Find the number of electrons I am not understanding,why do we have to subtract the charge from the atomic number while finding the number of electrons.

Suppose an atom of element 'X'

Let's assume the charge on this atom is 0 - hence, the number of protons in the nucleus and the number of electrons revolving about the nucleus must be equal, which results in a net 0 charge on the atom.

Now, lets assume that another atom of 'X' has +1 charge over it.
Well, how could this happen?
There are two ways for this to happen.

1) Nucleus gains an addition proton (+1 charge)

2) Atom loses an electron(-(-1) charge)

Both of these cases result in the atom gaining +1 charge.

Case 1 is very difficult to achieve, hence its not a good idea to think about in any chemical reaction (in fact it is not one at all, see comments!), instead consider case 2.

Notice, losing an electron results in +1 charge over the atom - another way to think of this is that we have ruined the balance that existed between the number of electrons and number of protons.

So we could say the above sentence in your words, we subtract the total charge present on the atom from the atomic number (remember, originally when the atom had no net charge, atomic number = number of protons = number of electrons. This is no more the case now!) to get the number of electrons present in the atom.

This is a very basic form of conservation of charge.


Wasn't Rutherford's assumption of gold foil to be made of single layer wrong? My teacher taught about Rutherford's gold foil experiment today. Sir said that, Rutherford used gold foil of thickness $10^{-7}$ m and concluded atom to be of size $10^{-10}$ m. Sir also said that Rutherford assumed foil to be consisting of only single layer of gold atoms.

What I thought was, if Rutherford used gold foil of $10^{-7}$ thickness, which is greater than the size of atom [$10^{-10}$] which he had concluded. So, gold foil should be made of more than single layer of atoms. But this would be against his assumption. If we assume gold foil to be consisting of more than one layer of gold atoms. Then we will be in need to know the structure of gold, i.e to know whether it has atoms in ABAB.. or AAAA.. or ABCABC... type of arrangement. My book says that gold has ABCABC... or ccp arrangement, it mean that there is no vertical allignment of atoms, thus there will be no vertical allignment of nuclei. It makes significant changes to Rutherford's predictions with respect to deviations. Won't it?

You were confused about the result of the experiment. It did not measure the size of the atom, but the size of the nuclei. Nucleus is the area of concentrated positive charge in the core of the atom, and is only a very small fraction the size of the atom itself.

And the result, $10^{-15}\ \mathrm{m}$, was the size of the nuclei, while the size of gold atoms has the order of magnitude of ångströms ($1\ \mathrm{\mathring{A}}=10^{-10}\ \mathrm{m}$).

The measurement comes from the analysis deflection angle of alpha particles when hitting the nuclei. The one-layer-of-atom assumption leaded to Rutherford to use a mathematical model for one single collision to analyze the angular distribution. Because the nucleus is so small, even if there were a small number of layers, most alpha particles will still only have one collision and the result will be correct.

However, the alpha particle was passing through some hundreds of atoms and many particles actually were deflected multiple times. The calculation would be much more complicated. Even if he knew the number of layers of atoms and tried to do it the right way, he might not be able to arrive at the correct number. Of course people back then would not be able to know that, but nevertheless, the existence of nucleus is still evident and Rutherford did almost get the order of magnitude right.


Can the concentration of a sodium carbonate solution be determined using volumetric analysis? In the titration of sodium carbonate with hcl , isn't the first end point determined by phenol phthalene is equivalent to the volume of sodium carbonate

because my text book say thats the volume at the first end point is just half the sodium carbonate and the seconds end point is the whole sodium carbonate.

I know that at the first end point all the sodium carbonate all be converted to sodium bicarbonate but why shouldn't i stop here?!

The titration of hydrochloric acid with sodium carbonate is a common school practical, used to determine the concentration of the acid (standardisation).

In the first step, as you rightly point out, the sodium carbonate reacts with one equivalent of hydrochloric acid. The phenolphthalein shows the pH change associated with the formation of the sodium hydrogen carbonate (aka bicarb). You could use this value to calculate concentration, however, in order to ensure your answer was correct, you'd need to run this titration at least twice more, averaging your answers and discounting any anomalous titres.

$\ce{Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)}$

As the bicarbonate is also reactive with hydrochloric acid, you can do something known as a 'double titration'. After the first titration, you can immediately do a second titration, this time using a different indicator, such as methyl orange. By doing this, you get a second value. If your titration is good, the values for the first and second titrations should be equivalent.

$\ce{NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) }$


How do orbitals overlap in difluorosilane? In $\ce{SiH2F2}$ four orbitals of silicon $(\mathrm{3s},$ $\mathrm{3p}_x,$ $\mathrm{3p}_y,$ $\mathrm{3p}_z)$ overlap with two hydrogen $(\mathrm{1s})$ and two fluorine $(\mathrm{2p}).$
I want to know how these orbitals are paired for overlapping.
Is it based on strength of overlapping?

A simple model for bonding in this molecule is taking $\ce {Si}$ to adopt $\ce {sp^3}$ hybridisation, as suggested from the tetrahedral geometry of the molecule. Subsequently, we can refine our model by applying Bent's rule (see more about it here), which tells us that $\ce {p}$ orbital character concentrates towards more electronegative substituents. Since the electronegativity of the elements has the following order: $\ce {F}$ > $\ce {H}$ > $\ce {Si}$ with Pauling electronegativity values of $\ce {3.98}$, $\ce {2.20}$ and $\ce {1.90}$ respectively, the $\ce {3p}$ character of $\ce {Si}$ would concentrate towards $\ce {F}$ while $\ce {3s}$ character concentrates towards $\ce {H}$. Thus, we would have $\ce {Si}$ using $\ce {sp^{3+x}}$ orbitals to form the bonds with $\ce {F}$ and $\ce {sp^{3-x}}$ to form the bonds with $\ce {H}$, where $x$ is some small value between $\ce {0}$ and $\ce {1}$.


Under what circumstances that Liquid-Liquid extraction would be good to use? Under what circumstances that LLE would be useful ?
Are the 2 top reasons -
1. Difference between the boiling points of the components are very low or similar.
2. For Temperature sensitive compounds as there is no heating which may cause thermal decomposition.

are there any other reasons ?

The above reasons to use LLE are focused on the compound thermal similarities and instabilities.

But it does not take into the account the amount of compounds. While distillation separation can be great for major components, it may fail for minor one, and definitely for trace ones.

The LLE is frequently used for separation and content enrichment of trace amounts, in wide range of grams to nanograms per litre.

Using the approach of chemistry of complexes, it may also lead to great separation of non-volatile compounds/ions/elements.

LLE is e.g used for mutual separation of  nuclear fission products and not spent fuel by the PUREX method.


How to choose proper active space in method such as CASPT2 and CASSCF? I am confused about other research paper which use CASPT2 method. I am not clear about how to choose active space. What can I refer to? (tutorial reviews, papers, or any other things)

Choosing the active space in any calculation with a multiconfigurational reference (CASSCF, RASSCF, GASSCF, CASPT2, NEVPT2, MRCI, the list goes on) requires more chemical intuition than more commonly-used "black box" quantum chemistry methods, such as DFT. Here, chemical intuition means preexisting knowledge about 1. the system under investigation, and 2. the properties being calculated. For example, a minimal active space may be sufficient to qualitatively capture static correlation in a system, but the active space might change when targeting excitation energies, and might be different again when calculating spin-orbit coupling constants. More specifically, it is understanding what orbitals are most important in composing your wavefunction and have the largest influence on molecular properties of interest. Unfortunately, this can also involve quite a bit of trial and error.

Some very general recommendations on the workflow for performing multiconfigurational calculations are in the numbered list in this answer (yes, a shameless plug).

Here is a more complete list of references I've found useful in my own research:

[1], [2] are slides from the group that develops GAMESS, one of the premier packages for multiconfigurational calculations. The second set of slides in particular has examples that highlight the chemical motivation for different active spaces depending on the molecule and the property (internal rotation in ethylene, internal rotation in $\ce{H2C=NH}$, dissociation of $\ce{H2C=O} \rightarrow \ce{CH2} + \ce{O}$, isomerization of bicyclobutane to 1,3-butadiene, and others).
[3] (paywalled) is a more formal overview of the above. Excellent.
[4] is a set of very detailed and theory-focused slides from the original authors of DALTON. I link to them only for completeness; there aren't many practical examples here.
[5] (paywalled) is a comprehensive review article from the authors of COLUMBUS, with many examples. Excellent.
[6] (paywalled) is an article from the authors of COLUMBUS on the application of multiconfigurational methods to geometries of small molecules. It has good explanations on the choice of active space based on electronic configurations.

In addition, there are many other online resources that are more tutorial oriented: [7], [8], [9], [10]. These may appear to be program-specific, but the chemistry is transferable between different programs; only small technical details in how to set up calculations should be different.

Finally, some programs have nice examples or helpful hints in their documentation. DALTON (PDF), Molcas, and ORCA (PDF) come to mind.

References

[3]: Schmidt, M. W.; Gordon, M. S. The construction and interpretation of MCSCF wavefunctions. Annual Review Of Physical Chemistry 1998, 49, 233-266. DOI: 10.1146/annurev.physchem.49.1.233
[5]: Szalay, P. G.; Müller, T.; Gidofalvi, G.; Lischka, H.; Shepard, R. Multiconfiguration Self-Consistent Field and Multireference Configuration Interaction Methods and Applications. Chemical Reviews 2012, 112, 108-181. DOI: 10.1021/cr200137a
[6]: Shepard, R.; Kedziora, G. S.; Lischka, H.; Shavitt, I.; Müller, T.; Szalay, P. G.; Kállay, M.; Seth, M. The accuracy of molecular bond lengths computed by multireference electronic structure methods. Chemical Physics 2008, 349, 37-57. DOI: 10.1016/j.chemphys.2008.03.009


Determining the value and units of the rate constant Consider this initial-rate data at a certain temperature for the reaction described by
$$\ce{2NOBr(g) -> 2NO(g) + Br2(g)}$$
\begin{array}{cc}\hline
\ce{[NOBr]_0 (M)} & \mathrm{Initial \,rate \, of \,Br_2}\, (\pu{M/s}) \\ \hline
0.600 &1.08 \times 10^2 \\
0.750 & 1.69 \times 10^2\\
0.900 & 2.53 \times 10^2\\ \hline
\end{array}
I'm not understanding how to find initial rates from the data given and then use that to find the rate constants. I also don't really understand how this information can give me the orders of reactions, for example when does the number double, stay the same, or quadruple, etc.

