der Waals type.


Phosphate group acidic or not? I would like to ask whether or not the "Phosphate Functional Group" itself is acidic. I understand that the compound Phosphoric Acid is very acidic, with several hydrogens. However the Phopshate itself lacks hydrogens to donate. Please clear up any misudnerstandings I am having right now. Thank you so much.

No, the phosphate functional group, $\ce{P(O)(OR)_3}$, itself is not acidic. Only protic phosphates are acidic. Phosphates that aren't phosphoric acids (i.e. $\ce{P(O)(OH)_{x}(OR)_{3-x}}$) are actually Lewis bases. The carbonyl oxygen is able to coordinate metals and other Lewis acids.


Why put zinc solution in a galvanic cell? In a galvanic cell with Cu and Zn electrodes, I was taught that each electrode is immersed in a solution containing it's own positive ions. I understand why you need copper ions, so they can gain electrons, but why do you need zinc ions - the reaction is going to produce zinc ions anyway, and the negative ions from the salt bridge will migrate to neutralize them, so I don't see why you need zinc ions at the first place. Is there anything I'm missing in how the cell works?

It's tempting to look at the whole cell here and ask: hmm what's going on? But try taking a reductionist approach and look at each electrode system one at a time, in isolation.

Let's consider a basic model.If we immerse an electrode of $\ce{Cu}$ metal into a solution of $\ce{Cu^2+}$, the following can be thought to take place:

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

Where the ions in the solution are bombarding the electrode surface, and are being taken up by gaining two electrons from the metal, whilst the metal is giving up ions from its surface, keeping hold of two electrons. At equilibrium, this forward and backward bombardment occurs at the same rate.

You might say to me: okay but we can't actually probe this system with a voltmeter to say that this is going on, because this would require a circuit to be made with an external system, and hence would perturb any effect we wish to observe. I'd then say to you: you're correct but bear with my argument, just noting what I've said.

By a similar token, in our other isolated beaker we have:

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

Now, the whole idea behind an electrochemical cell is to extract chemical energy by systems which are not at equilibrium. If a system is at equilibrium then no energy will be available. The equation which gives the free energy available from an electrochemical cell is (derivation not given):

$$\Delta G=−RT\ln K_\mathrm{eq}=−nFE_\text{cell}$$

Where $R$ is the ideal gas constant, $T$ is the parameter of temperature, $F$ is the Faraday constant (the amount of charge carried by a mole of electrons) and $E_\text{cell}$ is your available electrochemical energy.

So for these two individual single-electrode systems at equilibrium, can we extract chemical energy? Well no, because Gibbs at equilibrium is zero, and hence so is $E_\text{cell}$.

What then do we do?

Complete the circuit between the two, and if the potential between two half-reactions is sufficient to warrant the moving of electrons, a dis-equilibrium will be created until the reaction has completed (which is indeed what we get for a system of zinc/copper). Once the electron transfer between zinc and copped has been achieved i.e. equilibrium has been achieved, then we can no longer extract any useful chemical energy. Our cell is dead!

So in summary: the dual metal electrode/same-metal ion solution is needed for your initial half-reactions, which you then perturb by completing your circuit. This perturbation of the individual half-reactions results in our being able to extract useful chemical energy by shifting the equilibrium until a new equilibrium is established.


Negative Ohm reading from a Electrochemical reaction? The purpose of my experiment is to determine which salt is the best electrolyte. The setup is a tarnished spoon (Ag2S) is placed on top of thin plastic pieces that separate it from a container coated with aluminum foil. A solution of 1.0 M Salt and .2 M baking soda is heated and then poured into the container.  The following electrochemical takes place 3 Ag2S(s) + 2 Al(s) → 6 Ag(s) + Al2S3(s), the redox reaction being  Al + 3 Ag+ → Al3+ + 3 Ag. A multi-meter set to 200 ohms is attached, the positive end to the foil, and the negative end to the spoon. The picture included is my set up.

My logic was I could take the resistance measurement, take its reciprocal, and then find the conductance. I thought whichever salt would give me the best conductance would be the best electrolyte. However, I've only gotten readings of Negative ohms. My readings have been -190 ohms and then it steadily increased. The problem there can't be a negative conductance. From research I've found this, "conductivity is negative, it means that the resistance is negative, right ? This may occur if you have created a generator in your experiment, for example through a non-desired oxydoreduction phenomenon between two metals. If a current is produced in your experiment due to such reaction between 2 metals of different oxidation potentials, and if you material acts as en electrolyte, your apparatus may interpret it as a negative resistance."

How could I use the negative Ohms reading? Should I be collecting data in a different way? Help!

How could I use the negative Ohms reading?

You're using the meter the wrong way. A meter is set up to have a particular low voltage and measures how much current flows. Since your electrochemical cell has a voltage and creates its own current the meter reading goes whacko.

  Should I be collecting data in a different way?

