text:
Why copper(I) is unstable in aqueous medium?
I am expecting that $\ce{Cu+}$  attains a $\mathrm d^{10}$ configuration by losing one electron from s-subshell. Since it has fully filled d-orbital, it should be stable. But it is found that it is unstable and tends to disproportionate in aqueous medium or with moist air to form $\ce{Cu^2+}$ where  $\ce{Cu^2+}$ has incompletely filled d-orbital ($\mathrm d^9$ configuration). Few examples from Doc Brown's Science Revision Website — The chemistry of copper:
\begin{align}
\ce{\overset{+1}{Cu}_2O(s) + H2SO4(aq) &-> \overset{0}{Cu}(s) + \overset{+2}{Cu}SO4(aq) + H2O(l)} \tag{R1} \\
\ce{\overset{+1}{Cu}_2O(s) + 2 H+(aq) &-> \overset{0}{Cu}(s) + \overset{+2}{Cu}^2+(aq) + H2O(l)} \tag{R2}
\end{align}
Why is this so?

You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t.
Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^2+}$ ion has a greater charge density than the $\ce{Cu+}$ ion and so forms much stronger bonds releasing more energy.
The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu+}$ ion loses an electron to become $\ce{Cu^2+}$ which can then release this hydration energy.  A nearby $\ce{Cu+}$ ion is the most facile reduction target for the removed electron, which is why $\ce{Cu(s)}$ is also formed.
Hence, $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in aqueous medium.

curiousbrain, I added a sentence to your answer -- please review and edit/rollback if you don't like it.

@Brian thanks for adding the part. It should be part of the answer according to the question. Thanks a lot! :)

While I agree generally with curiousbrain’s answer, I don’t think that the charge density alone is the culprit.
Rather, $\ce{Cu+}$ is a $\mathrm{d^{10}}$ ion which therefore has no real preference for any ligand shell — much like zinc(II). All 10 d-electrons will always populate antibonding orbitals with respect to the $\ce{M-OH2}$ coordinate bond, weakening them and creating a highly labile ligand sphere.
On the contrary, $\ce{Cu^2+}$ has a rather well defined strongly Jahn-Teller distorted octahedral ligand sphere. While the lower $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{z^2}$ orbitals are antibonding[1] and fully populated, the very strongly antibonding $\mathrm{d}_{x^2 - y^2}$ orbital is only populated by a single electron, strengthening the copper-water interactions and making the ligand sphere less labile. This may well be a reason why the displacemen of an electron is favourable in aquaeous media.

Notes:
[1]: If one looks at the orbital scheme of a typical octahedral complex, additionally includes π interactions between ligands and metal, and finally also considers the Jahn-Teller distortion, it becomes evident that all metal-centred orbitals are antibonding to a certain extent. However, $\mathrm{d}_{x^2-y^2}$ is much more strongly antibonding than all other d-orbitals.

Soft base ligands stabil7ze copper(I) over copper (II), and in many cases these produce more ligand field splitting than hard bases like water. So I am not sure this argument holds up.

@OscarLanzi This has been a while but off the top of my head I think the explanation would be that the bonds between thiourea and copper(I) or chloride and copper(I) would be stronger or 'more co-valent' leading to more energy gained by lowering the donating electrons' orbitals while the d electrons' slight destabilisation is less loss. For the harder ligands, the donating bond is centred more on the ligands meaning the energy gain is less and the d orbital shenanigans matter more, hence why the water complex is tipped in favour of copper(II).

As usual, there are a number of partial explanations, all of which are true in themselves but all need to be considered to get the true picture -- or so I interpret this right here, right now! ;)

Quite the contrary, copper(I) is actually relatively stable compared with most fourth-period transition metals in the +1 oxidation state. As discussed here, copper in the +1 oxidation state can be stabilized, even in water solvent, by complexing with a soft base such as thiourea or chloride ion. That does not work with the +1 oxidation state in earlier transition metals in the series.
What happens with most fourth-period transition metals is an imbalance between ionization energy and solvation energy: the second ionization energy is small enough so that the additional solvation energy of a +2 ion versus a +1 ion by water outweighs it, therefore the +1 ion goes on to +2 (or more, especially early in the transition series).
With copper, however, the filled $d$ valence shell whose electrons only partially shield each other from the nuclear charge raises the second ionization energy of the copper. We still favor going to the +2 oxidation state when copper is complexed by water. But a ligand with more covalent bonding character and weaker electrostatic attraction, meaning a soft base, might now allow the +1 oxidation state to remain stabilized given copper's slightly higher second ionization energy. Then we get a stable copper(I) complex with the soft base, which in some cases is also soluble in water.
In the above, copper is contrasted with earlier transition elements in the fourth period. What about a comparison with zinc and later metals in the period? In these cases, the second ionization energy again drops off, until we reach the metalloidal elements arsenic and selenium, because the second ionization no longer comes from the $3d$ subshell (it comes from $4s$ or $4p$). Thus the post-transition metals in Period 4 also tend to go on to a +2 or higher oxidation state while copper, given the right environment, can be kept at +1.


text:
Shouldn't all of group 16 form double bonds?
I notice that out of all of group 16 oxygen is the only element that can form a double bond. Why is that because shouldn't all of group 16 be able to form a double bond and be diatomic because they all have six valence electrons?

Consider $\ce{CS2}$

I removed your picture - it was terrible and misleading. Also tags were improper.

@permeakra Why isn't it diatomic then?

@Mithoron Thanks, but then why aren't the double bond molecules also diatomic

It isn't proper place to describe bonds or multiatomic molecules, you should know basics when asking such question

I notice that out of all of group 16 oxygen is the only element that can form a double bond.

This premise is not true. Sulfur in fact forms double bonds in many organic compounds.  Selenium and tellurium both form double bonds with triphenylphosphine as triphenylphosphineselenide and triphenylphosphinetelluride.
The larger chalcogenides do have difficulty in forming double bonds in organic molecules due to lower bond strength, higher reactivity, preference of molecules for oxygen, and steric bulk.  They can however indeed form double bonds; it's just not as common.

Then why aren't they diatomic?

I suppose in the gas phase they could be.  In the solid phase though Sulfer is a 8 membered ring and selenium and tellurium are crystalline, but may for polymeric glasses.

Sulphur atoms are larger then oxygen atoms. This means that 3p orbitals in sulphur can't overlap efficiently or enough to create a pi bond (which is the "second" bond in the double bond) between two sulphur atoms. Oxygen, however, is small enough that its 2p orbitals can overlap to form a stable pi bond. The size of the atoms only increase down group 16, and so anything larger than oxygen will not form stable double bonds.

When you say "second bond"  what do you mean, because there is only one bond right?

"Larger" alone does not quite answer the question. If S atoms were just like enlarged O atoms, then $\ce{S2}$ would be just like enlarged $\ce{O2}$ and would feel just as well. This is not the case, though.

The pi bond in the double bond

But isnt there only one bond?

@RuchirBaronia A double bond is made up of a sigma bond and a pi bond. The pi bond results from overlap of p orbitals. Because p orbital overlap would be poor in sulphur, the pi bond cannot form, and so the single sigma bond is all that remains.

But each atom shares only one electron in a double bond, right? So  you're saying that the size of the atom impacts whether or not that one atom will be shared?

@RuchirBaronia if it is localised in a p orbital then yes it can. Overlapping p orbitals are what cause Pi bonds,

@Ruchir Baronia a double bond is made up of two bonds, as Georgeos Hardo said. So, each atom shares two electrons – hence, a total of four electrons are shared in a double bond. The size of the bonding atom can inhibit the ability to form one of the two bonds making up a double bond, i.e. the π bond. Thus, they only form single bonds.

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text:
How durable is a silanized surface?
I have two questions regarding silanization:
1) How durable is a silanized glass surface? (i.e. How long a silanized surface can be kept without any change in its wettability or contact angle?)
2) What are the optimum conditions for keeping a silanized surface while it's idle and dry? (e.g. maintenance temperature, humidity, etc. Should it be kept in a desiccator or not necessarily?)

Try looking up rainx

If done properly, reactions with silanes or siloxanes are effectively permanent, unless removed by serious mechanical abrasion (e.g., scraping the material below the coating) or some sort of etching underneath.
The silanization process involves creating new covalent $\ce{Si-O}$ bonds, which are quite strong.

The silane surface itself should be durable. However, the contact angle may change as material adsorbs on top of it. For example, a colleague has shown that hydrocarbon contaminants change the wettability of many surfaces
I don't think any particular storage is necessary. We store in a vacuum desiccator to minimize the effect of contamination, but as long as you clean the surface, the properties should be consistent.

Therefore, I can keep my silanized glass spheres in a closed-cap bottle for a long enough time and probably wash them before I need to use. BTW, I know that with silanization by Octadeyltrichlorosilane, I can reach to CA~110 deg for glass. Do you know what is the Maximum contact angle people could reach on glass, and with what solution? Thanks

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text:
Determining the mass of sodium sulfate that forms when reacting sulfuric acid with sodium hydroxide
I have come across simple problem in the chemistry book which I cannot solve. The question is below. I would appreciate if you have any solution to this.
Question:

$3\ \mathrm{g}$ of $\ce{NaOH}$ are mixed with $4.9\ \mathrm{g}$ of $\ce{H2SO4}$. How much sodium sulfate $\ce{Na2SO4}$ will form?

(The answer is $3.55\ \mathrm{g}$)
Note: I have calculated it as $5.33\ \mathrm{g}$ of $\ce{Na2SO4}$ don’t know where I am mistaken. It doesn’t seem to be a typo in the question because I couldn’t solve other similar problems in other books wither. There is something I always miss.

You seem to be assuming that all $\ce{Na}$ would go to $\ce{Na2SO4}$, and the unused $\ce{H2SO4}$ (which is in excess) would stay that way, i.e. as $\ce{H2SO4}$. This is not quite so.