Alright, so in the reaction

$$\ce{2NOBr(g) -> 2NO(g) + Br2}$$

the rate law is

$$\text{rate} = k[\ce{NOBr}]^x$$

where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.

By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get

$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$

$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$

The $k$'s cancel out.

$$2.25 = 1.5^x$$

$$x = 2$$

The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.

$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$

$$\begin{align}\text{rate after concentration} &= \left(2[\ce{NOBr}]\right)^2\\
&= 2^2[\ce{NOBr}]^2\\
&= 4[\ce{NOBr}]^2\\
&= 4 \times \text{rate before doubling concentration}\end{align}$$

A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.


Nitric Oxide Dimerization This is the molecular orbital diagram for nitric oxide:

Now, we know that nitric oxide can dimerize. However, why does nitric oxide dimerize as to form an $\ce{N-N}$ bond? Why not an $\ce{O-O}$ bond? Wouldn't the lone electron be more likely "possessed" by the oxygen since oxygen is more electronegative (if we were to step away from molecular orbital theory for a second).

Is this a Coulombic effect? Oxygen, by nature of being more electronegative, attracts more electron density to itself. This in turn would weaken any hypothetical $\ce{O-O}$ bond due to electron-electron repulsions.

Something is not correct with your MO diagram.  Here is a correct MO diagram for NO

Note that in this diagram the oxygen atomic orbitals are lower in energy than the nitrogen AOs.  This makes sense since the AOs of the more electronegative element should always be at lower energy than the AOs of the more electropositive element (in your diagram the relative energies of the oxygen and nitrogen AOs are somehow reversed).  The energetic ordering of the starting AOs is important as we'll see in a minute.

The closer in energy a MO is to an AO from which it was formed, the more character of that AO it shows.  So let's examine the nature of the single electron in the $\ce{\pi_{2p}^{\ast}}$ orbital since this electron will be involved in forming the new bond when $\ce{NO}$ dimerizes.  This MO is formed from both the nitrogen and oxygen p atomic orbitals, but it is closer in energy to the nitrogen p AO.  Therefore, we would expect this electron to have more nitrogen and less oxygen character (in other words it will reside more around nitrogen than oxygen).  Because this electron has greater density on nitrogen, dimerization to form an N-N bond is more likely than dimerization to form an O-O bond.


How were the elements discovered before modern technology? I recently was in need of zinc for an experiment. So I resorted to obtain it from some dry cell. But I wondered if it was pure zinc or was it $\ce{ZnO2}$ or any other compound of $\ce{Zn}$. It then occurred to me that I had knowledge of existence of pure $\ce{Zn}$ and its compounds, but how would its discovery have happened without the prior knowledge of its existence?

How did scientists discover the elements without knowing of their existence and unique properties?

Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?

Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?

In many cases, they weren't - the ancient... maybe we can't call them scientists. Let's call them "natural philosophers." The ancient natural philosophers (up until about 1700) only knew about a handful of elements. This is part of the reason that it took so long to put the periodic table together.

At the time, an element was something that was not reducible to smaller components. The ancient philosophers tended to conflate properties of things with this idea of fundamental components, and as a result, things like earth, air, fire, and water were thought of as elements. When we say they "knew about" modern elements, what we mean is that they knew of the substances - they did not necessarily know that they were chemical elements in the sense that we define them today.

That modern concept of an element defined by the number of protons in the nucleus wasn't developed until around 1913. The wikipedia article on the history of chemical element definitions gives a good overview of how the definition has evolved over time.

The time period you are probably most interested in is that between the "ancient" era (pre-history up until the mid-1660's or so) to the beginnings of the modern era (say early 1900's). During this period, chemistry was evolving from alchemy into a modern science. The systematic investigation and communication of results was occurring, and a large body of known but poorly understood chemical "recipes" was available to chemists. Using a range of reactions and isolation techniques, they basically just randomly tried things (maybe not completely randomly) until they reached a point where it seemed like something could not be decomposed further, or until they identified a new substance. In many cases, they would discover the new element, but not be aware that it was an element - in other words, they could isolate and identify a substance, but because our collective understanding of chemistry was still very limited, they had no way of knowing that it was an elemental substance.

Wikipedia's list of elemental discoveries gives details about the discovery of each element - although you should keep in mind that "discovery" in this case means "identified as a new substance," not necessarily "identified as a chemical element."


In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital? s-p mixing in $\ce{N2}$ causes the $1\sigma _g$ orbital to decrease in energy, the $2 \sigma _g$ orbital to increase in energy (to higher than the $1 \pi _u$) yet the $1 \sigma _u$ is left unchanged by the interaction. Why is this?

My lack of understanding may be as a result of the fact that I am unclear as to what exactly it is that is mixing. Clearly they are the s and p orbitals but is this within the atom or is the s orbital on one atom interacting with the p on the other?

Some clarification as to why the $1\sigma _g$ orbital to decreases in energy, the $2 \sigma _g$ orbital to increases in energy would also be much appreciated.

Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals.  Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance.  With that understanding, let's step through your diagram and questions.

It might be useful to take a look at the $\ce{N2}$ MO diagram here.  It's just like your diagram. but they break down the steps and it might be a bit less confusing.

In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals.  If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.

$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds.  In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized.  We form this bond by

Mixing the s and  a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.
Same thing happens on the other nitrogen atom
As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)

Now for pi bonds,

The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom.  This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital

With that background, let's now look at your questions

  s-p mixing in N 2   causes the 1σ g   orbital to decrease in energy,
  the 2σ g   orbital to increase in energy (to higher than the 1π u  )
  yet the 1σ u   is left unchanged by the interaction. Why is this?

I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered).  The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.

  Clearly they are the s and p orbitals but is this within the atom or
  is the s orbital on one atom interacting with the p on the other?

Hopefully I answered this above.  I tried to distinguish between atomic and molecular orbitals.  If I haven't been clear, post back.

  Some clarification as to why the 1σ g   orbital to decreases in
  energy, the 2σ g   orbital to increases in energy would also be much
  appreciated.

Hopefully what I wrote above clarified this as well, let me know if it didn't.


Why does cobalt have no negative charge? I would like to know why cobalt cannot have a negative charge (or at least why a negative charge for cobalt isn't typical). I am not sure where I have gone wrong in my reasoning.

The electronic configuration for cobalt is:
$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7$

Clearly, you could give cobalt a $+2$ charge in order to get a more stable ion as follows

$\ce{Co^{2+}}$: $[\ce{Ar}] 3d^7$

And from there, you could go further and remove $2$ more electrons from the $d$-orbital to become even more stable like this:

$\ce{Co^{4+}}$: $[\ce{Ar}] 3d^5$

If this logic is correct, then could I apply this same thinking but for adding electrons? Such that

$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7\to$ $\ce{Co^{3-}}$: $[\ce{Ar}] 4s^2 3d^{10}$

Since the $d$-orbital has been filled, the atom should have become more stable. Is it true?

Thus, am I able to argue that Cobalt has $3$ main ion charges: $+2, +4, -3$?

Cobalt can be found in various oxidation states including negative ones such as Co(-I) in NaCo(CO)4. In this instance however the excess electron density on the cobalt center is relieved by backbonding with the CO ligands and therefore such a low formal oxidation state of the Co atom is stabilized. On the other hand the free Co$^-$$^1$ ion cannot hold the extra electron first due to the low electronegativity of Co as TAR86 hinted out in his comment and second due to the fact that the extra electron is placed in a d orbital which is not very  penetrating and thus shielded by all other electrons. Hence, it is unlikely to encounter compounds such as [Na]$^+$[Co]$^-$ where there are no ligands on the cobalt to stabilize the excess electron density. The opossite applies when you remove electrons from Co (and any other transition metal): the atom becomes positively charged and the nucleus attracts the remaining electrons even more effectively (Higher Zeff). Therefore, positive oxidation states are more common.


Color difference among oxovanadium species Why dioxovanadium $\ce{[VO2]+}$ is pale yellow, whereas orthovanadate $\ce{[VO4]^3-}$ is colorless?

Both of them should be colorless if we ignore charge transfer for a moment, because in both cases vanadium is at +5 oxidation state.

Now considering the charge transfer is the reason for the pale yellow color of dioxovanadium species, why orthovanadate does not show charge transfer band?

The colour will depend on several factors. You've identified the oxidation state (+5 in both cases) as one.
Other factors are:

Ligands
Complex shape

With $\ce{[VO2]}^{+}$, the actual complex ion is $\ce{[VO2(H2O)4]}^{+}$ (Taken from ChemGuide). So our ligand is water, and the shape is tetrahedral (or square-planar).

The $\ce{[VO4]^{3-}}$ ion usually only exists at high pH in solution but polymerises at lower pHs. These polymeric ions can have colour (e.g. $\ce{[V10O28]^{6-}}$ is orange).(Link for colour / polymerisation)
However, I could find no details as to the ligands (I would assume that they would be hydroxide ions at high pH) nor shape.

If we are talking about the solid, then the same applies... does the solid have any water of crystallisation. (Think copper sulfate... blue when "wet" and white when "dry").


How do equilibrium shifts affect solids? Say we have the following reversible reaction: $$\ce{NaOH{(s) }<=> Na^+{(aq) }$+ $OH^{-}{(aq)}} +10.6 \mathrm{kcal}$$

If we add $\ce{OH^-}$ and equilibrium shifts to the left, does that affect the amount of $\ce{NaOH{(s)}}$ present or does it remain constant?

I'm confused because I remember that pure solids and liquids don't affect equilibrium value, or does that only apply when we're talking about adding pure solids and liquids..not the effect on them by an equilibrium shift?

When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of product relative to reactants and the rate of the reverse reaction will increase relative to the forward reaction until equilibrium is reestablished. This means that the ions will recombine into a crystal lattice and form a precipitate. So, to answer your first question, no, the amount of $\ce{NaOH_{(s)}}$ does not remain constant; more of it will be formed if additional ions are added to a solution already at equilibrium.

The reason why pure solids are not factored into equilibrium expressions is that they are not in fact part of the solution. Any excess precipitate, irrespective of the exact quantity, has no impact on the composition of the solution at equilibrium because it exists outside of the solution by definition. If equilibrium is disturbed due to changing conditions, then more precipitate can form, or some of the existing precipitate may dissolve back into solution. Under any given set of controlled environmental conditions, however, the product of the concentrations of the dissolved species remains constant at equilibrium, which is precisely why the equilibrium will shift if conditions or concentrations change.

