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In molecular orbital theory when two different atoms are combined, how do you tell which one is higher energy?

So in the picture above, see that there is an Atom A and Atom B and they combine. Notice that Atom A is higher in energy than Atom B (as indicated by how the 2s and 2p lines are drawn higher). Suppose I say that this is the diagram for NO. How would I tell which atom and which is O? What determines which atom has higher energy in the orbitals?
Thanks.

Welcome Pallas to Chemistry Stack Exchange. Generally the more electronegative atom will be lower in energy. So in your example atom A would be nitrogen and atom B would be oxygen.

Think about ionisation energies as well. A higher IE must mean that the AOs are lower in energy (since it requires more energy to remove electrons).


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Schrödinger's equation
Schrodinger's equation has been successfully solved in one-electronic systems, such as the ionic molecular hydrogen $\ce{H2+}$.
What is the exact reason that this equation can not be applied for multielectron atoms?

It surely can be, and is applied. It can't be solved in elementary functions, but that's another story.

http://chemistry.stackexchange.com/questions/39795/why-cannot-the-schr%C3%B6dinger-equation-be-solved-exactly-for-systems-in-which-more 2) http://chemistry.stackexchange.com/questions/34606/analytical-solution-of-the-schr%C3%B6dinger-equation-for-ab-systems 3) http://chemistry.stackexchange.com/questions/4438/how-was-the-quantum-3-body-problem-solved

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Radical Bromination of a 1-isopropyl-4-methylcyclohexane
I'm considering a problem whereby 1-isopropyl-4-methylcyclohexane with the indicated stereochemistry shown undergoes radical bromination at the indicated position. I understand there are other products produced. However, if the radical initiator (not shown) extracts the hydrogen at the 4 position, then the bromine radical attacks, there will be two distinct products. However, I am having trouble determining if these molecules are chiral? Are they diastereomers to each other?

Also, as a follow up, how would you name these (reagent + two products) to indicate their stereochemistry?

No, they aren't chiral because of symmetry plane, they are cis/trans isomers

Mith is right. But that doesn't mean they're not diastereomers though :)

They are not chiral because $\ce{C}-4$ is only considered to be bonded to 3 unique groups because the ring is completely symmetrical through $\ce{C}-1$ and $\ce{C}-4$.
As for naming these compounds, the reagent is simple: cis-1-isopropyl-4-methylcyclohexane (or cis-1-(1-dimethylethyl)-4-methylcyclohexane if you're a stickler for the rules).
The products are a little tricker. Let's start with the basics: it is 1-bromo-4-isopropyl-1-methylcylohexane. Now to account for the steric relations of these substituents, label the reference substituent (labeled r) followed by a hyphen and the substituent locator number and name. The relative configuration of other substituents are then reported as cis (c) or trans (t) to the reference substituent.
This means that in the order that you present the products, their names are r-1-bromo-t-4-isopropyl-1-methylcylohexane and r-1-bromo-c-4-isopropyl-1-methylcylohexane.
[Source - (page 1060)]

Do you have any references for the naming of the products with the r/s c/t nomenclature? I was trying to work these out myself but hit a brick wall. The only info I had was from March's Advanced Organic Chemistry, but the discussion was incredibly brief.

"if you're a stickler for the rules" - "isopropyl" is (probably) the preferred IUPAC name for $\ce{-CHMe2}$.

Yes, it is the preferred name. Maybe I should have said if you're a stickler for systematic names :)

@orthocresol The preferred prefix for $\ce{(CH3)2CH-{}}$ is “propan-2-yl”. The prefixes “isopropyl” and “1-methylethyl” may be used in general nomenclature. Besides, the stereodescriptors cis and trans are not used in preferred IUPAC names, but they may be used in general nomenclature.

@Loong in the source that I linked, cis-inositol is listed as a a preferred IUPAC name. I think cis-trans nomenclature is preferred less than other stereodsicriptors, but when they are not an option cis and trans are in fact used.

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MO shapes for dx2-y2 sigma bond
How do the Bonding and Anti-bonding MOs of dx2-y2 sigma bonding with another dx2-y2 look like ? I mean to ask that if the sigma bond is along the x-axis then is the shape of the lobes along y-axis altered during the formation of the MOs.


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

It is a big universe. There has to be at least one such "molecule" floating around in deep space somewhere. You certainly couldn't make a pressurized tank of it.

PubChem database N4O10 (molecular formula) search shows that there is no known compound with this formula.
While $\ce{NO}$ (“mononitrogen monoxide”) and/or $\ce{NO2}$ (“mononitrogen dioxide”), having odd number of electrons, will dimerize or combine to form
$\ce{N2O2{: } 2NO -> ON-NO}$
$\ce{N2O4{: } 2NO2 -> O2N-NO2}$
$\ce{N2O3{: } NO + NO2 -> ON-NO2}$
(note the $\ce{N-N}$ bond),
$\ce{N2O5}$ is formed by the condensation of nitric acid $\ce{HNO3}$:
$\ce{ 2 { }O2N-OH ->[-H2O] O2N-O-NO2 }$
(note the $\ce{N-O-N}$ bond), thus can be considered nitric acid anhydride. And unlike polyphosphoric acid formation from $\ce{H3PO4}$, there are no remaining hydroxy groups for further condensation.
With non-standard formal approach, $\ce{N2O5}$ already is a dimer of hypothetical franctional-oxygen-atom-count “mononitrogen(V) oxide” $\ce{NO_{2.5}}$.

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General notation for one of the d-orbitals
What is the general notation to represent the d-orbital with $l=2$, $m_l=0$, i.e. the orbital normally referred to as $\mathrm{d}_{z^2}$. To elaborate more, this orbital can be ordered in various ways in a crystal environment and can be written as: $\mathrm{d}_{3z^2-r^2}$, $\mathrm{d}_{3x^2-r^2}$ or $\mathrm{d}_{3y^2-r^2}$. So is there a general notation to represent all three, like $\mathrm{d}_{3A^2-r^2}$?

The $d_{z^2}$ orbital points along the $z$-axis. I believe you choose the $z$-axis to point wherever you want that orbital to point (not the other way round).

Yes, but in some crystal structures there is an orbital ordering along different local axes, like in LaMnO$3$. So there will be a $d{3z^2}$, $d_{3y^2}$ and $d_{3x^2}$. I just wanted to know how to refer to all three types collectively.

Generally you are free in your choice of coordinate system to describe a certain problem and calling the d-orbital with $l=2$, $m_l=0$  either $\mathrm{d}_{3z^2-r^2}$, $\mathrm{d}_{3x^2-r^2}$ or $\mathrm{d}_{3y^2-r^2}$ depends on this choice but by convention the $z$-direction is usually chosen to be the system's preferred direction, i.e. it points along the main symmetry of the system.
Let's move to $d$ orbitals in crystals: Usually, metal $d$ orbitals are not "properties" of the crystal itself but "belong" to the local metal centers. Thus, the $z$-direction along which a $\mathrm{d}_{3z^2}$ orbital points refers to the preferred direction of the metal ion in its local environment, e.g. in a Jahn-Teller distorted Perowskite $\ce{ABO3}$ the $\mathrm{d}_{3z^2}$ orbital of metal ion $\ce{B}$ always points along the direction of the elongated/compressed $\ce{B-O}$ bonds within the local $\ce{BO6}$ octahedron or quadratic bipyramid. So, even though in $\ce{LaMnO3}$ you have differently tilted $\ce{MnO3}$ octahedra that have different preferred directions and thus differently oriented $\mathrm{d}_{3z^2}$ orbitals, you still refer to all of them as "$\mathrm{d}_{3z^2}$" and everyone you talk to about this will know which orbitals you mean and where they will point in the different local octahedra in the crystal. That is pretty much the status quo.
 $\hspace{2em}$

Now, what you want to do, is talk about the $\mathrm{d}$ orbitals not in the context of their local coordinate system but in the context of a more global coordinate system related to the whole crystal, e.g. declare a certain crystal plane to be the $x$,$y$-plane and name the $\mathrm{d}$ orbitals according to this choice to let their name reflect their orientation within the plane.

Of course, you can do this and it has been done here pretty much as you suggested  where the authors chose notations like "$\mathrm{d}_{3\sigma^2 - r^{2}}$ orbital ($\sigma = x, y , z$)" or "$\mathrm{d}_{3(x, y)^2 - r^{2}}$ orbitals". But be aware that this is a kind of notation that needs to be explained beforehand when you use it because it refers to an unconventional choice of coordinate system as a frame of reference for the $\mathrm{d}$ orbital labels.
TL;DR: Yes, you can use a general notation like you suggested but make sure to explain which coordinate system you refer to because otherwise you might cause confusion.


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Rearrangement of a bicyclic dienone under acidic conditions

Find the major product of the following reaction.

Attempt:
First step would be the formation of oxonium ion. After that, there are two possibilities. Either the upper double bond or the lower bond will shift and create a carbocation (A and B)

In product A there are two possible migrations. The better migrating group is the carbon which is a part of the second ring. But it can't migrate because of formation of a bridged carbocation. So methyl group migrates.

If B carbocation is formed, then the better migrating group can attack the carbocation.

My teacher told that the product A is wrong because a weaker migrating group is acting while in B, the stronger migrating group is acting. Is this correct? Because B involves ring contraction and more steps.

Incidentally, the structures labeled A and B are resonance structures, i.e., the same compound.

Your analysis of the possible products is excellent. I would consider both of the products to be major. Waring and coworkers studied this problem with a slightly different substrate (an extra methyl group at the 1-position, next to the carbonyl on the 'bottom' of the molecule). Under aqueous acidic conditions (sulfuric acid), the corresponding products were obtained in ~25% (mechanism A) and ~40% yield (product B). Only when run with sulfuric acid in acetic anhydride did mechanism B truly dominate (~90% vs 5% mechanism A).
There is actually a third pathway, where the methyl group shifts from ring juncture to ring juncture, and then goes through a mechanism similar to B. For the dienone in the question, the product is indistinguishable from mechanism B, but for the dienone studied in the publication, this gave another product in ~30% yield. I suspect that for the dienone in the question, the total yield of product B would be ~70%. That makes it appear as if mechanism B is dominant, but it is really two similar pathways operating that happen to give the same product.

I will add figures later if desired.

Please add figures especially for third pathway

For an update, see: https://chemistry.stackexchange.com/questions/116043/8a-methyl-1-2-3-4-4a-8a-hexahydronaphthalen-4a-ylium-carbocation-rearrangement/116067#116067


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Why is a face-centred cubic unit cell not regarded as equivalent to a body-centred tetragonal unit cell?
When studying crystal lattices, I decided to learn which combinations of unit cells and lattice systems lead to the 14 Bravais lattices. I learnt that, for example, no centred triclinic unit cell exists.
Later while searching the Internet and Stack Exchange, I found an answer saying that an end-centred cubic cell can be transformed into a (smaller) tetragonal unit cell as shown in th picture below.