Rather than working with funny fractions, let’s use decimals.
$$m(\ce{H2SO4}) = 4.9~\mathrm{g}\\
m(\ce{NaOH}) = 3~\mathrm{g}\\
M(\ce{H2SO4}) = 98.09~\mathrm{g \cdot mol^{-1}}\\
M(\ce{NaOH}) = 40.00~\mathrm{g \cdot mol^{-1}}\\
M(\ce{Na2SO4}) = 142.05~\mathrm{g \cdot mol^{-1}}\\
~%
\\
n(\ce{H2SO4}) = 50~\mathrm{mmol}\\
n(\ce{NaOH}) = 75~\mathrm{mmol}$$
Remember when doing stoichiometry to correctly consider your reaction equations. In this case:
$$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2 H2O}$$
If we want to do this, we see that we don’t have enough $\ce{NaOH}$ — it is present only in substoichiometric amounts. We would need $100~\mathrm{mmol}~\ce{NaOH}$ for a full reaction with $\ce{H2SO4}$:
$$\frac{n(\ce{NaOH})}{n(\ce{H2SO4})} = \frac{2}{1}\\
n(\ce{NaOH}) = 2 n (\ce{H2SO4})\\
n(\ce{NaOH}) = 2 \times 50~\mathrm{mmol}= 100~\mathrm{mmol}$$
If we now assumed that all $\ce{NaOH}$ would participate in the formation of $\ce{Na2SO4}$, we would arrive at $37.5~\mathrm{mmol}~\ce{Na2SO4}$ or $5.4~\mathrm{g}\ \ce{Na2SO4}$. But that would leave free $\ce{H2SO4}$ lying around. Instead, consider a stepwise process:
$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}\\
\ce{NaHSO4 + NaOH -> Na2SO4 + H2O}$$
We see that we need to use the first $50~\mathrm{mmol}\ \ce{NaOH}$ to generate $50~\mathrm{mmol}\ \ce{NaHSO4}$; with $25~\mathrm{mmol}\ \ce{NaOH}$ remaining. These $25~\mathrm{mmol}\ \ce{NaOH}$ can react with the $\ce{NaHSO4}$ to generate $25~\mathrm{mmol}\ \ce{Na2SO4}$ and leave $25~\mathrm{mmol}$ unreacted $\ce{NaHSO4}$.
$$n(\ce{Na2SO4}) = 25~\mathrm{mmol}\\
m(\ce{Na2SO4}) = n \cdot M = 25~\mathrm{mmol} \times 142.05~\mathrm{g \cdot mol^{-1}} = 3.55~\mathrm{g}$$


text:
What are the strategies to deal with fluorescence from oil in Raman Spectroscopy?
I have a sample (for the purpose of this question let's call this Sample A) consisting of oil (edible olive oil) mixed with a contaminant (Contaminant B) from the Polycyclic Aromatic Hydrocarbons family.
I have tested pure forms of Contaminant B with gold nanoparticles and I am able to see great signals. (Surface Enhanced Raman Spectroscopy)
When I try to perform SERS on Sample A I notice that the peaks are suddenly gone. I suspect the following:

The signals from the PAH is being shadowed by the fluorescence of
the oil
The signals from the PAH is being absorbed by the oil
molecules, leaving no observable peaks
The oil molecules is
preventing the PAH molecules from being in close physical proximity
to the gold nanoparticles for SERS's enhancement

Question ONE: Which of these is true?
Question TWO: How can I solve this problem?
I can only think of:

Photobleaching (not really practical for the purpose of my paper)
Chromatography on Sample A (tried Acetone and Methanol but to no avail)
Adding an agent that could eliminate fluorescence from the oil molecules (I have no idea what could achieve this)
Adding an agent to digest or destroy the oil in Sample A leaving just the PAH in the remaining solution (again, no idea what this could be)

To answer, it would be helpful to have more technical details about your measurements.  What wavelength of (laser?) light are you using for your Raman detection?  When you say "the peaks are gone", is it because the baseline has gone up?  Or because the baseline is the same and the signal is lower?

Also, what concentration of Contaminant B did you use in your studies of the "pure form" of B?  100%?  What concentration range do you expect for B in Sample A?  Parts per thousand?  Parts per billion?  Does that expected range overlap with what you used in your "pure form" studies?

@CurtF. 785nm laser. Peaks are gone = peaks are not there anymore but the baseline is not higher.
In my pure form of study of Contaminant B I was able to pick up signals all the way from solid to 1 ppm. Sample A contains 100 ppm of Contaminant B

When you made solutions of pure Contaminant B to study, what solvent did you use?

@CurtF. We followed this paper: https://www.scribd.com/doc/294198642/Characterization-of-polycyclic-aromatic-hydrocarbons-using-Raman-and-surface-enhanced-Raman-spectroscopy
To answer your question: Methanol

As per my understanding SERS is not always reproducible ~


text:
How to determine the iron content in spinach from a titration with permanganate solution?
I used titration to find the mass of iron in a $5~\mathrm{g}$ sample of spinach after leaving the spinach to soak in tap water for five different lengths of times. I added $\ce{H2SO4}$ to the solution and then titrated it with $\ce{KMnO4}$.
These are my results:

I found the following total reaction occurs:
$$\ce{5Fe^2+ + KMnO4 + 8H+ -> Mn^2+ + 4H2O + K+ + 5Fe^3+ }$$
I am now trying to do my calculations to find my answer. I first found the amount of substance of $\ce{KMnO4}$ needed to neutralize my solution and then what I think is the initial amount of iron ($\ce{Fe^2+}$). I am stuck as I am not sure how to continue to find the actual mass of iron on the right side of the equation. Do I use stoichiometry? Is finding the $\ce{KMnO4}$ value even necessary?

Yes, use the stoichiometry. You have the KMnO4 concentration used, so you can calculate the iron mass.
For example:
If you used 0,01 mol of permanganate in the titration you have 0,05 mol of iron, this based in the stoichiometry.

Besides the question you asked, I wonder if you're getting just a reaction with iron. It seems that you were titrating a solution in which spinach leaves were soaking. There is all sorts of organic compounds in spinach that could be oxidized with permanganate. For instance spinach contains the oxalate anion. Normally for an inorganic analysis for trace metals the sample would be ashed, or the whole leaf chemically oxidized.

As @MaxW stated above , im my experience to get the best representative sample of the iron in spinach I burnt the spinach to ash in the crucible to free the iron in the sample from the biological molecules it is bound to. This also enables the Fe+3 ions to react with the titrating reagents efficiently. HCL was then added to the ash and potassium thiocyanate was added and mixed. The red iron thiocyanate molecule was produced and the concentration was determined via spectrophotometer readings and a calibration curve previously prepared from a control solution 0.0001% Fe(SCN)3 .

After doing more research I found that with my experiment set up I could only find the amount of ferrous oxalates in my sample of spinach. I used stoichiometry to find this value but now have to do further experiments to distinguish between the iron and the oxalate...

Hey what are the 5 different amount of times you used??

have you tried theoretical stoichiometry

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text:
How can three p orbitals per atom bond in tin (II) sulfide?
Tin (II) sulfide, a mainly covalent compound, bonds using its p orbitals only (the 5s orbitals are considered to be part of the core in divalent tin). Its structure is made up of Sn–S–Sn–S chains, which are further stabilized by Sn–S bonding across the chains:

How would VB or MO theory explain the fact that the tin molecule bonds three times, when only two of its p orbitals are half filled (Electron configuration: [Kr]5s24d105p2), while the third is empty? The observed bond angles of ca. 90° (96°/88°) would indeed suggest plain p orbital bonding, as suggested by this paper, but how is this possible?
And how could sulphur bond three times (Electron configuration: [Ne]3s23p4) with similar bond angles, when one p orbital is full?
The only thing that comes up to my mind is a dative covalent (co-ordinate) bond, in which sulphur provides both p electrons, thus permitting coordination. But then, the tin atoms would have a negative charge attached to them (due to having more electrons around than protons) and the sulphur atoms a positive one. Wouldn’t this be energetically unconvenient, considering that the compound has ca. 9% of ionic character, which depends on the tin ions being positive and the sulphur ions negative?
Addendum:
The dative bond assumption clashes against the bond lengths observed in tin (II) sulfide. The paper I cited above states that the Sn–S bond length on the chains are 2.68 Å long, while those between the chains are 6 pm shorter (2.62 Å). The IUPAC Gold Book says that dative bonds

are distinguished by their significant polarity, lesser strength, and greater length

when compared to covalent bonds. This is exactly the opposite of what happens in tin (II) sulfide.

I think your conclusion of a dative bond between sulphide anions and tin cations (assuming pure ionic bonding, which is clearly not the case if you study the image closely) is correct.
Note that the charges you are observing are nothing but formal charges. In a macroscopic picture, tin is still less electronegative than sulphur and the electrons would be considered sulphur’s e.g. when determining oxidation states. The mere fact that an electron pair with correct symmetry is adjacent to an atom is enough to stabilise said atom’s empty orbital.

And yet, the charge on the tin atoms is more negative with the dative bonds than it would be without them... So how is the bond convenient? Wouldn't single polymer chains be more thermodynamically stable?

@Corundum Think octet rule. That way, all three equivalent p-orbitals take part in some binding interaction; none is left alone in a non-bonding state. At least, that is how I would explain it quickly.

and what about the problem with bond length (explained in the addendum)?

@Corundum That requires me to do more thinking. I may be back later (or delete this answer; I’ll have to see).


text:
Active mass,Pressure and concentration
The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = 1\:\mathrm{mol/L}$. At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to0\:\mathrm{mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
The partial pressure is connected to the activity via
$$a(\ce{A}) = \frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=1\:\mathrm{bar}$ or traditional use of $p^\circ=1\:\mathrm{atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
$$\lim_{p\to0\:\mathrm{bar}}\left(\phi_{\ce{A}}\right)=1,\hspace{10ex} p(\ce{A})=y_{\ce{A}}\cdot{}p.$$
Now my question is since we can approximate activity to partial pressure as well as concentration therefore pressure must approximately equal to concentration. However we know that $p=cRT$, where $c$ is the concentration, hence we definitely cannot approximate pressure and concentration. So how is this possible?


text:
How to determine gerade & ungerade symmetry of a MO orbital?
J.D.Lee writes in his book Concise Inorganic Chemistry:

[...] An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes remains the same,  the orbital is gerade, and if the sign changes, the orbital is ungerade.

Now, it can be detected easily which MO, formed due to the overlapping of s-s or p-p orbitals, is gerade or ungerade as shown in these pics:

But what about the molecular orbitals formed due to the overlapping of s-p orbitals or p-d orbitals?
mo due to to overlapping of s-p orbitals:

mo due to overlapping of p-d orbitals:

If I rotate about the first orbital about an axis perpendicular to the inter-nuclear axis, the smaller lobe goes to the right only; the sign always remains same; just the position of the lobe has been changed. But in the first picture, only the bonding mo is gerade while the anti-bonding mo is ungerade; I'm not getting how by rotating along the axis perpendicular to the inter-nuclear axis, the sign of the antibonding mo changes making it ungerade.
Can anyone please help me how to apply the procedure for the molecular orbitals formed due to the overlapping of s-p & p-d orbitals?