For a more rigorous and mathematical explanation, you could read up on dynamic equilibria, equilibrium constants, and thermodynamic activity.


Why does calcium chloride cause a greater freezing point depression in water than magnesium chloride? So I saw this equation for the freezing point depression, when you add a solute: $\Delta T= K_\mathrm fmi$

Where,

$\Delta T =$ The freezing point depression
$K_\mathrm f =$ The freezing point depression constant (Cryoscopic constant)
$m =$ Molality
$i =$ van't hoff factor.

I read online that $\ce{CaCl2}$ would be better than $\ce{MgCl2}$, but I cannot understand why in the terms of this formula since $i$ for both of them is the same and $K_\mathrm f$ seems to be independent of the solute since I saw values of $K_f$ of water online. (without any solute given)

So, could someone answer why this happens? Because it seems to me that the $K_\mathrm f$ could be the only factor that would explain this. Specifically I would like to know if the charge or the electronegativity of the dissolved ions would make a difference.

As quantified in Fig. 17 of Manual of Practice for An Effective Anti-Icing Program (FHWA-RD-95-202):

By weight percent, MgCl2 depresses the freezing point of water to a greater degree than CaCl2, up to about 24%.  This is consistent with the freezing point depression formula, as Mg has less mass than Ca.

Beyond 24%, CaCl2 is superior.

This is because the eutectic point of MgCl2-water is at 21.6 percent, whereas the eutectic point of CaCl2-water is at 30%.

The freezing point depression formula is only valid at reasonably low concentrations.  It is certainly not valid beyond the eutectic point.


IUPAC name for a branched alkane What is the systematic name of this compound:

My try: 3-propylpentane.

You need to take the longest chain first, i.e. the basis of the hydrocarbon skeleton is hexane. It has a ethyl group at position 3, so the IUPAC name will be 3-ethylhexane.


Could compressed refrigerant be stored feasibly in residential setting? In Vapor-compression refrigeration, a refrigerant is compressed releasing heat (typically outside a building) and then decompressed in another location, absorbing heat.

A residential air conditioner works on this principle with a compressor outside the home and a blower over the decompression coils to circulate cold air.

Would it be feasible to compress ahead of cooling demand, storing the compressed refrigerant in a high pressure container for use at a later time of the day, thus shifting the power consumption to times of the day where there is less electrical demand?

By feasibility, I'm thinking of questions like:

How much storage volume would one need for a typical household usage over an afternoon of running air conditioning given a container type that would be available cheaply, ex: propane canisters?
What would be the sources of efficiency loss added by something like this?
etc.

I happen to have a bottle of refrigerant in my house and another in my car (besides the stuff already being stored in my AC units).

While this is definitely more of a physics question, or perhaps a question for the HVAC guys on the Home Improvement SE... to generally answer your question, there are many factors for cooling a home, like size and location, for example. But let's say your average 2-ton window-unit AC pumps 24,000 BTU's per hour... and 4000 BTUs requires 1 gallon of refrigerant per minute.... you need 6 gallons per minute (being released through an evaporator coil) to match a window unit AC.


Why do α-amino acids have a C-H bond at the α-carbon? I've just started studying biochemistry and I read that a general $\alpha$-amino acid looks like (ignore the ionisation for now):

My question is: why isn't the general $\alpha$-amino acid formula like $\ce{CRR'NH_2COOH}$ instead of $\ce{CHRNH_2COOH}$? Coming from a physics background, the latter feels more "natural" as it is more general.
Some possible reasons I've thought of (and disqualified):

Maybe the $\ce{C-H}$ bond dissociation energy is too high: I looked at a paper [1] but the bond energies seem to be not too high (relative to say $\ce{C-C}$ bonds.
Steric factors: I don't think this applies as having a small group, like say $\ce{CH_3}$, won't create a lot more steric repulsion relative to $\ce{H}$.

If the answer is along the lines of "it is defined this way because we actually see repeating units of these structures in proteins, so we naturally define the smallest subunit accordingly", then provide chemical grounds (if possible) to justify the fact that structures like $\ce{CRR'NH_2COOH}$ are not prevalent in proteins.
[1]: Moore, B. N., & Julian, R. R. (2012). Dissociation energies of X–H bonds in amino acids. Physical Chemistry Chemical Physics, 14(9), 3148-3154.

In the most general sense, the term α-amino acid does encompass doubly substituted structures like those you mention. After all, the name doesn't specify any particular substitution, or lack thereof, at the α-carbon. The name only really means that there is an amino group α to a carboxylic acid, i.e. one carbon away. Similar nomenclature is used to describe molecules where the amino group is 2, 3, ... carbons away from the carboxylic acid, also without any specification of substitution.

The disubstituted compounds you described, $\ce{H2N-C(R)(R')-COOH}$, can be synthesised in the laboratory. Pharmaceutical companies seem to have an interest in these compounds. Here is one example from a collaboration between Eli Lilly and IUPUI1 (there are many more - pharmaceutical companies like Pfizer2 even have some patents on these compounds):

Wikipedia actually has an example of a naturally occurring α-amino acid without a hydrogen at the α-carbon. Apparently, it is produced by some fungi:

However, the term α-amino acid is nearly always used to refer to the proteinogenic amino acids, which do happen to have a hydrogen at the α-carbon.
These proteinogenic amino acids were discovered many years ago. Generally, they are obtained by treatment of proteins with acid (which hydrolyses proteins into their constituent amino acids). Their chemical formulae were painstakingly identified by a series of decomposition reactions - back in the old days, there was no NMR. The history of the discovery of the proteinogenic amino acids is described in a review article3 from 1931(!) - which tells you how long ago these were discovered.
So, part of the reason why these aren't disubstituted is simply because we found them to not be so. However, I can understand if one complains of this conclusion being intellectually unsatisfying. So, here is some extra speculation.
One major step involved in the biosynthesis of amino acids is transamination. I don't have a list on me right now, but I think most amino acids go through such a step at one point in time.

The mechanism (described on Wikipedia, or in any biochemistry textbook) necessitates that one hydrogen is added to the α-carbon. [For the organic chemists out there, this is essentially an enzyme-mediated reductive amination.]
It turns out that, on top of this, it is also possible that such disubstituted amino acids are actually harmful. According to Weber and Miller, writing in the Journal of Molecular Evolution:4

Replacement of the α-hydrogen by a larger substituent, such as a methyl group, would also increase significantly steric hindrance around the amino and carboxyl groups. Steric difficulties have been encountered in the chemical synthesis of peptides with α-aminoisobutyric acid.
Also, the severely constrained stereochemical behavior of peptides of α-methyl-α-amino-n-butyric acid suggests that peptides composed of α-methyl substituted amino acids would not have the structural versatility of peptides derived from α-hydrogen substituted amino acids.
An additional factor involves the ribosomal peptidyl transferase, which would develop a specificity for either an α-hydrogen or an α-methyl of the amino acid. This would prevent the synthesis of proteins containing both types of amino acids.

References

Green, J. E.; Bender, D. M.; Jackson, S.; O'Donnell, M. J.; McCarthy, J. R. Mitsunobu Approach to the Synthesis of Optically Active α,α-Disubstituted Amino Acids. Org. Lett. 2009, 11 (4), 807–810. DOI: 10.1021/ol802325h.

Graham, S. R.; Mantell, S. J.; Rawson, D. J.; Schwarz, J. B. (Pfizer, Inc.) Amino acid derivatives. U.S. Patent 7612226 B2, Nov 3, 2009.

Vickery, H. B.; Schmidt, C. L. A. The History of the Discovery of the Amino Acids. Chem. Rev. 1931, 9 (2), 169–318. DOI: 10.1021/cr60033a001.

Weber, A. L.; Miller, S. L. Reasons for the occurrence of the twenty coded protein amino acids. J. Mol. Evol. 1981, 17 (5), 273–284. DOI: 10.1007/BF01795749.


Chirality of substituted adamantanes Is this compound chiral?

I can't tell where the chiral centres are because this polycyclic backbone is too confusing. I tried to draw a mirror image, but I got really confused and couldn't tell whether it was superimposable. And I don't have a molecular modelling set so I can't build a model.

Yes, this compound is chiral.

The polycyclic backbone is called adamantane. It has $T_\mathrm{d}$ symmetry, meaning that as far as chirality goes, it behaves like a perfect tetrahedron, somewhat like methane does.

It also has the interesting property that if you extend the C-F, C-Cl, ... bonds inwards, they will all meet at the same point. Those red dotted lines form a perfect tetrahedron.

Now, since the adamantane backbone behaves exactly like a methane molecule - why don't we just replace it with one? [n.b. this is not 100% rigorous but works here.]

Hopefully you'll agree that this is chiral. It basically behaves like it has a chiral centre right in the middle of the adamantane backbone.

If you really wanted to be sure, you could build a mirror image and verify that they are non-superimposable. (Hint: they are.)

Each of the four substituted carbons is connected to four different groups, and theoretically one would expect $2^4 = 16$ stereoisomers. However, because three of the four groups (the polycyclic backbone) are held in a rigid, fixed geometry, this number is reduced to 2.

There is more discussion of this class of compounds here.


What is the meaning of the '2' on some Compressed Gas Hazard signs in the lab? The main chemistry prep room at my institution (and also at the ambulance station) is labelled with the standard green compressed gas sign.  However, there is a small '2' included within the diamond.  None of the chemistry technicians, tutors or ambulance staff seem to know what it is referring to and I can't find any reference of it on the internet.  Would anyone here happen to know?

The related products of this online provider show that other hazards have different numbers but some also have two (e.g. flammable).  Is there a standard that is being referred to here?

The 2 refers to DOT (Department of Transportation) hazard class 2. This reference describes the different DOT hazard classes.

As summarized in the comments:

  Class 2 (compressed gas) is divided into three subclasses (2.1 - flammable, 2.2 - non-flammable, non-toxic and 2.3 - toxic). I imagine the diamonds for Class 2 don't contain the subclass number because there is a different diamond colour/symbology for each subclass (as opposed to for instance 5.1 (oxidising agent) and 5.2 (organic peroxide), which have the same symbol and colours).


pH of CH3COOH + HCl solution (shouldn't be a buffer) $\pu{0.05 L}$ of a solution containing $\pu{0.3 M}$ acetic acid $(\ce{CH3COOH}$, $K_\mathrm{a} = \pu{1.8e-5})$ are added to $\pu{75 mL}$ of a solution containing $\pu{0.02 M}$ hydrochloric acid ($\ce{HCl}$, a strong acid). Calculate the $\mathrm{pH}$.