Then I thought on and realised that a face centred cubic lattice can be transformed into a (smaller) body-centred tetragonal unit cell, and into an even smaller one, too.
So, why are some centred unit cells (for hexagonal, triclinic and rhombohedral lattices) considered non-existent, while others (e.g. face-centred cubic) are considered to exist even though they, too, can be reduced?

By that logic we should remove all centered unit cells, not just FCC, because each of them can be reduced to some primitive non-centered cell.

I don't know whether it is correct to say"... each of them can be reduced to some primitive non-centered cell."Moreover my question is justified

The logic I gave for my question about FCC have earlier been given in this site ,quite unsatisfactorily, so I decided to impose this question again as it has always confused me as to why other type of unit cells are not included in bravais lattice for example end centred cubic unit cell.

I think your question is quite unclear. Please not that the 'usual' picture of an fcc lattice is drawn with a 'conventional' unit cell, consisting of multiple atoms, and drawn to highlight the actual cubic symmetry of that particular Bravais lattice. Something like the Wigner-Seitz unit cell, while an appropriate reduced unit cell containing only one atom, and still showing the full Bravais lattice symmetry, is just harder to look at and directly see the cubic nature. I believe that you have difficulties in understanding space groups vs Bravais lattices vs point groups vs conventional cells.

@Sikander Don’t worry, since an edit to a question on hold automatically puts said question into a reopen review queue. But even after your edit your question remains unclear to me. Are you asking why crystallography uses some centred unit cells even though they can be reduced to primitive ones?

@Jan Yeah! this is my question.In fcc lattice one could also find in-centre tetragonal unit cell.

Only if all the unit cells in the crystal share a common structure (bcc, fcc..), it is said to have that particular structure. I think the flaw in your argument is that if you redraw the unit cells in this manner, you will not be able to reproduce the same throughout the crystal without losing the inherent symmetry.

In an FCC lattice  a=b=c while in a body centered tetragonal lattice a=b but c is unconstrained. The first example you gave should be that an end centered tetragonal is equivqlent to a primative tetragonal lattice. There is no end centered cubic lattice as that would violate the minimum cubic symmetry of 3 four fold symmetry axes. however the transition from FCC to a body centered tetragonal lattice would be displacive not transformational.


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Differences in Gibbs activation energies for rotation around a bond

Question
Why is there a difference in energy in the rotational barrier between these two structures?

My reasoning is such:

Due to conjugation between the O electrons and the pi- double bonds, the bond has actually single bond characteristics and is therefore weaker than a conventional double bond alkene. But that doesnt explain the difference between the molecules? Can someone please explain?

One possibility is that there is an unstable resonance structure you haven't drawn, where (going from the right-side one you drew) the pi bond between the left Oxygen and Carbon dissociates, putting a positive charge on Carbon. This would be better stabilized by having a more substituted carbon attached to it. However, I'm skeptical that alone is enough to explain this large a difference in energy. Especially since steric effects would favor B.

In both molecules, there is π delocalization across the $\ce{C=C}$ bond which extends across the two carbonyl groups on the alkene. In order to maximize the overlap of the p orbitals involved in this conjugated system, it means that this portion of each molecule has to be flat. It is not a problem for the conjugated system to be flat in molecule B, but in molecule C you have three bulky methyl groups near the conjugated system. Steric clashes in molecule C between the t-butyl group and the carbonyl group on the same side of the $\ce{C=C}$ bond mean the carbonyl will rotate out of the plane of the conjugated system to avoid "bumping into" the t-butyl group, lowering the overlap between the p orbitals in the conjugated π system and thus destabilizing the molecule.
Because molecule C does not experience as much favorable pi electron delocalization as  molecule B, molecule C sits at a relatively higher energy than molecule B. The transition states of the two molecules, however, are not so different in energy. The Gibbs energy of activation is defined as the difference in Gibbs energy between the transition state of a reaction and the reactants (where our "reaction" is a rotation about the $\ce{C=C}$ bond). As molecule C sits at a higher energy relative to molecule B, but the transition states are relatively similar, it makes sense that there is a lower barrier to rotation about the $\ce{C=C}$ bond in molecule C.

Hyperconjugation could also stabilize the compound B as compared to compound C...couldn't it?


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Presence of Hydroxide Ions in Chemical Reaction in a Basic Medium
I was studying NCERT book of chemistry about chemical properties of hydrogen peroxide. First there is a chemical equation under a subtopic named 'Oxidising action in acidic medium' which is...
$$\ce{2Fe^{2+} + 2H^+ + H_2O_2 \rightarrow 2Fe^{3+} + 2H_2O}$$
Note here that there are hydrogen ions in the reactant side because it is in acidic medium.
Then another subtopic named 'Oxidising action in basic medium' has reaction
$$\ce{2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^-}$$
Now in this equation there is no hydroxide ions as it should be because it is basic medium.
I am not asking about its oxidizing and reducing properties or how it works. I know that oxygen atom of $\ce{H_2O_2}$ has $-1$ oxidation state. I am just asking why isn't there any hydroxide ion in the reactants.
Sorry if it is a foolish doubt, thanks in advance.

Because you don't need OH- to form Fe3+.

@orthocresol why not?

maybe it has to do something with reaction rates. maybe if there is no hydroxide ion already present, the reaction will proceed very fast and can go through alternative path way to give something else as product. so maybe its a kinetic control here

@PunarbasuRoy then why this reaction should occur in basic medium as stated in the book?

kinetic control perhaps. to slow down the reaction. not always you would like a reaction to be very fast.

Nothing to do with rxn rates - just stoichiometry - look at the balanced equation you have, you can add OH- on the left if you want, all that happens is that another OH- is added on the right, if it's not involved in the reaction why bother adding it to the equation? it's like mixing aqueous AgNO3 and NaCl, you can write $\ce{Ag+ + Cl- -> AgCl}$, just because water is in the reaction system doesn't mean it has to be in the equation - would you write $\ce{Ag+ + Cl- + H2O -> AgCl + H2O}$?

The reaction consumes H+ either from H3O+ or from H2O [or in the second case from H2O2, it is an acid]. The choice of reductant is unfortunate because both hydroxides are insoluble in base, the ferric much more so. To maintain ions the reaction must be run in acid.  The large difference in solubilities most likely means Fe[OH]2 would be oxidized to Fe[OH]3; but must check the potentials to see.

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Reversible & irreversible adiabatic processes
Work done in reversible is more than that of irreversible
Suppose we are going from state $(P_1, V_1, T_1)$ to some other state say $(P_2, V_2, T_2)$ for reversible and $(P_2, V_2, T_2')$ for irreversible. Which should be greater, $T_2$ or $T_2'$?
My attempt
Let us say heat supplied is same i.e q for both reversible and irreversible. We already know that $U = q - W$
Reversible as work done is more so $U$ becomes less and in irreversible $W$ is less so $U'$ is more than $U$ i.e $U' > U$
As $U'$ is greater $T_2'$ should be greater. Is this correct explanation?

For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$.  Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$.  It doesn't matter what process is used to get from state 1 to state 2.  Only 2 intensive variables are required to specify the state of the system (assuming no phase change).  So $T_2=T_2"$.
Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.

This does not work because one the processes is irreversible which means there is some internal mode of changing the temperature. So if it were an expansion, there would be internal friction which would raise the temperature instantaneously making the overall change smaller. So the OP is right.

I stand by what I said.  The state of gas is established by specifying 2 intensive properties:  P and V, P and T, or T and V, where V is the specific volume.  If P, V, and T are the final parameter values for an adiabatic reversible expansion, you will not be able to identify an adiabatic irreversible expansion from the same initial state that matches more than one of these final parameter values in the final state.

I believe your answer is correct.
Because we are doing work in this case, we can imagine that this is simply a piston being expanded against the atmosphere.
As you have said, a reversible process corresponds to the maximum amount of work, which in this case also corresponds to the largest change in energy.
Because we are only interested in comparing the temperature $T_2$ and $T_2'$ because the pressures and volumes change by the same amount, we must ask ourselves if the temperature is going to increase or decrease?
Well, without any mathematical proof, I will say that while we are doing work, and thus giving out energy to the atmosphere, the temperature must decrease because our system must lose energy. Thus, while outputting work, the more work we do, the greater the change in temperature.
The above paragraph depends on the process being adiabatic though because if the process were not adiabatic, we would have to consider the relative amounts of heat flow and things could get weird.
But, we have said that $$W_{rev}>W_{irrev}$$thus, $$|\Delta T_{rev}|>|\Delta T_{irrev}|$$but, since temperature is decreasing as we give out energy, $$T_2<T_2'$$
So, you are correct unless we are both making a stupid mistake.
This answer makes sense though, because the for the reverse process, we would be compressing a piston and thus must get back as much energy as possible in the form of a temperature increase. With changes in pressure and volume being the same, we know immediately that the process which begins at the lower temperature and gets back to the original temperature $T_1$ has the largest change in internal energy which must be true for the reversible process.

I challenge you to offer one single specific example to back up what you are saying here, such that the final pressure and specific volume for the irreversible expansion are exactly the same as the final pressure and specific volume for the reversible expansion (starting from the same initial state).

If the adiabatic reversible and adiabatic irreversible expansions both end at the same (P2, V2) (as specified in the original question), what does the ideal gas law tell you about the final temperatures?

Hmmm.... I'm pretty convinced now that your answer is correct. I've been thinking about it quite a bit over the last few hours and I can't think of any reason why your answer is wrong besides that I feel like the irreversibility must lead to a decrease in  work output and hence the temperature shouldn't change as much. Could you point out what is wrong with my mathematical reasoning for my own sake? This is such a straightforward thing I can't believe I'm not understanding this.

The point I was trying to make is that the original problem statement was over-specified.  One cannot say that P2 and V2 are the same for both the adiabatic reversible path and the adiabatic irreversible path.  The problem statement would have been fine if the reversible path ended at P2,V2,T2 and the irreversible path ended at P2',V2,T2' (so that only the final volumes were required to be equal).  I think that otherwise your analysis would have been fine.

Oh!! Right that makes a lot of sense. I was a little confused by what you meant at the end of your answer but now it makes sense. Ok. Thanks for clarifying! I wasn't trying to be rude I was trying to get to the correct solution.


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Pharmaceutical grade phosphoric acid
I am looking for a specification for pharmaceutical grade phosphoric acid.