If measure current that would give you some idea. In a better controlled analytical chemistry experiment a special electrode pair with two flat plates is used so that the geometric configuration stays exactly the same.


How to determine the shape of hybridized atomic orbitals in VB theory? From diagrams, it's rather obvious how $sp$ orbitals are hybridized - the hybrids are just a composite of the $s$ and the $\pm p_{(x)}$ orbitals. However, $sp^2$ orbitals are not just composites of $s, \pm p_x, \pm p_y$, and the same goes for $sp^3$.
This incongruity leads to my question.
I am aware that the general shapes are trigonal planar and pyramidal, respectively. However, I recall learning of a more mathematical approach that used the wave function (I think?)
In my notes, I found this formula.
$$\Phi_i=\sum_{j=1}^nc_{ij}\phi_j, i=1,2,3...n$$
My understanding is that $n$ is the number of atomic orbitals (and thus also the number of orbitals). I believe $c$ is a constant, and $\phi$ is the orbital.
Later, for $sp^2,$ I have the following formulas:
\begin{align}
\require{cancel}
\Phi_1 &=c_{1,1}\cdot s+ \cancelto{0}{c_{1,2}\cdot p_x} + c_{1,3} \cdot p_y &&= c_{1,1}\cdot s+c_{1,3} \cdot p_y\\
\Phi_2 &=c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot p_y &&= c_{2,1}\cdot s+ c_{2,2}\cdot p_x + c_{2,3} \cdot -p_y\\
\Phi_2 &=c_{3,1}\cdot s+ c_{3,2}\cdot p_x + c_{3,3} \cdot p_y &&= c_{3,1}\cdot s+ c_{3,2}\cdot -p_x + c_{3,3} \cdot -p_y \\
\end{align}
I have some recollection that the wave functions zeroes out for particular quantum number values, but I can't remember how it all fits together with this equation.
I couldn't find this formula online, so I'm hoping that it might make sense to someone here.
Thanks for the help!

What we call "Hybridization" is really just a mathematical transformation of an approximated wavefunction. Your basic summation formula describes how to build some sort of customized wavefunction by forming a linear combination of some set of "basis" orbitals; you will find this equation over and over if you look up "basis set expansion". Now, of course you can't just use any random coefficients $c_{ij}$ of your liking for this custom wavefunction; you have to adhere to some physical rules. Most importantly, the overall energy that the wavefunction represents must not change. I won't go into detail about the constraints, but if you are interested, google "unitary transformation".
For orbital hybridization, you first have to find a set of basis orbitals from a physically sound starting wavefunction. If you only look at an isolated atom, then atomic orbitals are obviously a useful choice. From there, you can form combinations of those orbitals (by some criteria that really aren't uniquely defined) while adhering to the rules of the unitary transformation. In the end, you will arrive at your preferred hybrid orbitals that fulfill your chosen criterion as best as they can while keeping the wavefunction energetically equivalent to the original one. Some of the coefficients $c_{ij}$ in your equations may come out to be 0, typically for reasons of symmetry. For example, an $sp^2$ hybrid that is oriented along the $y$ axis of your coordinate system would obviously only have contributions from the $p_y$ orbital, not from $p_x$ and $p_z$. But the exact values of the cofficients will generally depend on the specifics of your problem, most importantly the definition of your coordinate system. So you really can't say per se which coefficients will drop out to be 0 and which won't -- it's one of those "it depends" situations.
General Thoughts:
Unfortunately, hybridization is a concept that is so misleading that I would consider its general use in chemistry to be outright physically incorrect. People invoke it to suggest and visualize bonding situations in molecules and to explain the resulting molecular shapes, but unfortunately always seem to forget some very basic undergraduate quantum chemistry in this process.
Hybrid orbitals are linear combinations of eigenfunctions to different energy eigenvalues of the underlying one-electron Hamiltonian; think $s$ and $p$ orbitals. As such, they are not solutions to the time-independent Schrödinger Equation anymore, which in turn means that they are time-dependent functions. This implies two things:
1.) Hybrid orbitals do not have a fixed shape; what you see visualized is just a single snapshot in time. The orbital shapes periodically "morph" back and forth and can at other points in time look nothing like those suggestive localized "bonding" orbitals anymore. In fact, the hybrid orbitals can even change places among each other over time if the symmetry of the system permits.
To better drive this point home, here are some (semi-quantitative) visualizations of the time development of the bonding and non-bonding orbitals in water.
2.) Hybrid orbitals do not have defined energies, since they are not solutions to the time-independent Schrödinger Equation. If you were to do any sort of "energy measurement" on a hybrid orbital, you would end up with a Schrödinger's Cat situation where you observe the energies of the constituent atomic orbitals with certain probabilities that are connected to the expansion coefficients $c_{ij}$.
In addition, there is a third problem. When the pure integer mixtures of s and p orbitals (e.g. $\mathrm{sp}^2$, $\mathrm{sp}^3$) were not sufficient to explain molecular shapes, people started to come up with weird fractional mixtures like $\mathrm{sp}^{1.5}$ or some such. How are those numbers obtained? - As stated above, from the wavefunction side this comes down to a "localization" procedure of the molecular orbitals; however, the criteria by which to do this are not strictly defined, and there exist multiple different schemes for it. So this raises the third problem:
3.) Hybridization is not stringently mathematically defined if one regards it as an orbital localization procedure.
So, the next time someone presents you with hybrid orbitals to explain e.g. why methane is tetrahedral, ask them about the energy and the electronic density distributions of those orbitals, and how they would determine the hybrid orbitals for e.g. chloromethane where the bond angles and energies are clearly different than in methane.