Can I simply find the $[\ce{H+}]$ released by the reaction of $\ce{CH3COOH}$ with water and add them to the $[\ce{H+}]$ released by $\ce{HCl}$, add the two concentrations and then calculate $\mathrm{pH}$ that way? Or does some reaction happen that I'm not aware of?

Or does some reaction happen that I'm not aware of?

There are three reactions, and you are probably aware of all of them:

$$\tag{1}\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

$$\tag{2}\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$

$$\tag{3}\ce{Ch3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$

The concentration of $\ce{H+}$ in those three reactions is the same because it all happens in the same phase. So what happens in reactions (1) and (2) influences the equilibrium of reaction (3).

  Can I simply find the [H+] released by the reaction of CH3COOH with water and add them to the [H+] released by HCl, add the two concentrations and then calculate pH that way?

No you can't because if you go back to reaction (3), you are no longer at equilibrium. To avoid trying to fix one equilibrium while messing up another, there are two strategies:

a) Put everything into a system of equations, and solve them in one swoop (preferred method if you are using an equation solver).

b) Start with the major species ignoring minor species and reactions that don't influence major species much. Then, move on to those other reactions and the minor species. This is preferred when you have to do the calculations on paper and don't need the exact solution.

The approximate answer

After mixing and ignoring all acid dissociation reactions, the concentrations are the following:

c(acetic acid) = 50 / 125 * 0.3 M = 0.12 M

c(hydrochloric acid) = 75 / 125 * 0.2 = 0.12 M

So here are the steps:

Let hydrochloric acid dissociate and get hydrogen ion concentration and pH
Check whether acetic acid dissociates appreciably
Check whether water dissociated appreciably

For the first step, we get [H+] = 0.12 M, and pH = 0.92. These are tentative because we did not let the other acids (water and acetic acid) dissociate.

For the second step, we are not at equilibrium yet (no acetate yet). However, the pH is very low compared to the pKa of acetic acid, so unless the pH changes a lot, it won't dissociate much. Let's try to calculate the acetate concentration assuming that the change in acetic acid and hydrogen ion concentration is negligible.

$$ [\text{acetate}] = K_a * [\text{acetic acid}] / \ce{[H+]} = \pu{1.8e−5} * 0.12 / 0.12$$

So we are not making a big mistake if we say the concentrations of acetic acid and hydrogen ions did not change much due to reaction 2. If we want, we can update the hydrogen ion concentration from 0.12 M to 0.120018 M.

For the third step, we do what we always do when the pH is substantially acidic. We just calculate the hydroxide concentration assuming the hydrogen ion concentration is not much affected by water dissociation. It comes out as $\pu{8.3e-14}$. If we want, we can update the hydrogen ion concentration from 0.120018 M to 0.120018000000013 M and a pH of 0.92075. (None of this makes sense because we only had 2 significant figures for the dissociation constant of acetic acid.)

Because we assumed a hydrogen ion concentration of 0.12 M for the second step and 0.120018 M for the third, even though this is not quite true, reactions (2) and (3) are not quite at equilibrium for the concentrations we calculated. The good thing is that in most cases, that does not matter.


Deducing orbital degeneracy in geometries apart from octahedral or tetrahedral How do you determine the crystal field splitting pattern when given the structure of a complex?

It's quite simple for structures like octahedral or tetrahedral, but I find that the reasoning behind the patterns for the trigonal bipyramidal structure, for example, is somewhat arbitrary.

Continuing the example of a trigonal bipyramidal structure, the $d_{xy}$ and the $d_{x^2-y^2}$ orbitals are supposed to be degenerate, but the $d_{x^2-y^2}$ fully overlaps with one of the ligands on the complex, while the $d_{xy}$ partly overlaps with two, and I don't think we have enough information to assume that the two "parts" add up to one.

Do I just have to accept that this is how it works (as in, is this just a "limitation" of the crystal field theory)? Or is there some aspect that I am missing/misunderstanding?

The easiest way to do it is to look at what other people before you have done.

If you look at the $D_\mathrm{3h}$ character table, you see that $(x^2 - y^2, xy)$ transform together as $E'$.

I don't think there is a way of looking at diagrams of the orbitals and concluding that they are necessarily degenerate. As you said, there isn't "enough information to assume that the two parts add up to one".

It isn't really a limitation of crystal field theory, either. Crystal field theory isn't about looking at how much orbitals overlap with each other and saying "oh, I think those look about the same, they're probably degenerate". If that was the case then in an octahedral complex you would not be able to tell why $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ are degenerate, since the former only really overlaps with two ligands and the latter overlaps with four.

The degeneracy is predicted by symmetry, the application of which is integral to, and implicit in, crystal field theory. There is no CFT without symmetry considerations.


Polarity and stability Why is  cis-1,2-cyclohexadiol less polar than trans-1,2-cyclohexadiol?  I know it has something to do with stability and chair conformations, but I'm not sure how it relates to polarity.  I know cis-1,2-cyclohexadiol has a more stable chair conformation than trans.

If I assume you refer to 1,2-cylohexanediols, the two isomers to consider were (cis), and (trans)

than there are three conformers to consider.  For the (cis)-configuration, there is only the (axial, equatorial) orientation of the two O-substitutents:

This contrasts to the case of the (trans)-configuration, where the two substitutents may be either in axial, or (likely preferred by thermodynamics) equatorial orientation:

From the later picture, taking into account i) the relative orientation of the two hydroxyl groups towards each other as well in respect to the cyclohexane moiety and ii) the (assumed) conformational preference for this form over the conformer with two axial oriented hydroxyl groups, I would assume the two individual vectorial contributions along $\ce{C -> O}$ are better lined up than in the instance of the (cis)-configuration to yield a larger (global) dipolar moment.


The dissociation of hydrogen on UO2 and PuO2 surfaces: Homolytic or heterolytic? I have seen that $\ce{H2}$ can dissociate on a metal-oxide surface by two methods:

Heterolytically, forming a proton-$\ce{O}$ bond ($\ce{OH}$ group) and a hydride-metal bond ($\ce{M-H}$) on a metal-oxide surface.
Homolytically, where two $\ce{OH}$ groups are formed (adsorbed protons) and the two electrons of the $\ce{H2}$ molecule are transferred to low-lying $\mathrm{d}$ or $\mathrm{f}$ states of the cations. This leads to a change in oxidation state and a reduction of the oxide.

However, the reasons for the different behaviours are not obvious to me. Assuming two comparative MO compounds have the same crystal symmetry, how does the electronic structure of the metal impact which process dominates?

EDIT:

I am particularly interested in the example given by: Differences in the Existence States of Hydrogen in $\ce{UO2}$ and $\ce{PuO2}$ from DFT + U Calculations

Here, the preferential formation of a hydride ion $\ce{[(UO_{2})_{n}]^{+}H^{-}}$ or hydroxide ion $\ce{[Pu_{n}O_{2n-1}]^{+}[OH]^{-}}$ has been shown. The authors give the following explanation:

  The differences in the existence states of atomic hydrogen in the two dioxides are proposed to be dependent on the nature of $\mathrm{5f}$ electrons of uranium and plutonium; that is, uranium $\mathrm{5f}$ electrons are more delocalized and more favorable to participate in chemical bonding than plutonium $\mathrm{5f}$ electrons.

Why should the higher degree of electron itinerary (which enables electrons to more easily form chemical bonds) favour hydride formation?

This seems to conflict with: Interaction of hydrogen with actinide
dioxide (111) surfaces.

In this paper and others, $\ce{PuO2}$ has the more reactive surface. Here, the formation of a hydroxide is more favourable on $\ce{PuO2}$, and less favourable on $\ce{UO2}$. Therefore, I am confused as to how this availability of electrons changes.

This reference$\ce{^{[1]}}$ shows what can happen even in a single oxide, in this case magnesia, with different metal environments.

Magnesia is, of course, strongly ionic, so the dissociation of hydrogen to form $\ce{O^{2-}-H^+}$ and $\ce{Mg^{2+}-H^-}$ on the surface might well be expected.  Putting the magnesia as a thin film on silver does not change this.  But on gold the hydrogen dissociates in a redox fashion, forming only $\ce{O^{2-}-H^+}$ as the electrons are injected into the gold.  The lowered Fermi energy of the gold causes this transfer as the electrons "fall" into the lower energy state.  It's an example of gold acting as an electronegative element, as it does in compounds such as $\ce{CsAu}$, ultimately caused by the effects of relativistic mechanics on the electron structure of gold.

In the case of bulk uranium and plutonium dioxides, electron donation or withdrawal must come from the oxide itself.  With uranium (IV), the 5f orbitals are at relatively high energy, and where they have electrons those electrons are readily available for bonding and for donation -- but conversely, empty 5f orbitals are relatively poor electron acceptors.  So we get metal hydride formation like the $\ce{Mg^{2+}-H^-}$ bonds in plain magnesia or magnesia/silver.

With plutonium in the same +4 oxidation state, enough increased effective charge and orbital contraction have occurred to make the plutonium 5f subshell a better electron acceptor, making the $\ce{O^{2-}-H^{+}}$ bonding mode more favorable.

References:

Hsin-Yi Tiffany Chen, Livia Giordano, and Gianfranco Pacchioni, "From Heterolytic to Homolytic H2 Dissociation on Nanostructured MgO(001) Films As a Function of the Metal Support", J. Phys. Chem. C, 2013, 117, 20, 10623–10629.

2.
Chemical Forms of Uranium". Argonne National Laboratory. Archived from the original on 22 September 2006. Retrieved 18 February 2007.  Link


What is the mechanism behind the formation of phosgene from tetrachloroethylene Tetrachloroethylene used to be a popular automobile brake cleaner. Unfortunately, when applied to hot brake rotors, it can turn into phosgene. This has resulted in many injuries and even fatalities.

  At temperatures over 315 °C (599 °F), such as in welding, tetrachloroethylene can be oxidized into phosgene, an extremely poisonous gas.

