text:
Difference between microcanonical and canonical ensemble
I feel a little unclarity about this. Canonical ensemble is pictured as many systems in heat reservoir of infinite capacity having $N$ (number of particles), $V$ (volume) and $T$ (temperature) constant whereas microcanonical ensemble is the analogous system having $E$ (energy) instead of $T$ fixed. From my simple viewpoint of introductory pchem, it seems obvious to me that heat capacity is constant hence how is it possible for canonical ensemble to have different energies while having same temperature. Feeling temporarily frustrated I grab McQuarrie's pchem it reads on the topic:

"Each system has the same values of N, V, and T but is likely to be in
  a different quantum state, consistent with the values of N and V"

A light shines into my lower mood state.
So we talk about very little changes where the concept of heat capacity must be taken with reserve? That is, it is possible that system sucks heat from this infinite heat reservoir which is used to promote a particle in the system to higher rotational state for instance, which causes the system to acquire a tiny bit of energy while temperature will not change at all? Temperature is understood as describing translational motion of particles, right? Is my idea of the difference between these to ensembles correct?

In a microcanonical ensemble the energy (sum of kinetic and potential E) is conserved. In a canonical ensemble the temperature (dependent on kinetic E) is conserved via coupling to an infinite heat bath. What exactly is the problem you're having?

Rotational and vibrational states are understood as form of potential energy?

Typically (at least how I learned this after three years of chemistry) the canonical NVT ensemble is created by considering many microcanonical NVE "ensembles" as microstates in this NVT ensemble. I quote from my lecture script for statistical thermodynamics:

An ensemble with a constant number $N$ of particles in a constant volume $V$ and at thermal equilibrium with a heat bath at constant temperature $T$ can be considered as an ensemble of microcanonical subensembles with different energies $\epsilon_i$.
  The energy dependence of [the] probability density conforms to the Boltzmann distribution. Such an ensemble is called a canonical ensemble.

So your NVT ensemble is many NVE ensembles at different energies. Their statistical weights (the probability of finding a microstate in that particular NVE state) are Boltzmann distributed. The way this distribution of NVE subensembles looks like then defines the temperature of the overarching NVT ensemble.
To illustrate this I have written a small Python program that calculates this distribution for different temperatures of the NVT ensemble. The probability density as a function of the energy of the 48 possible (this number was arbitrarily defined, in principle it could be infinite) NVE subensembles is given below:

As you can see, the probability of finding NVE microstates with low energies drops drastically as the temperature increases. You can think of this as the thermostat taking a NVE microstate at higher energy, removing some of that energy, and then putting it in a lower energy state. Or the other way around if you want to "heat up" a system.

I'm not sure whether I adequately addressed the problem you have. Please indicate whether this cleared up your confusion. If not, maybe try to clarify in the question post.

Concrete example: Suppose you have microcanonical ensemble where systems are composed of many vials of water. How do you manage to change their energy while keeping the temperature constant i.e. how do you increase only potential energy of this system?

I'll get back to this after I had a good night's sleep.

I meant canonical ensemble (sorry).

@wuschi I amended the answer. Did it become clearer? You don't change the energy of the subensembles/microstates. You change the way they are distributed in your ensemble.

I don't like the notion of a "NVE ensemble microstate" since it adds additional complexity for no reason. A microstate is usually defined as a single point in phase-space and naturally has a well-defined N, V and E. Since all microstates in an NVE ensemble have equal probability, one could exchange almost everywhere in your answer "NVE ensemble" by "microstate".

One should distinguish finite systems and systems in the thermodynamic limit. In the latter limit there are no fluctuations and all ensembles become equivalent. In this case, one can calculate e.g. the energy $E$ of the system from $N$, $V$ and $T$. This is the realm of classical thermodynamics.
For finite systems however, there will be fluctuations of energy when you regard a system with constant $N$ and $V$ which is in contact with a heat bath of temperature $T$. Note that in this case you cannot calculate the energy from knowledge of these three quantities. And vice versa, you cannot deduce the temperature from knowledge of a particular microstate of the canonical ensemble. So one should regard temperature not as a property of the individual microstates comprising this ensemble, but rather as a property of the heat bath or of the ensemble as a whole.
Regarding the heat capacity: I'm no expert in this and I guess that there exist definitions for it in the different ensembles. But to be on the safe side, I would recommend you to apply the concept only for macroscopic systems.
Regarding the speculation of OP: there should not be made a distinction between rotational and translational motion, both kinds are relevant for temperature. Furthermore, the heat bath will always excite all kinds of motion in your system. As I said, one should simply abandon the idea that two different microstates from a canonical ensemble with temperature $T$ both "have" this temperature individually.


text:
why are halogens considered deactivating groups while methoxyl groups are activating?
Something's wrong with how I think of this:
methoxyl has the partial negative charge on oxygen which means it's activating. However, halogens are more electronegative yet they're considered deactivating. Why?
On my lecture notes, I have one section that says
"if Q is an electron withdrawing group then attack on the ring is slowed because this leads to additional positive charge on the ring".
While another section says "all meta directing groups have either a partial or full positive charge on the atom directly attached to the aromatic ring."

Methoxy not methoxyl

methoxyl has the partial negative charge on oxygen which means it's activating. Where did you get this idea from? Having a partial negative charge on oxygen has nothing to do with methoxy groups being activating.

Halogens are considered deactivating groups due to the inductive effects of their overwhelming electronegativity withdrawing electrons away from the ring, thus slowing the overall reaction. However, halogens are also able to donate one of their lone pars to the ring through resonance which is characteristic of activating groups. Thus the observance of ortho and para directing. Ultimately, halogens inductive effects will trump resonance due to their electronegativity.
I forgot to add, that while oxygen can donate a lone pair through resonance it's electronegativity is not of the same magnitude as that of a halogen. So the alkoxy group would lead to a more electron rich aromatic system, thus it is a strong activator.

Of the halogens, only fluorine is more electronegative than oxygen.

Because fluorine has such a high electronegativity, its inductive withdrawing effect is much larger than any resonance electron-donating effect.
The other halogens have poor orbital overlap with the pi system of an aromatic ring, which precludes any appreciable resonance electron-donation (this is why you almost never see a C-X double bond). Therefore, their inductive capabilities lead to a net deactivation of the aromatic ring.
This is why an alkoxy group is more activating  than a halogen. Oxygen does have a higher electronegativity than does chlorine, bromine or iodine. However, oxygen's orbital overlap with carbon allows it to more readily donate into the ring's pi system. The larger halogens aren't able to do this (to an appreciable extent).

All the good points been drawn, the electron withdrawing ability of oxygen is reduced by the I+ effect of the alky group of the alkoxide, thus all the more favouring electron donation to the aromatic ring.

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Deriving an expression for dU as a function of V and T
I'm currently preparing for a physical chemistry exam tomorrow, and was given the following question for practice:

Derive an expression for $\mathrm{d}U$ as a function of $\mathrm{d}V$ and $\mathrm{d}T$.

I'm confused as to whether they want me to derive the equation below:
$$ \mathrm{d}U=\left(\dfrac{\partial U}{\partial V}\right)_T \mathrm{d}V + \left(\dfrac{\partial U}{\partial T}\right)_V \mathrm{d}T $$
or this
$$ U = q + w $$
given an infinitesimal change
$$ \mathrm{d}U = \mathrm{d}q + \mathrm{d}w $$
where
$$ \mathrm{d}q = C\mathrm{d}T $$
and
$$ \mathrm{d}w = -p\mathrm{d}V $$
hence
$$ \mathrm{d}U = C\mathrm{d}T + -p\mathrm{d}V $$
Essentially, I need help interpreting the direction I should proceed, and advice regarding the correctness of my methods.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved.

I realize now what you're looking for:
Consider $\mathrm{d}S(V,T)$ as
$$\mathrm{d}S(V,T) = \left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T$$ and place into the first law:
$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V = T\left[\left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V +  \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T\right] - p\mathrm{d}V
$$
Combine like terms of $\mathrm{d}V$:
$$
\mathrm{d}U = T \left(\frac{\partial S}{\partial T}\right)_{V} \mathrm{d}T + \left[T \left(\frac{\partial S}{\partial V}\right)_{T} - p\right]\mathrm{d}V
$$
and you have $\mathrm{d}U$ a function of $V$ and $T$.
It turns out that $\displaystyle T \left(\frac{\partial S}{\partial T}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{V} = \mathrm{C}_V$ and $\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T} = \left(\frac{\partial p}{\partial T}\right)_{V}$ by a second-derivative Maxwell relations of $A$ (below) with respect to $T$ and $V$, and a "prettier" version is:
$$
\mathrm{d}U = \mathrm{C}_V\mathrm{d}T + \left[T \left(\frac{\partial p}{\partial T}\right)_{V} - p\right]\mathrm{d}V
$$
(In the special case of an ideal gas, $pV = nRT$ and the term in brackets is zero - just write $p = nRT/V$ and evaluate.)

To derive $\mathrm{d}A$:
What you're looking for is a Legendre transform. You know
$$
\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V
$$
from the First Law of thermodynamics.
Hence the natural variables for $U$ are $S$ and $V$, not $T$ and $V$ (which is desired). So instead of $U(S,V)$ you want $?(T,V)$ where $?$ happens to be the Helmholtz free energy $A$, sometimes denoted $F$ by physicists.
To do the transform, you take your original function - the exchange variable times its differential form, i.e.
$$ A(T,V) = U(S,V) - S\left(\frac{\partial U}{\partial S}\right)_{V}$$
You know also that $$\left(\frac{\partial U}{\partial S}\right)_{V} = T$$ because
$$\mathrm{d}U(S,V) = \left(\frac{\partial U}{\partial S}\right)_{V}\mathrm{d}S +  \left(\frac{\partial U}{\partial V}\right)_{S}\mathrm{d}V = T\mathrm{d}S - p\mathrm{d}V$$
by total derivative formula for a function, in this case $U$.
So you have $A = U - ST$ or as commonly written, $A = U - TS$ so the differential form of $A$ is $\mathrm{d}A = \mathrm{d}U - T\mathrm{d}S - S\mathrm{d}T = T\mathrm{d}S - p\mathrm{d}V -T\mathrm{d}S - S\mathrm{d}T$, thus $\mathrm{d}A = -p\mathrm{d}V - S\mathrm{d}T.$ That is a function of $V$ and $T$.