Those MOs simply cannot be labelled as g or u. MOs of non-centrosynmetric molecules (i.e. molecules that do not possess a centre of inversion) cannot be labelled g or u. Look at the MO diagram of water (C2v) - none of the orbitals have g or u labels.

@orthocresol: So, do you mean to say those in the pic cannot be labelled g or u?

That's right, they cannot be labelled as such.

@orthocresol: But the author did label those orbitals as g & u :/

I think the point of all this is about MO orbital symmetry, nodes, bonding and antibonding. For these asymmetric examples, consider the orbital lobes between the nuclei if they are sigma- or pi-overlap, and if the orbital formed between them has genade or ungerade symmetry.

Related : How derive g and u symmetry labels for orbitals?

The better way to do it is to check what happens under inversion ($i$ or $\bar 1$). If the orbital stays the same, it is g, otherwise u.
However, as Orthocresol mentioned in the comments, checking that is only possible if the entire molecule contains inversion symmetry. Not all point groups do, and the vast majority of molecules does not (partly because $C_1$ is likely the most prevalent point group out there).
For example, consider transition metals’ d orbitals. In octahedral complexes, they are labelled $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In tetrahedral complexes, which do not have inversion symmetry, they are $\mathrm{t_2}$ and $\mathrm{e}$.

Hmmm... could you tell what point group is? Also, if there is no inversion symmetry in those orbitals, why did Mr. Lee labelled those orbitals as $g$ and $u;?$

@user36790 If you check out which symmetry features a molecule has, you realise that these happen to form a mathematical group, i.e. there’s identity, there’s an inverse element to each element and when adding two operations together, they exhibit commutativity. Only a handful of different groups actually exist, and since they help so much with symmetry considerations, they are taught in second or so year of chemistry studies. I’ll check out if I find a question here on chem.SE about them ^^

So, should I say the author is not right in marking those orbitals as $g$ and $u$?

Unfortunately, there is no ‘basic’ question about group theory here as far I can tell. Feel free to ask one ^^'

@user36790 Yes, it does not make sense labelling something as g or u if there is no inversion present in the molecule.

Thanks for the confirmation. He confused me for 2 days! I'm going to read about point group in order to understand it better. Thanks, again.

If opposite lobes have inversion centre i.e. the centre of symmetry with respect to  phase of wave functions then the orbitals are gerade otherwise ungerade.


text:
Is the name of this chemical 3-oxopropanoic acid?
Is the name of this chemical 3-oxopropanoic acid?
I included "oxo"- for the aldehyde group, and "-oic acid" for the carboxylate group.
However, I'm unsure whether this is the correct name since I cannot find it on the internet, while I can find almost all other names I've been trying to figure out.

Related: Nomenclature priorities with ethers, acyl groups and ketones

The name ‘3-oxopropanoic acid’ for the compound $\ce{OHC-CH2-COOH}$ given in the question is correct.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

(…)
  7. acids
  (…)
  15. aldehydes
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.
Systematic names of carboxylic acids are formed as follows:

P-65.1.2 Systematic names
Carboxylic acids are named substitutively using the suffix ‘oic acid’ or ‘carboxylic acid’ and the prefix ‘carboxy’ to describe carboxy groups that cannot be included as the principal characteristic group or to express the carboxy group in the presence of a higher principal characteristic group.
P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; (…). Except for (…), systematically formed names are preferred IUPAC names; (…)

Therefore, the preferred IUPAC name (PIN) of the unmodified carboxylic acid is ‘propanoic acid’. (Note that the retained name ‘propionic acid’ may be used in general nomenclature.)
Aldehydes can be systematically named in various ways. Monoaldehydes derived from alkanes are preferably named substitutively using the suffix ‘al’. In this case, however, this method cannot be used since the carboxylic acid group has priority to be cited as a suffix. Therefore, the aldehyde group has to be expressed by a prefix as follows:

P-66.6.1.3 In the presence of a characteristic group having priority to be cited as a suffix or when present on a side chain, a $\ce{-CHO}$ group is expressed by the preferred prefix ‘oxo’ if located at an end of a carbon chain, or, otherwise, by the preferred prefix ‘formyl’.

Thus, the PIN for the complete structure is ‘3-oxopropanoic acid’.

The corresponding name that is formed by using the prefix ‘formyl’ is ‘formylacetic acid’. Such names are not acceptable as PINs but may be used in general nomenclature.
Furthermore, a special rule for aldehydic acids applies to the given compound:

P-65.1.6.3 Aldehydic acids
When a dicarboxylic acid has a retained name (…) and when one of its carboxy groups is replaced by a formyl group, $\ce{-CHO}$ (…), the resulting structure is called an aldehydic acid and, in general nomenclature may be named by replacing the ending ‘ic acid’ of the name of the dicarboxylic acid by the ending ‘aldehydic acid’. Preferred IUPAC names for aldehydic acids derived from all dicarboxylic acids are constructed systematically. The prefix ‘formyl’ is used in preferred IUPAC names, except for a $\ce{-CHO}$ group at the end of an acyclic chain, which is designated by the prefix ‘oxo’.

For the compound given in the question, the PIN of the corresponding dicarboxylic acid is ‘propanedioic acid’.

This dicarboxylic acid also has the retained name ‘malonic acid’, which may be used in general nomenclature. Hence, according to P-65.1.6.3, the corresponding aldehydic acid may be named ‘malonaldehydic acid’ in general nomenclature.

By way of comparison, for the analogous example $\ce{OHC-CH2-CH2-COOH}$, all three corresponding names are explicitly mentioned in the Blue Book:

4-oxobutanoic acid (PIN)
3-formylpropanoic acid
succinaldehydic acid

Probably worth noting that many biochemists would call this compound "malonic semialdehyde", as it is derived from the common biochemical/metabolite malonic acid by reduction of one (but not both, thus the "semi") of the acid moieties to an aldehyde.

3-Oxopropanoic acid (if carbon of aldehyde is numbered, use "oxo" as prefix) is correct IUPAC-name for the given chemical compound.
Formylethanoic acid (if carbon of aldehyde is not numbered, use "formyl" as prefix) is also correct IUPAC -name.

A compound typically has one (in digits: 1) correct systematic IUPAC name.

If we are using formyl prefix and parent chain has two carbon only of an acid and "1" is considered, it is wrong. Suppose, in case of  "2-formylpropanoic acid" and "3-formylpropanoic acid" both are different compound and correct. "1-formylpropanoic acid" is incorrect IUPAC and even also this compound doesn't exit.

@Jan even if people don't know the correct concept, they make the correct answer and concept prove wrong. This thing is wrong on stackexchange.

My point was: (1) There is exactly one correct systematic IUPAC name per structure. (2) Of course, that does not mean the others are bad or discouraged or allow for misinterpretation. However, you should clearly show that the other name is a non-systematic IUPAC-ish name. (3) I don’t know what you mean with your second comment …


text:
What is the product of the reaction of carbon with sulfuric acid?
I don't know if $\ce{C}$ or $\ce{CuO}$ react with $\ce{H2SO4}$. Some sources say that:
 $$\ce{C + H2SO4 -> CO2 ^ + 2SO2 ^ + 2 H2O}$$
(It seems to be similar to $\ce{Cu}$ reactions with $\ce{H2SO4}$).
Is the reaction possible? If the answer is yes, in what conditions does the reaction take place?
And I was pretty sure that $\ce{CuO + H2SO4 -> CuSO4 v + H2O}$
But my chemistry book says that none of the reactions is possible.
I'm little bit confused, some help would be apreciated.

Carbon reacts with sulfuric acid to produce carbon dioxide and sulfur dioxide gas along with water. Sulfuric acid should be a concentrated, heated solution.
$$\ce{C + 2H2SO4 ->[\Delta] CO2 + 2SO2 + 2H2O}$$
More insight about this reaction can be found here.
Copper(II) oxide, a black solid, and dilute sulfuric acid react to produce copper(II) sulfate, giving a characteristic blue colour to the solution. From this solution, blue copper(II) sulfate pentahydrate crystals may be obtained if desired.
$$\ce{CuO + H2SO4⟶CuSO4↓+H2O}$$
It is a general "insoluble metal oxide is reacted with a dilute acid to form a soluble salt" type of reaction. It is a standard class experiment.
More insight about this reaction can be found here.


text:
Reaction of Hydrogen and Dinitrogen Tetroxide?
What might be the possible products of a reaction between hydrogen and dinitrogen tetroxide ($\ce{N2O4}$)? They are fuel and oxidizer respectively in rockets.

And what do you think they are?

I have no Idea.

In future please try to have any idea, otherwise question can be downvoted and/or closed.

A likely reaction between dinitrogen tetroxide and hydrogen would be as follows with nitrogen and water being the products
$\ce{N2O4 + 4H2 → N2 + 4H2O~(I)}$
$\ce{\Delta H_{rxn}=[0+(4 x -241$.$ 8)-9$.$2 - (4 x 0)] ~=~ -976$.$4~ kJ}$
For the record, dinitrogen tetroxide and hydrazine (usually in combination with methylated hydrazines) is currently used as a rocket fuel.  The equation describing this reaction is
$\ce{N2O4 + 2N2H4 → 3N2 + 4H2O~(II)}$
$\ce{\Delta H_{rxn}=[(3 x 0) +(4~ x ~-241$.$ 8) - 9$.$2 - (2 x 50$.$6)] ~=~ -1077$.$6~ kJ}$

are traces of NOx possible in either case?

Sure, $\ce{N2O4}$ can produce $\ce{NO}$ directly.

@F.Sherrif or redissociates into $\ce{NO2}$.


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can magnesium chloride and iodine react together?
If everything is in a mixture of alcohol and water, would it be possible for magnesium chloride and iodine to react and produce something?