What is the scientific term to describe the operation of a bong? What is the scientific term for the separation of dust particles from air by suction through a specialized flask in a way similar to the operation of a bong?

The thing that led me to this was a vacuum cleaner that uses water to separate dust and other particles from the air by vacuuming the air through a water bowl, and the only thing that is similar to that operation is the bong. It is like sparging, but the gas coming in will contain particles and those particles are the ones that remain in the liquid, and that same gas is the one coming out, filtered to some degree from those unwanted particles.

I think the term describing the initial process of separation of the solid phase (dust particles) from the gas phase via bubbling is a wet scrubbing process.

On an industrial scale, it's more efficient to spray liquid phase, whereas on a laboratory scale bubbling in a compact glass gas scrubber or a gas washing bottle is more convenient.
In order to increase efficiency of the lab apparatus, it's preferred to

keep the surface between gas bubbles and liquid phase as large as possible;
increase time of contact.

This is usually achieved by producing many smaller bubbles by using a spiral tubing (larger particles), frit of long columns filled with Raschig rings.

Figure 1. Left: Büchi miniScrubber, a glass gas scrubber (Source: Büchiglas — Gas scrubber). Right: Corning PYREX® 125mL Gas Washing Bottle with Extra Coarse Fritted Disc (Source: Thomas Scientific)

Eventually collected insoluble dust particles tend to settle out from the solution; this a basis for sedimentation, another separation technique.


Enthalpy change for exothermic and endothermic reactions I understood the concept behind endothermic and exothermic but finding it hard to understand it numerically.Let me describe my confusion through example.

Suppose for a reaction,
reactants require energy=$\rm40~J$
which on forming product give energy=$\rm50~J$ (so yes it is exomthermic)
where change in enthaply=$\rm10~J$

But as per the equation:
"Enthalpy change=Enthalpy of products- Enthalpy of reactants."
since enthalpy at constant pressure is Total heat content.
So I should have

Change in enthalpy=[U(internal energy)-50]-[U+40]=-90J

Could You explain it where I am getting this wrong? Is my understanding of enthalpy correct?

According to the definition,
 Enthalpy = Enthalpy(products) - Enthalpy(reactants), this should be followed.

When you are saying that 40 J are used up it means that 40 J are used in transforming the reactants to an activated complex(reaction intermediate) and when you say that 50J are released then it means that 50J is liberated when this activated complex is transformed to the product.

Assume that the energy of the activated complex be 0, then the energy of the reactant must be -40J(because 40J are required to convert reactant to the activated complex) , and the energy of the products should be -50J( because energy is released in converting the complex to the product)

Enthalpy(products)=-50J

Enthalpy(reactants)=-40J

Therefore,
 Enthalpy = (-50)-(-40) J = -10J.


How to meausre ion interferences in solution? Is there a way to quantitatively measure the interference of different metal ions in solution? I plan to measure concentration of Pb(II), Fe(II), and Cu(II) in solution together using a spectrophotometer.

I know that I can put multiple samples of two ions together to test for the interference trends/effects, but can I do this for all three? e.g. measure interference effects of both iron/copper on lead? Or are there too many variables to measure concentrations?

If not feasible, are there other accessible ways (besides mass-spec) to measure multiple concentrations of all three at once?

EDIT: I already plan to measure concentrations using reagents, specifically Hach TnTplus 850 and 858, and Hach CuVer 1, as they seem easy to get and use. My purpose is to accurately measure concentrations of one ion when in the presence of two other interfering ions.

What you can do is to add a reagent which forms a highly coloured complex with one of the metals, this will allow you to make a measurement which is good for one of the metals. You may need to do three different experiments on three seporate aliquats from the same solution.

For example for the iron, I would suggest that you use 2,2'-bipyridine, or 2,2'-bipy as it is sometimes known. This forms a very deep red complex with ferrous iron. This would allow you to measure the iron.

You would need a series of calibration samples with different amounts of iron, please do be aware that the matrix can sometimes interfere with this measurement. I have found that oxalic acid does inhibit the formation of the iron tris bipy complex.

Maybe you should read this paper which is about 2,9-dimethyl-4,7-diphenyl phenanthroline. This is a reagent which has been reported to be good for copper determination.