What is the mechanism behind the oxidation to phosgene? I know that these things probably happen--alkene bond is broken and oxidized--but is there a detailed mechanism?

Given that the reaction occurrs at high temperatures, the reactive species is likely singlet oxygen, which reacts much like an alkene$^{[1]}$. Electron rich alkenes without allylic hydrogen, like tetrachloroethylene, react with singlet oxygen to form 1,2-dioxetanes$^{[2]}$.

A $[_\pi2_s + _\pi2_a]$ cycloaddition, though a poor reaction geometry for two alkenes, is made more accessible by the small, linear oxygen molecule. This, as well as peroxides' predisposition to nonplanar conformations, makes the transition state hardly any more strained than the dioxetane itself. Once formed, the dioxetane quickly decomposes to give two molecules of phosgene. To give you an idea of their stability:

  A small sample of the [dioxetane] product, on warming to room temperature, melted and exploded.$^{[3]}$

$^{[1]}$ Wikipedia, Singlet Oxygen, Reactions, Organic Chemistry
$^{[2]}$ Singlet Oxygen: Applications in Biosciences and Nanosciences Volume 1
$^{[3]}$ Stereospecific Formation of 1,2-Dioxetanes from cis- and trans Diethoxyethylenes by Singlet Oxygen


Solubility of glucose On eating powdered glucose I felt a cold sensation in my mouth. Then I thought it would be a result of the reaction between saliva and glucose. Then I thought that this would imply that the reaction would be endothermic. I don't find myself to be pretty sure and need to know whether I am right or not. Why would the reaction be endothermic in spite of the fact that glucose is readily soluble in saliva?

The dissolution of glucose in an aqueous solution is an endothermic process.  As stated in a comment, a reaction need not be exothermic to be spontaneous. Don't forget about the entropy term.

Simply put, the glucose dissolved endothermically in your saliva causing a heat transfer from your mouth to the glucose/saliva solution, resulting in a slight decrease in the temperature of your mouth, making it cooler than it was prior do dissolving the glucose.


I can't find reaction enthalpy ∆H for water formation reaction $$\ce{1/2 O2(g) + H2(g) → H2O (l)}$$

The tabulated value is $\Delta H = \pu{−285.8 kJ/mol}$

I'm using the bond energies values for each molecule and for some reason, I am getting a different result, not even close to $\pu{-285 kJ/mol}$.

Bond dissociation energies:

$\ce{H-H} = \pu{436 kJ/mol}$

$\ce{O=O} = \pu{499 kJ/mol}$

$\ce{O-H} = \pu{464 kJ/mol}$

My attempt to calculate from bond dissociation energies:

(500.1/2) + 436 → 464.2

250 + 436 → 928

($∆H$ found is positive)

Here is the tally of bonds:

One $\ce{H-H}$ bond broken
One half $\ce{O=O}$ double bond broken
Two $\ce{O-H}$ bonds made

Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $\ce{O-H}$ bond depends on what else is attached to the oxygen. For H-H and O=O, this is no problem because there is only one molecule, respectively, with that kind of bond.

$$\Delta H = \pu{436 kJ/mol} + \pu{249.5 kJ/mol} - \pu{928 kJ/mol} =  \pu{-242.5 kJ/mol}$$

We got the right sign and are not too far off in magnitude.

Why is the physical state important?

In solids and liquids, there are intermolecular interactions that contribute to the enthalpy. This is why the enthalpy of sublimation and boiling is positive - it takes energy to break the intermolecular forces. Here are the values for enthalpy of formation for water in the two states at room temperature:

$$\Delta H_\mathrm{f}(\ce{H2O(l)}) = \pu{-285.8 kJ/mol}$$

$$\Delta H_\mathrm{f}(\ce{H2O(g)}) = \pu{-241.8 kJ/mol}$$

And here is the enthalpy of vaporization

$$\Delta H_\mathrm{vap}(\ce{H2O(g)}) = \pu{43.99 kJ/mol}$$


Se+P White or red P? I read this in an old chemistry book. It says:

  When selenium is dropped into melted phosphorus, it dissolves rapidly, and the compound sinks through the liquid phosphorus in red streaks.

My question is whether the phosphorus is red or white?

Liquid phosphorus is white phosphorus. It melts at 44.1 °C.
Red phosphorus does not melt. It burns at 200 °C and sublimes in an argon atmosphere at 280 °C, without melting.


Computing the commutator of the Fock and density matrices in AO basis (Hartree-Fock theory) I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute

$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$

However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is

$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$

as this is equal to zero to within $1\times10^{-14}$. But how to derive this?

Derivation

An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions

$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$

Acting the commutator on $\phi_i$ and expanding it to AO basis yields

$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha}  - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$

Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition

$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$

Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\sum_\beta C_{\beta i}^* \chi_\beta^*$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$

$$
[\mathbf{F}, \mathbf{D}] = \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha}  - \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$

from in braket notation becomes

$$
[\mathbf{F}, \mathbf{D}] = \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle  - \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$

At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. Further, due to the orthonormality of the MOs, then I can imagine that the summation terms only survive when $\alpha = \beta$. But I'm not sure how to derive this properly.

Start from the AO Hartree-Fock equation and its adjoint
$$\mathbf{F}^{AO}\mathbf{T}=\mathbf{S}\mathbf{T}\epsilon \text{ and } \mathbf{T^\dagger}\mathbf{F}^{AO}=\epsilon\mathbf{T^\dagger}\mathbf{S}$$

where $\mathbf{T}$ is an $N\times n$ matrix that is essentially the occupied block of $\mathbf{C}$ ($n$ is occupied, $N$ is total orbitals). We use this $\mathbf{T}$ matrix because it has the convenient property that

$$\mathbf{T}\mathbf{T}^\dagger=\mathbf{D}^{AO}$$

Now, we can multiply the HF equation by $\mathbf{T^\dagger}\mathbf{S}$ on the right and multiply its adjoint by $\mathbf{S}\mathbf{T}$ on the left, which gives

$$\mathbf{F}^{AO}\mathbf{T}\mathbf{T^\dagger}\mathbf{S}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$

$$\mathbf{S}\mathbf{T}\mathbf{T^\dagger}\mathbf{F}^{AO}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$

Subtracting the first equation from the second and yields the desired commutator relationship.

$$\mathbf{F}^{AO}\mathbf{D}^{AO}\mathbf{S}-\mathbf{S}\mathbf{D}^{AO}\mathbf{F}^{AO}=\mathbf{0}$$

The notation I use here is based on a similar derivation given in Chapter 6 of McWeeny's Methods of Molecular Quantum Mechanics, 2nd edition.


Calculating concentration of OH⁻ ions Calculate concentration of $\ce{OH-}$ ions in $\pu{0.66 mol L^-1}$ solution of $\ce{NH4+}.$

I think that there might be something missing, as the only law I have for this kind of problems is

$$[\ce{OH-}]^2 = K_\mathrm{b} × C_\mathrm{b},$$

but is there a way to calculate $[\ce{OH-}]$ by using concentration of base only $(C_\mathrm{b} = \pu{0.66 mol L^-1}).$

Your problem seems to have no solution without further information. You may be interested in the the general method for any acid /base calculation is summarised below.

When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ).

The following calculations apply for each or any of these solutions.

The full calculation for problems of this type starts by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$

$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$

where the concentrations are those at equilibrium.

There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and

$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$

(The activity of water is 1 so is ignored in this equilibrium ).

We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionized species.

The total concentration of HA is

$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$

and for the base

$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$

The e subscripts are now dropped for clarity. (Note that some authors use mass and charge balance instead to work out the concentrations.)

These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation that can be used to solve all problems of this type;

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$

which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions.

Rewriting eqn 2 gives

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - \mathrm{[OH^-]} }{c_a - \mathrm{[H^+]} + \mathrm{[OH^-]}  }$$

assuming that $\mathrm{[OH^-]} \gg \mathrm{[H^+]}$ then

$$K_A=\mathrm{[H^+]}\frac{c_b  - \mathrm{[OH^-]} }{ \mathrm{[OH^-]}  }$$

As $K_w=K_AK_B$ then

$$K_B = \frac{\mathrm{[OH^-]^2}}{c_B-\mathrm{[OH^-]}}$$

and making a further approximation, if $c_b >\mathrm{[OH]^-}$ then $K_Bc_B\approx \mathrm{[OH^-]^2}$, which is the equation you quote. Therefore there seems that your problem cannot be solved without more information.


Those basis functions are removed because they are considered redundant. There is a completely uncontracted shell with the same exponent and angular momentum, so they can be removed without much loss of accuracy.

N     0
S   9   1.00
   9046.0000000              0.0007000
   1357.0000000              0.0053890
    309.3000000              0.0274060
     87.7300000              0.1032070
     28.5600000              0.2787230
     10.2100000              0.4485400
      3.8380000              0.2782380
      0.7466000              0.0154400
      0.2248000             -0.0028640        <=====
S   9   1.00
   9046.0000000             -0.0001530
   1357.0000000             -0.0012080
    309.3000000             -0.0059920
     87.7300000             -0.0245440
     28.5600000             -0.0674590
     10.2100000             -0.1580780
      3.8380000             -0.1218310
      0.7466000              0.5490030
      0.2248000              0.5788150        <=====
S   1   1.00
      0.2248000              1.0000000        ***Same exponent

Unfortunately, this is not documented on the old BSE. Here is the reference:

T. Hashimoto, K. Hirao, H. Tatewaki
'Comment on Dunning's correlation-consistent basis set'
Chemical Physics Letters v243, Issues 1-2, pp, 190-192 (1995)
https://doi.org/10.1016/0009-2614(95)00807-G

There was a period of time when the option was not working on the new BSE website (https://www.basissetexchange.org) but has been fixed (I hope).

Note: I am the maintainer of the new Basis Set Exchange :)


Why does SYBR green emit more signal when bound to DNA? Principle of SYBR Green

  SYBR Green I (SG) is an asymmetrical cyanine dye used as a nucleic
  acid stain in molecular biology.

  (wiki)

This molecule is commonly used in qPCR to meassure the amount of double stranded DNA. The fluorescent dye SYBR Green I binds to the minor groove of the DNA double helix. In solution, the unbound dye exhibits very little fluorescence (left figure below), however, fluorescence is greatly enhanced upon DNA-binding (right figure below).

(image source)

Question
Although its common use I cannot figure out why SYBR green fluorescence more when bound to DNA.

