text:
Nomenclature for two cyclohexane rings linked by a single bond

From my current basic organic chemistry knowledge, I suppose it is 1,1-dimethylcyclohexane-4-methylhexane. I considered one of the hexane ring as parent chain. But I'm looking for the correct answer with appropriate explanation for it.

The relevant sections on ring assemblies of two identical cyclic systems in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually read as follows:

P-28.2.1 Ring assemblies with a single bond junction
Assemblies of two identical cyclic systems joined by a single bond are named by one of two methods:
(1) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding parent hydride enclosed in parentheses, if necessary. Parentheses are used to avoid confusion with von Baeyer names;
(2) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding substituent group (for names of substituent groups, see P-29), enclosed in parentheses, if necessary.
Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
(…)

P-52.2.7.1 Preferred IUPAC names for assemblies of two or more identical cyclic systems joined by a single bond are formed using the names of parent hydrides rather than the names of substituent groups, (…). For two-component assemblies, locants of one ring are unprimed; locants of the second ring are primed. Locants, including the locants 1 and 1', are necessary in preferred IUPAC names to indicate points of attachment of rings or ring systems (see P-28.2.1 ).

Therefore, the preferred IUPAC name for the compound that is given in the question is formed according to Rule (1) as 4,4'-dimethyl-1,1'-bi(cyclohexane).
The corresponding name according to Rule (2) is 4,4'-dimethyl-1,1'-bi(cyclohexyl). This name may be used in general nomenclature.

Chemspider claims  4,4'-Dimethyl-1,1'-bi(cyclohexyl) was the  [ACD/IUPAC Name].
The parent structure are 2 linked cyclohexane rings.
User @orthocresol found

P-52.2.7.1 of the Blue Book, "Preferred IUPAC names ... are formed using the names of parent hydrides rather than the names of substituent groups"; hence cyclohexane rather than cyclohexyl)

The outdated 1993 version had this rule

R-2.4.4.1 Assemblies of two identical cyclic systems are named in one of two ways:
(a) by placing the prefix 'bi-' before the name of the corresponding substituent group (see R-2.5) enclosed in parentheses, if necessary (additive operation); or
(b) by placing the prefix 'bi-' before the name of the corresponding parent hydride enclosed in parentheses, if necessary (conjunctive operation).

According to that rule, both bi(cyclohexyl) (a) and bi(cyclohexane) (b) were valid. However, I do not have access to the to the current version of the blue book at the moment and do not know, whether this rule still exists that way. Would be great, if someone could comment on this. I will  continue using bi(cyclohexane).
Regarding parentheses, user @orthocresol found

P-28.2.1 "Parentheses are used to avoid confusion with von Baeyer names."

or, as described in an Wikipedia article:

Fused and bridged bicyclic compounds get the prefix bicyclo

Since the rings are not fused, the carbon atoms are numbered separately in both rings, i.e. 1 to 6 in one ring, and 1' to 6' in the other.
The numbering starts in each ring at the carbon atom which is linked to the highest molecular group, i.e. the C that links to the other ring. This gives the 1,1'-bi(cyclohexane) part.
Using this numbering, you end up at the C4 and C4' as the number for the carbon atoms that are linked with a methyl-group each. Thus, the name is prefixed with 4,4'-dimethyl-.
Putting this all together, a correct IUPAC name should be 4,4'-dimethyl-bi(cyclohexane).

Side note:
According to PubChem the IUPAC name is 1-methyl-4-(4-methylcyclohexyl)cyclohexane.
In this case, they ignored the ring assembly and used only one cyclohexane ring as a parent and treated the other one as a substituent.

Well, IUPAC nomenclature is pretty arcane, so only Loong may know the truth here, but it's usually "bicyclohexyl" without parenthesis.

@Mithoron I'd look up the IUPAC rules on parenthesises, if you think the source is wrong there.

@Mithoron "Fused and bridged bicyclic compounds get the prefix bicyclo"  (from https://en.wikipedia.org/wiki/Bicyclic_molecule) Thus, I'd say the parenthesises are necessary to make the distinction and to indicate this isn't a  bicyclo, but two cyclo  rings

There were only mine comments, so no ping needed. I think chemspider may be weak, but it's not my "specialisation". Check out IUPAC rules or wait until Loong finds it... Do as you wish.

ChemSpider is not a good source for IUPAC names. The PIN is 4,4′-dimethyl-1,1′-bi(cyclohexane) (cf. P-52.2.7.1 of the Blue Book, "Preferred IUPAC names ... are formed using the names of parent hydrides rather than the names of substituent groups"; hence cyclohexane rather than cyclohexyl). Furthermore, the first letter of the name should not be capitalised unless at the beginning of the sentence (excluding things such as stereochemical descriptors e.g. R/S/E/Z or atom locants e.g. N-).