Prediction of surface atoms in molecule from its graph If we are given a molecular graph i.e. al the atoms involved and their connectivity, how can we make reasonable prediction if the atom would lie in the surface or in the inner part. On way might be check it degree (how many atoms it is bonded to), higher the valency higher it surface accessibility.

I will most of the time have small molecules (drug molecules) which have in average 50 atoms.
My broader interest is to find out which atoms directly affect the activity of the molecule more than the other. Probably, atoms lying at surface have more role to play. Any other suggestions aimed at this identification would be helpful.

The Quantum Theory of Atoms in Molecules (QTAIM) provides a fairly unambiguous answer. First, I would like demonstrate the concept of an Atomic Basin. A region of space belongs to a given atom if the steepest ascent path through the electron density terminates at that atom.  Consider a plane that contains the nuclei of the water molecule:

and the atomic basins could also be represented in 3D:

These atomic basins are the QTAIM definitions of ATOMS in MOLECULES. Note that there are gradient paths for each atom that extend to infinity, hence each atom could be considered an "external atom."  (This is a quite trivial case, of course.)

Now consider formamide in this configuration:

The atomic basins of form amide look like:

You can see that the carbon atom is fairly "inaccessible" in the plane of the molecule, but it does have some extent "peeking" out of the plane of the molecule (here, the maroon-colored basin, sorry for change-up in color):

This is a completely general concept, and so can be extended to any scenario. Here is a molecular graph of thiol on a silver cluster (meant to mimic a surface) (note the bond paths unambiguously show which atoms are bonded to what, even for C-Ag, S-Ag, H-Ag interactions where these "cooked-up" rules of hybridized bonding and "valency" are totally out of their league):

Plotting the atomic basins is just an exercise:

The final answer: you can define "interior atoms" as QTAIM atomic basins that do not have steepest ascent trajectories starting at infinity that terminate at that atom, such as an interior silver atom in the above example. If you are conceptually separating a system into a substrate and an adsorbate, then you could say "exterior" atoms of the substrate include the ones that have bond paths connecting the two.

I must caution that as two or more atoms approach each other, their atomic basins change their shape.  (Of course they do!)  And so one must consider the possibility that atomic basins of "exterior" atoms move out of the way and lead to exposure of what was, in isolation, an interior atom.

(PS:  the atomic basins of peripheral atoms actually extend to infinity.  Here, I have applied cutoffs so that one can actually plot the molecule in a finite region of space.]


Nomenclature of complex benzene-based substituents For a complex alkyl substituent, (at least the ones I have come into contact with thus far) you name it as a chain numbering from the first non-parent carbon and placing the name in square brackets: e.g. isopentyl can also be named [3-methylbutyl]. What is the convention for more complex benzene-based substituents? Specifically, can you still use the more common names like phenol and toluene as a parent for your name if say you bond C5 of 2,3-dimethylphenol to your parent and what suffix will you use?

Intro

Your statement, that complex substituent names are placed in square brackets, is incorrect. These are to be placed in parentheses (round brackets). Correct preferred IUPAC name (PIN)[1] example:

  2-(3-methylbutyl)phenol (PIN)

Square brackets are used when the substituent name already contains parentheses, e.g.:

  2-[(2R)-2-methylbutyl]phenol (PIN)

Retained aromatic compounds allowed for substitution

You can see that the names above are based on phenol, which is one of retained aromatic functional parent compounds, with allowed substitution, in preferred names.

The other ones are aniline and benzoic acid (see IUPAC rule P-34.1.2.[1]). E.g.:

  2-hydroxybenzoic acid (PIN)
  (not 2-carboxyphenol, as $\ce{-COOH}$ group is preferred over $\ce{-OH}$)

Retained alkyl-substituted aromatic hydrocarbons (not allowed for substitution in PINs)

There are also retained alkyl substituted aromatic parent hydrocarbons (P-22.1.3):

  toluene (PIN)
  methylbenzene

and

  1,2-xylene (PIN)
  o-xylene
  1,2-dimethylbenzene

(and the two other xylenes).
But none of them is allowed for substitution in preferred names, e.g.:

  1-chloro-4-(chloromethyl)benzene (PIN)
  α,4-dichlorotoluene

Xylenes are not allowed for substitution even in the general non-preferred names:

  1,2-bis(bromomethyl)benzene (PIN)
  α-bromo-2-(bromomethyl)toluene
  (not α,α′-dibromo-o-xylene)

(Yet another retained name mesitylene for 1,3,5-trimethylbenzene (PIN) is allowed only for general names, not allowed for substitution.)

Complex aromatic substituents

Now if a complex substituent is made from substituted phenol, aniline or benzoic acid, the substituent name is no longer based on these names. They can be thought as suffixes, but now we have another higher-priority “suffix”, -…yl. E.g.:

  3-hydroxy-4,5-dimethylphenyl acetate (PIN)
  (not 2,3-dimethylphenol-5-yl acetate)

(In case you didn't hear ‘phenyl’ yet – it's not to be confused with ‘phenol’, it's the only, notorious, name for “benzenyl”, $\ce{-C6H5}$)

References:

Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)


How to distinguish between carbon-carbon double bond lengths? Which of the following compounds should have the longest carbon-carbon double bond?