Will you confirm that you are looking for the actual USP specification for phosphoric acid?

here are tests for phosphoric acid to meet the specification http://www.pharmacopeia.cn/v29240/usp29nf24s0_m64570.html


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Calculating the wavelength of an emitted photon of a Ne^9+ ion
I'm supposed to calculate the wavelength of an emitted photon going from the 6p orbital to the 2s orbital. The first part of this problem will be to use the equation
$$E=-{Rh\over n^2}$$
in order to calculate the energy.
That's all easily done when considering a hydrogen atom. Our chemistry professor, however, has told us to consider the ion $\ce{Ne^9+}$.
Therefore, my question is this: What's the difference (if any) when calculating the energy when it comes to this specific ion? What part does the nucleus play in this?
Also side note(Just to make sure): There is no difference when the professor refers to the orbital numbers right? I'm only supposed to consider the principal quantum numbers, correct?

The equation is already wrong. That equation only applies for the hydrogen atom. For a hydrogen-like species, i.e. an ion with only one electron like $\ce{Ne^9+}$, you need to consider the effect of the nuclear charge $Z$ - see https://en.wikipedia.org/wiki/Rydberg_formula#Rydberg_formula_for_any_hydrogen-like_element

Have you considered in class the change in wavelengths going from hydrogen to deuterium? As noted by @orthocresol, Ne$^{9+}$ is a single electron, hydrogen-like system, except that the nuclear charge and mass are not that of a proton. This does change the levels of the Rydberg states, but in a way usually covered in a straightforward way by texts or at good on-line resources.

Check this page.


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Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?
The molecule $\ce{[PdCl4]^2-}$ is diamagnetic, which indicates a square planar geometry as all eight d electrons are paired in the lower-energy orbitals.
However, $\ce{[NiCl4]^2-}$ is also $\mathrm{d^8}$ but has two unpaired electrons, indicating a tetrahedral geometry.
Why is $\ce{[PdCl4]^2-}$ square planar if $\ce{Cl}$ is not a strong-field ligand?

The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference.
Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. If it were to adopt a square planar geometry, the electrons will be stabilised (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired, are now paired:
$\hspace{30 mm}$
(source: Wikipedia)
and on top of that, there are additional steric repulsions introduced by moving the ligands closer together. (This is, to some extent, a throwback to VSEPR theory, which simply says that the ligands would rather adopt a tetrahedral arrangement because of their mutual repulsion.)
We can label these two factors as $\Delta E$ (stabilisation derived from occupation of lower-energy orbitals) and $P + S$ (pairing energy + steric repulsions) respectively. One can see that:

If $\Delta E > P + S$, then the complex will be square planar;
If $\Delta E < P + S$, then the complex will be tetrahedral.

This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter $\Delta_\mathrm{O}$, also called $10~\mathrm{Dq}$ in older literature, plays the same role as $\Delta E$ does above. (However, unlike $\Delta_\mathrm{O}$, there is no simple graphical way to represent $\Delta E$ on a diagram since multiple orbitals are changed in energy.)
The two main factors that differentiate the Ni(II) complex from the Pd(II) complex are therefore:

The $\mathrm{4d}$ orbitals of $\ce{Pd^2+}$ are more radially diffuse (i.e. bigger) and therefore form stronger overlap with the orbitals on $\ce{Cl-}$, as compared to the $\mathrm{3d}$ orbitals of $\ce{Ni^2+}$. This leads to larger splitting of the $\mathrm{d}$ orbitals - or the relevant MOs (I'm too lazy to check their symmetry labels) - and hence a larger $\Delta E$.
Again because the $\mathrm{d}$ orbitals are more diffuse, the pairing energy $P$ is smaller in the palladium complex (it basically costs less energy to stuff them into the same orbital).

Together, these two factors ensure that practically all $\mathrm{4d}$ and $\mathrm{5d}$ $\mathrm{d^8}$ $\ce{ML4}$ complexes adopt a square planar geometry, even if the ligand is not a strong-field ligand. Other examples of such square planar complexes are $\ce{[PtCl4]^2-}$ and $\ce{[AuCl4]^-}$.
In fact, I am digressing here, but the same factors also cause the octahedral complexes to be almost invariably low-spin. For example, the $\mathrm{d^5}$ complex $\ce{[Fe(ox)3]^3-}$ has five unpaired electrons, indicating a high-spin configuration, $\mathrm{(t_{2g})^3(e_g)^2}$. However, $\ce{[Ru(ox)3]^3-}$ has one unpaired electron, which implies a low-spin configuration $\mathrm{(t_{2g})^5}$.

On  a somewhat unrelated note: I find it interesting that this effect is also present in d9 transition metals. The Jahn-Teller effect is like if this orbital energy inversion happened but not quite all the way... like it was getting halfway to becoming square planar. And what you get is a distorted octahedron where two "axial" ligands have extra long bond lengths compared to the other four "equatorial" ones... like it was halfway to becoming square planar.

@orthocresol Are the steric repulsions included in the splitting diagramms? I mean do they just change the absolute value of the energy of the system but not the difference $ΔE$?


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Heat Dissipation of Chemical Reaction
When a chemical reaction dissipate heat to the surrounding, the molecules that participate in the chemical reaction will have high kinetic energy and will transmit some of the kinetic energy to the surrounding and the remaining kinetic energy will increase their temperature.
When we say that heat is dissipated to the surrounding we are talking about "some of the kinetic energy" right? Also when we write that heat dissipated is equal to $\Delta H$, it is included the remaining kinetic energy?
Can somebody confirm whether my understanding is correct or clarify where I went wrong if it is not?
Sorry, but I am far from chemistry field.

By DH do you mean $\Delta H$?. If that is the cause, the answer is not always. Heat dissipated is known as q (heat transfer) which is only equal to $\Delta H$ when pressure is constant.

To expand on Nanoputian's answer, $-\Delta H$ is the amount that has to be removed (dissipated) to hold the temperature constant at its initial value.

Let establish some thermochemistry terminology first:
Work: the total energy transferred by a system to another. There are several types of work. In thermodynamics, the most common one is pressure volume work, given by: $w = -p_{ext}\Delta V$
Internal Energy: the total kinetic and potential energy of a system
Enthalpy ($H$): is the total energy of a system. It also can be thought as the amount of energy that is required to make all the bonds in the system plus all the energy that is required to make room for the system. Hence it should be clear that it is given by the formula: $\Delta H = \Delta U + \Delta {(pV)}$
Heat Transfer (q): the amount of heat energy (kinetic energy) that is transferred from a system given by the formula: $q = \Delta U -w = \Delta U + p_{ext}\Delta V ~~$ (when only pressure volume work is done)
In your question, you are actually referring to $q$ (heat transfer), not $\Delta H$. This is because $\Delta H$ also accounts for the change in pressure and volume.
However under certain conditions, $q$ and $\Delta H$ can be equal. If the external pressure is constant (which is a common case as most experiments are done in the open atmosphere which has a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p_{ext}\Delta V = q$$


text:
Nuclear Fallout
If there was a nuclear explosion and years later I come out of a fallout shelter like in the popular game series Fallout, what elements would react to the radiation in a way that could harm humans and what happens to humans when they interact with these substances?

While this is interesting question, it too broad and imprecise. Situation would depend on lots of factors.

IMO it would be much different question. I think earlier was better but you should narrow it down

We can discuss it in chat

@Mithoron I think I'll leave this question as is and ask a separate and more thought out question later. I realized the question is completely different now but I'm still interested in what answers to this question would be. Maybe if stackoverflow added a tree of topics instead of separate tags it would be possible to convert broad questions into topics with an array of subtopics and/or questions inside.

The whole SE model is to have one clear question, and provide answers to that question. Multiple queries in one question do not fit the model.

As for the core question, the 'harmful' elements (isotopes) are either generated directly in the nuclear explosion, or are part of the directly generated isotope decay chains (i.e. daughter products). If I recall properly, Cs, Sr, and I are major issues for humans since they are taken up by the body resulting in internal rad dose.

One of the best examples would be Bikini Atoll.

Today, the key residual fallout radionuclides of potential radiological concern include cesium-137 and strontium-90 and, to a lesser extent, plutonium isotopes and americium-241.
The most important pathway for human exposure to residual fallout contmaination is ingestion of cesium-137 contained in locally grown food crop products such as coconuts, Pandanus, and breadfruit

According to the CDC's Radioisotope Brief: Cesium-137

External exposure to large amounts of Cs-137 can cause burns, acute radiation sickness, and even death. Exposure to Cs-137 can increase the risk for cancer because of exposure to high-energy gamma radiation. Internal exposure to Cs-137, through ingestion or inhalation, allows the radioactive material to be distributed in the soft tissues, especially muscle tissue, exposing these tissues to the beta particles and gamma radiation and increasing cancer risk.


text:
How do we know if a compound can be oxidized?
I'm studying organic chemistry, and I have to identify compounds that can be oxidized in some exercises I am doing.
For instance, an $\ce{R-CHO}$ can be oxidized further, but an $\ce{R_{3}-COH}$ cannot, and presently I can tell this only if I remember it by heart.
How do I know if a compound can be oxidized? Is there an algorithm/method to follow?

At this stage, I would recommend memorising. The mechanisms for redox reactions are often not trivial and sometimes unknown - it'll be difficult to come up with a systematic way to predict which compounds can be oxidised/reduced that isn't just a disguised form of "memorising".

I think your best bet other than memorizing right now is paying attention to the definition of "oxidation". What does oxidation mean? So, what do the compounds need to have in order to be oxidizable? Note that this is rather qualitative and can put you in trouble when you're not expecting it.

Oh and just by the way - since it's probably good to expand your chemical vocabulary - $\ce{RCHO}$ is called an aldehyde and $\ce{R3COH}$ a tertiary alcohol.

Every compound that contains carbon and is not $\ce{CO2}$ or a (hydrogen) carbonate can be oxidised.


text:
Dilution with 3 Different Percentages
I'm trying to solve this problem.

How many grams of a solution at $25\%$ do you have to add to $100\ \mathrm{g}$ of a solution at $40\%$ to obtain a solution at $30\%$.

My Attempt
It's a bit more than a simple diluition. So far I've calculated the mass of the solute in $100\ \mathrm{g}$ at $40\%$: $(40\times 100\ \mathrm{g})/100 = 40\ \mathrm{g}$
Then the mass of the solution at $30\%$ so: $(40\ \mathrm{g}/30)\times 100 = 133\ \mathrm{g}$
$133\ \mathrm{g} - 100\ \mathrm{g} = 33\ \mathrm{g}$ (of solvent you need to add)
Now, how does that $25\%$ fit in all this? Can anybody point me to a useful formula or an example? Thanks!