Magnesium chloride and Iodine with form an equilibrium.  This is a bit of a simplification of the reactions but the dissociated chlorides and free iodine will reversibly react to form iodide and triiodide ions and chlorine.
$$\ce{Cl- + 1/2 I2 <=> 1/2 Cl2 + I-} $$
$$\ce{MgCl2 + I2 <=> MgI2 + Cl2} $$
$$\ce{I- + I2 <=> I3-} $$
$$\ce{MgI2 + 2I <=> Mg(I3)2} $$
The amount of products of this reaction though will not be very appreciable, but yes a very literal sense magnesium chloride and iodine will react.

$\ce{<=>}$ gives you the correct arrows: $\ce{<=>}$

what do you mean by not be very appreciable? Could you quantify it generally?

That would be pretty negligible. On the same grounds you may claim that any two compounds react with each other, which also would be technically true in some sense.

Looking at it macroscopically, the first two equilibria are so far on the reactants’ side, that one would say ‘no reaction is observed’.

(-1): Disagree. See my explanation and source.

Would someone like to provide a reference here.

For the record, elemental iodine does not displace chloride!
Per an educational source, for example, to quote:

These reactions will take place in one direction only. The reaction of a halogen X2 with a different sodium halide (NaY) will occur only if X2 is more reactive than Y2. If X2 is less reactive than Y2, the reaction shown in Equation 1 will not take place.

However, in the real world, does this mean that in the presence of a transition metal impurity (say iron in the form of ferrous as found in tap water) is it likely that aqueous $\ce{MgCl2}$ is unreactive with an iodine presence?
To answer, I will explore some advanced fenton-type chemistry (which could be subject to recycling with light).
Consider the reaction between ferrous and HOI formed in the first step with the reaction of iodine and water:
$\ce{I2 (s) + H2O (l) <=> H+ + I- + HOI}$
$\ce{Fe(2+) + HOI -> Fe(3+) + .IOH-}$
And, at pH over 5 (assuming chemistry paralleling that of the chlorine-based $\ce{.ClOH-}$ radical anion):
$\ce{.IOH- -> .OH + I-}$
$\ce{Cl- + .OH -> .Cl + OH-}$ (Source)
$\ce{.Cl + .Cl -> Cl2}$
But, at pH under 5:
$\ce{.IOH- -> .I + OH-}$
implying likely no chlorine creation.
Note, per photolysis:
$\ce{I- + hv -> .I + e-}$
which could recycle iron ions:
$\ce{Fe(3+) + e- -> Fe(2+)}$.
Also, any ferric iodide formed is unstable and reverts to ferrous, which creates a cyclical clock reaction system (discussed here).
Bottom line, in my assessment, except with favorable impurities, pH and light, it is not likely that any free chlorine, as an intermediate, would be formed in a system of MgCl2 and I2.

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


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How to choose the best method and basis set for a calculation in computational chemistry?
In Quantum chemical calculations, it is not always the case that the more complex (correlated) the method or the basis set the more accurate the results.
For example, it's sometimes found that MP2 results better fit the experimental results than that of CI calculation.
In other instances diffuse functions negatively impacts the accuracy of the calculation.
Are there any general rules for the choice of a method and the basis sets?
What is the degree of accuracy of a calculation with no experimental reference ?

This might be of interest for you: http://chemistry.stackexchange.com/q/27302/4945

The experimental equivalent of your question would be: what kind and level of impurity am I going to accept in my experiment?
In terms of basis sets: in principle you keep increasing the complexity of your basis set until you reach convergence in terms of electronic energy. What you consider to be converged depends of course on your preferences.
With regard to the choice of method. Again, this depends on what you consider to be accurate enough. In heterogeneous catalysis, typically density functional theory calculations are performed using a PBE functional. This is accurate to about 5-15 kJ/mol which is fine in this field. This type of accuracy would be completely unacceptable in other fields. If high accuracy is required, people typically use post-Hartree-Fock methods such as MP2 or higher or CI.
My recommendation in picking a method and a basis set is the following:

Decide what you consider accurate enough.
Look into the literature if there are calculations being performed on the same or equivalent type of system and analyse their results.
Experiment using different methods on a model system. Look how the energy converges with respect to basis set and methodology. Pick the method that you think is best, both in terms of computational speed and accuracy. (it is OK to pick a couple of methods)
Perform the bulk of your calculations using this method.
If you are going to report your calculation. Clearly explain the reasoning why you picked a certain method. The potential readers of your report really appreciate to hear your motivation.

thank you ifilot .... so in case of no experimental reference i have to choose an equivalent computed reference

Exactly. Just critically read through their results. Try to reproduce a couple of these results and then try to extrapolate their method to your system.

I don't agree with the comparison to the impurity level. Less impurities is always better while in computational chemistry that kind of order might not exist, especially if you compare different density functionals.

To add onto this answer, you can always consider additional levels of theory (e.g. a few density functionals, for instance). Nobody would complain about providing more data. So, if you do this, feel free to publish whatever you run in the SI (explaining any anomalies in the actual text). I also would like to specifically agree with Point 5. More often than not, there is no justification for the level of theory used even if the authors had some underlying reason. It is extremely helpful for the reader if you briefly support your decision, even if it's just a matter of citing similar papers.


text:
What's the reaction here?
I have found a gif photo on the internet and I am very curious what is happening there? What reaction is the reason for that?

I kinda like the GIF, but using dangerous, reactive reagents such as concentrated $\ce{H2O2}$ to perform a rather violent reaction without any safety glasses?  Bad idea.

$\ce{KI}$ with $\ce{H2O2}$ with a tiny bit of dishwashing liquid. It is a rather violent redox reaction wherein $\ce{H2O2}$ is decomposed to $\ce{H2O}$ and $\ce{O2}$. You can actually see that $\ce{I2}$ is being formed by the brownish-yellowish color in the beginning of the reaction.
There is a lot of information about this reaction. I performed this reaction quite a few times myself for my students. It is a lot of fun. Some people call it making elephant toothpaste. :-)

authors: COSMOS 環宇海外移民服務


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What is the significance of Law of Multiple Proportions?
I cannot understand how the Law of Multiple Proportions is significant, or how does it further improve over Law of Conservation of Mass or Law of Definite Proportions. From Wikipedia:

If two elements form more than one compound between them, then the
  ratios of the masses of the second element which combine with a fixed
  mass of the first element will be ratios of small whole numbers.

Now I must say that I don't understand anything from the "small whole numbers" part, since it is not rigorously defined. For example, $31454315/43546546$ may be a ratio of small whole numbers for me but not for you.
Nevertheless the following is the algebraic interpretation of the law, with substituting the "ratios of small whole numbers" part with: "a real constant", since the ratios of two whole numbers is always a real constant.
Compound 1: Has $m$ mass of component $A$ and $x$ mass of component $B$.
Compound 2: Has $m$ mass of component $A$ and $y$ mass of component $B$.
By the Law of Definite Proportions, the following are correct:
$x/m = c_1$
$y/m = c_2$
where $c_1$ and $c_2$ are real constants. This is because Law of Definite Proportions states that "a chemical compound always contains exactly the same proportion of elements by mass".
Dividing $x/m$ by $y/m$ gives:
$x/y = c_1/c_2$
Since the ratio of two real constants is another real constant, we can express this as:
$x/y = c_3$
Since a real constant can always be expressed as the ratio of two whole numbers, this gave us the Law of Multiple Proportions.
As I have shown above, the Law of Multiple Proportions is a corollary of the Law of Definite Proportions.
So again: What is the significance of the Law of Multiple Proportions? Isn't it just an application of the Law of Definite Proportions? How does Law of Multiple Proportions improve over Law of Definite Proportions and Law of Conservation of Mass? How does it contribute to the first scientific atomic theory?

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

Is my question out of the scope of the community?

First read tells me no.

These laws all date back from the very early days of chemistry and are only taught at school to give pupils a way to grasp parts of the subject more easily. Except for the law of conservation of mass, which some professors/supervisors love stating frequently, they are hardly mentioned in universities outside of science history classes.
That said, as I understood them, they refer to slightly different cases. The law of definite proportions should be understood to read:

No matter how much mass of two elements will react with each other, the product will contain the masses of its elements in a certain ratio and this ratio only.

Taking an easy example, if I reacted $46~\mathrm{g}$ of sodium and $16~\mathrm{g}$ of oxygen, I would get $62~\mathrm{g}$ of $\ce{Na2O}$ where the ratio of sodium to oxygen is $46:16$. If I had instead reacted $92~\mathrm{g}$ of sodium with an unlimited amount of oxygen, I would have gotten $124~\mathrm{g}$ of the same product which contains sodium and oxygen in a $46:16$ ratio. No matter how much sodium I let react with how much oxygen, the product will contain the definite proportion of sodium to oxygen $46:16$.*
The law of multiple proportions now states that some elements are special. For example, iron can react with chlorine to give two different chlorides: $\ce{FeCl2}$ and $\ce{FeCl3}$. For each of the two you can, as you realised set up a law of definite proportions which often gives you a rather weird rational ratio. However, when comparing those two with each other, you find that the mass of chlorine in $\ce{FeCl3}$ is 1.5 times as much as the mass of chlorine in $\ce{FeCl2}$ — corresponding to a $3:2$ ratio. And it is these very small (usually single digit) whole numbers that are actually meant.
Taking a different example, the relative atomic masses of sulphur and chlorine are $32.07$ and $35.45$, respectively. So a sulphur dichloride would have mass ratios of $32.07:70.9$ — far from any small number, but still a rational ratio. Sulphur tetrachloride exists, too, with another pretty unclean ratio. However, the ratio of the chloride contents is $1:2$ — isn’t that a nice simple ratio?

*: This example is not technically correct. Burning sodium gives $\ce{Na2O2}$, sodium peroxide. But I wanted nice and whole numbers and oxygen and sodium have nice and whole mass numbers.

And those "small whole number" ratios lead to the idea that there must be discrete (rather than continuous) building blocks for the matter around, which can be explained with the idea of atoms right? I mean AFAIK this is the idea. But I am arguing that getting whole numbers in a ratio has not meaning whatsoever. Because even if we get, say 1.27/5.65, we can multiply the numerator and denominator by 100 and get 127/565, which is again a ratio of whole numbers. It is a ratio of "small" whole numbers maybe for me, but maybe not for you. That's the part that I do not understand. How does ...

... Law of Multiple Proportions have any improvement over Law of Definite Proportions and Law of Conversation of Mass to propose the first scientific atomic theory?