*I first doubted whether I should have posted this on biology stackexchange due to its relevance in this field, however I think this question covers way more chemistry than biology.

In dyes such as this rotation about bonds that can be involved in the conjugated part of the molecule contribute to radiationless transitions causing the excited state to transfer its energy to  a triplet or back to the ground state rather than fluorescing. When intercalated between base pairs as you suggest in your diagram, the molecule is held more rigidly and so radiationless transfer is reduced and fluorescence increased.

In molecules with a dipole it may additionally be the case that in a polar solvent the stabilisation due to the solvent polarity brings the initially formed excited state closer to a crossing to a charge-transfer state and this non-radiative process will lower fluorescence yield. Intercalated in the DNA the polarity will be lower than in water and again this will result in more fluorescence.  If your molecule changes wavelength in polar vs non-polar solvents then this is indicative of this process.


Thermodynamic vs Kinetic Sulphonation of Naphthalene The left hand isomer is clearly the kinetically favoured product and the right hand isomer is the thermodynamically favoured product. However, I have no idea why. Can you explain?

In contrast to many other electrophilic aromatic substitution reactions, aromatic sulfonation is reversible, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate.  This makes sulfonation a nice method to 1) control where a second substituent will be introduced or 2) serve as a protecting group (block second substituent introduction) - and then when you are done, you simply de-sulfonate.

Now to your question

If we look at the intermediates leading to the 2 isomers we see that we can draw 2 resonance structures that preserve aromaticity in one ring for the transition state leading to naphthalene-1-sulfonic acid, but only one resonance structure where the other aromatic ring remains intact for the 2-isomer (draw the resonance structures to convince yourself).

(image source)

Therefore the transition state leading to the 1-isomer will be lower in energy and naphthalene-1-sulfonic acid will be formed fastest - it is the kinetically favored product.

However, naphthalene-2-sulfonic acid is a few kcal/mol more stable than its 1-isomer. This is due to an adverse 1,8 steric interaction present in the 1-isomer.

(image source)

Because the sulfonation reaction is reversible, if we run it long enough at high temperature (favoring reversibility), eventually we will wind up with the thermodynamically preferred (more stable) product, the 2-isomer.


Acids used in Friedel-Crafts alkylation I am learning adding alkyl groups to benzene using Fridel-Crafts alkylation:

I know that it is very common to use $\ce{AlCl3/AlBr3}$ as Lewis acid catalyst in this reaction. But I am just wondering whether I can use a Bronsted acid, e.g., $\ce{H2SO4}$, as the catalyst or not. For example, in a $\mathrm{E1}$ or hydration reaction, $\ce{H2SO4}$ is added to alkene or alcohol, respectively, forming a carbocation, which is the same ultimate result when $\ce{AlCl3}$ is used (make the alkyl group more electrophilic).
I am thinking $\ce{AlCl3}$ is the only way if I am adding primary alkyl halide since in  $\mathrm{E1}$/hydration reactions, an actual primary carbocation is impossible to be formed (too unstable), while it exists in a complex with $\ce{AlCl3}$ (not entirely a carbocation, so can be formed).
SO, what is the problem with secondary/tertiary alcohols if I use the Bronstad acid? If there is a problem, way? Both Lewis and Bronsted acid ways will lead to carbocation rearrangements (I have not learnt the acylation).

Brønsted Acid has been used in Friedel-Crafts alkylation before (for example, Ref.1). According to this reference, when toluene use as the solvent, the F-C alkylation reaction of toluene is fruitful for tertiary and secondary alkyl bromides, tosylates, and alkenes (which give both carbocations upon protonation). However, the only primary bromide, which worked in this alkylation are benzyl and cinnamyl bromides. Other active primary bromides such as allyl and crotonyl bromides gave ditolyl products (e.g., crotyl group substituted first, then underwent double bond protonation followed by hydride shift to give secondary benzyl carbocation, which then reacted with the second toluene nucleus to give 1,1-ditolylbutane). Keep in mind that these reactions worked only with activated phenyl nucleus. For example, chlorobenzene gave only trace of products even with higher temperatures.
Later, this reaction is successfully used in Tandem-Friedel-Crafts alkylation reactions to synthesize hitherto unknown perylene derivatives (Ref.2):

References:

Mathew P. D. Mahindaratne, Kandatege Wimalasena, "Detailed Characterization of p-Toluenesulfonic Acid Monohydrate as a Convenient, Recoverable, Safe, and Selective Catalyst for Alkylation of the Aromatic Nucleus," J. Org. Chem. 1998, 63(9), 2858–2866 (https://doi.org/10.1021/jo971832r).
Mark A. Penick, Mathew P. D. Mahindaratne, Robert D. Gutierrez, Terrill D. Smith, Edward R. T. Tiekink, George R. Negrete, "Tandem Friedel–Crafts Annulation to Novel Perylene Analogues," J. Org. Chem. 2008, 73(16), 6378–6381 (https://doi.org/10.1021/jo800558c)(PDF).


If for a chemical reaction the standard gibbs free energy is negative then : If the standard gibbs free energy for a reaction is negative then the forward reaction will take place and the reaction will not be reversible. Is it true ? I have a problem understanding what is gibbs free energy and its significance in a chemical reaction.

Thanks in advance !

All reactions are in some sense equilibrium reactions. The Gibbs free energy is just a convenient measure of energy that we can use to understand more about our system. In particular you may be familiar with a few equations for the Gibbs energy:
\begin{equation}
\Delta G =\Delta H-T\Delta S,
\end{equation}
\begin{equation}
\Delta G=-RT\ln k
\end{equation}
\begin{equation}
\Delta G =-nFE
\end{equation}
For chemists, these three equations are the most commonly used and depend on what exactly you are investigating, but there are plenty of others too.

If we consider the second of these equations we note that $k$, the equilibrium constant, appears. We can see that as the equilibrium constant increases (and since $x$ follows $\ln x$ in limits) we note that the change in Gibbs energy becomes more negative the larger the equilibrium constant becomes. In other words the more products you have relative to reactants, the more negative the change in the Gibbs energy is, we call this spontaneous.

We could also take the third equation and note that $E$ is the reduction potential for a redox couple ($X^{n+}+e^{-}\rightarrow X^{(n-1)+}$). The more positive the potential for the reduction, the more spontaneous the reaction is.

Hope that helps you, oh and I'm assuming you mean reversible in the reaction sense of equilibria not in the thermodynamic sense of a reversible process?


Oxidation state in transition metals Why does scandium has +2 oxidation state unstable? It only has +3 oxidation state stable.
Scandium has outermost electronic configuration as 3d¹4s². Why it does not lose two 4s e- only instead of all three?

Scandium actually does have oxidation states lower than +3.  Wikipedia reports a double chloride $\ce{CsScCl3}$ where scandium is clearly in the oxidation state +2.

The stability of oxidation states in transition metals depends on the balance between ionization energy on the one hand, and binding energy due to either ionic or covalent bonds on the other.   In the case of scandium the third ionization energy is low because all three valence electrons are held rather loosely, being in diffuse orbitals that are shielded from most of the nuclear charge by the argon core.  Thus it's easy to get from the +2 to the +3 oxidation state because of the greater binding energy enabled by the latter.  Later first-row transition elements like iron have more tightly bound $3d$ electrons and thus higher third ionization energies, making the +2 oxidation state more favorable.


What fraction of air gets expelled upon heating an empty flask? A student forgot to add the reaction mixture to the round bottomed flask at $\pu{27 ^\circ C}$ but instead he placed the flask on the flame. After a lapse of time he realised his mistake and using a pyrometer, he found the temperature of the flask was $\pu{477 ^\circ C}$. What fraction of air would have been expelled out?

By solving $$\frac{V_1}{V_1 + V_2} = \frac{T_1}{T_2}$$ I found $$\frac{V_2}{V_1} = \frac{3}{2}.$$ But according to my book it is $\frac 35$. What exactly is this ratio?

Your first equation should be:
$${V_1\over V_2} = {T_1\over T_2}$$

We want to know what $V_2$ is in terms of $V_1$. Here, I call $V_2$ final volume. Arranging the equation and using the values for temperature in kelvin:
$${\rm final\;volume} = {V_1 \times 750\over 300}= {5\over 2}V_1$$

To get the amount that was expelled, we subtract the initial volume ($V_1$) from the final volume:
$${\rm amount\;expelled} = {5\over 2}V_1 - V_1 = {3\over 2}V_1$$

The fraction expelled is then:
$${{\rm amount\;expelled}\over{\rm final\;volume}} = {3\over 5} $$


Chemical Formulas of Acids and Bases How can we identify bases and acids by their chemical formulas?

Take for example this reaction. I know that HNO3 is an acid and KOH is a base, but if I didn't know that how would I be able to identify that using the chemical formulas

$\ce{HNO3(aq) + KOH(aq) \rightarrow KNO3(aq) + H2O(l)}$

I'm going to assume you are in high school, thus talking about Brønsted acids/bases.

A Brønsted acid is a compound that can donate a proton $\ce{H^+}$ whereas the base can accept a proton. For some chemicals it can be easily seen that they can donate a proton, like in the example of $\ce{HNO3}$ you give. There is obviously a proton in there so it makes sense that it is an acid. For chemicals with $\ce{O^-}$ in their structure (like the $\ce{KOH}$ in your example which dissociates into $\ce{K^+}$ and $\ce{OH^-}$) it is also clear that they can accept a proton and thus are obviously a base.

Now the difficult part is when a molecule can do both: these are called amphiprotic substances, the most famous being water $\ce{H2O}$, but there are many more of these, think about $\ce{H2PO4^-}$, it both can accept and donate a proton. In these situations molecules can indeed be considered both a base and an acid and the way they behave depends on the other molecules they come into contact with. If the other chemical is a stronger acid, as measured by the acid dissociation constant then the current molecule will act as base and vice versa.


Reversibility of an acid's reaction with water depending on its strength This is how my chemistry professor compared reaction of a weak and a strong acid with water:

  $$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$
  In this case $\ce{H2O}$ breaks the bond between $\ce{H}$ and $\ce{Cl}$ in $\ce{HCl}$ which results in he formation of $\ce{H+}$ and $\ce{Cl-}$ ions. The $\ce{H+}$ ion combines with $\ce{H2O}$ to form $\ce{H3O+}$ (also known as hydronium). Since adding water to an acid causes its hydration, which is an exothermic process, the reaction also results in the emission of heat. This reaction favors going to the right hand side of the equation.
  $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$
  In this case, the size of the acid is greater than the size of $\ce{HCl}$. It means that the process of hydration will be slower. This will lead to a comparatively less number of $\ce{H+}$ ions being combined with water. So, considering Arrhenius' definition of acids, $\ce{CH3COOH}$ is a weaker acid than $\ce{HCl}$. Also, this reaction is going to be reversible.