Isn't this really a maths problem? You've calculated how much solvent you need to add when you're actually required to add a 25% solution. You need to generate a simultaneous equation based on the solution mass and solute mass relationships.

OK, for what it's worth:
Let $x_f$ be the final solution mass and $x$ be the mass of the 25% solution.
The dissolved mass can be defined by
$$0.3x_f~=~40+0.25x$$
The solution mass is:
$$x_f=100+x$$
Substitute $x_f$ into the first equation and solve. You should get the 25% solution mass to be added and then solve the easy solution mass equation to check your answer.


text:
Meaning of "Diffusion Rate"
A friend is trying to understand a paper which discusses the diffusion of proteins through milk. Rather than listing a diffusivity (with units [Length]^2/[Time]), the paper lists the 'diffusion rate' of the proteins through the liquid, in units of mm/hr (i.e. [Length]/[Time]).
We can't make heads or tails of this. I was wondering about some sort of 'average front propagation velocity', but I'm not sure how that would be defined, as a diffusion front's velocity can change over time, and depends on the geometry, concentrations, etc.
Has anyone run into this terminology before?
Thanks

In the attached pdf (page 7, part C) the diffusion of a dye is measured as the liner distance the drop of dye solution penetrates into a test tube of gelatin. http://www.instruction.greenriver.edu/kmarr/biology%20211/Labs%20and%20ALEs/B211%20Labs/B211%20Labs/4_Lab%204_DiffusnOsmosis_B211_Fall2009.pdf // There must be some sort of experiment where the diffusion was measured in some sort of linear manner...

Could you give information on the paper? It might make the question easier to answer for anyone with access.


text:
Strange valence numbers?
I'm trying to understand why ChemReference sometimes lists the valence electrons as a number I don't expect. Take oxygen, for instance. It should have six valence electrons, however the site lists two. Why?
I hear that, for transition metals, the d-orbitals count towards the valence. Is there another rule like that I need to know about that might explain Oxygen?

I suspect software bug might be the answer here. I can't think of any definition of valence electrons (there are some different takes on the concept, but not for light elements) by which Oxygen would have 2 valence electrons, let alone how it could have 2 and Nitrogen would still have 5. But perhaps there is a definition by which it makes sense.

I just saw a bug report in which he refers to those valance as the "common valence". Never heard of that before. Have you?

I have not. It seems this is what he is referring to: https://en.wikipedia.org/wiki/Valence_%28chemistry%29#Common_valences . That's pretty obscure terminology, and I'm not sure how it relates to the values being given. Perhaps what he's getting at is the common number of bonds formed by the element. That wouldn't make sense for N, though.

Look closely on that site. It's not 'valence electrons' but 'valence'. There is a difference between the valence electrons and valence.
Valence electrons are the number of free electrons in an atom that can take part in making the bond. There are 6 such electrons in the oxygen atom as you stated.
While the valence or valency is the number of bonds that can be made by the atom.
That's why valency of oxygen is 2

Interesting. Why aren't those numbers the same? If there are six electrons in a valence shell, aren't all six free to make bonds?

there is also the octet problem if all those 6 electrons participate in the bond formation. A neutral oxygen atom needs 2 electrons to complete its octet and he get it by making two bonds.

So the way to calculate it would be to take the number of valence electrons an element has and then figure out which is less... adding electrons to get to 8, or subtracting electrons to get to 8, and then that number is the valence?

@Stradigos Yes but it only works until the octet rule is followed.


text:
Is it possible to separate/remove sugar from common drinks, such as juice?
Is it possible to remove sugar from a drink (for example, juice) using common household equipment?

I suspect what you're ultimately asking is whether there is a way to "unsweeten" a drink, i.e. a way to selectively remove sugars from a complicated aqueous solution? That would probably preclude the usage of any sort of technique which involves strong/prolonged heating, as it could change the flavour of the juice.

One way that technically would work in some rather constrained conditions is to add a large amount of pure ethyl alcohol to the juice, in which sugar is poorly soluble but most flavourful compounds would still stay dissolved, then filter off the precipitated sugar and evaporate the excess alcohol (if you want...)

@Nicolau Saker Neto thanks, you understand me correctly.

@NicolauSakerNeto how to evaporate it ?

It's unlikely you'll be able to evaporate all the alcohol, especially not quickly. Since ethyl alcohol is somewhat more volatile than water (pure ethanol boils at 78 ºC), you could heat the juice mildly for a while to remove a portion of it. You could also bubble air through the juice for a prolonged period, though that could oxidize some of the flavour compounds, which I think may cause some souring. If you happen to have a good vacuum pump and a rotary evaporator in your house, that would be most effective, and also very cool.

Also, as I mentioned before, this process only works technically. The resulting juice would have a fairly high alcohol content. Also, I really need to point out that pure ethanol for domestic use is almost always sold as denatured ethanol, which is not drinkable. Denatured ethanol usually contains a very bitter substance added to it (often toxic, too), and you don't want to drink any of that.

Yeah, the real answer to the question "Can I make this beverage less sweet?" is 'no, not really'.  Just try to make it less sweet next time. But if you ask a bunch of science nerds if you can separate the water from a sugar solution, then we will think of (impractical) methods to do so.

I've seen similar question about lactic acid - these aren't questions for chemists but cooks or bartenders.

@RS But the vitamins are irrelevant if the rest of your diet is good and the sweetness is what make the drinks taste good but is, in most circumstances, actively bad for your health (many experts even include natural fruit juice in the "too much sugar to be good for you" group).

Why not add yeast to your juice and allow it to ferment?  This will depend on the juice having pH and sugar concentrations; look at commercial wine making kits which provide the proper equipment and recipes. If you conduct the fermentation properly without introduction of air, the sugar will be converted to alcohol, which later can be boiled off (see above) to give the sugar-free version of your juice. Otherwise, with air present, the fermentation will go on to form acetic acid and give the juice a vinegary taste.

What can I use to precipitate the acid ?

I don't know of any way to precipitate acetic acid from dilute solution. You are far better off controlling the fermentation in the absence of air, so the acid does not form in the first place.

Technically @Kyle is correct (the best kind of correct), in that boiling the sugar water solution will drive off the water. But any candy maker will tell you that as the water boils off, the temperature of the boiling solution increases, and you just have a more and more concentrated solution. If you keep adding heat then the remaining sugar will burn (my poor caramel!!)
By boiling you can drive off most of the water, but not all.  Evaporation at room temp is a way to get rid of all the water, and it can happen faster under application of a vacuum.
I imagine there are also high-pressure microfilters that you can use to separate them as well.

Nicolau Saker Neto is closer to what I want than @Kyle, heating will kill the taste.

Well, adding a large amount of ethanol to an overly sugary liquid definitely is one way to make it more palatable.....

Just add yeast under anoxic conditions and wait a few weeks, then fractional distillation to remove the ethanol. Also if reverse osmosis removes salt, it should also remove a larger compound like sucrose.

Yes, this is a physical separation. Since sugar ($\ce{C12H22O11}$) has a boiling point of around $\pu{205^\circ C}$ and water boils at $\pu{100^\circ C}$, you can evaporate the water by heating your drink over a stove at any temperature between $\pu{100^\circ C}$ and $\pu{205^\circ C}$ (preferably about 3/4 between, closer to the sugar boiling point).
Depending on your drink you may have to do some other separations (boiling orange juice won't remove pulp). You should be left with sugar in your pot.

It is possible to remove sugar partially by decantation.
Sugar solution is nothing more that a dispersion of the sugar molecules in water due to hydrogen bonding. However, the solution is not perfectly stable since the molecule of sucrose is quite large, therefore we can let it sit on the table for a few hours where most of the denser sugar sink and rest at the bottom. Slowly we can decant the less populated solution into another vessel to have a less concentrated form of juice.

It's not a suspension: it's a solution. Solutions don't separate on standing.

Matt is right.. A solution won't separate by sitting.


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Is aluminium hydroxide a strong base?
Is $\ce{Al(OH)3}$ in a very dilute solution deemed as a strong base? Are there any actual experimental data for it?

$\ce{Al(OH)3}$ is all but insoluble in water (though you may call that "a very dilute solution", if you like), and most certainly is not a strong base.

No, it is not a strong base.
It is a weak acid with pKa of 5
See DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION at the page numbered 143 (page 12 of the file).


text:
Reaction rate constant conversion
I'm flustered, I'm not into chemistry and need to convert $\mathrm{(nM\cdot s)^{-1}}$ to $\mathrm{(M\cdot s)^{-1}}$.
Is it correct to multiply with $10^9$?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I am unsure what your question is supposed to be; turning nano-molar-seconds into mols? That doesn’t really make sense to me …

I'm designing a model of the EGFR pathway with simmune-software-tool and have to extract all kind of information from journals/papers about association rates and dissociation rates of the proteins in the pathway.

Wouldn't the conversion of asssumed nM into M imply dividing by $10^9$ instead of the multiplication mentioned in the OP?  Alternatively, a multiplication with $10^{-9}$ instead of $10^{9}$ ...

You have a rate constant for a second order reaction.  You want to convert from $(\mathrm{nM}^{-1} \cdot s^{-1})$ to $(\mathrm{M}^{-1} \cdot s^{-1})$.  So you have to write it out using fractions, like
$$ \mathrm{nM}^{-1} s^{-1}  = \frac{1}{\mathrm{nM} \cdot \mathrm{s}}\times \frac{10^9 \mathrm{nM}}{\mathrm{M}} = 10^9 \mathrm{M}^{-1} s^{-1}$$
So yes, you multiply the rate by $10^9$


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Which of the four balloons will shrink?
Question:

Four gas balloons A, B, C, D of equal volumes containing $\ce{H2}$, $\ce{N2O}$, $\ce{CO}$, $\ce{CO2}$ respectively prickled with a needle and immsersed in a tank containing $\ce{CO2}$. Which of them will shrink after some time?
a) A
b) B
c) C
d) Both A and C

My attempt: I think that the ballons filled with the gases which will react with $\ce{CO2}$ will be the ones that will shrink. Now $\ce{H2}$ reacts with $\ce{CO2}$ to form $\ce{CH4}$, so the amount of $\ce{H2}$ will decrease and the balloon will shrink. Also I know $\ce{CO}$ cannot react with $\ce{CO2}$, so the answer has to be (a). I need hep determining the correct solution.

This is a rather unclear question (your test question, not your question so much). If they're pricked, I would expect gas to be released. If it weren't for that detail, I would assume this was about ideal gas vs. nonideal gas. Are you sure that bit was in the question? You wrote "pickled" but I am suspecting it's a typo. Also, the options are A/B/C not the gases.