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text:
What kind of bonds can form between benzene and transition metals in π-complexes?
I'm trying to draw the six π-orbitals (3 bonding, which are filled, and 3 anti-bonding) of a benzene ring when it complexes with metal d-orbitals and which metal orbitals have the correct symmetry ($\sigma$, π, $\delta$) to overlap with each of them. I know that metal orbitals will only form a bonding interaction with ligand orbitals if they have matching symmetry but I am not sure which ones will interact.
I think a π bond is formed between $e_{1g}$ of benzene and $d_{xy}$ and $d_{yz}$. Also a $\sigma$ bond is formed between $a_{2u}$ of the benzene and $d_{z^2}$ and $a_{2u}$. No δ bond is formed.

In theory benzene and a transition metal could form σ, π, δ, and even φ bonds as the symmetry allows this.
Here are the orbitals of benzene, calculated at the PBE0/def2-TZVPP level of theory in the D6h  point group with Gamess.

The (valence) orbitals of chromium on the same level of theory in the same point group are as follows:

The linear combination of these orbitals would yield orbitals that have the above mentioned bond symmetries. We will ignore for now that they will most likely not occupied in a complex like bis(benzene)chromium.

benzene a2u + Cr s → σ
benzene a2u + Cr dz² → σ
benzene e1g + Cr dxz → π
benzene e1g + Cr dyz → π
benzene e2u + Cr dxy → δ
benzene e2u + Cr dx²-y² → δ

If we look at f-orbitals (that is of course not very feasible for chromium), we could construct another bond type:

benzene a1g + Cr fx(x2−3y2) → φ

The necessary orbital looks like this:

Of course which of these bonds will form or not depends on the metal in question, the occupation of the orbitals and more. After all, there are also p orbitals which can overlap to form σ and π bonds.

Colour code: Occupied orbitals are shown in blue and orange, while virtual orbitals are shown in yellow and red. Phases have been assigned randomly. The total density of the molecule uses a violet contour.


text:
Comparing IR stretching bands in molybdenum complexes
Why would the infrared stretching frequency of $\ce{CO}$ (which is about $2140~\mathrm{cm^{-1}}$ in gaseous form) be $1982~\mathrm{cm^{-1}}$ in $\ce{[Mo(CO)_6]}$ and even lower in $\ce{[Mo(CO)3(η^6$\text{-}$Mes)]}$ ($\ce{Mes} = $1,3,5-trimethylbenzene)?
I think it may be due to the π* orbital being used in the $\ce{CO}$ ligand which reduces the bond order and lengthens the bond. The stretching frequency is proportional to the bond strength so if the bond gets weaker then the frequency is lowered.


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Significant figure rules when plugging a value into an equation or formula
I have not been able to find an answer to what may be a silly question so I am asking it on here as a last resort.
If I am to plug in say a number like 14.0 Celsius into the formula that solves for temperature in Fahrenheit.  So I would do 9/5 times 14.0 to get 25.2.  When I then add the 25.2 to 32 does the rule for addition matter here?
Would the answer be 57.2 or 57?
Is there a general rule for when I am plugging in a measured value into a formula that is given to me?  I believe that only the measured values are significant and all numbers in the formula are exact with infinite significant digits.

The conversion between unit is in principle exact, hence any uncertainty in the measured value is the same in the converted value. Related: http://chemistry.stackexchange.com/q/15862/4945

If the conversion is exact, all uncertainties carry over. Arguing the same for significant digits is very tricky and can lead you on erroneous paths. It is thus easier for you to convert the uncertainty that is present in the originally measured value into a new uncertainty for the converted value and then infer from that the number of significant digits.
Let's assume that your measurement is $\vartheta = 14.0 \pm 0.1 ~ \mathrm{^\circ C}$.
The conversion to Fahrenheit is
$$ \phi(\vartheta) = \left(\frac{9}{5} \frac{\vartheta}{\mathrm{^\circ C}} + 32 \right) ~ \mathrm{^\circ F} \,,$$
which we can straightforwardly apply to the mean of the measurement which gives us $57.2~\mathrm{^\circ F}$.
To do the error propagation under the assumption that the errors are independent, we do the following:
$$\sigma_\phi = \sqrt{\left( \frac{\partial\phi}{\partial\vartheta} \right)^2 \sigma_\vartheta^2} = \frac{\partial\phi}{\partial\vartheta} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \cdot 0.1~\mathrm{^\circ C} = 0.18~\mathrm{^\circ F}$$
The term under the root would in principle be a sum, but here we only have one variable with an uncertainty associated with it. For more details, look up this Wikipedia article on error propagation.
Anyway, our newly converted value is correctly expressed as
$$ \phi = 57.20 \pm 0.18 ~ \mathrm{^\circ F} \,.$$

This mostly answers the question. Since 14.0 Celsius has three significant figures then when converted to Fahrenheit we'd generally write three significant figures also, so 57.2 F would be the better conversion. But that obviously belies the exact situation. The 14.0 Celsius temperatures value is measured +/- 0.05 degree which is not exactly the same as +/- 0.05 F which is indicated by 57.2 F value. You'd have to make a judgement call if the error conversion is necessary too.

Point being the rule of thumb is to use same number of significant figures in the answer as the smallest number of significant figures used in the any one of values used in the calculation.

This is new to me.  How would I choose between the error conversion and not using it?  So the error being +/- 0.1 why do you put +/- o.o5 degrees MaxW?

So the error must be converted to the new units as well always? What does it mean to assume that the errors are independent?

Usually for our purposes the errors are independent. If they are not (and that can sometimes be the case) then you need to use more advanced modelling of the error than the one I gave here.


text:
Enthalpy change with temperature
Recently I stumbled across a book that describes enthalpy change with temperature, specifically for a product of combustion process (e.g. CO as a result of solid carbon combustion). The book says that enthalpy of formation at $298$ K is $-110.5~\rm\frac{kJ}{mol}$, meaning that this is the amount of energy that gets released during combustion.
But in a very next paragraph there are two opposite calculation for enthalpy at 2000K: one says it is $-53~\rm\frac{kJ}{mol}$ (less energy get released during combustion), and the other it is $+167~\rm\frac{kJ}{mol}$ (the energy must be brought into system)?
Can you please clarify which number is correct and does energy goes out or into the system?

Well, perhaps you could summarize the 'two opposite calculations' to give us a hint about them and why they might be different?

What it says is that the enthalpy at 2000 K (value of enthalpy in the book: -53 kJ/mol or +167 kJ/mol?) is the enthalpy of formation at 298 K (value = -110 kJ/mol) + change in enthalpy from 298 K to the 2000K (value = 57 kJ/mol). My guess is that authors ment this is enthalpy for a case where CO2 had to be first produced through the combustion at referent temperature 298 K (resulting in negative enthalpy of -110 kJ/mol) and then heated to 2000 K by adding 57 kJ/mol extra energy from enviroment.  Resulting -167 kJ/mol would be just an error and -53 kJ is the right value. Does it sound ok?


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Why are hydrogenation reactions exothermic?
I learned that all reactions that yield hydrogen are endothermic (such has reforming) and reactions that use up hydrogen are exothermic (FCC cracking, hydrogenation, etc.). But why is this so? I know what exothermicity/endothermicity means, but I haven't found a specific explanation for the correlation with hydrogen generation or consumption yet.

In hydrogenation entropy is lowered so heat must be generated - otherwise change in enthalpy also wouldn't be favorable and reaction wouln't happen

@Mithoron I like this line of reasoning.

Let's take a look at an example of a hydrogenation reaction, namely the hydrogenation of ethene to give ethane
$$\ce{CH2\bond{=}CH2\ (g) + H2 (g)-> CH3\bond{-}CH3 (g)}$$
And now let's determine the enthalpy change of this reaction. By Hess's Law, the enthalpy change of the reaction is essentially the same as the enthalpy change of breaking all the bonds of the reactants and forming all the bonds in the molecules of the product. We can see that

1 $\ce{H\bond{-}H}$ bond has been broken
1 $\ce{C\bond{=}C}$ bond has been broken
1 $\ce{C\bond{-}C}$ bond has been formed
2 $\ce{C\bond{-}H}$ bonds have been formed

The other bonds remain unchanged by the end of the reaction
The following is the list of bond enthalpies of the aforementioned bonds, obtained from the Internet

$\ce{H\bond{-}H}$ bond: +432 kJ/mol
$\ce{C\bond{=}C}$ bond: +614 kJ/mol
$\ce{C\bond{-}C}$ bond: +347 kJ/mol
$\ce{C\bond{-}H}$ bond: +413 kJ/mol

Breaking bonds gives a positive enthalpy change since it is an endothermic process, and vice versa for bond formation. We can now estimate the enthalpy change of the reaction
$\Delta$$H$ = 2(-413) + (-347) + 614 + 432 = -127 kJ/mol
As indicated by the negative value of the enthalpy change, the reaction is exothermic.
In fact, the enthalpy changes of hydrogenation reactions of other alkenes are also similar (though more substituted alkenes may have a lower magnitude of heat of hydrogenation, but still the reaction is exothermic). You may notice that for other alkenes, the way you calculate the enthalpy change is the same since they involve the same bond breaking and bond forming processes aforementioned (any substituents bonded to the carbons with the $\ce{C\bond{=}C}$ bond remain unaffected throughout the reaction and will still be present in the product being bonded to the carbon).
To simply put, the energy released from bond forming is larger than the energy required (absorbed) for bond breaking, and that is why you get net energy released from the reaction.

Only one bond is broken, as in H-H bond in hydrogen gas. But two bonds are made, that are C-H bonds. One extra bond is made. As we know bond making is exothermic, so more energy due to that one bond made will be released than absorbed.

You're breaking a C-C pi bond as well.