@Utku Well, of course, you can derive one from the other as you showed. But sometimes the milestone is merely finding the connection. 127/565 would probably not have lead to any helpful theory in any short time, while 2/3, 1/2, 1/3 actually help finding a theory.


text:
Why does a dynamic equilibrium exist?
We know that at equilibrium Gibbs free energy is minimum. We also know that at equilibrium both forward and reverse reactions occur simultaneously, and we also know that for a reaction to be spontaneous Gibbs free energy should decrease in that direction. So if at equilibrium we have minimum Gibbs energy then both forward and backward reactions will be non spontaneous, since they are characterised by an increase in Gibbs free energy, therefore there should be no forward and backward reactions and hence, no dynamic equilibrium. Then why does a dynamic equilibrium exist?

Roughly speaking, all reactions occur all the time, no matter what the Gibbs energy is. We call a reaction "spontaneous" when it goes faster than its reverse reaction (and yes, this is the case when the Gibbs energy decreases, and not otherwise). There is no such thing as static equilibrium. (Again, this is a bit of oversimplification, but I think it will do.)

Zero change in Gibbs reflects an overall zero change in entropy.  If you consider collision theory, even though the reaction may be at constant T there's still a Boltsmann distribution of energy so some collisions will lead to successful reactions.  At equilibrium the rate of successful collisions in each direction is the same, so successful collisions in each direction cancel each other out overall.

Crossposted to http://physics.stackexchange.com/q/227487/2451

I'm voting to close this question as off-topic because it's cross-posted on physics.SE.

@Todd Minehardt No I asked this question first!

The user who crossposted is not the same one as the user who asked the question here. I call plagiarism on physics, not crossposting. Voting to leave open.

@JM97 According to English typography (and most others), no spaces are placed before commas. That includes spontaneous, comma.

@Jan , yes that's right but I was editing only to attract viewers by doing some activity.

Please do not do that. Especially not on purpose after there already have been edits. It is very bad style and could lead to serious consequences. If you feel that your question did not get enough attention, continue collecting reputation and ultimately place a bounty on it (see in the [help]). You need 75 reputation to do that and it is the encouraged way to give a question more exposure.

You mixed up two different levels: the microscopic level, where the forward and backward reactions occur and the macroscopic level, where the Gibbs energy (or whatever depending on the system) is minimal. The macroscopic level is just an average, broad view on what is happening in miscroscopic level.

To understand the confusion here you need to take two different points of view about what is happening in a reaction: the molecular view and the statistical view.
The statistical view ("Gibbs free energy is a minimum") is an average of all the molecular activity in the reaction. It describes the net, bulk behaviour of the reaction. At equilibrium, there is no net direction of change. But this tell us little about what is happening to the individual molecules or atoms in the reaction.
The molecular (or atomic) view is about what is happening with the individual things that make up the reaction. At any temperature above absolute zero there will be plenty of things happening at the molecular level. There is plenty of energy around and the molecules will bang into each other a lot. Moreover, not all the components will have the same energy (some molecules are slow moving, others fast: the distribution of energies will follow a Boltzmann distribution). Sometimes there is enough energy for the products of the reaction to fall apart; sometimes there is enough energy for the reactions to react to give the product. Equilibrium is reached when, at a given temperature, the rates of both reactions are the same not when everything stops at the molecular level. Only when we look at the overall averages in the bulk reaction do we see no net change: at the molecular level things are always happening.
It is worth noting that the molecular view gives more insight than the purely thermodynamic view which only considered bulk averages. Some reactions at the molecular level happen fast, others slowly. Often in chemistry it is the kinetics that determine the product not the thermodynamics and thinking about the molecular point of view often explains the observed situation in the bulk.
It is quite important and insightful in chemistry to be able to switch from one view to the other as you will get a much richer interpretation of what happens in real reactions if you can.

I concur with what @matt_black writes. The thermal energy will always cause reactants to form products and product to form reactants at all points in a reaction, but only at equilibrium will the rates (not rate constants) be equal. If one could observed just a few molecules, at equilibrium, then the number of reactants and products would vary up an down as time progressed. This is due to the random nature of the energy supplied via molecular collisions with solvent  and distributed according to Boltzmann. Single molecule fluorescence experiments using a confocal microscope do allow such measurements.
When studied over a long time the forward and backwards rates (from single molecule measurements) would be equal. In a thermodynamic calculation we assume truly vast numbers of molecules, so any fluctuations occurring at the molecular level are unobservable.


text:
Constructing a Walsh diagram of AH2
According to Wikipedia, for the $\ce{AH2}$ Walsh diagram:

and the Walsh diagram in Shriver and Atkins Inorganic Chemistry, the $a_1$ orbital sinks in energy going from 90 to 180°. Furthermore, the $b_2$ orbital at 90° is higher in energy than the $a_1$ (lone pair).
However according to the second, newer diagram for a AH2 molecule on the same page:

the $a_1$ is rises in energy and the $b_2$ is definitely lower in energy than  the $a_2$ (or $a_1$ lone pair). This matches also my own self-made diagram, using Hartree-Fock 6-31G*, calculated through Spartan. Other pages use the same Walsh diagram.
What is going on here?
Shrivers textbook explains the rise of the $a_1$ orbital through decreased H-H overlap and reduced involvement of the $\ce{X~2_{pz}}$.

Just fyi for passers-by, here's the diagram in the 5th edition of Shriver & Atkins: http://i.imgur.com/VTcN1up.png

I'd say the diagram from Atkins "Inorganic Chemistry" is in error.
I went to what I consider the "definitive source" for such Walsh diagrams: Orbital Interactions in Chemistry by Albright, Burdett and Whangbo.
Here's their diagram for $\ce{H2S}$ (the linear on the left, the bent on the right):

They don't readily indicate the method they used for the calculation, but on this energy scale the difference between the $1\sigma_g$ and $1a_1$ is minimal. They show that the $1a_1$ is lower in energy since the interaction with the central p orbital will be stabilizing.
As to the $1b_2$ rising above the $2a_1$ in the bent structure -- the diagram from Albright, Burdett, and Whangbo agree that there's no "crossing."


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When is there hybridization and when is there no hybridization?
When is there hybridization and when is there no hybridization? Do all structures have hybridization or only some specific structures have it and why?

Hybridization is there when you are thinking about it, and vanishes when you stop thinking about it. Hybridization is not a physical phenomenon, really.

What do you mean? I know that hybridization is a model / theory, when considering this model, should I assume that all compounds have hybridization or only some?

@bean you could've attached a few examples, like $\ce{NH3}$ and $\ce{PH3}$, think about what will happen when bond angles become ~90° (ie: P-H).

As Ivan already mentioned, hybridisation is not a physical phenomenon. Physically, the most accepted theory at the moment is that you have nuclei and the electrons roam freely in between them, forming standing waves (orbitals). These standing waves generally spread across the entire molecule unless it is a standing wave of a core electron.
However, it is non-trivial for us chemists to come up with a proper description of these molecular orbitals. Hence, we use a certain set of assumptions. For example, the LCAO approach for molecular orbitals used in some calculations assumes a set of hydrogen-like orbitals on every atom and linear combines them to create the most stable set of molecular orbitals. Since that is still rather hard to do, chemists have also derived a pen-and-paper only approach that does this on a per-atom basis (called hybridisation) and then checks what kind of orbitals would form which bonds — so that’s two steps of linear combinations following each other.
But all of this is only our thought experiment to better describe the physical picture of a molecule. Nature does not need hybridisation or linear combinations. Nature just ‘does it’ and comes up with a final solution. So whenever hybridisation ‘is involved’, it is always man thinking about how to look at the molecule.

Once you have taken that for granted, the question may remain:

But when do I need to take hybridisation into account for orbital considerations?

There’s a set of simple rules:

Elements of the third and higher periods will remain mostly unhybridised unless they require hybridisation for any specific issue ($\ce{SiH4}$ cannot be explained without $\mathrm{sp^3}$ hybridisation on silicon).
Halogens and noble gases will never be hybridised unless they are hepta-/ octacoordinated.
Generally assume a hybridisation for all elements of the second period. Use the one that best explains the geometry. For oxygen, assume at least one unhybridised p-orbital.
For a nitrogen with three single bonds, remember that the hybridisation will always switch between $\mathrm{sp^3}$ and $\mathrm{sp^2 + p}$. This is the nitrogen inversion.
Hydrogen needs no hybridisation.

Nice Jan!  I like to use the hybridization "language" to try and describe physical reality.  For me it is a simple and convenient language. A quick comment on your rule involving nitrogen - can't nitrogen also be sp hybridized as in a nitrile?

@ron Yes, and plain sp² in imides, azobenzene and related; I was thinking only about $\ce{NR3}$ cases where there are three single bonds.


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Can the green spots in these coins be copper chloride?
My father collects coins as a hobby and lately has observed some green spots in some of his coins so he told me to ask here to see if anyone knows what it is.

The coins are made of silver (999 parts out of 1000 according to the description) so at first he thought that it might be some silver oxide but the silver oxide is dark brown and not green.
He looked it up in the Internet and he has read that it might be copper chloride but it doesn't seem right to him due to the purity of coin.
Is this feasible? Can the spots be copper chloride? If not, what can they be?

Unless for some reason you prefer to breathe chlorine at home (which would be very peculiar indeed), I find the copper chloride hypothesis quite improbable. More likely, this is basic copper carbonate, a typical component of patina formed on copper as it gets oxidized by air.

Indeed, as Ivan noted, it is most likely copper patina. Silver is usually obtained as a byproduct in copper refinning, so a copper impurity is likely.

This phenomenon is relatively common among jewellers and silversmiths and they call it verdigris. This verdigris can be copper oxide, copper acetate, copper carbonate and even copper chloride. The former options are more likely, but copper chloride can actually be a possibility if you live close to the sea.
Perhaps relevant to your father: verdigris can be cleaned by applying a mildly acidic solution to it, such as lemon juice or vinegar. The formation of verdigris can be prevented by keeping the metal dry as this kind of corrosion is typically formed by contact of the metal with moisture.


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Why does the carbonyl group in an acid anhydride have two stretching frequencies?
The carbonyl stretching frequencies for an acid anhydride are approximately $1820~\mathrm{cm^{-1}}$ and $1760~\mathrm{cm^{-1}}$. These are both higher frequencies than a simple ketone owing to the electron withdrawing capability of the adjacent group. However, why are there two distinct infrared stretching frequencies?

The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride.