$\ce{H2}$ will not react with $\ce{CO2}$ for two reasons: (1) such a reaction requires pretty high temperatures and (2) $\ce{H2}$ is inside the balloon, while $\ce{CO2}$ is outside. And if we consider that a gas can flow out of a pierced balloon (as our common sense says),  then it will be the case with all four of them, so all balloons will shrink.

pickled balloons?  Like this?

I have a feeling this is a poorly dressed up question asking about the relative effusivity between the gasses.

The relative molecular mass ($M_r$) of the gas determines the rate of diffusion of the gases from a region of higher concentration to a region of lower concentration.
The smaller the $M_r$ of the gas, the greater the rate of diffusion (movement of gas molecules).

$M_r$ of $\ce{CO2}$ = 40

$M_r$ of $\ce{H2}$ = 2 (Balloon A)
Since hydrogen gas has a smaller $M_r$ than carbon dioxide gas, more hydrogen gas diffuse out of the balloon than carbon dioxide gas into the balloon per unit time. Hence, this balloon will shrink.
$M_r$ of $\ce{N2O}$ = 44  (Balloon B)
Since nitrous oxide gas has a greater $M_r$ than carbon dioxide gas, more carbon dioxide gas diffuse into the balloon than nitrous oxide gas out of the balloon per unit time. Hence, this balloon will inflate
$M_r$ of $\ce{CO}$ = 28  (Balloon C)
This balloon will shrink (same logic as $\ce{H2}$).
$M_r$ of $\ce{CO2}$ = 40  (Balloon D)
Since the balloon and the tank both contains $\ce{CO2}$ (same $M_r$ of gases), there is no net diffusion of gases in/out of the balloon, thus the balloon will not inflate/shrink (theoretically).


text:
Elimination of HCl from isomeric chlorocyclohexadienes

Which of these is most easily dehydrohalogenated?

a. 5-Chlorocyclohexa-1,3-diene (1)
b. 3-Chlorocyclohexa-1,4-diene (2)
c. 1-Chlorocyclohexa-1,4-diene (3)
d. Both 1 and 3

I encountered this question a while ago.  The given answer is option (b) (compound 2). But the dehydrohalogenation of compound 1 produces benzene which has high heat of formation. So why isn't that the answer?

They both form benzene. Does that change your analysis?

@jerepierre Can you elaborate a bit?

Can anyone please explain?

That's a fair question,
Assuming that you use a strong base for dehydrohalogenation you will be deprotonating the most acidic $\ce{H}$ resulting in formation of a carbanion which has the highest stability in compound number 2
You can see that the carbanion formed in compound numer 1 will have resonance with only one of the double bond and hence will be less stable then the carbanion in the compound 2 which has resonance from both the double bond.
Mechanism can be as follow :-

Also;

I thought that extended conjugation is a better stabilizing effect than cross conjugation?


text:
How does the ionic strength formula work? (simple version)
The ionic strength formula provides a way to estimate the Debye screening length. I’m having some trouble understanding the structure;
$$I = \begin{matrix}\frac{1}{2}\end{matrix}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2}$$
It seem a bit tricky to find the derivation for this, but all I’m really looking for is an intuitive answer to a couple of questions;

Firstly; why is there a factor of a half out the front?
Secondly; why do we square the charge number of the ion?

I am a physicist, so please explain any chemistry concepts like I’m 5. Thank you for your patience.

Firstly; why is there a factor of a half out the front?

To avoid double-counting. For a salt such as $\ce{NaCl}$ comprising a monovalent cation and a monovalent anion, the ionic strength is defined such that it should be equal to the molar concentration of the salt, as opposed to the sum of both the anion and cation concentrations.  Therefore, since such a salt has two constituents, each constituent contributes half of the total ionic strength.

Secondly; why do we square the charge number of the ion?

For this there's not much choice but to turn to the Poisson-Boltzman equation, which provides a differential expression for the electric field inside the electric double-layer. (Sorry, 5-year-old version of Jekowl!) Below is the version I have in my graduate transport book, Eq. (15.4-6):
$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\exp{\left(-{z_iF\over RT}\Phi\right)}}
$$

$\Phi$ is the electric potential, $\mathrm{J} \over \mathrm{C}$ or $\mathrm{V}$
$F$ is Faraday's constant, $96,485\ {C \over \mathrm{mol}\ q}$
$\epsilon$ is the electrical permittivity, $\mathrm{C}\over \mathrm{V\ m}$
$z_i$ is the valence of ion $i$, $\mathrm{mol}\ q \over \mathrm{mol\ ion}$
$C_{i\infty}$ is the concentration of ion $i$ far from the surface of interest, $\mathrm{mol} \over \mathrm{m^3}$
$R$ is the ideal gas constant, $8.314\ {\mathrm{J} \over \mathrm{mol\ K}}$
$T$ is temperature, $\mathrm{K}$
$q$ is the elementary charge

If the magnitude of the electric field is small such that $\left|z_iF\Phi/RT\right|\ll 1$, the exponential can be expanded as a Taylor series:
$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\left(1-{z_iF\over RT}\Phi+\dots\right)}
$$
If you then pull everything inside the parentheses, ignoring signs the second term becomes:
$$
C_{i\infty}z_i^2{F^2\over \epsilon RT}\Phi
$$
The second-order dependence of $z_i$ is readily visible in the above.

Poission-Boltzmann is safe for physicists, because you can derive it from statistical mechanics :D

First the whole notion of solution activity  is simplfied by the Debye–Hückel theory which describes dilute solutions in terms of the individual
ions. The solution activity is needed to predict the colligative properties properties of solutions like:

Relative lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

The 1/2 factor is so that an electrolyte solution like NaCl (cation and anion only have one charge each) in dilute solutions reduces to it molar strength even though it creates two ions in solution. (The molal strength  would be used in concentrated solutions.)
To understand the square term you'll need to read about the mathematical development of the Debye–Hückel theory.


text:
Lattice structures, Metallic Bonding
How and under what circumstances can cations form a lattice structure?
Would they not repel each other?

Cations alone can never form a lattice structure. As you correctly pointed out, they would repel each other.

Is this a problem with metallic bonding, or what?

Thanks, They are surrounded by localized electrons as per what i found  out.Yes it is a problem with metallic bonding


text:
Stoichiometric calculations for the combustion of octane
The equation for the burning of octane:
$$\ce{2C8H18 + 25O2 ->  16CO2 + 18H2O}$$

How do i find the number of moles of carbon dioxide which is produced when one mole of octane burns ?
How do i find , at what volume , at STP , is occupied by the number of moles determined in the answer to my first question ?
If the relative molecular mass of carbon dioxide is 44 , what is the mass of carbon dioxide produced by burning 2 moles of octane?

This is a homework problem, so I'll not spell the answer outright, but try and give as much as I would if you came to me in office hours.

Your equation tells you what the ratio of moles of $\ce{C8H18}$ to moles of $\ce{CO2}$.  For every two moles of octane burned, how many moles of $\ce{CO2}$ are produced?  Now, what would it be for only one mole of octane?
What have you tried so far on this?  You have the ideal gas law (PV = nRT), and you know the P, the n from the first question, and the T - so just solve for the V.
So you know one mole of $\ce{CO2}$ masses 44 grams, so just multiply this by the number of moles of $\ce{CO2}$ produced by 2 moles of octane (see my hint on point 1 above).


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link between acidity and [H3O+]
Suppose I have an aqueous solution (1) of $\ce{CH3COOH}$. The chemical species in the solution will be $\ce{H2O, H3O+, CH3COO-}$ and, since it's a weak acid, some $\ce{CH3COOH}$ will remain.
Suppose I have another solution (2) containing initially water and $\ce{HCl}$. The chemical species in the solution will be $\ce{H2O, H3O+}$ and $\ce{Cl-}$ since it's a strong acid. Suppose the initial quantity of $\ce{HCl}$ was calculated so that the obtained concentration of $\ce{H3O+}$ in solution (2) is the same than in solution (1). But in solution (1), in addition to the $\ce{H3O+}$ ions, we have some molecules of another acid, the remaining $\ce{CH3COOH}$. Then the $\mathrm{pH} = - \log [\ce{H3O+}]$ is the same for (1) and (2), while (1) contains more "acidic" molecules.
How is pH a measure of acidity then ? It seems to me that if for practical purposes I had to use one of these solutions and wanted to avoid protons being exchanged, I would not consider these two solutions equally.

Please use the \ce{...} environment for chemical formulae. Type $\ce{SO4^2-}$ for $\ce{SO4^2-}$ or $\ce{CH3COO-}$ for $\ce{CH3COO-}$. It not only makes typing much easier but also uses an upright font. Check out the [help] and this meta-post for more information.

$\mathrm{pH}$ is a measure of acidity of a solution, not of a substance. One of the measures for the latter would be the substance's $\mathrm{pK_a}$. If you look at the formula for the $\mathrm{pK_a}$, you can see that a type of normalization is going on; you divide the amount of protons and deprotonated acid in a solution of the acid by the amount of still protonated acid. If you really want to compare substance acidities using $\mathrm{pH}$, you also need to know the concentration of the acid, which ultimately allows one to calculate the $\mathrm{pK_a}$, even if only implicitly.

I was considering pH as a measure of acidity of a solution in my question. What puzzles me is that two solutions having the same concentration of  $\ce{H3O+}$, but one having on top of that some $\ce{CH3COOH}$ in it are considered equally acid, as far as pH is concerned.

$\mathrm{pH}$ is a measure of acidity but defining acidity as the concentration of dissoluted protons in solution. This is quite a bit different from defining acidity as the amount of acid in solution.
Taking your example, the appropriate concentration of $\ce{HCl}$ would be something like $1~\mathrm{mM}$ while the corresponding acetic acid concentration is maybe $1~\mathrm{M}$. (These numbers are most certainly not correct, and maybe even the dimension is off by one or two orders of magnitute, but they can still serve well as an example.) So you could say ‘but there’s one thousand times as many acid molecules in the acetic acid solution!’
When considering the actual effects of an acid, the concentration of the original acidic species is irrelevant. Vinegar is, for example, usually $5~\%$ acetic acid and can be added to food (tough guys might even drink it) — I wouldn’t try that with $5~\%~\ce{HCl}$ since that contains a much higher active proton concentration.
What you can do, though, is consider the higher concentration of weak acids a type of buffer: Dilute acetic acid and the $\mathrm{pH}$ will hardly change while diluting $\ce{HCl}$ will have an immediate effect on $\mathrm{pH}$. So a weak acid can supply the same weak source of protons for a longer time.
And even though real-world analogies should be avoided in any case since they often give a wrong impression: If you jump down the staircase of a high building two steps at a time there will be likely no effect (small effect but buffered — acetic acid) just like if you only jump the two steps once (small effect unbuffered — strongly diluted $\ce{HCl}$), but if you try to jump the whole staircase (large effect unbuffered — $\ce{HCl}$ in the same concentration as acetic acid) you won’t remain unaffected.