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Relation between various bond angles of Nitrogen compounds
I was thinking what could be order of bond angles of NH3, NF3, N(CH3)3 and N(C2H5)3.
Considering  NF3 and NH3:
Since there is backbonding between 2p-2p orbitals of N and F, there will be a partial double bond character between N and F and therefore, since repulsion between lone pair and double bond is more than that between lone pair and single bond and thus bond angle in  NF3 will be less than NH3
But, the size of F is very small, and F has lone pairs while H does not. If the N-F bond acquires partial double bond character, the lone pair of other F atoms should repel the bond making the F-N-F angle larger? Would the two effects cancel each other making the bond angle of  NF3 equal to  NH3
It is known that bond angle of  NF3 is less than NH3. Why? How do I decide which factor will dominate over other?
Now since N(CH3)3 and N(C2H5)3 have bulky groups, the groups will repel each other making the bond angle greater than NH3.
But between N(CH3)3 and N(C2H5)3 , which will have greater bond angle? I suppose C2H5 will repel other C2H5 groups more than CH3 groups.
Again there is another factor to be considered here. There are more numbers of bonds in C2H5 and shape is tetrahedral which might (there is a possibility) bring the CH3 connected to the C (which is connected to N) closer to N hence increasing repulsion between the group and N and thus decreasing C2H5-N-C2H5 bond angle.
Again, why does this not happen?
Can anyone explain the correct order between these molecules?

I would assume that there is no back-bonding in $\ce{NF3}$. There are no accessable empty orbitals on nitrogen that you can back-bond into.

$\ce{NH3}$ and $\ce{NF3}$ are easier to analyse and predict since there is only one set of atoms around the nitrogen — not three groups of four or three groups of seven as for trimethyl- and triethylamine, respectively.
Note that you did not (but should!) include bond lengths in your analysis. It is important to remember them. For ease, here they are as taken from Wikipedia (along with the bonding angles):
$$\textbf{Table 1: Bond lengths and angles}\\\textbf{of nitrogen compounds}\\
\begin{array}{lcc}
\hline
\text{compound} & d(\ce{N-X}) & \angle (\ce{X-N-X})\\
\hline
\ce{NH3} & 102~\mathrm{pm} & 107.8^\circ\\
\ce{NF3} & 137~\mathrm{pm} & 102.5^\circ\\
\ce{NMe3} & 146~\mathrm{pm} & 110.9^\circ\\
\hline
\end{array}$$
Starting with the difference between $\ce{NH3}$ and $\ce{NF3}$ — oh wait, I already answered that once. The gist is both compounds would love a $90^\circ$ bond angle but cannot due to too small bond lengths — the outer atoms would clash into each other, figuratively speaking. Therefore, the bond angles are expanded (at a loss of energy) to reduce this steric strain (gain of energy) up to an optimal sweet spot. Since the $\ce{N-H}$ bond is shorter than $\ce{N-F}$, the fluorines are comfortable with a slightly smaller bond angle as they are further away from each other (even though they are larger).
The difference when going from fluorine atoms to methyl groups (or ethyl groups) — as highlighted above — is the additional set of atoms. These cause further crowding, even though the slightly larger carbon atom (with respect to fluorine) allows for an even longer bond length. The steric strain of the additional hydrogens means that the bond angle must be extended even further to accomodate for all these.
Not much changes when going from trimethyl to triethylamine — except that the steric strain is larger still, and that therefore the bond angle is probably increased slightly again. One source I found including a crystal structure including triethylamine did not, however, give notably different bond angles when compared to trimethylamine, so this point may be moot. Note, however, that the angles reported are already larger than the perfect tetrahedral angle, indicating that the lone pair has an even lower s-proportion which would destabilise the overall molecule. Only by reducing steric strain can this low s-proportion be explained in any way.
As I already commented on the question: Back-bonding will not happen since there are no accessable empty orbitals on nitrogen that could accept a back-bond. Try drawing a back-bond in a Lewis formula without exceeding nitrogen’s octet to see that it is impossible.

You're right about no backbonding. My friend (whom I consider to be all knowing) told me that the reason would be backbonding so I wrote it without giving it much thought. I should've tried to think it out myself rather than believing him blindly. And that is a really good explanation and I forgot about bond lengths too. Thanks a lot.

You are forgetting about core electrons, when compare H and C/F

@permeakra I’m curious: How do they affect the picture aside from what I already wrote?

@Jan No idea =). I suspect that exchange repulsion is a thing here, though.


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How to calculate spectral overlap integral using spreadsheets?
I need to calculate the spectral overlap integral for the emission spectrum of coumarin 334 and the absorption spectrum for rhodamine, using spreadsheets (MS Excel).
Following the theory of Fluorescence Resonance Energy Transfer (Förster), I have the definition of the spectral overlap integral:
$$
J(\lambda) = \frac{\int_{0}^{\infty}f_\ce{D}(\lambda) \epsilon_\ce{A}(\lambda) \lambda^4\,\mathrm{d}\lambda}{\int_{0}^{\infty}f_\ce{D}(\lambda) \,\mathrm{d}\lambda}
$$
I also have the data for emission spectrum of the donor (coumarin 334) and the absorption spectrum of the acceptor (rhodamine). However, I am not sure how to proceed and actually extract the spectral overlap. Since the spectral overlap is a function of wavelength, will I end up with a "spectral overlap spectrum"? And how do I get the absorption coefficient of rhodamine? I have been given the maximum absorption coefficient (at wavelength of maximum absorption), but how do I get the molar absorption spectrum, $\epsilon_\ce{A}(\lambda)$?
Should I calculate values at a "per-wavelength" basis, or should I fit functions and integrate those?

This question would be better asked on Stack Overflow as a question about how to do numerical integration in Excel.  I think it would attract better answers.  If you are confused about the symbols and nomenclature being used in the Forster equation, then you could revise your question to be about that.

This isn't a pure spreadsheet inquiry. // In general you'd need to do the calculations on a "per-wavelength" basis. Trying to find a function to fit the spectra would be very messy. // You'd need to have both spectra at the same concentration. Not sure how to manipulate spectra (or if they can be) since you haven't provided concentration information.

So I ended up normalizing the emission spectrum to 1, and converting wavelength to cm. Because of the linear relationship between absorbance and the molar absorption coefficient, I used the maximum coefficient to find all other coefficients. Then I multiplied, at each wavelength, the normalized emission intensity, the absorption coefficient, and the wavelength^4, and summed all values. I ended up with a very small number, ~2.5E-12, which seems much smaller than I expected.

@Yoda the form of this equation doesn't make sense to me. J can't be a function of $\lambda$ if all of the $\lambda$ dependence is integrated out on the right hand side.

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Optical isomerism regarding diastereomers
Are diasteriomers always optically active?
If not then why?
Though they are chiral then also.

@Mansi Jain because they might have elements of symmetry.

If you carefully studied your book, you would have found an answer.

Some definitions (by IUPAC):

Diastereomer: A stereoisomer that isn't an enantiomer.
Stereoisomer: An isomer that has the same bond connectivity, but arises due to differences in the spatial arrangements of atoms.
Enantiomer: Non-superimposable mirror image.

So, diastereomers do not necessarily have to be optically active. Consider cis- and trans-1,2-difluoroethene.

They are stereoisomers (connectivity is the same, but the fluorines are on opposite sides) which are not enantiomers (they aren't mirror images of each other), which makes them diastereomers.
Neither are chiral because they are superimposable on their respective mirror images. Equivalently, the presence of the plane of symmetry (which is the plane of the molecule) makes both isomers achiral.

I don't think this is correct. They are stereoisomers, which are not enantiomers, but that does not necessarily make them diastereomers. Usually a diastereomer would exhibit two chiral centres or more, but doesn't necessarily have to be an enantiomer e.g. mesotartaric acid

A diastereomer is just the same compound with one group twisted about a stereocentre,right? That does not necessarily mean diastereomers are those stereoisomers which are not enantiomers? This confused me.

The IUPAC definition of a diastereomer is any stereoisomer that is not an enantiomer. http://goldbook.iupac.org/D01679.html @Beerhunter I could use meso-tartaric acid as an example too but the alkenes were simply easier. I may edit if I have the time later.

cc @Quark comment above Your definition is narrower than the correct definition. Compounds which differ in the configuration of some of their chiral centres are diastereomers, but diastereomerism is not restricted to that

@orthocresol Which other compounds are diastereomers then?

@Quark The alkenes above. Geometrical isomerism is stereoisomerism that is not enantiomerism - that makes it a form of diastereomerism. If you insist on having some chiral centres, here they are: these two pairs of compounds are both diastereomers. http://i.imgur.com/tuXqkTi.png

@orthocresol I thought this answer was sufficient, but perhaps add a definition of 'stereoisomer' to the answer, since there seems to be confusion on that point.


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

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about it.

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

Right, but doesn't quite go far enough. In aqueous solution the phosphate ion itself is actually basic, not acidic. Thus $$ \ce{PO4^{3−} + H2O -> HPO4^{2−} + OH^{-}}$$

OP asked about the phosphate functional group. If asking about the phosphate (tribasic) anion, then what MaxW said is correct.

Curious but pK2 of phosphate is 7.198 according to Wikipedia which would make the $\ce{HPO3^{2-}}$ slightly basic too. but all the second ionization of all(?) $\ce{R-PO(OH)2}$ compounds seems to be acidic. (I'd wonder if there isn't some weird molecule where the second ionization pKa would be more than 7 in aqueous solution.)

I believe you meant $\ce{HPO^{2-}_4}$.

absolutely. thanks for that correction and it should be $\ce{R-O-PO(OH)2}$.


text:
exchanging H2O in buffer for D2O - pH vs pD?
I have TBS pH 7.4 buffer that I am exchanging all the water out for $\ce{D2O}$. How do I find out what the pD is after all the water is gone? I understand that pH meters don't work for deuterium, and it seems like deuterium also changes the spectral properties of the chromophores in pH paper (or in this case pD paper)
Also, if I want to best mimic the alkaline conditions in $\ce{D2O}$ as $\ce{H2O}$, should pD final = pH initial?

I'd expect that a pH meter would work, but you'd need to "soak it" in D2O and you'd need a different calibration curve. // Interesting question though!

Are you asking how you measure the final pD (to see if you are finished), or how you can determine (for instance, calculate) it?

The articles by Glasoe,[1] Covington et. al.,[2] and (much more recently) Krężel and Bal[3] suggest in their intros that pD values can be obtained from a pH calibrated electrode by adding a correction constant of about $0.4$. This was determined by comparing the measured pH when the same amount of acid or base was dissolved in $\ce{H2O}$ and $\ce{D2O}$.
References:

P. F. Glasoe, Citation Classic - use of glass electrodes to measure acidities in deuterium-oxide. Current Contents/ Physical Chemical & Earth Sciences 1979, 16, 12-12. A1979HZ34800001 (pdf) Original paper: P. K. Glasoe, F. A. Long, J. Phys. Chem. 1960, 64,:188-190. Doi:10.1021/j100830a521
A. K. Covington, M. Paabo, R. A. Robinson, R. G. Bates, Anal. Chem. 1968, 40 (4), 700–706. DOI: 10.1021/ac60260a013
A. Krȩżel, W. Bal, J. Inorg. Biochem. 2004, 98 (1), 161-166. DOI: 10.1016/j.jinorgbio.2003.10.001 Available at researchgate.net.