Now, my questions are as follows:

Why the first reaction favors going to the right hand side of the equation? I think that it has something to do with heat as the heat is lost to the surroundings and the energy level on right hand side decreases. I am still not sure.
Why is the second reaction reversible? I have no idea of why it is reversible so I can't provide my opinions.
Why does water causes ionization of acids? This question might look dumb to some but I am wondering what special water has got that it ionizes acids (or maybe other species too).
Why is there a release of heat? I have read that breaking of bonds is an endothermic process.

$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$

This reaction owes its ability to go to completion to the great stability the ions it forms when dissolved in water ($\ce{H+}$ and $\ce{Cl-}$). This ionization is caused by stronger attractive forces between the water and the atoms in a molecule (or ions in a salt) than between the atoms within the molecule have themselves, an is an endothermic process. In the dissociation of salts like $\ce{KNO3}$, the solution becomes notably cooler. When an acid dissociates, however, the hydration of $\ce{H+}$ to $\ce{H3O+}$ releases a great deal of heat.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$

This reaction also results in the formation of $\ce{H+}$ ions, though I wouldn't say that $\ce{CH3COO-}$ is less stable because it is larger than the $\ce{Cl-}$ ion. Consider that $\mathrm{p}K_\mathrm{a}(\ce{HCl})=-7$, $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=4.75$, and $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=0.75$. In case you are unfamiliar, the $\mathrm{p}K_\mathrm{a}$ is a way of expressing how much an acid dissociates in water, and a drop in $\mathrm{p}K_\mathrm{a}$ by $1$ means an acid releases $10$ times as many $\ce{H+}$ ions. $\ce{HCl}$ releases $10^{11.75}$ times as many $\ce{H+}$ ions as $\ce{CH3COOH}$, but when acetic acid's $\ce{C-H}$ bonds are swapped out for $\ce{C-F}$ bonds to make $\ce{CF3COOH}$ (an even larger molecule), its acidity increases by $10^4$. This shows that ion stability has more to do with bond strength and the inductive effect than ion size.

Both of these reactions are reversible. This just means that the forward reaction is occurring as well as the backward reaction (when occurring at the same rate this is known as equilibrium), but the first reaction lies more towards the formation of the products due to its greater stability. All reactions proceed in the backward direction, unless they are irreversible. Though the dissociation of $\ce{HCl}$ is often considered irreversible due to how shifted its equilibrium is, it is not truly irrversible because it has the potential to reform a molecule of $\ce{HCl}$. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, which does not happen in this case.


Why does p-dichlorobenzene have zero dipole moment while hydroquinone does not? At first I thought that both the OH groups would cancel out their dipole moments in hydroquinone. On reading further, I realized that because of the two different planar structures possible, it has a non-zero dipole moment. My textbook says the dipole moment of para-dichlorobenzene is zero. But by the same logic we used for hydroquinone, shouldn't this have a non-zero dipole moment as well?

To facilitate this discussion, here are the two structures.

You may remember from the beginning of organic chemistry classes that free rotation can occur around single bonds. This is mostly taught for $\ce{C-C}$ bonds but true for any σ symmetric bond. On the other hand, rotation is restricted around (partial) double bonds which rely on exactly one π system such as the phenyl ring.

This means that all $\ce{C-O, O-H}$ and $\ce{C-Cl}$ bonds in the above image are freely rotatable. Rotating either $\ce{C-O}$ bond of hydroquinone will move the corresponding hydrogen out of the phenyl ring plane. (In more scientific terms: the dihedric angle will no longer be $0^\circ$ or $180^\circ$.) Therefore, planarity may be lost. Also, if both hydrogens point to the same side, inversion symmetry is lost. In both cases, a nonzero dipole moment will be observed, as highlighted in this related question.

In para-dichlorobenzene’s case, you can also rotate the $\ce{C-Cl}$ bond. However, the result is entirely philosophic in nature, as only the positions of atomic nuclei are relevant for symmetry considerations and such. However you rotate that ball at the end of a stick, the end result will be the same. There is no way in which you could reduce symmetry. Hence the molecule has zero dipole moment.


Which of the following reaction will have greater enthalpy and why? Which of the following reaction will have greater enthalpy and why?
$\ce{C (g) + 4 H (g) -> CH4(g)}$ (reaction enthalpy = $x\ \mathrm{kJ/mol}$)
$\ce{C (graphite) + 2H2 (g) -> CH4 (g)}$ (reaction enthalpy = $y\ \mathrm{kJ/mol}$)

I thought that since $\ce{H2}$ is in its aggregation state it would have enthalpy of $0\ \mathrm{kJ/mol}$ and hence $y\gt x$.

Maurice has stated why equation 1 is much more exothermic than equation 2. A visualization of the processes may help you understand the details. Equation 1 involves the reaction of a carbon atom with four atoms of hydrogen to form methane in the gas phase. The bond dissociation energy (BDE) of the hydrogen molecule is +104 kcal/mol. Accordingly, the heat of formation (ΔHfo) of the hydrogen atom is half this value or +52 kcal/mol. The ΔHfo of the carbon atom from graphite is +171 kcal/mol. Both hydrogen and graphite are taken as 0 kcal/mol at the Standard State. The right hand side of the chart shows that the sum total of the necessary atoms is +379 kcal/mol above the Standard State. This is your starting point for equation 1. To form methane in the gas phase the reaction will be exothermic by -379 kcal/mol plus the ΔHfo of methane (-18 kcal/mol) for a total of -397 kcal/mol for equation 1. On the other hand, equation 2 is nothing more than the ΔHfo of methane as -18 kcal/mol below the Standard State as shown in the left hand diagram. This diagram also shows the stepwise BDE's of C-H bonds starting from methane. The same value for equation 1 is obtained, -397 kcal/mol.


Is it possible to find the volume of condensed water by temperature change without knowing the initial pressure of the flask? I found this problem in a practice sheet. The original source doesn't indicate any authorship or whatsoever.

It states the following:

  A sealed container of $\pu{6.24 L}$ in capacity has humid air at $\pu{30 ^\circ C}$ with a $RH=90\%$. The container temperature is then lower to $\pu{27 ^\circ C}$. Find the volume of condensed water. The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$. You can consider that the volume of condensed water is insignificant.

The problem doesn't specifically mentions any other assumptions but I'm assuming that the gas is behaving ideally and since it is referring to water. I will need the density of water, which is $\pu{1 \frac{g}{mL}}$ and the formula weight of water, which is $\pu{18 g mol-1}$.

The strategy which I thought was that the moles of air will remain the same when the vessel is cooled but the total moles will change before and after, as the moles of water initially will be different when the vessel it is cooled.

Then to find the volume what I need to find is that difference in moles and use the formula weight of water and the density to get volume.

However this is a problem because, how can I characterize the vessel if I'm not given an initial pressure?

I'm aware that:

$$RH=\frac{\textrm{partial pressure}}{\textrm{vapor pressure}}\times 100$$

Therefore I can say that the partial pressure for water is:

$$0.9\times \pu{32 mmHg} = \pu{28.8 mmHg}$$

But what I need is the total pressure in the vessel. Is it possible to find it?

If I could find the total moles, what I could do is find the moles of air by difference as follows:

$$n_{air}=n_{t}-n_{water}$$

These moles will not change.

The moles of water could be found by using the ratio of the partial pressure of water to that of the total is equal to the moles of water divided by the total moles.

$$\frac{n_{water}}{n_{total}}=\frac{p_{water}}{p_{total}}$$

But as it can be seen, I'm stuck. I did not included numerical calculations because if I were to use the equation of the ideal gas:

$$PV=nRT$$

What would I had use as the total pressure? Therefore I'm trapped. Is there any way to find what it is being asked?

Edit:

Following the indications by the answer I did the following:

The number of moles of water vapor at $\pu{30 ^\circ C}$ can be found from using the equation of ideal gas:

$\left(0.9\times 32\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(30+273)K$

$n_{water}=0.009505\,mol$

Then the number of moles of water vapor at $\pu{27 ^\circ C}$:

$\left(27\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(27+273)K$

$n_{water}=0.009\,mol$

Then:

$\Delta n_{water}=0.009505-0.009=0.000505\,mol$

From this quantity the grams of water and volume can be obtained using density of water and the formula weight.

$V=0.000505\,mol\times\frac{18\,g}{1\,mol\,\pu{H_{2}O}}\times\frac{1\,mL}{1g}=0.009090\,mL$

Which would be the volume of the condensed water. Judging from the ammount obtained it makes sense that it is a very tiny volume.

This problem is sloppily worded and, technically, it is not possible to give a definitive answer based on the information provided.  But the missing information is not what you think -- you don't need to know the air pressure in the flask.  What's missing are two key pieces:  The degree of humidity ("humid" doesn't mean "saturated") and the volume of water in the flask.  If the flask contains no water, and air that, while humid, is sufficiently below 100% $RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$ (i.e. you'd just be increasing the $RH$ from some number <100% to a value that, while larger, is still <100%). Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. Thus here is what I think the author meant to ask:

  A sealed container of $\pu{6.24 L}$ contains an equilibrated mixture of air and liquid water at $\pu{30 ^\circ C}$. The container temperature is then lowered to $\pu{27 ^\circ C}$. Find the volume of water that condenses. Assume the volume of liquid water is insigificant relative to the volume of the container.  The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$.

As to how to approach the question, that's straightforward:  You know the partial pressure of the water vapor at each temperature.  You also know the total volume of the air.  Given this, you can determine the number of moles of water vapor at each temperature (using the ideal gas law).  The difference is the number of moles of water that condenses, from which you can then determine the volume of the water that condenses.


Hydrophilic vs water soluble? Is the term 'hydrophilic' a synonym of 'water soluble' or can you have a molecule that is water insoluble but hydrophilic (or water soluble and not hydrophilic)?

Hydrophilic substances are not necessarily water soluble, and the two terms are not synonymous.