Source: Introduction to Spectroscopy, Pavia and Lampman

You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=O group is lengthening as the other is shortening.  You can see an example of this kind of spectral pattern in the example below.

Source: Introduction to Spectroscopy, Pavia and Lampman

This explains why you get two frequencies for C=O bond in acid anhydrides, whilst ketones/aldehydes etc. only give a single IR frequency for the C=O bond.

I'm just curious that if both C=O are shortening, would that give 1760 cm^-1 ?

Its a continous motion of shortening and lengthening, so yes

Is there any simple explanation why the asymmetric vibration corresponds to a higher energy, I would have expected it to correspond to a lower energy.

Just to add on to NotEvans's answer:
The stretching frequencies of compounds can theoretically be predicted by group theory and symmetry considerations - some introduction can be found here. In general, for "simple" molecules such as $\ce{CO2}$ ($D_{\infty\mathrm{h}}$) and $\ce{XeF4}$ ($D_{4\mathrm{h}}$), the bonds do not stretch individually, but rather their vibrations are coupled. A diagram of the stretching vibrations for the two compounds mentioned above will illustrate this (there are other non-stretching vibrational modes which are omitted):

(source: Orbital Interactions in Chemistry 2nd ed., Albright, Burdett & Whangbo)
In general this arises because the $\ce{C=O}$, or the $\ce{Xe-F}$, bonds are related by symmetry. In ketones, there aren't any other $\ce{C=O}$ groups next to it, so the $\ce{C=O}$ bond tends to vibrate individually, which allows us to speak specifically of "the $\ce{C=O}$ stretch" in IR spectroscopy.
However, in anhydrides, there are two carbonyl groups next door to each other, and while they are not always exactly equivalent by symmetry, they still display coupling to each other, as @NotEvans described. It's hardly limited to anhydrides, though. In acetylacetone, where the two $\ce{C=O}$ groups are equivalent:

(source: Introduction to Spectroscopy 5th ed., Pavia et al.)
Another extremely easy-to-spot example of coupling is in primary amines $\ce{RNH2}$ or primary amides $\ce{RCONH2}$ - you will observe two distinct "$\ce{N-H}$" stretches, which are actually symmetric and antisymmetric stretches. Here's an example of a primary amine, $\ce{BuNH2}$:

Compare that with a secondary amine, $\ce{Bu2NH}$, which only has one $\ce{N-H}$ bond:

(source: Pavia et al.)


text:
Why is the carbonyl IR frequency for a carboxylic acid lower than that of a ketone whilst an ester is higher than a ketone
The approximate stretching frequencies are as follows:

Acid: $1710$, Ketone: $1715$, Ester: $1730$ (all $\mathrm{cm^{-1}}$)

This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid. However, I am finding it hard to reasonably justify why. Presumably this is also linked to the relative electrophilicity of the ester and the acid.

First things first, it's important to realise that functional groups may have IR frequencies over quite a broad range. I've provided the table below, from which you can see that actually there is a great deal of overlap depending upon the actual 'environment' of the functional group in question. Ketones and acids, for example, cover a very similar portion of the spectrum.

Stretching frequencies of common carbonyl groups (Taken from Introduction to Spectroscopy, Pavia and Lampman.

That said, the difference between the frequency range of an acid vs an ester is an interesting one, as on the surface of it they are electronically quite similar (in both cases the carbonyl is next to an electronegative oxygen).

@RobChem: This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid.

In your statement, you may have gotten a little confused, remember that the stretching frequency increases with bond strength. Therefore using your data we can say that the ester is stronger than the ketone whilst the acid is weaker.
We can explain the ester being stronger by considering the effect of the electronegative oxygen withdrawing electron density:

This argument however, fails to explain why the carboxylic acid has a stretching frequency below that of the ketone, since, as was already said, they both have the oxygen adjacent to the carbonyl.
One rationalisation for this is that the carboxylic acid group doesn't exist in isolation but rather interacts with other carboxylic acids in a hydrogen bonding interaction, which weakens the C=O bond:

This hydrogen bonding concept is universal, for instance in methyl salicylate, the C=O bond has dropped to well below that expected for an ester:

Silverstein et al., Spectrometric Identification of Organic Compounds writes that the C=O stretch in a monomeric carboxylic acid is ~$1760~\mathrm{cm^{-1}}$, which of course supports what you already said.

Sorry. I do not follow: We can explain the ester being stronger by considering the effect of the electronegative oxygen withdrawing electron density:

You wrote This argument however, fails to explain why the carboxylic acid has a stretching frequency below that of the ketone, since, as was already said, they both have the oxygen adjacent to the carbonyl. I think you meant to write ester instead of ketone.

the predominant effect of the oxygen of an ester is inductive electron withdrawal, so the resonance contributor with the single bond contributes less to the hybrid. The carbonyl group of an ester, therefore, has a C-O double-bond character than does the carbonyl group of a ketone, so the former is stronger and harder to stretch.
bond in a carboxylic acid has a partial double-bond character that is due to resonance electron donation.


text:
Why do all radioactive decay series terminate at lead?
Thorium series:

Uranium series:

Actinium series:

Why do all the radioactive decay series terminate at lead isotope? Why not hydrogen?
(All pictures were taken from Wikipedia.)

I'm no expert, but I thought they just kept on decaying until they reached a stable isotope, and lead just happens to be one

Well, terminating at hydrogen would imply that no elements beyond hydrogen are stable, which would imply a very different universe. Lead just happens to be the first reached stable nuclei.

You might benefit from studying the first diagram here while noting that most of the nuclides above and to the right of the intersection $Z = 82$, $N \approx 126$ decay towards lead, but lighter unstable isotopes generally do not.

I believe I read somewhere that at sufficiently high energies, lighter elements tend to fuse toward iron.

There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of this division (0, 1, 2 or 3, corresponding to the $4n$, $4n+1$, $4n+2$ and $4n+3$ mass number decay chains respectively) stays constant under alpha, beta or gamma decay. There are decays which change the remainder (neutron emission, proton emission, spontaneous fission, etc) and therefore allow hopping between the main decay chains, but for simplicity these can be approximately ignored for the isotopes which are not too unstable or too heavy.
Only three out of these four main decay chains "stop" at lead, so already the statement in the question is incorrect. The $4n$, $4n+2$ and $4n+3$ chains "stop" at $\ce{^208_82Pb}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$, respectively. The $4n+1$ decay chain reaches $\ce{^209_82Pb}$, but this lead isotope is quite short-lived ($t_{1/2}=3.25\ \mathrm{h})$ and decays further to $\ce{^209_83Bi}$ where it "stops" (or at least seemed to until 2003).
Note that I said the chains "stopped" at these isotopes. That's because in reality they don't. It just happens that $\ce{^208_82Pb}$, $\ce{^209_83Bi}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$ are all very long-lived isotopes, with half-lives billions of times greater than the current age of the Universe. This means that any continued decay is severely bottlenecked at these points, so for almost all practical purposes, the decay chains stop there. However, rigorously speaking, if you were really patient there would be further decays in the sequence. In fact, theoretically the heaviest isotope which is not susceptible to any known mode of spontaneous radioactive decay (except proton decay) is an isotope of zirconium, $\ce{^92_40Zr}$, so the decay chains actually go on well past lead or bismuth. It just takes an immense amount of time for the decays to happen.

Seems that proton decay has not yet been proved.  Overall +1.

Short Answer: They don't.
As the comments above indicate, the decay chains will terminate at a stable nuclei. For the series you mentioned in your question, lead has the first stable isotope.
A really useful tool for looking at different decay chains is available from Oak Ridge National Laboratory's Risk Assessment Information System.
You can pick from lots of different isotopes and see the decay chains. For example, $\ce{^142_56Ba}$ gives $\ce{^142_58Ce}$ as the final stable product. $\ce{^14_6C}$ goes to $\ce{^14_7N}$, etc. Many terminate with lead, but many don't.
As for hydrogen? Keep in mind that for large nuclei, fission and other nuclear processes that decrease the number of protons are usually favored, but for small nuclei, processes that increase the number of protons are often favored (e.g., beta decay) - changing the neutron-proton ratio and generating more stable nuclides.


text:
Why do the boiling and melting points decrease as you go down group 1 and vice versa for group 7?
I used to think that because an alkali metal needs to lose one electron to complete its outer shell, when the atom increases in size (atomic radius), the electron would be easier to lose as the attraction between it and the nucleus would be lower, resulting in increased reactivity as you go down the group (the opposite being true for halogens as they need one more electron and so need to retain their electrons).
Now the trouble I am having is that why would the melting point also increase, what does the ability to lose an electron have to do with the intermolecular forces between several atoms?

The group 1 elements are the so-called alkali-metals. The bonding between the atoms is caused by the interaction of the nuclei with the delocalized electrons. With increasing number of electrons and protons, the atomic radii get bigger and hence this interaction becomes weaker as the average distance between nuclei and electrons increases as you go down in group 1.
The group 7 elements are the so-called halogens. They exist under normal circumstances in their molecular form ($\ce{F2}$, $\ce{Cl2}$ and so on). In contrast to the group 1 elements, the dominant intramolecular force here is London dispersion (or van der Waals forces if you prefer). This attraction is caused by the correlated motion of electrons. With increasing amount of electrons, there can be more correlated motion and hence there is a stronger interaction between the molecules and an increasing melting / boiling point when you go down in group 7.

Thanks, I think I get it. So for what you are saying is that for group 1 elements, the forces between the electrons and the atoms influence the intermolecular forces (forces between individual molecules), thus resulting in varying boiling/melting points (due to energy needed to overcome forces etc)?

Elements can condense to form a solid or a liquid because there is an attractive force between them. For the group 1 elements, the dominant force is here the interaction between the nuclei and the electrons. When more and more heat is put into these systems, these intermolecular forces can no longer keep the elements together and the system undergoes a phase change.


text:
Molecular mechanics force fields for metals?
I know of several general molecular mechanics methods suitable for metals, including Dreiding and UFF.
I know it's difficult to get many things right for such classical treatment of inorganic / organometallic molecules, because oxidation state, spin state and other factors can greatly influence geometries and energetics.
But I'm surprised that so few attempts have been made to improve on these methods. Are there no other methods published since 1992? I'm looking for force fields that cover a substantial part of the periodic table (or at least most metal-ligand interactions) and are completely published - proprietary methods don't count.

Paging @F'x - I believe you have some expertise here..