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Why is cyclopentadiene anion is aromatic but cycloheptatrienyl anion is not?
I just can't seem to figure this out. The cyclopentadienyl anion (1) is aromatic, but the cycloheptatrienyl anion (2) is not, though its cation (3) is. Why?

That happens because of Huckel's rule of aromaticity. The cycloheptatrienyl anion has 8 electrons which translates as 4n electrons, not the 4n+2 as implied by Huckel. If you are about to construct the MOs for both cycles, you'll find that putting 4n electrons will rise in a diradical molecule, not stable as it already sounds. A quick trick for doing that is using Frost's circle. All you need to do is inscribe the correspondent polygon in a circle and make sure that you have a vertex touching the circle as low as possible. The vertices will give you the qualitative energy of the MOs.

Ground state O2 is a diradical and is pretty stable. I'd say it's just a form of a Jahn-Teller distortion - a nonlinear molecule with incompletely filled degenerate orbitals.

Agree. Indeed pseudo-Jahn-Teller effect is present in molecules like cyclobutadiene and distorts the geometry from square to a rectangular. It may also happens in cycloheptatrienyl anion.

"That happens because of Huckel's rule of aromaticity. "No, it doesn't happen because of this rule. The rule is a tool to predict aromaticity/antiaromaticity in very simple systems.

Aromaticity exists when there are $4n+2$ (i.e. 2, 6, 10, 14...) electrons in a planar, cyclic pi system. Planar, cyclic pi systems that have $4n$ (4, 8, 12, 16...) electrons are antiaromatic.
In the case of the cyclopentadienyl anion, there are 6 electrons in the pi system. This makes it aromatic. The cycloheptatrienyl anion has 8 electrons in its pi system. This makes it antiaromatic and highly unstable. The cycloheptatrienyl (tropylium) cation is aromatic because it also has 6 electronics in its pi system.

Cycloheptatrienyl anion(tropylium anion) has 8 pi electron system, therefore it must be antiaromatic but the extra lone pair on the one carbon would cause that carbon to be become sp3 hybridized and put those extra elctrons in one of the sp3 orbitals. This would make it non-planar and non-aromatic.

It's quite simple, really. Aromatic rings have all their relatively stable electronic orbitals filled and unstable ones empty.  The 4n+2 rule just says that a conjugated ring will have an odd number of stable pi orbitals to be filled, that is 2n+1 orbitals to be occupied by two electrons apiece for some whole number n.
When you try to put 4n (or 4n+4) electrons into the ring as in cycloheptatrienyl anion, you have filled and empty states at the same energy level; that's what the Frost circle tells you.  Sometimes molecules do that, but they generally are not as stable as what you get with all the filled states well below all the empty ones.  The latter is what the proper 4n+2 electron count does.


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What is the temperature at which lava evaporates?
According to this article ->  http://ntrs.nasa.gov/search.jsp?R=19660041998
Lava evaporates under natural conditions, but apparently, only a little bit.
I'd like to know the temperature at which lava could fully, or mostly, evaporate at 1 atm.
I know this is extremely difficult, if not impossible, to be accomplished under standard pressure with realistic tools, but I would still like to know, if you please.

Boiling points of $\ce{SiO2}$ and $\ce{Al2O3}$ would be a decent approximation.

You would likely need to consider the sub oxides as well in the thermodynamics. The gas phase will be multi-species.

There are many components of lava, such as molecular sulfur and water, as well as silica and alumina. The sulfur is frequently mined where it condenses around a fumarole: http://blog.michaelyamashita.com/?p=595

$\ce{SiO2}$ has b.p. 2950 deg  C, $\ce{Al2O3}$ 2977 deg C. So by natural Earth surface processes like never, in lab with a high power laser, the particular value does not matter much.

The problem with lava is that it isn't a single compound and evaporation leads to decomposition
A reasonable first guess of the temperature where lava might evaporate is the approximate, known boiling points of alumina or silica (as Ivan Neretin's comment mentions). These are both close to 3,300 K (or C as the difference isn't much at such high temperatures).
There are problems with this estimate, though. Most lava isn't a pure compound but a mixture of one mineral with varying compositions or even a solution of many minerals of varying composition. Both of these considerations might alter both melting and vapourisation significantly. Worse, the vapour will not, in general, have the same composition as the liquid due to decomposition (minerals are often complex macromolecular mixtures containing large networks of silicate and aluminium oxides in varieties of infinite chains, panes or 3D structures which start to break up when liquid and certainty when in the vapour).
But this has been studied. This paper by NASA ("Vaporization and Thermodynamics of Forsterite-Rich Olivine and some
Implications for Silicate Atmospheres of Hot Rocky Exoplanets") studied the vapour pressure and composition of the common mineral olivine (a very common  iron-magnesium silicate mineral). The vapour contains Fe, Mg, FeO, SiO, O and O2. This complicates any analysis. The experiments consisted of measuring the vapour pressures and composition of the vapour components above melted olivine at temperatures around the melting point.
However the paper provides a vapour pressure equation for the mineral that suggests the boiling point is about 3,700 K which is not outrageously different from the known values of SiO2 and Al2O3.
Other papers like this one "Critical vapourization of MgSiO3" (motivated by attempts to understand hot conditions after planetary collisions) have estimated critical liquid/vapour temperatures compatible with the estimate above.
So, a definitive answer is not possible as lavas are not all the same and the exact results are hard to measure directly. But estimates on some possible lava components and data from some pure compounds related to lava minerals  have been done and suggest temperatures of vapourization of between 3,000 and 4,000 K at atmospheric pressure.


text:
What is exactly ortho effect in Reimmer-Timmer reaction?
In the Reimmer-Timmer reaction cho group is added to only at ortho position not at para position , I got the answer for this as ortho effect. So what is exactly ortho effect.

I think that the coordination step is responsible for the ortho selectivity. The phenoxide anion has a large negative charge on the oxygen atom, so the electron-poor carbene will tend to go there and the conjugated attack will result in the ortho product.


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What is the electronic configuration of Cr (+1) ion
I believe it should be $\ce{d^5}$ , but i found out that it is $\ce{d^4s^1}$. Is it so?
I have not found any explanation for the following. So why the strange configuration?

Neither of those is the electronic configuration of $\ce{Cr+}$. You might be asking for what the configuration ends in.

@Ϻ.Λ.Ʀ. It's perfectly fine to talk of a $\mathrm{d^4s^1}$ configuration. Obviously it's not the absolutely technically correct description but the unwritten parts are implied. If we had to write [Ar] every time we wanted to discuss the configuration of a TM it would get very boring.

I assume we are talking about the free ions in the gas phase.
I checked my notes and it seems that the electronic configuration of $\ce{Cr+}$ should be $[\ce{Ar}]\mathrm{3d^5}$, the reference being a rather old book: Phillips & Williams' Inorganic Chemistry (1966). I tried to look in slightly newer textbooks, but I couldn't find any information on the monovalent ions.
I would not claim to be 100% certain on this, but it can certainly be rationalised by the increase in $Z_\mathrm{eff}$ in the $\ce{Cr+}$ ion as compared to neutral $\ce{Cr}$ (one less electron means less shielding and greater effective nuclear charge). That would serve to stabilise the 3d orbital more than it stabilises the 4s orbital, and therefore favour the $\mathrm{3d^5}$ configuration over the $\mathrm{3d^44s^1}$ configuration.
In general, the electronic configurations of transition metals and their ions are a balance of delicate factors. It's not easy to rationalise the electronic configurations fully and there are several factors that will affect the configuration adopted, e.g. 1) nuclear charge and relative energy of 3d and 4s orbitals, 2) relative e-e repulsions in 3d and 4s orbitals, 3) exchange energy.
Some interesting discussion on the topic can be found here:

J. Chem. Educ., 1978, 55(1), 2
J. Chem. Educ., 1989, 66(6), 481
J. Chem. Educ., 1994, 71(6), 469

you can also check here http://physics.nist.gov/PhysRefData/ASD/levels_form.html  3d^5 is lower by 1.48 eV


text:
Is there a chemical reaction that speeds up when the temprature gets lower?
We were talking about chemical reaction speed on our chemistry class, and I asked my teacher if there is a chemical reaction that speeds up when the temprature gets lower?
I think this is not everyday chemistry and my teacher didn't have an example for one.

Well, no I guess, since that's what the Arrhenius equation tells ya. You can't tell anything like "are there any positive charges that attract other positive charges?" so you can't say "Is there a chemical reaction that speeds up when the temprature gets lower?"

It seems like a bit of a cheat, but would enzyme-catalysed reactions count? Above some temperature, enzymes denature and become less effective, so as you cool the system below that critical temperature, it would go faster?

Related - Geoff's answer here http://chemistry.stackexchange.com/questions/18924/there-are-no-chemical-cliffs/21629#21629

+M.A.R "You can't tell anything about "are there any positive charges that attract other positive charges?" so you can't say "Is there a chemical reaction that speeds up when the temperature gets lower?" Those two are not the same thing. The Arrhenius equation works in only a few simple reactions and is not always applicable.
So there are many reactions whose rate increases with a dip in temperature.

That's a good question.
A reaction slowing down on a rise in temperature seems counter intuitive but it's not so always.
The Arrhenius equation works for only a few limited elementary reactions and the transition state theory is a much better explanation of rate laws and their variation with temperature.
The dependence on temperature of the reaction rate cannot always be analysed purely using the Arrhenius equation. It can be better understood by knowing the mechanism of the reaction.
For instance
$$\ce{2NO + O2-> 2NO2}$$
Has its first step as
$$\ce{2NO <=> N2O2}$$
And the second step as:
$$\ce{N2O2 + O2->2NO2}$$
The first step is exothermic in nature.
And the equilibrium shifts to the left on increasing temperature based on the Van't Hoff equation of gaseous equilibrium.
So the effective rate constant of the overall reaction decreases on increasing the temperature.
Thus, we see a reaction whose rate slows down with the rise in temperature.
There are in fact many such reactions.
To give another example, I could take an enzymatic reaction in a living cell and show that the rate of reaction falls down drastically with a rise in temperature. Ever wondered why your body rises its temperature during a fever? The aim there is to slow down the reactions in the cells of the invading germ and hence kill it.
From these examples, one could conclude that the famous Arrhenius equation works in only a few limited(and ideal) cases.