Is it correct to say that "acidity is defined as the concentration of dissoluted protons in solution" because we can neglect the presence of the other, non dissoluted protons still attached to the acid ? The argument in that case would be that if the extra $\ce{CH3COOH}$ were to matter in any reaction based on proton exchange, they would have done it already with the water they are in ?

@aidaGoG Well, it depends on what you want to do with your protons. If acidic protons are merely catalytic, it doesn’t matter at all. If you need stoichiometric protons, you need to add the appropriate concentration of the acid (e.g. 2 equivalents of $\ce{CH3COOH}$), not the appropriate concentration of free protons. And in that case you would keep the overall free proton count at a specific time more or less constant while maintaining a constant flow of new ones.


text:
Calculate Work Done for Reversible and Irreversible Adiabatic process
How to calculate it.
Work done for adiabatic process is (P1V1 - P2V2)/(γ-1) but this is also the work done for reversible process. How?
And How do we calculate for irreversible?

Related: Reversible and Irreversible adiabatic expansion

For an irreversible process, the pressure of the gas within the system is typically non-uniform spatially, and so there is no unique value that can be used in conjunction with the ideal gas law to calculate the work.  But, at the interface with the surroundings, the force per unit area exerted by the gas on the interface always exactly matches the pressure of the surroundings, which is usually referred to as $P_{ext}$.  So, if $P_{ext}$ can be controlled externally as a function of the volume, one can calculate the amount of work done by the gas on the surroundings.  In many homework problems encountered in thermodynamics, the value of $P_{ext}$ is often specified as being constant during an irreversible process.
In addition to the effects of pressure and temperature nonuniformities in a gas experiencing a rapid irreversible expansion or compression, there is also another physical mechanism acting.  This is related to the viscous behavior of a gas that is deforming.  In the case of a rapid irreversible deformation, the gas pressure depends not only on the gas volume, but also on the rate of change of volume.  The behavior is analogous to that of an ideal spring and damper in parallel (as in the front suspension of a car).  The force exerted on a spring/damper combination during a rapid stretching or compression depends both on (a) how much the combination is stretched, but also on (b) how rapidly it is stretched (because of the effect of the damper).  A very crude approximation to the behavior of an ideal gas that is experiencing a rapid (irreversible) rate of deformation is provided by the equation:  $$P=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}=P_{ext}$$ where k is a positive quantity related to the viscosity of the gas.  So the gas pressure on the piston face depends both on the amount of volume change and on the rate of volume change.  The equation indicates that, during rapid expansion (i.e., dV/dt > 0), the pressure exerted by the gas on the surroundings is less than that given by the ideal gas law, while, during rapid compression (i.e., dV/dt < 0), the pressure exerted by the gas on the surroundings is greater than that given by the ideal gas law.

Won't $\int P_{ext}dV$ give works done in irreversible expansion?

This equation applies to both irreversible and reversible expansions.  In a reversible expansion, the pressure of the gas is uniform throughout, and is not affected by the rate at which the gas is deforming (since the rate is negligible).

$\Delta E=W$ in adiabatic process.
As we know that, $\Delta E=Q+W$ and $Q=0$ in adiabatic process, so work done $W=\Delta E$.


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Difference between thermodynamic terms heat and enthalphy
What is the difference between heat and enthalphy?

What’s the difference between apples and grapes?

Heat is defined as the energy that flow into or out of a system because of difference in temperature between the system and its surrounding. Give special attention to the word 'flow'. It is not the energy that is stored in the body in the form of temperature. It is the energy that is transferred due to temperature difference.
Now, Enthalpy is the energy stored in the body at constant pressure. And here the energy is stored.

To elaborate on this, H is determined exclusively by the thermodynamic state of the system at a certain temperature and pressure (i.e., it is a physical property of the system), while the heat that flows into the system between state A and state B is determined by the process path.

@ChesterMiller Thanks, I forgot about it while answering.


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Relation between enthalpy and Cv
MY ATTEMPT
We already know that $\Delta U=nC_V\,\Delta T$
which is for constant volume
Now we know that $\Delta H = \Delta U + p\,\Delta V$
so $\Delta H = nC_V\,\Delta T + 0$ (as at constant volume $\Delta V = 0$)
$$\Delta H=nC_V\,\Delta T$$
But I have never such an equation anywhere.
Please verify.

For an ideal gas, your equation for $\Delta U$ applies not only to constant volume processes, but to all processes.  It's just that, in experiments to measure Cv, it is very convenient to use a constant volume path because, under those circumstances, no work is done, in which case $\Delta U$ is just equal to the amount of heat added.  So this gives you a direct measurement of Cv.
As far as $\Delta H$ is concerned, the correct equation is $\Delta H=\Delta U+\Delta (PV)$.  But, from the ideal gas law, $\Delta (PV)=nR\Delta T$.  So what does that give you for $\Delta H$?


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How to identify the anode and cathode in a concentration cell?
How do I understand which is the anode and which is the cathode in this problem?

We have a concentration cell with hydrogen electrodes. It's made up of two half-cells. One half cell contains [BaCl(aq)]=0.5 M, the other one has [BeOH(aq)]=0.5 M. Draw the cell diagram and calculate e.m.f at 25ºC

So I tried this:
$$\ce{Pt}(\ce{H2} \, 1~\mathrm{atm})| \ce{H+,\,Ba^{+2}} \,(0.5 M)||\,\ce{Be^{+2}}\, (0.5 M)|\ce{H2}(g,1~\mathrm{atm})|\ce{Pt}$$
I don't know how I should think to write the correct diagram. Can you please help me?

You should first assume one is the cathode and the other is the anode. Then you can calculate the theoretical emf - if the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode. If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode.

If I apply Nernst' eq I'd get the potential of each half cell, right? How do I do that? For instance, for the first half-cell is it correct to write $E_{(1)}= E^ 0 - \dfrac{0.0591}{n}, \log{\left( \dfrac{[0.5M]}{[1,atm]}\right)}$, with n=2 because Barium has +2e-?

This is a personal preference but I prefer to write out the full cell equation first, then apply the Nernst equation to the whole equation - so you have $E = E^\circ - (RT/nF)\ln Q$ where $Q$ is the familiar reaction quotient. 1 atm is the standard pressure for gases, so you can leave out those terms with 1 atm. The only terms that contribute to ln Q are the concentrations. If you want, you can also do the same thing with the individual half-cells and although I'm not 100% sure myself (because I don't do it this way ), your equation looks correct.

Ok, but in the exercises so far I've used $\log_{10}$ instead of $\ln{(\cdot)}$. Is n=2 in this scenario? Also, if Q= 0.5/0.5=1 then $\log1=0$, correct? I feel there's something missing here. Should I be considering which half-cell is more acidic (more $\ce{H3O+}$ ions in it) in order to find out who's the anode?

The ln and log10 forms are equivalent. No, Q probably isn't 0.5/0.5. You can't be reducing $\ce{Ba^2+}$ and $\ce{Be^2+}$ simultaneously - something has to be oxidised - it's either going to be $\ce{H2O}$ in the $\ce{BaCl2}$ solution, or $\ce{OH-}$ in the $\ce{Be(OH)2}$ solution, depending on which electrode is the cathode - either way, the concentration of the oxidised species is not going to be 0.5. There is no quick and easy way to see which is the anode and which is the cathode. You have to work out the theoretical emf for one way and if it doesn't work, you do the other way.

@Karsten Theis, This question is interesting because I was thinking of writing a J.Chem.Ed article on this topic for several years from a historical perspective. The first question is how do we know, if we are provided with a cell and asked to identify which electrode is the cathode and the anode from a physics perspective without requiring Nernst's law. For that we need to go back to the 18th century physics. If you are interested we can discuss this by email.

Let's restate the problem: we have two hydrogen half-cells with other stuff in them. Hydrogen electrodes will have voltages related to their pH, because that is the definition of a hydrogen electrode. Then it goes haywire: does one half-cell contain BaCl2 and the other contain Be(OH)2 at 0.5 M - certainly not BaCl and BeOH. In fact, Be(OH)2 is amphoteric and soluble at only 2 ppm, so even estimating the pH of that cell would be a blind guess; you could guess the BaCl2 solution to be about 7.0, but might be a tiny bit different.
I suspect the question is not one that has actually been created and measured, but is rather a made-up question with no dependence on reality, to ask how to visualize  two hydrogen electrodes in pH 7 (the BaCl2 solution) and pH 13.69 (pH of 0.5 M OH), which could not occur because of the low solubility and amphotericity of Be(OH)2. So the whole cell could be written:
Pt(H2,1 atm)|H+(aq,10e-7 M)||(H+(aq,10e-13.69 M)|Pt(H2,1 atm)
or, just consider both half cells to be at pH 7 and there's no overall cell voltage (my preference).

Whoever originally designed the question has no idea of real electrochemical cells or even basic chemistry because not only the formulae are wrong in the original question but Be(OH)2 has limited solubility as you pointed out. Since the title reads concentration cells, both cells $must$ have the same ions. The question is messed up by typos. However the idea is nice, how to identify anode and cathode $experimentally$?


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why can't I get my absorbance values to come out right on my oxybenzone uv-vis assay test?
I am attempting to assay a raw material sample, oxybenzone, against oxybenzone, USP using UV-vis. The requirement for acceptance is NLT 99% mg C14H12O3/mg raw material sample. The method instructs a standard preparation of 10ug/mL in CH3OH, and then a sample preparation by 100 mg sample oxybenzone in 100 mL, pipette one mL into another 100 mL flask, dilute to volume with CH3OH. I used these same steps to prepare my standard solution as well. Scan and use A value to calculate mg C14H12O3 in the sample using 10(C)(At/As) with At being A of the sample and As being A of the standard, C being concentration of the standard solution in ug/mL
assuming I know my sample should pass, my results make no sense.
As=.8341
At=.6304
C= 10
I used methanol as the blank per the method.
I am sure I must have made a mistake somewhere, but I am stuck. The only thing I can think of is if I made a mistake preparing the standard solution and it was too concentrated perhaps A would go up, but I don't think I did. I weighed 100 mg oxybenzone USP standard, diluted to volume, removed 1 mL, diluted to 100 mL again, same as I did for the sample solution prep.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

The method, and your result, both make sense to me: it indicates that the test sample is 75.57% pure, although as you point out the result is not consistent with the assumption that the sample should pass the requirement for acceptance. However, why would you run the test if you knew ahead of time that the material would pass?
I would recommend that you repeat your test 2 more times from scratch - that way you would know if there was an error in your technique. If you get the same result (within experimental error), then you would have to admit that it is hard to argue with the data.