  $\ce{CH2=C=CH2}$
  $\ce{H3C-CH=CH2}$
  $\ce{(H3C)3-C-CH=CH2}$
  $\ce{(H3C)2C=CH2}$

I know that single bond > double > triple length-wise, but how do you distinguish between double bonds?

You can draw hyperconjugation structures to compare bond length. The more α-hydrogens the alkene has, the more hyperconjugation structures it has, and the higher is the degree of delocalization of the π-bond and therefore the longer is the bond. When double bond character decreases, bond length increases. Therefore the answer should be (d).


Chalcogens' hydrides as acids? I've noticed that chalcogens never form binary acids, and instead end up like water, with very little acidity. Why is this? Or am I wrong and there's a counterexample, if so please give it.

To start off, all chalcogens are known to form binary hydrides of the form $\ce{H2X}$ which can then behave as binary acids, however weak or strong, by the equation $\ce{H2X <=> HX^- + H+}$.

Now to compare the strength of the acids. The most straightforward way of doing this is comparing gas phase acidities, which removes the complication of solvation effects. For this, we look at the proton affinities of the conjugate bases $\ce{HX^-}$. The definition of a substance's proton affinity is the $\Delta G$ when it is protonated in the gas phase (always an exergonic reaction for neutral species and anions), so gas phase acidity is simply the inverse reaction, with the opposite sign for the $\Delta G$. According to these two sources, the gas phase acidities of the hydrogen chalcogenides are:

$$\ce{H2O_{(g)} <=> HO^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta  G = +1635\  kJ\  mol^{-1},\ K^{25ºC} = 2.5\times 10^{-287}$$
$$\ce{H2S_{(g)} <=> HS^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1477\  kJ\  mol^{-1},\ K^{25ºC} = 1.2\times 10^{-259} $$
$$\ce{H2Se_{(g)} <=> HSe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1417\  kJ\  mol^{-1},\ K^{25ºC} = 4.1\times 10^{-249} $$
$$\ce{H2Te_{(g)} <=> HTe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G \simeq +1386\  kJ\  mol^{-1},\  K^{25ºC} \simeq 1.1\times 10^{-243} $$

As you can see, when going down the group, deprotonation becomes less endergonic, so the binary hydrides definitely become stronger acids, even if against what one might expect from electronegativity considerations. Calculating the equilibrium constants allows a more direct comparison. Don't worry that the constants are all extremely small, what matters to us is the ratio of the equilibrium constants. Thus, in the gas phase, $\ce{H2S}$ is a whopping 28 orders of magnitude more acidic than $\ce{H2O}$, while $\ce{H2Se}$ and $\ce{H2Te}$ are 38 and 44 orders of magnitude more acidic than $\ce{H2O}$, respectively. It is certainly noteworthy that $\ce{H2O}$ is a much, much weaker acid than its heavier homologues.

With the data presented, we ask, why does acidity increase down the group? There are two main factors presented in different sources. One of them is that the chalcogen-hydrogen bond becomes weaker for heavier chalcogens, and this is ascribed to the poor overlap between the tiny $\ce{H}\ 1s$ orbital and the increasingly larger chalcogen orbitals which weakens the covalent bond and presumably makes ionization easier. This can be seen from the dissociation energies of the $\ce{X-H}$ bonds ($\ce{O-H} = 428\  kJ\  mol^{-1}$ | $\ce{S-H} = 344 \  kJ\  mol^{-1}$ | $\ce{Se-H} = 305\  kJ\  mol^{-1}$ | $\ce{Te-H} = 268\  kJ\  mol^{-1}$). The second factor is the size of the chalcogen atoms. The heavier atoms are larger, meaning they spread the same charge in the conjugate base $\ce{HX^-}$ over a larger volume, which makes reprotonation less favourable. This latter explanation seems to be the most common.

Up to now we've been talking about acidity in the gas phase, but now let's add the solvent, specifically water. The presence of a solvent stabilizes ions immensely relative to gaseous ions (especially protons), and water is a particularly good solvent for ions, so all the dissociation equilibria will be shifted massively to the right compared to the gas phase. Let us compare the acid dissociation constants in water:

$$\ce{H2O_{(l)} <=> HO^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ k_w^{25ºC} = 1.0\times 10^{-14}$$
$$\ce{H2S_{(aq)} <=> HS^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.0\times 10^{-7} $$
$$\ce{H2Se_{(aq)} <=> HSe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.3\times 10^{-4} $$
$$\ce{H2Te_{(aq)} <=> HTe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \   K_a^{25ºC} = 2.3\times 10^{-3} $$

Some things to note:

All the binary chalcogen hydrides still increase in acidity after solvation when going down the group. This is not necessarily always the case. A classic example of relative acidities changing on solvation is the dimethylammonium ($\ce{Me_2NH2^+}$) and trimethylammonium ($\ce{Me_3NH^+}$) cations; in the gas phase the former is more acidic, but in aqueous solution the latter is more acidic as the $\ce{Me_3NH^+}$ ion is not as well solvated.
As I mention in the comments, aqueous hydrogen telluride ($pK_a=2.6$) is about as acidic as aqueous phosphoric acid ($pK_{a_1}=2.15$). Compare also with a common weak acid such as acetic acid ($pK_a=4.76$), a borderline strong acid such as nitric acid ($pK_a \simeq -1$), and a recognized strong acid such as hydrochloric acid ($pK_a \simeq -6$).
The $K_a$ of water is a little tricky, since it is both the reactant and the solvent. Its value depends on the definition used. I quoted the $k_w$, but many people use a $pK_a$ of 15.74 which implies $K_a = 1.8\times 10^{-16}$. The difference is not too important here.
$\ce{H2O}$ is a much stronger acid than expected from gas phase data, relative to its heavier homologues. This is ascribed to the exceptionally strong aqueous solvation possible only for $\ce{OH^{-}}$ among the listed conjugate bases, stemming from strong hydrogen bonding interactions, which helps weigh the equilibrium to the right side.


E - Z configuration of cyclic substituents The answer is coming out as Z but given is E.
I’m not sure where I’m wrong at this point as I moved along one chain at a time and wanted to follow heaviest atom as earliest as possible !

Welcome to Chem SE, A Singh: You have done a nice job laying out the digraphs for the cyclopropene (A) and cyclohexene (B) rings. The double bond in A is accessed early in both directions but only in one direction in B does access to the double bond occur early. Which means that A>B and the double bond is of the (E)-configuration. But I offer you the more rigorous treatment in the diagram below.As you move outward from C1 and C4, sphere by sphere, compare sphere 2, C1 has [2,3,H] while C4 has [9,5,H]. This is a tie. Now move to sphere 3 (marked in red). The C1 path reveals [3,(3),H;2,(2),H] while the C4 path displays [6,(6),H;8,H,H]. Comparing atom to atom, C1 has one more carbon than C4 (See bold (2) and H above). Therefore, the cyclopropene ring (A) has priority over the cyclohexene ring B. The double bond is of  the (E)-configuration given that Cl>C. Recall that all "real" carbons are equal to duplicate carbons (in parentheses) but all duplicate atoms are attached to three atoms of atomic number zero. In some cases, one has to go this far to make a decision.


When is Raoult's Law valid for solid-liquid solution? Raoult's Law for solid-liquid solution is stated as such in my books that it is applicable for all non-volatile solutes.
However different solutes must have different attractive or repulsive forces and in such a case, all of them will probably hamper vaporisation in different ways.
Then how come can we simply define the law as:
$$P_\text{vapour}=P_\text{vapour}^0\chi_\text{solvent}$$
where
$$\chi_\text{solute}+\chi_\text{solvent}=1$$
Here the equations do not provide any information about the solute.

Is the law valid for only a class of solutes?

It is valid for both volatile and nonvolatile solutes, as it refers to the partial vapour pressure, not total vapour pressure.
Saying that, note that  it applies only on mixtures with  ideal behaviour, as there are many more or less significant  positive and negative deviations from the law. These deviations relate to preference of intermolecular bonding to the same, or to the other molecules.
Typical examples are ethanol water mixture forming an azeotrop of the minimal boiling point,  or hydrochloric acid forming an azeotrop with the maximal boiling point.


What is the reason for the exceptional stability of the cyclopropylmethyl carbocation? Can someone explain this to me by drawing resonance structures for the cyclopropylmethyl carbocation please?

Also one more question, is the tricyclopropylmethyl carbocation more stable than tropylium ion?

It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the $\pi^*_{\ce{C=C}}$ orbital while the carbene $\mathrm{sp^2}$ orbital interacts with the $\pi_{\ce{C=C}}$ orbital, so this 'virtual' double bond behaves somewhat like a normal double bond.

On the other hand, the structure of the cyclopropylmethyl cation is downright weird. It is well known that both cyclopropylmethyl and cyclobutyl derivatives give very similar product mixture under $\mathrm{S_N1}$ hydrolysis conditions, resulting in both cyclopropylmethyl and cyclobutyl derivatives (see, for example, J. Am. Chem. Soc. 1951, 73 (6), 2509–2520). This is commonly described by the conjugation in following manner (here, the 1-cyclopropylethyl cation is depicted):

Here bonding of $\ce{C-4}$ with $\ce{C-1}$ and $\ce{C-2}$ can be roughly described as an interaction of the vacant $\mathrm{p}$-orbital of $\ce{C-4}$ with filled orbital of $\pi$-bond between $\ce{C-1}$ and $\ce{C-2}$. It does not matter much, where the original leaving group was - at  $\ce{C-2}$ or $\ce{C-1}$. Since the positive charge is more or less symmetrically distributed between three atoms and the small ring is somewhat relieved of its geometrical strain (both cyclopropane and cyclobutane are very strained molecules, not only due to angle strain, but also considerable steric interactions between hydrogens), the cation has remarkable stability.