But isn't this a special case of a complex reaction which is rather artificial in the sense that the overall reaction rate decreases because a reactant is depleted? It doesn't say anything about how the reaction rates of the elementary reactions in this network change with temperature. If one focuses on elementary reactions and elementary reaction rates, can we observe a speed up when temperature decreases? Are there cases where the free energy of activation decreases when the temperature decreases?


text:
Since PbCl2 is insoluble in water, how is it that lead leached from pipes due to Cl- reaction can become ingested?
I have reviewed a discussion concerning the salt content in the Flint River being directly connected to lead being leached from civic water pipes in Flint, Michigan. Wikipedia has a site dedicated to $\ce{PbCl2}$ and line 1 states that this salt, once formed is insoluble in water, unless that water is heated. Is there a secondary process that results in the lead returning to a form, which can be ingested — after it has been leeched from the wall of a pipe?

It is not insoluble. It is poorly soluble - enough to poison somebody, though.

Supporting Ivan's comment, see this:-http://chemistry.stackexchange.com/questions/42698/why-exception-is-noted-in-the-solubility-of-lead-salts

All in all the water chemistry is more complicated than you're supposing. // "Insoluble" is a rather poorly defined term in chemistry. As chemists we typically mean that you can't dissolve 1 gram of solute in 100 ml of water. But for lead we're concerned about 0.000000010 grams per ml of water (10 parts per billion).

Lead(II) chloride is not the sitting-at-the-bottom-like-a-brick type of insoluble. Rather it is the well-ma~ybe-I’ll-dissolve-a-little type of insoluble. So enough lead chloride will be dissolved — as Ivan mentioned, enough to poison somebody. Its solubility is given as $10.8~\mathrm{g/l}$ on Wikipedia. There are much less soluble salts like mercury(I) chloride ($2~\mathrm{g/l}$), barium sulfate ($2.4 \cdot 10^{-3}~\mathrm{g/l}$) or silver(I) iodide ($3 \cdot 10^{-6}~\mathrm{g/l}$). Note that for mercury(II) sulfide a solubility is not even given, it is merely termed insoluble (and is the sitting-at-the-bottom-like-a-brick type of an insoluble compound).
All values were taken from Wikipedia and are given at $20~\mathrm{^\circ C}$.


text:
Reaction of nitro compounds with strong alkali?
Why does a primary and a secondary nitro compound behave as an acid in the presence of strong alkali whereas tertiary nitro compounds do not?
Can anyone explain it with reactions?

How many $\alpha$ hydrogens are there on a tertiary nitro compound?

there is no alpha hydrogen in tertiary nitro compound

There is your answer then.

In order for a compound to act as an acid, it must have a proton which can be abstracted. As you correctly pointed out, tertiary nitro compounds have no $\alpha$ hydrogens and therefore they cannot be deprotonated at the $\alpha$ position.
Of course, they could be deprotonated elsewhere, but if the other groups are just alkyl groups then this isn't really feasible as they have such a high $\mathrm{p}K_\mathrm{a}$.

Organic nitro compounds can tautomerise to give its aci-nitro form. As we know tautomers exist in a dynamic equilibrium, the enol form of nitro compound is called its aci-nitro form, which is present in a small amount in the equilibrium mixture. Aci-nitro form is the acidic form of a respective nitro compound, which is able to liberate an acidic hydrogen when needed.
Thus, when an alkali metal is added to a nitro compound, they react with the remaining aci-nitro form (which gets used up) and thus the equilibrium shifts right, producing more aci-nitro compounds. This results in complete dissolution of the alkali in the nitro compound, provided that the given amount of nitro compounds is able to synthesise required amount of aci-nitro form to completely dissolve the alkali.
The role of alpha hydrogen in the nitro compound is to take part in tautomerisation:

Please edit your post for grammar, punctuation, and style; use MathJax for typesetting; and avoid ethereal links.


text:
Reaction speed of Na2CO3 with vinegar vs that produced by NaHCO3 with CH3COOH
I'm working on a project requiring the easy production of a gas not corrosive to "normal" household materials (e.g. carbon dioxide). I settled on vinegar (acetic acid) and baking soda or washing soda, as it is very easily available to most people. Looking at the chemical formula for both sodium carbonate and sodium bicarbonate, I can assume that roughly the same amount of gas will be produced by the reaction of both with vinegar.
However, is there any difference at all in reaction speed? I don't care about the end product as long as it can be washed down the drain.

Sodium carbonate, $\ce{Na2CO3}$, will not create gas when mixed with an equimolar  portion of acetic acid (i.e. $1\  mol\ \ce{Na2CO3}:1\ mol\ \ce{CH3COOH}$). The reaction would proceed as follows: $$\ce{Na2CO3} + \ce{CH3COOH} \ce{->} \ce{NaHCO3} + \ce{CH3COONa}.$$ Therefore, you would need another equimolar portion of acetic acid to protonate the bicarbonate ion, according to this reaction: $$ \ce{NaHCO3} + \ce{CH3COOH} \ce{->} \ce{CH3COONa} + \ce{H2O} + \ce{CO2_{(g)}}.$$ This should illustrate that using sodium carbonate simply causes you to use an extra portion of acetic acid to make the same amount of gas.
Thus, sodium bicarbonate and acetic acid will produce gas in less overall time than sodium carbonate and acetic acid. When starting from sodium carbonate, the reaction will proceed through both reactions above. When starting with sodium bicarbonate, only the second reaction must proceed. Technically, the rate of gas evolution will be the same once gas formation begins (assuming equal starting molarities). However, the sodium carbonate reaction wouldn't produce any gas until the first reaction had completely finished, taking more time overall.


text:
Equivalent of Szabo and Ostlund book for DFT
Szabo and Ostlund book Modern quantum chemistry [1] is extremely useful to understand Hartree-Fock and post-Hartree-Fock methods. Not only it explains the theory behind such methods, but it is also oriented towards a practical implementation. Following this book I was able to implement an Hartree-Fock code from scratch in order to perform Born-Oppenheimer molecular dynamics.
Since I already work with software packages based on density functional theory (Gaussian09 for calculations in vacuum and QuantumESPRESSO for solid state systems), I would like to implement a simple DFT code in order to better understand what is going on inside the "black box". However, I would like perform this task using a plane wave basis set: this is somewhat more involved than using Gaussian basis sets since it involves periodic boundary conditions and pseudopotentials. In addition, I will have to take care of fast Fourier transforms everywhere and of the approximation of the exchange-correlation functional.
Supposing that I will be happy with the local density approximation, there is a book out there similar to Szabo's book for DFT? Notice that I am not interested in the basics of DFT but I am searching for a good book explaining in details the practical implementation using PW.

Here some comments on the books I already considered:
Martins [2]: This book contains a lot of informations and is oriented to solid-state electronic structure, therefore it explains plane waves and pseudopotentials. However I think that it works well as a reference but not as first reading to understand in details the concepts.
Giustino [3]: Very nice book with the goal of teaching DFT to undergrad students. For this reason is too simplistic and does not go into technical details.

[1] A. Szabo and N. S. Ostrlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Doever Publications, 1996.
[2] R. M. Martins, Electronic Structure: Basic theory and practical methods, Cambridge University Press, 2004.
[3] F. Giustino, Materials modelling using density functional theory, Oxford University Press, 2014.

I'm not confident enough to post this as an actual answer (might be too basic), but the CPMD folks cite v2, part 5 of Handbook of Molecular Physics and Quantum Chemistry, "Density Functional Theory: Basics, New Trends and Applications." Unfortunately, it looks terrifically expensive, though it's available in a few European libraries.

Assuming LDA with no exact exchange and pseudopotentials the job is plain and simple search of minimum of a functional. Assuming plane-wave, the function in question is decomposed into a set of plane waves, so the job is minimaztion of a function over the decomposition coeffecients.

Have you had a look at this question? Maybe you could comment on the sources that are referred there, to make it easier to not give duplicate answers.

@Martin-マーチン I don't think the question is useful to me... In the comments you read " you are not even supposed to understand how DFT calculations are done step by step" or "everything you need to know about SCF is in Szabo & Ostlund". I would like to go a step further and implement a "simple" DFT code from scratch. S&O was extremely helpful to implement the HF method in a Gaussian basis set and I would like to find something similar for DFT. The books on which I studied DFT are extremely superficial and only explain the basic theory...

@Brian Thank for pointing this out. I will try to give a look to this book, provided that I can access a copy from my University.

Check out The ABC of DFT by Kieron Burke  here. It's free and comes in at 104 pages. Also, see Kieron's group literature page for other gems here.

I can recommend the book of Yang and Parr, although I am not sure it deals  with plane wave bases at all. In principle, a DFT code is not really different from a HF code. In fact, you can use exactly identical codes, just replacing the Coulomb and exchange integrals by numerical integration of some functional. E.g., if you choose $X\alpha$-LDA, than your exchange functional is
$$K = C_\alpha \left(\frac{3}{\pi}\right)^\frac{1}{3} \int \rho(\mathbf{r})^\frac{4}{3} \mathrm{d}\mathbf{r}$$
You can evaluate your density numerically, choose any numerical integration scheme and obtain the exchange energy. That works even, if you still calculate the Coulomb integrals the HF way. However, you wouldn't save much time since you still have the $\mathcal{O}(N^4)$ step in the Coulomb integrals, where $N$ is the number of basis functions. But once you integrate the Coulomb integrals numerically, you have basically a working DFT code.
There is a number of papers about choosing the correct grids and integration schemes for numerical integration. Once, I read one that compared the radial grids quite well, but sadly I don't remember the reference. But the buzz words would be e.g. Becke grid, (Treutler?-)Ahlrichs grid or simply Gauss quadrature for the radial integration (most of the grids are a form of Gauss quadrature, though). For the angular integration there is e.g. the Lebedev grid, but its calculation is a little more involved.
My suggestion, based on teaching experience, is to write first a HF code in a Gaussian basis. Then you pimp it with a numerical integration scheme and something like $X\alpha$. In the next step, you replace your Coulomb integrals. Lastly, you can switch from a Gaussian basis to the basis of plane waves. The last step only changes your matrix elements but not the algorithm (HF/DFT) itself. Since you first implemented a HF code for a Gaussian basis, you can always check your algorithm before going into uncharted waters.