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Ionic reaction between lithium and chlorine
I got a weird question in my chemistry book. It says Li and Cl react to form LiCl an ionic compound. Cl has .... electrons in its outermost occupied shell and the answer of the book is 6 electrons . How?

Book is wrong. Chlorine has 7 valence electrons.

Unless of course the book meant the p-shells only and the resulting ion. Which would be weird as hell.

@Jan - it would be p "subshell" then, although I wouldn't put it past a book that says chlorine has 6 valence electrons to mix up shell and subshell...

Maybe your book is wrong. It can't be. There has to be 8 electrons in the outermost occupied shell and to complete its octet. I searched over the google about it and found three links to cover my answer.

https://in.answers.yahoo.com/question/index?qid=20111018173152AAn9DHz
https://quizlet.com/38997670/chem-1305-exam-2-study-guide-flash-cards/ (question number 3)
http://slidegur.com/doc/1661637/ionic-bonds (slide number 18)

Chloride has 8 valence electrons, chlorine has 7. Btw, I don't think it's really necessary to cite sources for this kind of thing... but if you really want to cite sources, please try to choose more reliable sources. Even though people always like to say "Wikipedia is unreliable", it is not a bad start.

@orthocresol that's why i chose to take three sources. How much probability is there that all three of them would be wrong. I would have commented them but due to lack of reputation points, i couldn't.

You can can comment under your own post

@Mithoron Now i can comment anywhere..lol...i have just got the 'comment everywhere' privilege

That was fast :D You can also use chat now :)


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Computing two-electron integrals with an STO-3G basis set
I am trying to implement a restricted Hartree-Fock calculation using an STO-3G basis set, for fun. I managed to perform this calculation where only $\mathrm{1s}$ orbitals are present ($\ce{H2}$ and $\ce{HeH+}$) as explained in Szabo and Ostlund's book. In this book, authors give explicit formulas for overlap, kinetic, nuclear-electron, and electron-electron integrals for $\mathrm{1s}$ orbitals and they work correctly.
In order to generalize my calculation to systems containing $\mathrm{2s}$ and $\mathrm{2p}$ orbitals (for $\ce{H2O}$ and $\ce{N2}$), I used the general formulas I found in Cook's book for the electron-nuclear and electron-electron integrals. In this case, I obtain results that are slightly different from Szabo's book:
$$E_\text{tot}(\ce{H2O}) = -74.4442002765 \text{ a.u.}$$
instead of
$$E_\text{tot}^\text{Szabo}(\ce{H2O}) = -74.963 \text{ a.u.}$$
This is obviously problematic since orbital energies suffer from the same error and this leads to an erroneous ionization potential (0.49289045 a.u. instead of 0.391 a.u., a difference of approximately 63 kcal$\cdot$mol$^{-1}$).
Since I checked my code multiple times and I wrote the two-electron computation code from scratch twice, I was wondering if there is a typo in Cook's book. Is there is a good reference where I can find the (correct) formula to compute two-electron integrals of gaussian functions (in Cartesian coordinates) with arbitrary angular momenta? At the moment I am not looking for a very efficient (recursive) algorithm to perform this task, I only need an exact formula like the one proposed in Cook's book.
Sources:
[1] Szabo and Ostlund, Modern Quantum Chemistry, Dover, 1989.
[2] Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.

Can you put your code on github?  Difficult to answer the question without seeing your formulae.  Your result could be due to a floating point error ... I've found this when reproducing the computations of other scientists.

@user1945827 I was thinking about the same thing, but many calculations (see Geoff's answer) gives a value of -74.96(...). My code is already on GitHub (https://github.com/RMeli/Hartree-Fock) but I didn't gave the link because I didn't wanted too much help! ;) The routine computing the integrals is the ELECTRONIC in INTEGRALS.PY (the program is in Python because I can code fast, but I will migrate it to fortran when it works fine).

consider https://scholar.google.ru/scholar?q=computing+gaussian+function+integrals&btnG=&hl=en&as_sdt=0%2C5

@permeakra I spent much time using Google and Google Scholar. However, Google Scholar results are normally about very efficient algorithms to compute two-electron integrals. What I would like to implement at first is an exact formula (like the one proposed in Cook's book or in Szabo's book for 1s orbitals).

@R.M. This article is on the first page of the google scholar link from above.... I dunno, seems suspiciously like what your want. http://rspa.royalsocietypublishing.org/content/royprsa/258/1294/412.full.pdf

Actually there is a mistake in the analtical expression in Cook's Book. On his web page he has a pdf with the corrected verison
http://spider.shef.ac.uk/
Maybe this solves your problem, but I would also recommend to implement the Obara-Saika Scheme or rys-Quadrature since they are really much more efficient. If your are programming in Python, you might have a look at the PyQuante project, which implements all this stuff.
Concerning Obara-Saika you might also read about the Head-Gordon Pople Scheme. It is in principial an adapted version of Obara-Saika which reduces the FLOP count.

This DO solve my problem!!! Thank you!!! I will certainly look to Obara-Saika or other efficient schemes, but for the time being the goal is to implement BO-MD and not to have an efficient code! ; )

My advice is to implement the Obara-Saika recurrence formulae that are outlined in "Molecular Electronic-Structure Theory" by Helgakar, et al. I would stick with Cartesian functions, since a) they are easier and b) spherical harmonics don't matter for molecules anyway.
I did this years ago (in Mathematica) when I was in a similar place -- having completed Szabo and Ostlund and wanting an arbitrary "angular momentum" code.
The book will cost you a small fortune, but if you can borrow it through Interlibrary Loan, it is a phenomenal text.

I will certainly look at this book, thank you!

Unfortunately your suggestions on Boys function does not solve this problem (as I hoped). The error should be directly in the computation of two-electron integrals. I will look to Obara-Saija recurrence.

Check normalizations of you basis functions

Why you suspect about normalization?

Did Sij evaluate to 1?

Yes, Sii = 1 (I print 5 decimal digits)

My suggestion would be to use another existing code and run the calculation.
For example, if I do an HF/STO-3G calculation on $\ce{H2O}$, I get:
$$E_\mathrm{H_2O}=-74.9659011\:\mathrm{a.u.}$$
I don't have Cook's book on hand, so I can't look for the error, but I'd suspect some error as you do.

It will help if I write down explicitly Cook's formulas? I already checked on internet for other HF/STO-3G results and all of them start with -74.96(...) a.u., this is why I am concerned.

@R.M. Just because other more modern techniques are efficient does not mean they are inexact. Feel free to post the formulas, but I'd adapt a different method.

@Goeff I know there are very good (and efficient) techniques, but this will be the next step in my project. At the beginning I just want to code the simplest algorithm possible, even if it is inefficient. Cook's analytical solution only imply a series of loop, which is extremely easy and straightforward to implement (in addition, this is how I implemented the other integrals). However, if I will not find an analytical solution quickly, I will certainly spend some time to implement a more difficult technique.


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Metallic Aluminum and Dry Ice
We see $\ce{2Mg + CO2 -> 2MgO + C}$ , quite often in lab experiments.
Under which conditions can $\ce{4Al + 3CO2-> 2Al2O3 + 3C}$ be used - if chemically feasible?

Based on this (very old) paper - http://link.springer.com/article/10.1007%2FBF00740367#page-1 - the aluminum will ignite if you heat it past the melting point of aluminum oxide, 2060C.


text:
Do I need to include lattice energy when calculating the energy required to vaporize NaCl?
I understand that the lattice energy is the energy released when a crystal forms and that's the reason it takes the negative sign. But do we need to add this energy again when we are trying to vaporize an ionic compound ?
For example, let's assume we want to completely vaporize a mole of NaCl starting at room temperature, how much energy do we need ?
I don't know neither the specific heat nor the latent heat of vaporization of NaCl, but I understand we would need:
1- Energy to raise the temperature of mole of NaCl from 298 °K to its boiling point.
2- Energy equal to the latent heat of vaporization of a mole of NaCl.
I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.
Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol. So my question here is: do we need to add this lattice energy to the calculated steps above to account for all the energy required to vaporize a mole of NaCl ?

Well you left out steps...(1) Energy to raise NaCl to melting point (2) Energy to melt solid NaCl to liquid NaCl at its melting point  (3) Energy to raise liquid NaCl from melting point to boiling point (4) Energy to vaporize NaCl at its boiling point. // So the lattice energy is (2)

@MaxW, I wasn't sure if NaCl would turn to liquid first. But anyway, is't (2) the latent heat of fusion ?
Also, once an ionic compound is melted, this means that there are no more lattices to break, is that correct ?

Sorry to have mislead you. The lattice energy is defined as the energy liberated when gaseous ions are united to form the solid lattice. This would be at a constant temperature. So the whole heating it to some other temperature doesn't apply.

I'm sure that measuring lattice energy from ions is impractical so that the Born-Haber cycle is used.  https://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle // I upvoted you question in hope that someone who has studied thermodynamics more recently than I can answer your question. I'm a bit confused about the temperature dependence of the lattice energy.

@MaxW, its also important to find out what we will end up with once we vaporize the NaCl mole. Will we get Na+ and Cl-  or just molecules of NaCl in gas phase ?

For lattice energy it is from the gaseous ions to the solid.

No I mean if I heat up NaCl to boiling point and so it starts to vaporize, what will the vapor be ? gaseous ions or neutral molecules ?

I would guess that most wold be diatomic NaCl not the ions.

I think you are correct. I found a similar question here: http://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions

Lattice energy

Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol.

First, we have to understand the term lattice energy.
Here is the textbook explanation (Fleming: Physical Chemistry): The lattice energy is the energy required to separate the ions in an ionic lattice so that they are at infinite distance (but still ions). This would be difficult to do experimentally, but the value may be determined by using a Born Haber cycle. As the diagram below shows, we know the enthalpy of reaction for:
$$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$$
We can get the same energy in a thought experiment (path 2 in the diagram), first turning all species into gas, then into atoms, then transferring the electron and - finally - forming the lattice from the individual anions and cations. All these processes are experimentally accessible except for the last, so you can determine the lattice energy this way.

Source: diagram adapted from the cited textbook.
Enthalpy of vaporization

But do we need to add this energy again when we are trying to vaporize an ionic compound?