I think the cynical answer is that it's probably very hard to get funding to do something like this now. I'm looking for more technical answers. Thanks!

The modelling of metals using a 'simple' potential is relatively difficult. I have a suggestion for a force field, though it does not fit exactly all your criteria, but I would like to offer it anyway.
Adri van Duin has developed a force field called ReaxFF. This force field is non-proprietary and is readily available in the open-source package LAMMPS. This force field gives pretty good results and can be applied to any kind of elements, it can describe chemical reactions and is also valid for hetero-atomic interactions. That being said, the biggest challenge when using the ReaxFF force field is that you have to fit / train it to your system. This requires, for instance, a database of ab initio calculations that are representative for your system.
If you cannot find any force field, than this might be a interesting alternative to developing a force field completely from scratch.

I know of ReaxFF, but my understanding is that there isn't a generally-parameterized version like UFF. People end up tailoring the parameters for reactions of interest. So I think of ReaxFF as a "family" of force fields, rather than a general one. Is there a "base" parameterization for general use?

Your understanding is correct. :-) That is why I mentioned in my answer that it does not exactly fit all your criteria. The advantage of fitting a force field though, if you base it on ab initio calculations, is that you tend to get better results than using a 'general' force field.

I think most new parameterization uses high-quality calculations with schemes like ForceBalance. My point is rather that this seemingly productive area of research (e.g., "fast geometry cleanup") has quietly, while other areas of MM have gained significant advances.

You could have a look at VALBOND from Clark Landis, don't know how generally it was parameterized. LFMM from Rob Deeth is very good, but not many metals. I think old PCModel is still around, one of the better for metals. Allinger did a metal extension in MM3(94), with generalized parameters for the periodic table, but not much used.
I've created a number of specific metal force fields based on MM3* in MacroModel, but nothing general.
The easiest is usually a POS model with bonds from covalent radii, but no directionality.

Wow, great to get an answer from you! I know that UFF incorporates VALBOND concepts, as does ReaxFF. I'm aware of LFMM and similar ligand-field schemes, but as you say, they don't seem very general.


text:
What is the reason for enantiomerically pure ketone reaction under acidic conditions giving a 1:1 mixture of diastereoisomers?

(b) Treatment of the enantiomerically pure ketone C under acidic conditions leads to a 1:1 mixture of diastereomers

My thinking is that perhaps the explanation has something to do with the carbonyl group being prochiral and having two faces, each side generating one of two diastereoisomers?
I'm not very sure of this explanation though. What is the correct reasoning for this?

Oh the pain of alpha-chiral carbonyl compounds ^^'

Carbonyls are able to undergo keto-enol tautomerisation, this is accelerated in the presence of acid/base.
This phenomenon means that alpha-chiral carbonyls are generally regarded as configurationally unstable, as they are prone to racemisation, although in the absence of acid/base the extent to which the alpha-chiral carbonyl exists in its enol form is often negligible (the C=O, C-H combination is ever so slightly more favoured than C=C, O-H in a lot of cases).
The enol form of the carbonyl (the middle structure in my diagram below) is planar. When the enol tautomerises back to the ketone form, the proton may be transferred to either face of the planar structure, leading to both diastereomers being formed, as you describe.


text:
Why does tin form tin (II) compounds?
The stability of the +II oxidation state in lead has been widely attributed to the so-called “inert pair effect”: the 6s subshell is stabilized through relativistic contraction due to the near-c speed of its electrons. Now, Pekka Pyykkö states in his 1988 article (p. 585) that

In addition to the 6s2 inert pair discussed above, analogous compounds (such as GeO and SnO) are known for rows 4 and 5. While the relativistic 5s stabilization is not entirely negligible, the 4s stabilization due to the “d-block contraction” probably overweighs relativity on row 4.

How does the d-block contraction cause a change in valency? I know that due to poor shielding by d electrons, atomic radii across the fourth period don’t increase particularly. But how is it responsible for the increasing stability of the +II oxidation state in group 14 (before getting to lead, where relativistic effects play a significant role)?
Would I be wrong if I said that the cause for the increasing stability of the s electrons in germanium and tin was the existence of a full d subshell (absent in, say, silicon), which, however slightly, shields the nuclear charge for the p electrons better than for the s electrons? Or what is it that happens, exactly?
Finally, according to the orbital potential energies listed on page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.), the 5s electrons of tin aren’t more stable than those of silicon – their energy (–14.56 eV) is higher than the energy of the 3s electrons in the Si atom (–14.89 eV), even if only by a small amount. Compare this to germanium and lead, whose valence s orbitals are both more stable than the silicon’s 3s (–16.05 eV and –15.12 eV, respectively). How is it, then, that tin forms divalent compounds while silicon doesn’t? Is Pyykkö correct in associating the Sn valency with 5s stabilization?

An excellent question, you poked my interest. Indeed, tin's outer shell electrons are less bound than that of silicon. However, energy of formation of tin-element bonds is generally lower than for silicon. My guess is that this is because of d-subshell interference: its interference lengthens covalent bonds, preventing effective orbital overlap, thus tin compounds are shifted towards ionic spectrum, while silicon compounds are mostly covalent.

Indeed, it seems that the bond length is the reason, but I don’t think it is related to the d-subshell (see my answer below).

Drago wrote a paper that discussed this exact issue: J. Phys. Chem. 1958, 62 (3), 353–357 The conclusion is of course the same as what you have written. It arises from a combination of the stabilisation of the s subshell, and the difficulty of forming covalent bonds.

I feel very ashamed to answer my own question, but having found a possible answer, I don’t see why I shouldn’t share it with the community.
I would like to start by bringing in this table of orbital energies (from the appendix I linked to in the question) as a reference point for further explanations:
$$\begin{array}{ccc}
\hline
\text{Element} & \text{Energy of outer s orbital / eV} & \text{Energy of outer p orbital / eV} \\
\hline
\text{Carbon} & -19.43 & -10.66 \\
\text{Silicon} & -14.89  & -7.78 \\
\text{Germanium} & \mathbf{-16.05} & -7.54 \\
\text{Tin} & -14.56 & -7.01 \\
\text{Lead} & \mathbf{-15.12} & -6.81 \\
\hline
\end{array}$$
As you can see, the energies of the p orbitals get higher and higher as you go down the group, while the energies of the s orbitals don't.
The reason why, in general, orbitals rise in energy when going down a group is that even though the nuclear charge is bigger (which has a stronger effect than the rise in the quantum number $n$), the increased screening effect of added electrons prevails over this rise.
Germanium has such a low 4s electron energy because of the so-called d-block contraction. Since it has 10 added protons due to the existence of the d-block, and the d electrons don’t shield this increased nuclear charge well, the s electrons are stabilised. The p electrons don’t feel this effect as much because they are shielded more than the s electrons (less penetration), thus experiencing the rise in the nuclear charge less, while being shielded additionally by the stabilised s electrons (and presumably by the d electrons as well, to a lesser extent).
The lowered energy of the 6s orbital of lead is the result of the equivalent f-block contraction, amplified by the relativistic contraction of the s shell. The p electrons don’t have a noticeable change for the same reason they didn’t in germanium.
Now, tin has d electrons too, so it will experience the d-block contraction – but to a far more limited extent, since the change in nuclear charge is not as big in percentage. To illustrate this concept, it is useful to bring in this formula:
$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$
where $E$ is the energy of the electron (in the hydrogen-like atom!), $Z$ the nuclear charge and $n$ the principal quantum number. Since it refers to hydrogen-like atoms, this formula obviously doesn’t bring in the screening effect. To see why the d-block contraction doesn’t have such a strong influence on the tin atom, we can use this formula to calculate how the non-screened, non-relativistic energy grows from Si to Sn:
$$\begin{array}{cc}
\hline
\text{Element} & \text{Non-screened energy of outer orbital / eV} \\
\hline
\ce{Si} & -296.178 \\
\ce{Ge} & -870.4 \\
\ce{Sn} & -1360 \\
\hline
\end{array}$$
As you can see, the percentual change from Si to Ge is much higher than the change from Ge to Sn – thus, the increase in nuclear charge from Ge to Sn doesn’t have an enormous effect on the electron energy, while the screening effect between Ge and Sn grows steadily: the orbital energy is lowered in the many-electron atom. Hence, the Sn s shell is stabilised, but to a lesser extent.
Now, why does Sn form divalent compounds so easily, when Ge and Si have lower s energies and don’t? There’s a passage in Greenwood’s Chemistry of the Elements (2nd ed), pp 226-7 that explains it:

The term “inert-pair effect is somewhat misleading since it implies that the energy required to involve $n\mathrm{s^2}$ electrons in bonding increases in the sequence $\ce{Al} < \ce{Ga} < \ce{In} < \ce{Tl}$. […] This is not so […]. The explanation lies rather in the decrease in bond energy with increase in size from Al to Tl so that the energy required to involve the s electrons in bonding is not compensated by the energy released in forming the 2 additional bonds.

Indeed, that is what happens: the $\ce{Sn-X}$ bonds are longer than the $\ce{Si-X}$ bonds, and thus have smaller energy, as permeakra pointed out in his/her comment. However, I believe this is not due to the interference of the d-subshell, but simply to the larger Sn nuclear charge. This increased nuclear charge amplifies the repulsion to the nuclei of the other species (say, oxygen), thus lengthening the bond. Since the s electrons have a very low energy compared to the p ones, they do not participate in this bond – they cannot overlap.
In short, tin forms divalent compounds because:

Its s subshell is more stable than the p subshell, because of screening differences;
Its bonds are longer, due to the increased nuclear charge: thus, the s electrons don’t bond.

The combined effect is smaller in germanium compounds (Ge is more tetravalent), because even though the s shell is more stabilised, the bonds are shorter; it is stronger in lead compounds (Pb is more divalent) because the 6s shell is more stabilised than the Sn 5s shell and the bonds are even longer.


text:
How does this brominated intermediate lose its methyl group?
The problem is such:

So far my working has been:

I'm concerned about the last step where the ester forms a ring. The methyl group doesn't look like a plausible leaving group, but it does not appear in the product so how can I get rid of it?

@A.K. Sorry! I was having difficulty uploading the image.. Please have a look now!

I think your top drawing is incomplete.  Specifically, hydrolysis of the ester is required in order to first convert it to the carboxylate anion.  It is the oxygen in this anion that reacts with the bromonium ion to form the intramolecular lactone. To achieve this the bromination takes place in the presence of sodium bicarbonate which hydrolyzes the ester.  You can google "bromolactonization" or see this link for the analogous iodolactonization.