Two of the books that I found close to your needs (i.e. not a basic intro but a detailed implementation including programming and algorithm) are the following:

Ab initio molecular dynamics by Marx and Hutter (published by Cambridge Uni Press)   - This has some info on the implementation of algorithms in CPMD code. Some flow charts of the algorithms.
A chapter (Ch 6) in the book "Computational Physics" by Thijssen (also by Cambridge Uni Press) comes with a short computer program (downloadable from the companion site written by the author of the book).


text:
Is there any software package for quantum chemistry that includes CAMB3LYP?
I want to do calculations on systems with photoinduced electron transfer, and I've read that the Coloumb Attenuating Method is a modification of the functionals that makes calculations including long-range electron transfer more accurate. Unfortunately I can't seem to find CAM in Gaussian. Is there some other software I should look at?

CAM-B3LYP is implemented in Gaussian. Please look again.

CAM-B3LYP is also implemented in ORCA (free, but for non-commercial use only).

It seems I have an old version of the Gaussview program I use to generate my input files. I will just have to write the input file manually instead.

CAM-B3LYP is present in Gaussian, Q-Chem, GAMESS, NWChem, ORCA, DALTON, DIRAC, and perhaps other major software packages, either as cam-b3lyp or camb3lyp, however keywords are entered.
Notably it isn't available in TURBOMOLE as of version 7.1.

It should be noted, that no DFT calculation is good without calibration.

Of course. Additionally, there are better long-range-corrected functionals than CAM-B3LYP (at least in theory), but that's a topic for another thread.

... and Dalton (example for $1 ~\gamma$, $2 ~ \gamma$, and $3 ~ \gamma$ absorption) and Dirac (example).

Not all of QM program use adirect keyword of functional. Some of them you've to mix an ingredient of % functional, says in NWCHem. In that, you aint use keyword 'cam-b3lyp' directly, you need the combination of keywords like

xc xcamb88 1.00 lyp 0.81 vwn_5 0.19 hfexch 1.00
  cam 0.33 cam_alpha 0.19 cam_beta 0.46


text:
Can you artificially create molecules with a machine?
Is it possible to create molecules with machines? For example: you shoot one oxygen-atom $(\ce O)$ on one diatomic hydrogen $(\ce{H2})$ to create one water molecule $(\ce{H2O})$.

Crossposted from http://physics.stackexchange.com/q/232530/2451

You kinda can, but that would be prohibitively expensive.

Surely you can create any molecule with automation. MIT have done it. Is there another example as water isn't artificial

Sounds like a replicator from Star Trek.

This sounds to me like molecular beam chemistry. Within a vacuum chamber, you can create and collimate a focused "beam" of moving atoms or molecules and direct it to hit a stationary target or another beam. You can use spectroscopy to study the species that are formed. The idea is to look for and characterize reactive intermediates or transition states between starting materials and products. But if you were to create a familiar species such as water in this manner, it is not "artificial" water, but the real thing.
Maybe there is a physical chemist out there that can elaborate further on molecular beams.

First Thanks! I think the "artificial" is misleading, I meant it so: "Create Molecules Artificially" ...without the Chemical Reactions!


text:
Are there any colorful, odorless gases?
Gases are frequently described as "colorless, odorless, and tasteless."  I am wondering if this is redundant, since I can't think of many gases that have a detectable color to begin with, much less ones that have a color but no odor.
Are there any gases that are colorful and odorless at standard temperature and pressure?  In particular, gases that impart enough of a color to white light that one could look into a room filled with the gas and detect its presence visually?
(In the limit I suppose one could argue that a standard atmosphere is "colorful" since over many miles one can see blue scattering, but the question here is whether it is ever useful or helpful to describe a gas as "colorless.")

$\ce{O2}$, $\ce{N2}$, noble gases, $\ce{SF6}$, lower hydrocarbons and freons... see, the list of odorless gases is pretty short, and it just so happens that all of them seem to be also colorless.

@IvanNeretin - interesting: I hadn't thought to ask the question that way.  Now I'm trying to think of other odorless gases.  $\ce{CO}$ and $\ce{CO2}$ famously make the list, but I can't think of others right now.

Yes, I forgot $\ce{CO}$. As for $\ce{CO2}$, to call it odorless would be a bit of understatement.

There are a few colorful gases but they also possess strong odors, such as the halogens. Iodine vapor in particular is a vivid purple. Even though it's "b.p." is listed as ~180 C, iodine is quite volatile at STP (just carefully sniff some crystals ;-).

$H_{2}$, $D_{2}$ $HD$ and other three isotopic forms with tritium : all are colorless and odorless !

I googled far and wide but couldn't find anything. The closest I could find was the gas phase of nickel tetracarbonyl (43 Celsius). It doesn't have a powerful odor but it does have a smell.

There are plenty that are smelly and colourless, or are odourless and colourless but I would agree that there are none that are coloured and odourless. Surely it is just coincidental if any of these pairs occur because the chemistry involved in vision and smell are totally different. (Whatever the chemistry of smell is it does not involve photons!) If you were less restrictive in your conditions then most gases will be 'coloured' in the sense that they have electronic absorption in the near ultra-violet and also infra-red at longer wavelengths. Perhaps some insects would see gases such as N$_2$O or O$_3$ as coloured. Also if you were to allow vapours then many compounds can be considered if heated, for example azulene (an isomer of naphthalene) is blue and has a small but measurable vapour pressure if warmed slightly. I don't think I want to smell any, however. The ideal of filling ones nose and lungs with aromatic or dye vapours that then condense is not an attractive one!

Azulene actually smells quite nice, kinda like crayons

I would say, there's a solid no, here.
Unless, y'know. You'd be willing to smell some fluorine, which, although, technically odorless, that's only because it tends to destroy the nerve receptors that actually handle smell. (Even this isn't technically true. Fluorine is smellable, to some extent, but only off of certain rocks.)
Something else that would be technically colorful, I suppose, would be iodine vapor. This is actually a very, very pretty violet color. However, this is neither nontoxic nor odorless. Nor tasteless. In fact, it's pretty toxic if you breathe it in.
Same goes for the rest of the halogens, really.
The problem with gases and color, is where color actually comes from and how you perceive it. Photons bounce off of whatever, the thing it bounced off of absorbs a few spectrums of light and reflects others, and you perceive the result of this process as color. (This is not true for light emitters. With light emitters, it emits the spectrum you perceive as color.) The issue is, there's not much for light to bounce OFF of, when it comes to gas. It's just so diffuse in the air, so there's not much for molecules to bounce off of.
Even when in liquid form, they're still really tiny molecules. Only oxygen is a faintly blue color. The halogens form fairly opaque liquids, but I have no idea why you'd want liquid halogens running around.
As for whether it's useful to describe a gas as colorless and odorless: Yes. In fact, it's extraordinarily useful for issuing warnings to anybody. Telling people that something that can potentially kill you, and then proceeding to tell them they can't see this thing at all, tends to invoke a heightened state of awareness for their own safety. Which is the goal, after all.

As @IvanNeretin suggested in the question comments: There may be very few odorless gases, and it may be that only two of them are potentially lethal.  But the question was a little more abstract: Yes, it might require a large volume of any gas to reflect its color.  So pick an arbitrarily large container -- e.g., a cubic mile if necessary -- when enumerating odorless gases that reflect visible light at STP.

Cubic mile? Even then, you're unlikely to get any gasses that truly produce any color. Our atmosphere produces a color because of the way the light scatters, and that's over 50 miles worth of air going straight up. But the few I suggested (i.e. The Halogen Gasses) are visibly colorful in an ampule full of it. As far as odorless gasses? I think you're outta luck.


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How the chemical reaction is happening?
I have got this IMAGE peculiar chemical reaction and  tried to find reason behind it. But I couldn't. Can you write the reason?

It's hydrophobic sand, I think. Apparently called 'Magic Sand' : https://en.wikipedia.org/wiki/Magic_sand.

It isn't really a "chemical reaction" since there isn't a product made from reactants.


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What's the composition of gun smoke and is it dangerous?
Background
I've been conducting an informal (i.e. mental) risk assessment of visiting gun ranges. There are a couple of obvious sources of hazard:

Injury by bullet
Hearing loss
Hot casings flying around
Gun malfunction

The first one is addressed by the fact that you should assume every gun to be loaded until personally convinced otherwise, and to always keep it pointed at the target area when handling the weapon. The second and third points are addressed by safety measures such as wearing ear and eye protective equipment.
The fourth one is kind of a wildcard, because you would assume that people who go shoot guns in a range are professionals with training. This may not always be the case, and statistics of that should be taken into account in the final verdict of the risk assessment, but I will ignore it for the time being.
The Question
Suddenly, I was struck by the idea that guns also may pose a respiratory hazard. The smoke that is formed has a distinct smell, possibly combined from the lubricant on the gun and the combusted propellant of the bullet. What exactly is this "gun smoke" made of, and should I be concerned for my health in this regard?
An image to show what this gun smoke can look like (Source: Arizona Sonora News):

Bonus Question
There is an additional level of concern when handling even larger payloads than a simple 9 mm bullet, but the documentation may be restricted and as such I do not require an answer to the following question: Are there any requirements for military personnel operating cannons and other large artillery to protect their airways from the fumes?

There are two much more significant risks that you missed. First the danger of driving to and from the gun range. Second, many handgun ranges are indoors. The most significant long term risk is from lead dust.