Similar effects are common in the chemistry of small bicyclic systems, with norbornane derivitives being the chosen test subjects for decades with the 2-norbornyl cation being probably the most well-known example. March's Advanced Organic Chemistry, 7th ed., Section 10.C.i discusses such nonclassical carbocations in great detail, with the cyclopropylmethyl system being described on pp 404–406.

With further addition of multiple cyclopropyl groups, however, the full conjugation becomes sterically hindered, so extra groups beyond the first have less of an effect. Of course, the stability of these cations is far below that of the tropylium cation, which has very little strain and also possesses aromatic character, distributing the positive charge over seven(!) carbon atoms.  In fact, the stability of the tropylium system is so high, that even the cyclooctatrienyl cation (also known as the homotropylium cation) adopts a similar structure.


Changes in physical properties in homologous series : Solubility In the text that I'm reading it is stated that :

  As the molecular mass increases in any homologous series, a gradation in physical properties is seen. This is because the melting points and boiling points increase with increasing molecular mass. Other physical properties such as solubility in a particular solvent also show a similar gradation.

My question is regarding the last line; what kind of gradation in solubility is observed? Increase or decrease?

My question is regarding the last line, what kind of gradation in solubiltiy is observed? Increase or decrease?

Both

Solubility is not one single property, but a collection of related properties. Solubility in water is different from solubility in ethanol, which is different from solubility in benzene.

The cause of the gradation of (all of these) properties is a systematic change in intermolecular forces. The number and strength of these forces increase as molecular mass increases, which increases the boiling points and melting points. For solubility, we care about the interplay between the solute and the solvent, so we also need to know the type of intermolecular force (polar vs. nonpolar) in addition to the total strength.

For the homologous series of alcohols below, I have listed their water solubilities (mined from UC Davis's ChemWiki and from Wikipedia articles). Notice that as the number of $\ce{CH2}$ units increases, the water solubility decreases. This decrease occurs because, while the strength of nonpolar intermolecular forces (London-Dispersion) increases with each added $\ce{CH2}$ until the nonpolar interactions overpower the polar interactions (dipole-dipole and hydrogen-bonding) of the alcohol functional group. Water is a polar solvent, and with each added $\ce{CH2}$, the alcohol molecules begin to have stronger attraction to themselves than to water, and the solubility decreases.

Solubility of the homologous series of linear n-alkanols in water

Methanol $\ce{CH3OH}$ - miscible (infinite)
Ethanol $\ce{CH3CH2OH}$ - miscible (infinite)
n-propanol $\ce{CH3CH2CH2OH}$ - miscible (infinite)
n-butanol $\ce{CH3CH2CH2CH2OH}$ - 80 g / L
n-pentanol $\ce{CH3CH2CH2CH2CH2OH}$ - 22 g / L
n-hexanol$\ce{CH3CH2CH2CH2CH2CH2OH}$ - 5.9 g / L

At the same time, the solubility of this homologous series of alcohols increases in nonpolar solvents like hexane and benzene. Those data are not readily available, but a quick experiment will verify it. The increased amount of nonpolar interactions in the alcohols increases their affinity with nonpolar solvents.


Can there exist any element which has an 8th principle shell? I am a high school student . While studying the atomic structure a question raised in my mind. We know that an atom can have the highest limit of 7 bohr energy shells or principle energy shells ; K,L,M,N,O,P,Q respectively. My question is, can there exist any element that consists more than 7 energy shells having more electrons? Or , can there exist any element that has an 8th energy shell of R shell?

That is certainly possible, and in fact very likely. Nuclear physics predicts an island of stable elements around an atomic number of 120 $^{[1]}$, with the two heavier of the three elements predicted to be most stable having an $\ce{8s^2}$$^{[2]}$ and an $\ce{8s^2 8p^1}$$^{[3]}$ configuration, respectively.


Why does boron trifluoride form boron tetrafluoride ion instead of boron pentafluoride ion? From ChemWiki by UC Davis:

  As a side note, it is important to note that $\ce{BF3}$ frequently bonds with a $\ce{F-}$ ion in order to form $\ce{BF4-}$ rather than staying as $\ce{BF3}$. This structure completes boron's octet and it is more common in nature.

Wait, what? Boron has three electrons. Bonding with four fluoride ions gives it four electrons — these are covalent bonds (shared electrons). That makes for a grand total of 7 valence electrons. Did the author mean $\ce{BF_5^2-}$?

As you point out, the $\ce{B}$ in $\ce{BF_3}$ is sharing a total of 6 electrons, 3 from the boron and 3 from the three fluorines.  Now add a fluoride anion that brings 2 more electrons to share.  This now makes for a total of 8 electrons around the boron.