@ron yes thanks for pointing it out. My working contains the next few steps of the reaction mechanism, but i had the most difficulty with the first steps as pictured

A methyl cation definitely won’t leave by itself. However, you can consider bromide ions floating around in solution. These bromide, being nucleophilic can attack the methyl group in an $\mathrm{S_N2}$ manner, because any positively charged oxygen is a good leaving group. Therefore, the side product would be bromomethane.
Other than that, your mechanism is looking good.

i was wondering what CHCl3 does in the reaction?

@justbehappy Be a solvent, I guess. (There shouldn’t be anything it does to participate.)

I propose:

Hydrolysis of ester to acid through basic hydrolysis
Halogenation of the double bond
Having in mind the leaving groups, you get the desired cycle.

The first set of conditions provided by the o/p don't permit the ester to be 'reduced' (though the link provided is for basic hydrolysis).    @ Jan's solution seems more plausible, though I agree with @Ron that it is much more common to hydrolyse the ester first before closing the ring.


text:
What is the reason behind neglecting & putting the overlap integrals $0$ in Simple Hückel Theory?
This is an excerpt about Simple Hückel Theory from Elements of Physical Chemistry by Peter Atkins:

The first step that Hückel took was to ignore the $\sigma$-bonding framework and focus solely on the $\pi$ electrons. That is, he assumed that the atoms had taken up the positions they have in the actual molecule, then calculated the properties of the $\pi$ orbitals that matched that framework.
[...] we show in Derivation 14.3 that the Schrödinger equation for the orbitals $\hat H \psi = E\psi\;,$ then becomes the following pair of simultaneous equations for the coefficients: $$(H_{AA}- E)c_A- (H_{AB}- ES)c_B= 0\\ (H_{BA}- ES)c_A+ (H_{BB}- E)c_B= 0$$ ... . These equations are called the secular equations.
Hückel then made further approximations. First he neglected all the overlap integrals and set $S= 0$ wherever it appear. .....

I didn't quite get why he made this approximation. There must be some reason for neglecting the overlap integral, I suppose. Also, Atkins didn't mention why Hückel only made calculations for $\pi$ orbitals only; does his equations not work for $\sigma$-bonding?
So my questions are:
$\bullet$ What is(are) the reason(s) behind making the approximation of putting all the overlap integrals $0\;?$ After all, making $S= 0$ would mean there is no overlap.
$\bullet$ Why did Hückel work only on $\pi$ orbital? Doesn't his calculation work for $\sigma$-bonding?

What is(are) the reason(s) behind making the approximation of putting
  all the overlap integrals 0?

I am not 100% sure here, but using such a simplification, the mathematics becomes significantly easier. In principle, Hückel theory would still work if there would be overlap. In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system. You do not have to back-transform your solution as is the case in the Hatree-Fock method.

Why did Hückel work only on π orbital? Doesn't his calculation work
  for σ-bonding?

Only $\pi$ electron molecular orbitals were included because these determine the properties of the molecules that Hückel was studying (conjugated molecules). The $\sigma$ electrons were ignored. This is known as $\sigma$-$\pi$ separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. As a consequence though, the Hückel method is in principle limited to planar systems.

Taking $S=0$ implies an orthogonal basis set of AOs & this indeed makes the maths easy. But if there is no overlap between any of the constituting AOs, then how can MO form?

Even without overlap, by the diagonalisation of the Hamiltonian matrix you obtain eigenvectors and -values. Those vectors represent the molecular orbitals which are linear combinations of the atomic orbitals. For example, in the ethylene system you obtain $\psi_{0} = \frac{\phi_{0} + \phi_{1}}{\sqrt{2}}$ and $\psi_{1} = \frac{\phi_{0} - \phi_{1}}{\sqrt{2}}$ for the HOMO and LUMO, respectively.
This is for instance demonstrated on the Wikipedia page concerning the Hückel  method: https://en.wikipedia.org/wiki/H%C3%BCckel_method#H.C3.BCckel_solution_for_ethylene.

so, what does resonance integral being zero imply?

The resonance integrals are the (off-diagonal) integrals of neighbouring atoms. If those are zero as well, then you were correct that you cannot find any MO because every MO will then simply be an AO and all these MO/AO will have the same energy. (and your system is no longer conjugated)

In those examples, $S_{12}= S_{21}= 0$ meaning there is no overlap between atom 1 and 2; still $H_{12}= H_{21}\ne 0= \beta$; I'm not getting that- if there is no overlap which is evident from $s_{12}$ being zero, how can there be non-zero $H_{12}$ as it would mean there would be overlapping? I'm really not understanding how $S_{12}= 0$ and $H_{12}\ne 0$ go with each other.

Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij} = \delta_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle \phi_{i}|\hat{H}|\phi_{j} \rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a nonzero outcome.

Use \langle \rangle to get $\langle ;\rangle$ rather than <> to get $<;>;.$

(thanks for mentioning about the mathjax commands) To give an example of my previous comments; have a look at this website: http://www.ivofilot.nl/posts/view/29/Introduction+to+Electronic+Structure+Calculations%3A+The+variational+principle. If you look at example 2, you see that $\psi_{1}$ and $\psi_{2}$ are orthogonal to each other so they have zero overlap. Yet the off-diagonal elements of the Hamiltonian matrix are nonzero.

You are right at all your points but still I'm not getting the physical insight; I'll post it as a different question where I can reflect my problem more clearly. Thanks, nevertheless for the answer; +1.

It does not really have a physical interpretation. You should see it as something transcendental. Like a decoherence of the wave function if you will. That is the reason why diagonalization of the Hamiltonian matrix results in the correct energies of the electronic states: because you get rid of the off-diagonal elements. It is actually a very good question and deserves a more elaborate answer than what I can give in this comment.

It's probably worth mentioning that there is an extended Hückel method that includes both $\sigma$ and $\pi$ electrons.


text:
Can nonpolar molecules exhibit dipole-dipole forces?
Dipole-dipole forces occur when the positive part of a polar molecule is attracted to the negative part of a polar molecule. In a nonpolar molecule, there may still be polar bonds, it's just that the dipoles cancel each other out. So why can't there be dipole-dipole forces between nonpolar molecules with polar bonds? There are still positive and negative parts of the molecule, so there can be attractions between them.
For example, in $\ce{CO2}$, there are negative areas near the oxygens and positive areas near the carbon.

related http://chemistry.stackexchange.com/questions/14641/why-is-o-c-o-nonpolar

Short answer: there are many electrostatic interactions between two non-polar molecules.
Beyond monopole (full charges) and permanent dipole moments (polar molecules), there is a full multipole expansion for the electrostatic potential around any molecule. (This is technically true for atoms and ions too, but higher-order terms are really only useful for molecules.)
So there are electrostatic potential energy interaction terms for charge-dipole, dipole-dipole, dipole-quadrupole, quadrupole-quadrupole, etc.
These terms are important - the quadrupole-quadrupole interactions dictate the orientation of the benzene dimer and $\ce{CO2}$ dimer in your example.1
The problem is that most of these interactions die off very quickly. The quadrupole-quadrupole term is:1
$$E(r)=\frac{-\Theta_1\Theta_2}{4\pi\epsilon_0r^5}\times\Gamma(\theta_1,\theta_2,\phi)$$
So roughly $1/r^5$, compared to $1/r^3$ for dipole-dipole interactions, or $1/r^6$ for dispersion forces like induced-dipoles.
When such molecules are close, the quadrupole moments (and other multipole electrostatic terms) can dictate packing and distances, but are not as strong or as long-range as dipole-dipole or charge interactions.

So nonpolar molecules with polar bonds would have higher melting and boiling points than nonpolar molecules with no polar bonds right?

@swenger There are a lot of things that go into melting points and boiling points (e.g., internal rotations, etc.) but if you had two hypothetically equivalent non-polar molecules and one had polar bonds (and thus quadrupole interactions) it should have higher forces and higher mp and bp, yes.

@Geoff Hutchison Can polar molecules like water exhibit induced dipole moment interactions? We know that water molecules have permanent dipole moment but are there any changes in the dipole moment due to polarizability so we have an extra interaction?

In the case of non-polar molecules, dispersion forces or London forces are present between them. These forces are induced dipole - induced dipole interactions. As we know that in non-polar molecule, the whole molecule has zero dipole moment but bonds are polar.

When two non-polar molecules comes closer to each other. The negative part (electrons) of one molecule attract the positive part (nucleus) of another molecule. As a result, two dipoles are induced. Such dipoles are called induced dipoles and interaction is called induced dipole - induced dipole interactions.
Example - CO2 - CCl4 and He - Ne intractions

This doesn't actually answer the question about whether there are dipole-dipole forces in molecules like $\ce{CO2}$ and if not, why.

Dipole-dipole interactions are electrostatic interactions between the permanent dipoles of different molecules. These interactions align the molecules to increase the attraction.
In order to form a dipole-dipole attraction, there should be a dipole moment for the considered molecule.
For an example: Water ($\ce{H2O}$)
It has a large permanent electric dipole moment. It's positive and negative charges are not centered at the same point; it behaves like a few equal and opposite charges separated by a small distance.
The permanent dipole in water is caused by oxygen's tendency to draw electrons to itself (i.e. oxygen is more electronegative than hydrogen). The 10 electrons of a water molecule are found more regularly near the oxygen atom's nucleus, which contains 8 protons. As a result, oxygen has a slight negative charge (δ-). Because oxygen is so electronegative, the electrons are found less regularly around the nucleus of the hydrogen atoms, which each only have one proton. As a result, hydrogen has a slight positive charge (δ+)
If we considered a NON POLAR molecule with POLAR BONDS as you mentioned; such as $\ce{CCl4}$ or $\ce{CO2}$
Molecules often contain polar bonds because of electronegativity differences but have no overall dipole moment if they are symmetrical. In the molecule tetrachloromethane ($\ce{CCl4}$), the chlorine atoms are more electronegative than the carbon atoms, and the electrons are drawn toward the chlorine atoms, creating dipoles. However, these carbon-chlorine dipoles cancel each other out because the molecular is symmetrical, and $\ce{CCl4}$ has no overall dipole movement.
Though $\ce{CO2}$ have polar bonds,it does not have a dipole moment, so it can not form dipole-dipole interactions.