Toxic smoke from propellant combustion has been known about for over 500 years..
The propellants used today are Nitrogen based, thus nitro glycerin, nitrocellulose and nitroguanidine . These produce gases containing oxides of nitrogen, carbon oxides, carbon monoxide, hydrogen cyanide, ammonia and several other gases..
The oxides of nitrogen can be inhaled deeply into the lung thus causing damage as deep as the Alveoli, thus causing bronchiolitis obliterins resulting in a "Pulmonary Edema". This can happen 5 years after exposure..
If you show signs within 24 hours of exposure then oxygen and Arterial Blood gas tests are the solution to treating the condition and knowning the effect. The symptons show just like influenza but no sign of virus.
This is known as an Acute Toxic Smoke Inhalation Injury.
WW1 and WW2 this was studied to reduce the effect on gun crews on both land and sea. It is still a problem today with Armoured vehicles and Artillery and one of the reason the Navies use unmanned turrets.
Look up this file and you will find quite a bit of information about this condition.
Occupational Health - The Soldier and the Industrial base
I had first hand experience of this injury in 1988 and am still alive to tell my story, I had a "Pulmonary Edema" in 1991 and survived that, so I am in the 2% of survivors from that period. In the mid 1990's they discovered how to treat a pulmonary edema thus it has increased the survival rate.
There is thousands of pages of information on this subject, just look for "Propellant exhaust gas exposure" or "Toxic gas inhalation" that will bring up more information.
The military threshold limits are higher than civilian but a tank will expose the crew to 1,300 the TLV for less than 5 minutes if the fume extractor is working. In my case it was not!
http://www.public.navy.mil/surfor/documents/p_5041.pdf  Chapter 10
http://www.cs.amedd.army.mil/FileDownloadpublic.aspx?docid=a1d49fb8-3a16-4489-9408-1089f7044b4c
https://permanent.access.gpo.gov/lps101453/www.bordeninstitute.army.mil/published_volumes/occ_health/OHch10.pdf
There are 3 links for you to go to...
Enjoy reading it...
Oh just as an idea, if you had 1,100 .308(7.62mm) rifles fired in a 100 m3 room, its would be the equivalent of 5x L7 105mm tank rounds...
Enjoy..

It would help if you could include a source of (link to) the file you are referencing and also to summarise its contents if that is not already part of your answer. If you do quote entire paragraphs, please put the > symbol in front of them for block quote formatting.

Potential health issues associated with the immediate exposure to toxic propellant gases have been addressed above by Webber GK.
However, a review of the biochemistry literature does suggest that there can also be long-term consequences from radical-induced illness to the lungs (in particular, see this source, as an example) which may also be a concern here.
As a foundation, I start by looking at the decomposition products likely generated from the discharge of weapons. Starting with gunpowder, per a source:

Gunpowder...consists of a mixture of sulfur (S), charcoal (C), and potassium nitrate (saltpeter, KNO3). The sulfur and charcoal act as fuels while the saltpeter is an oxidizer.[1][2]
A simple, commonly cited, chemical equation for the combustion of black powder is:
$\ce{2 KNO3 + S + 3 C → K2S + N2 + 3 CO2}$
A balanced, but still simplified, equation is[126]:
$\ce{10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2}$

I suspect an unmentioned intermediate is sulfite which is converted to sulfate with air contact.
Next, per Wikipedia on Potassium nitrate, comments on thermal decomposition:

Between 550–790 °C (1,022–1,454 °F), potassium nitrate reaches a temperature-dependent equilibrium with potassium nitrite:[16]
$\ce{2 KNO3 ⇌ 2 KNO2 + O2}$

Likely more prevalent, however, is the use of smokeless powder alternatives, which includes nitrocellulose, $\ce{C6H7(NO2)3O5}$ where the more elementary decomposition product could add NH3, NO2 and even some HCN to the list.
Also, the primer employed in bullets is impact-sensitive lead styphnate ($\ce{C6HN3O8Pb}$). As such, I would expect its decomposition to yield  CO, CO2, NO, NO2, Pb and water vapor. Also upon the interaction of water and nitrogen dioxide, for example, a possible acid mist:

$\ce{•NO2 + •NO2 -> N2O4}$
$\ce{N2O4 + 2 H2O ⇌ HNO2 + HNO3}$

The direct action of $\ce{NO2}$ on $\ce{NH3}$ may also be possible in the presence of $\ce{NO}$ at increasing temperatures.
Interestingly, cited gases $\ce{N2, NO, CO2, SO2, NH3, and O2}$ are colorless. However, the interaction of ammonia and acidic-formed mists could create a smokescreen (in the current discussion, ${NH3}$ with $\ce{H2SO3}$ or $\ce{HNO2}$ or $\ce{HNO3}$ or $\ce{HCN}$…).
To this visible gun smoke, I would add incomplete combustion products including C (as soot) and small amounts of Pb. Also, with the use of KNO3, several potassium salts (sulfide, sulfite, nitrite, sulfate). A largely confirming source given previously, to quote:

This chapter primarily considers the risks of exposure via inhalation, the skin, and the eyes to the major combustion products of commonly used solid propellants—carbon monoxide, ammonia, hydrogen chloride, oxides of nitrogen and sulfur, and lead—in army weapons systems.

However, a rifle’s barrel does wear down due to nozzle pressure and friction generated heat. As such, I would further add trace amounts of iron and chromium metal particles to the outgassing as well, where the presence of transition metals in the lungs could be critical to further reactions.
In particular, with  respect to chemistry sourcing radicals (which could produce in time health issues), I start with Josef Prousek’s book "Fenton chemistry in biology and medicine" by, to quote reaction (15) on page 2330, a general depiction of Fenton-type reactions, to quote:

For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],
$\ce{Fe(II)/Cu(I) + HOX → Fe(III)/Cu(II) + •OH + X-}$ (15)
where X = Cl, ONO, and SCN.

Assuming HOX could also represent, in the current case with NO2 creation, HONO formation.
Now, a corresponding possible reactions in the current context:

$\ce{Fe(II)/Cr(II) + HONO → Fe(III)/Cr(III) + •OH + NO-}$

where $\ce{NO-}$ is more correctly depicted as hyponitrite anion, $\ce{N2O2(2−)}$.
Further, in the human body, the presence of ascorbic acid is known to provide some recycling of ferric (and likely chromium/other transition metal) ions back to ferrous to keep a Fenton-type reaction active (as a reference, see, for example, 'Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions' by Edgar Vidrio, at al.
The cited work by Vidrio notes oxygen and solvated electrons could lead to the creation of the superoxide radical anion, capable of reducing Fe3+ back to Fe2+, thereby promoting cyclic activity in the reaction system:

$\ce{O2 + e- ⇌ •O2-}$
$\ce{H+ +  •O2- ⇌ •HO2}$
$\ce{•O2- + Fe(III) ⇌ O2 + Fe(II)}$ (a reversible reaction)
$\ce{Fe(II) + H+ + •HO2 -> Fe(III) + H2O2}$
$\ce{Fe(II) + H2O2 + H+  -> Fe(III) + •OH + H2O}$

Where the possible creation of the powerful hydroxyl radical is problematic from a health perspective as it can extract a hydrogen atom from any organic (forming water), including DNA, potential leading to health issues.


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What happens to the third hydroxyl group in a phosphodiester linkage?
A phosphodiester bond occurs when exactly two of the hydroxyl groups in phosphoric acid react with hydroxyl groups on other molecules to form two ester bonds.
But what happens to the third hydroxyl group; i.e. why is it an oxygen and not hydroxyl?

At physiological pH it is simply deprotonated, i.e. it becomes $\ce{-O-}$ instead of $\ce{-OH}$. That's why DNA is said to have a negative charge.
I realise that this image on Wikipedia omits the negative charge, which is strictly wrong (the negative charge is probably implied because writing it out every time would clutter the diagram). A proper biochemistry textbook will show that it is deprotonated. Here's an image from Berg & Stryer's Biochemistry, 7th ed:

Here's one from Alberts et al. Molecular Biology of the Cell, 6th ed:

Whether you draw the negative charge as being delocalised across the two oxygens (as Berg & Stryer have done) or being localised on one oxygen (as it is in MBotC) is not relevant.

This is definitely helpful, am I to understand that the two oxygen "groups" not participating in bonding each have a resonance/bond order of 1.5? This is definitely not mentioned in my cellular biology book, and it seems relevant. Also, if the third hydroxyl group is deprotonated, does that technically make DNA a conjugate base?

Yes and yes! Technically the bond order is 1.5 but it's really not that relevant in biology. And deoxyribonucleic "acid" is in fact not an acid - it's actually the conjugate base of the acid.


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What are good lubricants for Bunsen burners?
We've been running into a lot of problems the last several years with our laboratory Bunsen burners becoming jammed and/or corroded.  I've just come to find out that our lab tech has been lubricating the burners with glycerol, which seems wholly inappropriate to me. I guess she figured that if it worked for inserting glass tubing into rubber holes, it would work for anything. To me, something like a WD-40 would seem more appropriate based on household experience, but I'd like to hear if anyone else has found something for this exact application that works well for them.
I'm currently experimenting with white lithium grease.  An engineer friend also recommended Permatex anti-seize lubricant.  These seem to work well on a very short term, but we'll have to see how it holds up throughout the next several weeks.
Anyone have a favorite lubricant for something like this?

You're right glycerol is hygroscopic which makes it inappropriate as a metal grease.

We've been running into a lot of problems the last several years with our laboratory Bunsen burners becoming jammed and/or corroded. I've just come to find out that our lab tech has been lubricating the burners with glycerol.

Does the issue itself seem to have been caused by the glycerol?
Admittedly, we don't use many Bunsen burners nowadays as they're seen as being fairly unsafe in the kind of organic lab I work in where we handle large amounts of solvent, however we do have a well vented hood with several burners in.
The ones we do have are very old indeed (pre 50's) and still work fine without the need for any maintenance (i.e. greasing).
Bunsen burners themselves are incredibly simple pieces of kit, if you do have issues with corrosion (i.e. because of an overly humid environment or harsh chemical vapours), the most sensible thing would be to disassemble, remove the corrosion and re-assemble.  The use of any kind of grease should be un-neccesary, especially as very few of them (other than very specialised products) can withstand long term exposure to heat without degradation.

I think WD40 seems too fluid for this application. A more viscous, greasy lubricant seems more appropriate to me. Also, lithium grease can release flammable vapors based on this MSDS. I'm not sure if that is just the aerosol or if that applies to the residue after application, as well.
I thought Dow-Corning's high vacuum grease was nonflammable and good for high temperatures, but I don't see it listed as being nonflammable. Dow's Krytox (Amazon) seems appropriate, though. It specifically says nonflammable and good up to 200 °C. Depending on what part of the Bunson burner you're applying it to, this may be reasonable. I'm assuming it's the base, where it twists. If that is the case, this area shouldn't get too hot. I'm always able to adjust mine with bare hands while it is burning.