You problem is not well-defined. What are the initial and final temperatures, what is the pressure? What does it mean to vaporize solid NaCl or any ionic compound, i.e. what is the product? If we boil sodium chloride at 1 atm pressure, it produces mostly NaCl monomers in the gas phase, see https://chemistry.stackexchange.com/a/14560. If you are interested in this process, you could write it as:
$$\ce{NaCl(s) -> NaCl(g)}$$
So to get the required energy, you could first separate all ions (lattice energy) and then calculate the energy of forming individual pairs of sodium and chloride ions (assuming the interaction is purely ionic).
$$\ce{NaCl(s) -> Na+(g) + Cl-(g)}\tag{1}$$
$$\ce{Na+(g) + Cl-(g) -> NaCl(g)}\tag{2}$$
It should be obvious that process (1) cost much more energy than what you get out of process (2) because you are going from 6 nearest neighbors to just one. If you look up the atomic distance of $NaCl(g)$, you could estimate the enthalpy of (2) pretty accurately and get a numeric answer.

[Comment from https://chemistry.stackexchange.com/a/14177]: Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid NaCl crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates. – Nicolau Saker Neto Jul 8 '14 at 16:39

According to the comment above, the heat of vaporization should be about half of the lattice energy (with opposite sign).

I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Because enthalpy is a state function, it does not matter if you choose the actual (realistic) path or take a different path. You are right that if you step through what actually happens, you need to know heats of fusion and vaporization as well as heat capacities of all phases for the relevant temperature ranges.

The energy to vapourize one mole of NaCl is the negative of its lattice energy, assuming you are starting at 25 degrees C.
Can be calculated via born habor cycle-
The Born-Haber Cycle can be reduced to a single equation:
Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities

But the lattice energy is a huge number compared to the latent heat of vaporization. For example, Al2O3 has a lattice energy of -15600 kJ/mole, so if you are correct, then it would take 15.6 MJ of energy to vaporize 102 grams of Al2O3. When comparing this to iron, you will notice that it takes about 450 kJ/mole to or about 820 kJ to vaporize 102 grams of iron. Can you see the difference ?

You are also assuming that once NaCl is vaporized, the gas will be made of its gaseous ions and not NaCl molecules. Could you please explain why this will happen ?


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Why does the phosphate ion have resonance structures?
My first-year chemistry teacher depicted $ \mathrm{PO_{4}^{3-}}$ as exhibiting the resonance structure inset below this question.
However, I can't figure out what diagramming rules (which is what she was teaching) she used to deduce her answer. I thought that the double bonded oxygen would be single bonded and that it would have six unbonded electrons like the other oxygens.

Why should one oxygen be different than any of the others?
What diagramming rules imply that  $ \mathrm{PO_{4}^{3-}}$ should be depicted as my teacher did?

Actually your teacher omitted two paths. There should be connections between the diagonals of the square shown. This forms a "complete graph" in mathematics.

There is also at least one resonance structure missing, where the charges are separated and the molecule formally obeys the octet rule. Resonance structures with an extended octet are also not really accurate.

@MaxW Are most resonance structures complete graphs?

A much better representation of phosphate is the one given below (You need to add three lone pairs to each oxygen atom which I was too lazy to do):

It does not expand phosphorous’ octet, which calculations have shown is unlikely.
It does not require phosphorous’ 3d-orbitals which are energetically too far removed to participate in hybridisation in a noteworthy manner.
It gives you four equivalent $\ce{P-O}$ bonds a priori without needing to resort to mesomery.
It gives all oxygen atoms the same charge a priori and also gives phosphorous a charge in better agreement with experiment/calculation.
The experimental bond lengths are better explained as single bonds rather than having a bond order of $\frac{5}{4}$.

This can all be summed up as ‘phosphorous obeys the octet rule much like every other main group atom (doublet rule for hydrogen/helium).’
But you probably want answers to your questions on the decidedly minor mesomeric representations your teacher drew. Well:

Why should one oxygen be different than any of the others?

Those four mesomeric structures are entirely equal, just like in benzene. To get a ‘picture closer to the truth’ you would need to put them all on top of each other and look through them. You would see that every $\ce{P-O}$ bond has the same (minute) chance of being a double bond and thus no oxygen is different from the others.

What diagramming rules imply that $\ce{PO4^3-}$ should be depicted as my teacher did?

Well: none, really. If you are going to stick with the double bond representation, the order of the four structures does not matter as long as they are all there. Also, you can go from every single one to every single other one in a single step by appropriate electron shifts. So, as MaxW commented, you really should add the two missing diagonal arrows to be correct.

What are the calculations which you talk about? I haven't seen many calculations showing octet rule 'more important' than hypervalency-formal charge...; In any case, +1...

Why should one oxygen be different than any of the others?

A very important thing to understand when representing a molecule with its resonance structures are that they do not exist, at least not individually. What the Lewis depiction of a molecule does a semi-poor job of showing is electron delocalization. This is why we have to represent a molecule that shows delocalization with its resonance structures. It is not the case that one oxygen is double bonded while the rest of them are single bonded, in fact all the $\ce{P-O}$ bonds are identical in length, and each have the strength of $\frac{5}{4}$ of a $\ce{P-O}$ single bond.

What diagramming rules imply that $\ce{PO_4^{-3}}$ should be depicted as my teacher did?

Again, this is a Lewis structure. The general rules for this should be in the link I provided. As for what you need to know about phosphorus, or any element below the second period really, is that it has the ability to form more than the usual 8 bonds through hybridization of their $\mathrm{p}$ and $\mathrm{d}$-orbitals.
The most basic thing you need to know for this problem is this: a neutral phosphorus atom has 5 valence electrons. If it forms a single bond with each of the four oxygens, it will be left with one unbonded electron. To explain this in the Lewis structure, it is drawn as having a double bond with one of the oxygens.


text:
Why is the cyanide ion toxic?
As the title implies, what is the molecular basis of cyanide toxicity? I did some searching around at the CDC and it only states that it prevents cells from using oxygen. I also read how it could take as little as $131~\mathrm{ppm}$ to kill. How can a simple $\ce{CN-}$ ion do so much damage?

I think this question is much, much, much more interesting from a chemical point of view. I think it is well on topic here and decide not to close it. I hope someone will find the time to write up a better explanation than what is currently found on [bio.se].

Cyanide is a pretty good ligand for coordination compounds. The electron pair on carbon (which, incidentally, also carries the Lewis structure’s formal charge) is located in the HOMO — much like as in $\ce{CO}$, whose molecular orbitals can be found in this answer by Martin (replace oxygen with nitrogen to arrive at $\ce{CN-}$) — making it a good Lewis base and a good σ donor.
It is remarkable for being a strong-field ligand, i.e. one that often creates low-spin complexes even for 3d-metals (which generally prefer high-spin if there are no competing reasons for low-spin). While one could draw analogies to $\ce{CO}$ or even $\ce{NO+}$ whose π-antibonding orbitals are strongly Lewis acidic and participate in metal to ligand backbonding, it turns out that for cyanide the main reason is a high covalency of the metal–ligand-σ-bond due to the relevant orbitals being very similar in energy. Thus, cyanide is unique in that it both stabilises higher oxidation states by being an anion and low-spin complexes due to its binding modalities.
One might think, in analogy to $\ce{K4[Fe(CN)6]}$, that cyanide preferentially binds to haemoglobin’s iron(II) centre and thereby inhibits oxygen transport, but that is in fact not the case as this centre requires a less-oxidised iron(II). Oxygen (due to the resulting redox-mechanism) and carbon monoxide are much better ligands for haemoglobin.
What cyanide does strongly inhibit is cytochrome c oxidase, the last enzyme in the respiratory cycle which collects electrons from four cytochrome c’s and transfers them to an oxygen molecule to create two water molecules:
$$\ce{4 Fe^2+-cytochrome~$c$ + 4 H+ + O2 -> 4 Fe^3+-cytochrome~$c$ + 2 H2O}$$
This is done by a plethora of metal centres consisting of two haems, to cytochromes and two copper centres. I can’t tell you exactly which one, but there will be at least one which needs accessable iron(III) at some point in the catalytic cycle to which $\ce{CN-}$ can bind well and strongly inhibiting the catalysis of the enzyme.
Since the inhibited process is directly responsible for cellular respiration by transferring electrons to molecular oxygen (and being the only enzyme dealing with molecular oxygen in the respiratory chain), inhibition of this protein will cause the breakdown of the entire respiratory chain and lead to cellular suffocation.
Considering that an amount of $200~\mathrm{mg}$ can be toxic according to Wikipedia, and also considering that a human being contains some 10 trillion cells also according to Wikipedia, that’s still some 400 million cyanide ions per cell demonstrating the efficiency of binding and inhibition.
Note that popular antidotes also directly address the fact that cyanide is a strong ligand:

One antidote, hydroxycobalamin, presents the cyanide ion with a $\ce{Co^{III}}$ centre and a weakly bound, easily displaced hydroxido ligand. Hydroxide is displaced by cyanide creating cyanidocobalamin, a variant of vitamin B12 that the body has no problems dealing with. It is ejected via the kidneys. Injected into the circulatory system, it scavenges cyanides before they can enter cells.
Another antidote, also injected into the circulatory system, is 4-dimethylaminophenol (not to be abbreviated as DMAP to prevent confusion with 4-dimethylaminopyridine). It is an oxidising agent designed to oxidise haemoglobin’s iron(II) to methaemoglobin (containing iron(III)) thereby creating a much better acceptor for cyanide. Since there is a lot of iron(II) (a quick estimate with $8~\mathrm{mmol/l}$ gives $40~\mathrm{mmol}$ haemoglobin in human blood) in the blood for oxygen transport, small amounts of this can be sacrificed to capture cyanide before that reaches the respiratory cycle.

Cyanide is made up of a nitrogen triple bonded to a carbon with a lone pair, [C≡N]−. (Carbon has 4 valence electrons and Nitrogen has 5; pairing will result in a triple bond, a pair of electrons around the nitrogen, and an extra electron pairs with the 4th electron around the carbon). Carbon does not like to have a 5th electron, it is unstable, even the electron charge is shared between the Carbon and Nitrogen. The resulting negative charge (and triple bond) means that cyanide will be very reactive and want to form another bond.
As a result, cyanide is particularly deadly because it will bond with iron, (usually Fe3+ or Fe2+). In our body, iron is found in protein in the electron transport chain. When cyanide inhibits the electron transport chain by binding to iron, the cell can no longer produce energy.
Cyanide is also deadly when it bonds with hydrogen to create hydrogen cyanide, or HCN. HCN can be inhaled, where it will bond with iron, and result in cell death.

"The resulting negative charge (and triple bond) means that cyanide will be very reactive and want to form another bond." Not really - octet rule is satisfied in CN-.

As ortho said. Even more: cyanide is actually pretty stable. You can store it on the shelf for years.