On a molecular scale, "hydrophilic" is defined by the IUPAC Gold Book$^1$ as:

  'Water loving'. The capacity of a molecular entity or of a substituent to interact with polar solvents, in particular with water, or with other polar groups.

Note that the term hydrophilic does not necessarily refer to an entire molecule, and that interacting strongly with a molecule or a part of a molecule does not imply that it must be soluble in water.

Also, entire macroscopic surfaces can be said to be hydrophilic, meaning that they interact strongly with water but certainly are not soluble in water. In this context, the degree of hydrophilicity is measured by the contact angle, which can be simply stated as the angle formed when a droplet of water is placed on a surface, where a small angle (as seen from inside the droplet) means that the droplet has spread out over the surface, and that the surface is therefore hydrophilic.

So, although small, hydrophilic molecules are frequently water soluble, solubility plays no part in the definition of "hydrophilic" on any scale or context.

1) IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8


A way to calculate orbital energy of an electron Maybe this is just not included in my course, but I still wonder how to calculate the binding energy of an electron .at a certain electron orbital.

Intuitively,The ionization energy of an element should vary depending on which orbital the valence electron is in.

PS: I only know how to calculate ionization energy using principle level.

You are correct that the ionization energy of an electron depends on the orbital. There is no easy formula, because the electrons in atoms are governed by quantum mechanics and for multi-electron atoms, the equations are un-solvable algebraically.

That said, there are some reasonable tabulations, which have been verified by calculations and by spectroscopy, e.g., http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf (from a quick Google search).

Both ultraviolet photoelectron spectroscopy and x-ray photoelectron spectroscopy can be used to eject electrons from various materials, and you can thus identify elemental composition (and other properties) from the known orbital energy levels


What would be the product (B) in the following sequence of reaction? Here is how I approached this:
As we know, alc. KOH results in dehydrohalogenation of the compound which takes place as an anti elimination. And then, we're simply conducting halogenation of the compound which would result in a dibromo substituded product.

Where I'm facing problems is deciding the final orientation of the product. How will we represent the product in the fischer representation as the initial reactant is given in its fischer representation as well.

For reference, here's the product (B) given in my book:

This is a fun one, and a common question in this course.  First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved.  That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

Now we need to add Br+ to make a bromonium intermediate.  Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality.  The result, as shown in your answer, is a molecule with a mirror plane.  In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up.  This compound is a meso compound.  But ... what about the other isomer if the Br+ originally goes to the other side of the double bond?  Well, just turn your Fisher structure upside down. :)


Why exactly does high pressure (72 bar) nitrous oxide's pseudocritical¹ temperature drop so far when oxygen is added to make 50%v/v mixture? (Entonox) This answer to Can precipitation occur in states of matter other than liquids? mentions the unusual case of Entonox.
The effects of temperature on nitrous oxide and oxygen mixture homogeneity and stability (Litwin P. D. 2010 BMC Anesthesiol. 2010; 10: 19. doi: 10.1186/1471-2253-10-19) says:

Poynting found that the critical temperature and pressure of a vapour may be affected when it is mixed with another gas. The critical point of a substance is the temperature at above which a gas cannot be liquefied, no matter how much pressure is applied. Conversely, it is the temperature and pressure point at which condensation of the gas into liquid will commence. For nitrous oxide this critical point occurs at a temperature of +36.4°C and at a pressure of 72.45 bar (1050 psi) [12]. For a cylinder of a 50% nitrous oxide and 50% oxygen v/v mixture filled to 138 bar (2000 psi), the new critical temperature of the nitrous oxide (known as the pseudocritical temperature) decreases from +36.4°C to -6°C.

Answers to Can precipitation occur in states of matter other than liquids? tend to guide us away from thinking of the Poyting effect as being related to dissolution or solubility but my concern is that "Poynting effect" is simply a name for an observation, and by itself not a well developed scientific concept or explanation.
That Wikipedia article says:

In thermodynamics, the Poynting effect generally refers to the change in the fugacity of a liquid when a non-condensable gas is mixed with the vapor at saturated conditions.

and that linked subsection ends with:

This equation allows the fugacity to be calculated using tabulated values for saturated vapor pressure. Often the pressure is low enough for the vapor phase to be considered an ideal gas, so the fugacity coefficient is approximately equal to 1.

So if we start with tabulated basic measurements of vapor pressure and calculating fugacity f and then see an unexpected, significant increase in vapor pressure of A when introducing gas B to the vapor, we say "Oh, that's the Poynting effect!"
We might then quickly change the subject or excuse ourselves before "What's that?" or "Why does that happen exactly?" gets ask.
Question: This change has a name, but what is happening? Why does the high pressure (~72 bar) nitrous oxide's pseudocritical1 temperature drop so far below it's pure substance critical temperature (from +36.4°C to -6°C) when oxygen is add to achieve a 50% v/v mixture at ~138 bar? Is the mechanism at least analogous to dissolution in some ways? Or could it be considered a "liquification point depression" analogous to a freezing point depression?
The citation of a readily available reference or published work explaining "exactly why" this is thought to happen will be most helpful, and if in what ways the process is and is not like dissolution can be added, then that's a complete answer.

1As pointed out in comments, critical temperature is defined for a pure substance so "pseudocritical temperature" is used in the block quoted work accordingly. See Real gases; Real gas law or Interaction Parameters for Kay's Pseudocritical Temperature for some examples of the term in use.

Dissolution(solvation) is solvation of solute molecules by molecules of solvents. This decreases chemical potential $\mu={\left(\frac{\partial G}{\partial n}\right)}_{T,p}$ of the solute, comparing it (often hypothetically ) to the chemical potential of the solute at the same concentration in gaseous phase.
Different solvents cause different chemical potential drop, leading to different solubility. As dissolution equilibrium means the equal solute chemical potentials in the solid phase and the solution.
Chemical  potential and fugacity

It [fugacity] is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy[=chemical potential] as the real gas.

are linked by relation $\mu=\mu_0 + RT \ln{\frac{f}{f_0}} \tag{1}$
Molecules in gaseous phase move independently on each other, unless there is a strong chemical affinity, what is not this case. Pointing effect does not depend(*) on the used gas, as there is no molecular wrapping in gas.
Pointing effect does not directly affect chemical potential of vapor. But  pressure affects the chemical potential of liquid instead, leading to higher saturated vapor pressure at the same temperature. The vapor pressure therefore increases to balance chemical potentials of the  substance in both phases.
For liquids near their critical conditions, like for the case of nitrous oxide with or without oxygen, there are big changes of state variables like molar volume or density with small change of conditions like pressure and temperature. This leads to relatively big change of saturated vapor pressure with the change of pressure of the system.
Note that for the liquid at its critical temperature $T_\mathrm{c}$ and near its critical pressure $p_\mathrm{c}$, $$\lim_{p \to p_c}{\ \left(\frac {\partial V}{\partial p}\right)}_{T }= -\infty \tag{2} $$
to get a picture what these changes are like.
As we know from the thermodynamics, for stable composition and zero non-volume work:
$$\mathrm{d}G=-S \cdot \mathrm{d}T + V. \mathrm{d}p  \tag{3}$$
Therefore $${\ \left(\frac {\partial G}{\partial p}\right)}_{T }=V \tag{4}$$
$${\ \left(\frac {\partial \mu_\mathrm{liq}}{\partial p}\right)}_{T }=V_\mathrm{m, liq} \tag{5}$$
If we assume ideal vapor behavior, we can consider $p_\mathrm{vap} = f_\mathrm{vap}$ and
$$RT \cdot {\left(\frac{ d \ln{p_\mathrm{vap}}}{\mathrm{d}p}\right)}_T =  V_\mathrm{m, liq} \tag{6}$$
$$RT \cdot \ln {\frac{p_\mathrm{vap}}{p^{\circ}_\mathrm{vap}}} =  V_\mathrm{m, liq} \cdot (p - p^{\circ}_\mathrm{vap}) \tag{7}$$
$$ p_\mathrm{vap}  =  p^{\circ}_\mathrm{vap} \cdot \exp {\left(\frac {V_\mathrm{m, liq}}{RT} \cdot (p - p^{\circ}_\mathrm{vap})\right)} \tag{7}$$
(7) is a simplified saturated vapor pressure dependency rather for "normal" liquids with $T_\mathrm{c} \gg T$, when the liquid compressibility can be neglected. Otherwise, $V_\mathrm{m,liq}=f(p)$ would have to be included to the integration, as the liquid molar volume and compressibility start steeply grow toward the critical point. E.g. water vapour and liquid densities converge to about $\rho = \pu{0.21 g/cm}$ at the water critical point.
For illustration, we can take water at the boiling point with
$V_\mathrm{m} = M/\rho = (\pu{18.02 g/mol})/(\pu{0.9584 g/cm3})=\pu{18.8 cm3/mol}$, $T=\pu{373 K}$ a $p^{\circ}_\mathrm{vap}=\pu{e5 Pa}$. It can be seen the pressure effect can be neglected for small and medium pressure.
$\begin{array}{cc}\small \\
\text{pressure [MPa]} & \text{H2O vapor pressure [kPa]} \\
0.1 & 100.0 \\
0.2 & 100.1 \\
0.5 & 100.2 \\
1.0 & 100.5 \\
2.0 & 101.2 \\
5.0 & 103 \\
10 & 106 \\
20 & 113 \\
50 & 135 \\
100 & 183 \\
\end{array}$

(*) Special cases are systems with significant chemical interactions, including gas solubility in liquid. E.g. well soluble gases cause additional effect of the opposite sign than the Pointing effect. They decrease the saturated vapour pressure by decreasing solvent molar fraction and eventually its chemical potential in both phases by chemical interaction. E.g. $\ce{NH3(g + aq),H2O(l + g)}$.


How can meta-cresol be made, given that the hydroxyl group is ortho,para-directing? How is it possible for m-cresol (3-methylphenol) to be synthesized?

The methylation of phenol would lead to the ortho- and para-isomers instead.

Here is a synthesis of m-cresol starting from benzene.  It achieves the desired meta orientation of the final substituents by introducing a meta-directing substituent (the aldehyde group) early-on in the synthetic scheme. In the nitration step, the meta isomer is by far the major product (o:m:p 19:72:9), but a distillation or recrystallization would be needed (or you could carry all the isomers along and separate later, but you'd probably have the widest range in boiling points at this stage with the nitro group present). The only novel step is the double Wolff-Kishner used to reduce both the aldehyde and nitro groups in a single reaction.