My thought went to dry lubricants.
There are liquid and aerosol suspensions and dry powder dispensers.
Teflon or graphite might work well in your situation if contamination is not an issue.  The temperatures are not likely to be a problem.  Various silicone oils and greases are also known to handle higher temperatures (cf. vacuum greases in the other answer) but may also outgas and contaminate something one day.
Fitting a brass or other lubricating strip or wear pad in the moving parts may be appropriate for a functional restoration of a treasured vintage burner but replacing the burner may give you a decade of trouble free life if the glycerol lubrication habit is given up.


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There are three things in the world that smell like fish. But why?
As a philospher once said, there are three things in the world that smell like fish. One of them is fish.
What connects these three things that gives them this particular, characteristic smell? What chemical exactly do we smell?

Small alkyl amines, particularly trimethylamine.


text:
How do you figure out the positional and chain isomer possibilities of a compound?
I need to figure out the chain and positional isomers of C4H7Cl. I am not looking for someone to directly give me the answer, I just want to know what steps are needed to be able to figure them out for a given compound.
This is different to my prior question. Here I am asking for the technique, not the direct answer. Thanks.

Possible duplicate of What are some examples of chain and positional isomers of C4H7Cl?

No. Please re-read.

You should have edited your old question, don't ask second time about the same topic.

Would it have reappeared on the main question page if I edited it?

It surely would.

It would need to be reopened before answer could be posted - editing would need to validate opening.

http://chemistry.stackexchange.com/questions/7452/constitutional-isomers-of-cis-1-2-dibromocyclopentane

Is the parent hydrocarbon unbranched? That would be n-butane, $\ce{C4H10}$. Replacing one hydrogen atom by chlorine would lead to $\ce{C4H9Cl}$. Apparently, you're not having that!
Move a methyl group in n-butane, yielding 2-methylpropane, $\ce{C4H9}$. Here, replacing one hydrogen atom by chlorine gives $\ce{C4H8Cl}$. You're not having that either!
Does your target molecule have $\ce{C=C} $double bonds? Calculate the number of double bond equivalents in $\ce{C4H7Cl}$! I count one.

Work your way down from here.


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Can oxidizing agents act as comburents?
Comburent, to the best of my knowledge, refer to substances that catalyze combustion reactions without reacting itself or being flammable independently, such as potassium chromate. How would oxidizing agents assist combustion reactions?

Well, potassium chromate is a useful oxidizing agent, so you may have already answered your title question.

Sorry for not clarifying, I meant to say that potassium chromate is a strong oxidizer, I didn't know it was a comburent. Also, I don't understand how or why.

Comburent would seem to include materials which are themselves combustible. http://www.thefreedictionary.com/comburent


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How does a cathode half reaction work with different metal electrode and electrolyte
If you had a galvanic cell with the cathode half cell using silver as its electrode and Copper Sulfate as its electrode, will the copper (s) coat the silver electrode?

Do you mean copper sulfate as an electrolyte ans copper metal as an anode? If yes, then the answer to your question is yes, as silver is more noble metal than copper.

Sorry I meant silver as the cathode half cell electrode with the electrolyte of the cathode half cell being copper sulfate. When the electrons are given from the anode reaction, will the copper ions in the cathode electrolyte coat the silver cathode electrode?

The electrolyte for the two half cells has to be copper sulfate. The copper  metal must be the anode. In this case copper will coat the silver cathode.  We call copper the sacrificial anode.


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Is chloronitrobenzene and nitrochlorobenzene same?

Is it 2-chloronitrobenzene or 2-nitrochlorobenzene?
If we take nitro group as the substituent, it will be 2-chloronitrobenzene. But if we take chloro group as the substituent, it is 2-nitrochlorobenzene.
What if we start numbering the carbon atom from the carbon to which the nitro group is attached, it should be o-chloronitrobenzene. But no, carbon atom numbering starts from the group to which chloro group is attached.
Is it the fact that chlorobenzene forms ortho-para substituted product and it is named accordingly or according to some IUPAC rules?

Related: Priority when naming benzene derivatives

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of citation of substituent prefixes and the numbering of the corresponding locants are laid down in two different rules.
Simple substituent groups that are named by means of prefixes (such as ‘nitro’ and ‘chloro’) are arranged alphabetically.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the compound given in the question is x-chloro-y-nitrobenzene (not x-nitro-y-chlorobenzene).
The locants x and y are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings (e.g. chlorobenzene or nitrobenzene); but in substitutive names of disubstituted benzene, numerical locants have to be used to distinguish the 1,2-, 1,3-, and 1,4-isomers.
Note that the letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4-isomers of disubstituted benzene (e.g. o-chloronitrobenzene). According to current IUPAC recommendations, this usage is strongly discouraged and it is not used in preferred IUPAC names.
The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), the compound given in the question could be named as 1-chloro-2-nitrobenzene as well as 2-chloro-1-nitrobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-chloro-2-nitrobenzene rather than 2-chloro-1-nitrobenzene since chloro is cited first as a prefix in the name.


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Calculating Raman Intensities from Static Raman Activities
Whenever I want to calculate a Raman spectrum, my quantum chemical programs give me Raman scattering activities. As the common Raman spectrum is dependent on the incident wavelength of the used laser this seems quite logic, because the program cannot guess which laser the experimentalists could have used.
At least in Gaussian 09, for some methods, there is the possibility to calculate the Raman scattering activities for a certain incident light frequency through cphf=rdfreq.
As far as I know, the latter would give me the dynamic Raman activities and whenever I don't enter the incident light frequency I will get static Raman activities.
To get the intensities out of the activities, I was given the following equation, which I was told only holds for the dynamic activies:
$$I_k \varpropto \frac{S_k}{\tilde{\nu}_k} \frac{\left(\tilde{\nu}_\text{in} - \tilde{\nu}_k\right)^4}{1-\exp\left(-\frac{h c \tilde{\nu}_k}{k_B T}\right)} $$
But what if the calculation method that I use is not able to produce dynamic Raman activities (i.e. because it's not implemented) and I am only able to get those static Raman activities? How do I get Raman intensities from them?

You have two options in that case.

You can try to understand the frequency dependent (or dynamic)  property in the formulation for Raman activity. This property is a bit complex and is known as polarizability. For Raman spectra,  certain invariants of this tensor are involved in the intensity formulation. Be careful with this.

You can then find the formulae for activity first (by refering to the Gaussian manual and cheking referred literature there). After having a formula, then calculate the components required (including polarizability) individually, and finally obtain the dynamic activity.

You can calculate static activity very accurately and scale it using dispersion relations. Dispersion relations are relations which relate a dynamic property to static ones. (I am not sure if Gaussian has dispersion relation calculation.) In some cases you can calculate dispersion coefficients also and use them for scaling. Important is that you need to verify whether static activity can be scaled in such a manner. (I know that polarizability can be scaled but, not near absorption regions.)


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What temperature is required to burn pencil lead graphite?
I tried to burn a bunch of 8B pencil lead in a gas flame the other day. None of it actually caught fire, but being made of carbon (in the form of graphite) should cause it to burn, as graphite is essentially purified coal. What temperature is required to set it burning, and is extra oxygen required?

Well, it depends. I was easily able to ignite graphite by using point 5 - point 9 lead, using a simple concoction of alligator clips (Copper) holding up the lead, and connecting it to a mere 18 volts of energy!

8B (the softest lead) contains about 15 % clay.

@user40353 if it didn't continue glowing after removing the electricity then it wasn't really "burning", just oxidizing endothermically.

Nothing endothermic about graphite oxidation.

Curt F. I should have said "dissipating more energy than it liberates"

could it help to grind up the graphite then try burning it

You have two problems.
1.) pencil lead graphite is actually a graphite-clay composite.
2.) carbon does not sustain burning easily unless it is held at very high temperatures.
Pencil lead will not burn effectively for the first reason as clay is non-combustible and smothers any fire load. If you did have pure graphite to burn, a simple flame wouldn't be hot enough to combust it.  You need to sustain temperatures of around $1000\mathrm{-}2000~^\circ\mathrm{C}$ in order for pure carbon to burn.

Probably worth noting that the situation would change if you could enrich the air with oxygen.

@CurtF. Eh, not really. Graphite is heavily used in glassblowing/lampworking. Those flames are oxygen/propane flames and you can put chunks of solid graphite directly into the flame without any obvious combustion occurring. You might get a bit from carbon dust, but the block doesn't degrade at all.

Is a flame really an oxygen-enriched environment?  Also, http://my.rsc.org/video/318

Then how can coal be burned so easily?

@BadAtChemistry Coal is not pure carbon like graphite, but rather really heavy hydrocarbons like anthracene

It’s all in the bonds. It doesn’t matter what you have, if it is in a tight covalent bond structure it is going to take a lot of input energy to trigger oxidation. What many people don’t understand is that anything can burn unless it has already been fully burned. Other than that, you can burn it, no questions ask. Eventually, with a high input energy, graphite carbon will tear off its tight covanent structure and pick up some oxygen, but that might take a lot more energy than the reaction will release, meaning that you probably won’t get a self sustained combustion flame going. No cool fire, but definitely some toxic CO and CO2 and some very hot carbon!


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How to optimize the molecular geometry under PyMol?
I already now about the Optimize.py plugin, which seems to be working well, but I don't see an option to prevent it from changing double bonds to single ones. If Optimize isn't capable of such, how can I achieve a minimized molecule from a hand-built one?
As far as I know Optimize uses Open Babel. I could imagine that the problem is, that it doesn't export the state of bonds at all.

Disclaimer: I don't typically use PyMol, so I only made a quick look at the Optimize.py script.
You're right that the problem is with the bond orders. The Optimize.py script uses a PDB file as an intermediate representation that it sends to Open Babel for force field optimization and other properties.
The problem is that PDB does not support bond orders. So yes, you don't get double-bond information back, and I can't verify if Open Babel is getting bond orders from PyMol and Optimize.py.
I looked through the PyMol commands and it seems as if there's a way to write an MDL Molfile (which would have bond orders) but not read one - PDB is the only option for reading and writing in a PyMol script.
My suggestions:

Either push the PyMol developers to support other formats (e.g., Sybyl .mol2 or MDL Mol) in the scripting commands.
Use another program like Avogadro to draw and optimize the geometry and export to read into PyMol later.

(Disclaimer: I'm an Avogadro and Open Babel developer, but there are many programs that could be used.)

Unfortunately I have a "QGLFramebufferObject: Framebuffer incomplete, missing attachment" error with Avogadro, but not with Avogadro2 which works correctly

What OS do you use? (and what graphics driver?)

Linux with ati fglrx