Dow's Process for the manufacture of phenol When chlorobenzene ($\ce{C6H5Cl}$) reacts with aqueous ($\ce{NaOH}$) solution it first forms sodium phenoxide ($\ce{C6H5ONa}$), and then phenol upon reacting with $\ce{H+}$.

But my question is, why can't $\ce{Na}$ directly replace $\ce{Cl}$ from chlorobenzene by breaking $\ce{Na-O}$ bond instead of breaking $\ce{O-H}$ bond?

The mechanism of this reaction is that $\ce{NaOH}$ being a strong base with some extreme conditions abstracts  H from C adjacent to C containing Cl while Cl also leaves.

This creates benzyne as an intermediate.

Subsequent addition of $\ce{NaOH}$ forms the phenoxide ion.


How come when it rusts iron can undergo oxidation without the prescence of a another metal to undergo reduction? It seems to me rusting is an electrochemical process where oxygen, which has particularly high electronegativity, fills the role of cathode or is that an incorrect way to put it?

Rusting is a redox reaction.
$\ce{4 Fe + 3 O2 -> 2 Fe2O3}$

Redox reactions mean one chemical is being reduced and one is being oxidized. Reduction = electron gain, oxidation = electron loss. The reason it is called oxidation is because oxygen is so good at reducing other chemicals (it wants electrons). How do you tell what gets electrons in a compound? Oxidation numbers.

Oxidation numbers signify how many electrons each atom has. -n means it has n more than neutral state, +n means it has n less than neutral state. In molecules which have only one element, oxidation state is 0. So Fe/O2 both have oxidation state 0. There are some rules on oxidation states in compounds which are complicated, basically usually group 1 is +1, group 2 is +2, O is -2, and group 17 is -1. In this case each O in the end is -2, and Fe is +3. So each Fe lost 3 electrons.


How to rationalise the IUPAC name for tetrahedrane? I'd like some assistance in understanding the IUPAC name for tetrahedrane. I understand the basics of polycyclic ring naming however this example has me confused.

Let's start with bicyclobutane:

The proper name for this compound is bicyclo[1.1.0]butane. The compound is numbered starting at one bridgehead and working around. The numbers in brackets represent the number of atoms between the bridgeheads.

Now, tetrahedrane, which has an extra bridge between carbon atoms 2 and 4:

Once again, we list all of the bridges in the brackets, but we have an extra bridge, and it is not between the main bridgeheads (1 and 3). We list this extra bridgehead last and then use superscript locants to indicate which ring atoms this bridge connects. This extra bridge is between carbon atoms 2 and 4, thus $0^{2,4}$: tricyclo[1.1.0.0$^{2,4}$]butane.

For more examples, including applying the rules to crazier ring systems, here is the IUPAC publication


Basicity comparison of halides The order of decreasing basicity in the four halide ions is:

  $\ce{I- > Br- > Cl- > F-}$
  $\ce{Cl- > Br- > I- > F-}$
  $\ce{F- > Cl- > Br- > I-}$
  $\ce{Cl- > F- > Br- > I-}$

The answer is option 3) but i'm getting as option 1).

My reasoning is that these elements belong to the same group, and the atomic radius increases down the group. Bases are electron donors, and iodide is the largest, so it can donate electrons the easiest, hence it should be most basic. Why is this incorrect?

Define basicity. For Brønsted-Lowry, it's all about $\ce{H+}$ and the conjugate bases. In Brønsted theory, acceptance of a hydrogen ion is what a base does. If you didn't know before, you should know that the hydrogen halides acid strength is $\ce{HF < HCl < HBr < HI}$. So for the conjugate bases, you just flip the sign $\ce{F- > Cl- > Br- > I-}$.

You way overthought the question. You should also think about where you got this idea about "electron donation". (Lewis acid-base theory deals with electron pair donation/acceptance but there's a huge difference between electron donation and electron pair donation.) You claim for some species $\ce{X-}$, that if $\ce{X- -> X^0 + e-}$ easily, then $\ce{X-}$ is a stronger base (than some $\ce{Y- -> Y^0 + e-}$ which is less easy). (By 'easily', assume I mean in the thermodynamic (energetic) sense)

Well, take a look at that and ask yourself whether it is relevant to Brønsted acid-base theory. I say no, it's not. Whether it is true or not, it is not relevant to the base strength as defined by Brønsted-Lowry (at least it has no clear, direct relevance).


Can hydrides react with hydrocarbons? And why? I have read there exist so called "magic" (super strong) acids which can provide a proton to saturated hydrocarbons, producing cation and hydrogen molecule. Can hydrides act in the same manner (pick a proton to form free hydrogen molecule)? And why?

The pKa of sodium or lithium hydride is (solvent dependent) about 35, the pKa of Butyl lithium is around 50 so clearly H- is not strong enough to tear a proton off an unactivated alkane pKa tables here