... although it is not the preferred IUPAC name, 1,1′-bi(cyclohexyl) may still be used in general nomenclature. Regarding parentheses: P-28.2.1 "Parentheses are used to avoid confusion with von Baeyer names." cc @Mithoron

@orthocresol Thanks for checking! Do you know an online-resource that is reliable regarding IUPAC nomenclature?

ACD/I-Lab is generally the best (not 100% perfect), but needs a subscription.

@orthocresol I've finally made it -  incorporated your comments and looked up a few more things. Would be great, if you could check it!

The preferred IUPAC name according to latest nomenclature recommendations is 4,4′-dimethyl-1,1′-bi(cyclohexane). The alternative name 4,4′-dimethyl-1,1′-bi(cyclohexyl) is still acceptable, but is not preferred. PubChem's alternative 1-methyl-4-(4-methylcyclohexyl)cyclohexane is not acceptable. I think it would be good if you made this sufficiently clear in the text.


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How are bond lengths measured experimentally?
How is the bond length between any two atoms in a compound measured?
For example, in the case of graphite, the bond between its two layers is 340 pm. This number must have been measured, so what are experiments that can give lengths with that precision?

Xray crystallography is the most general method, but infra-red and Raman rotational spectroscopy has also been used for smaller molecules. There are some specialised NMR techniques that also measure distance but these are not routinely used as far as I am aware.

This is something to look up in a basic chemistry textbook, not by googling, especially not for scientific articles. If you haven't got a textbook, try wikipedia?

Possible duplicate of Has anyone even taken a picture of a molecule to confirm the geometry predicted by theory?

Transmission electron microscopy is one method of imaging atoms "directly".

@Mithoron That answer talks about shapes of molecules and not the experimentally measured bond lengths. How's it a duplicate then?

@GaurangTandon Bond lengths are part of geometry, along with angles.

@Mithoron "bond length" was only mentioned once in that linked question, so I was hoping that this question would have more detailing on that part...

The possible duplicate question is not useful, but the answers are slightly more useful. A good answer here would give much more detail than any of those answers.

A discussion of all the techniques would be useful as an answer. But it would have distinguish among the alternatives. Gaseous molecules require different techniques to crystals, for example.

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Why any non polar compounds dissolve in nonpolar substance like Benzene what is the reason behind it?
In case of ionic solvent the ionic solute dissolve in it because due to polar nature of water it break the bond between the ions through hydration energy and the ionic solvent dissolve in it.
So I want to ask that for what reason the non polar solute dissolve in non polar solvent ?
Means a basic framework like due to what reason it happens?
I searched on Google but I don't get slim and articles or content

This is something to look up in a basic chemistry textbook, not by googling, especially not for scientific articles. If you haven't got a textbook, try wikipedia?

That's your homework, sorry. Your question already contains a number of helpful terms to look up.

The simple answer is entropy. The explanation not so much.  If you consider solvation (a specific type of dissolving) to really be a melting process you can consider the resulting solution a eutectic/peritectic mixture.
To explain Figure 1 is the phase diagram of tin-lead compounds.  You could consider tin to be soluble in lead since at $300^\circ C$ lead is liquid and tin is solid but when tin is added to lead, it dissolves.  Or you could consider the mixture of the two to be a eutectic that lowers the finally melting temperature of both substances to $183^\circ C$ (at 61.9 wt% lead) which is commonly the case.  In fact if you were to make a phase diagram for any solute and solvent you would see similar behaviour.
Now sometimes the solvent forms a solid complex such as $\ce{CaCl2 *6H2O}$ which casuses a discontinuity. in Figure 2, point A is a eutectic point but points B and C are peritectic which is a result of the solid phase having a different equillibrium complex ($\ce{CaCl2*6H2O}$ vs. $\ce{CaCl2*4H2O}$.
That was some concept now for some math.  The entropy of mixing can be determined for a solution by: $\Delta S_{mix} = -R[x_a ln(x_a) + x_b ln(x_b)] $ where $x_a$ and $x_b$ are the molar ratios of the components  ($x_a = 1 - x_b$). Since $\Delta S_{mix}$ is always positive, $\Delta G_{mix}$ will always be negative as long as the heat of mixing ($\Delta H_{mix}$) is not too positive.
Figure 1: Eutectic Mixture Example:

Figure 2: Peritectic Example:

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Why is the oxidation of sulfur dioxide not termed as heat of formation of sulfur trioxide?
According to me, the following reaction's heat should be termed as heat of formation of sulfur trioxide:
$$\ce{SO2 + 1/2 O2 -> SO3}$$
However, in my textbook, they have that given it is not the formation reaction. Why is it so? And which is the formation reaction then?

There are quite a few different reactions that end up in SO3. Which of them should be called formation?

Thanks for editing my question. After reading the edited question, I have come up with an answer(not sure if I am correct or not). The heat of formation of sulfur trioxide, instead of being the heat of reaction of the above reaction, is the sum of the heat of reaction of a two-step reaction(because the formation of sulfur trioxide is a two-step reaction).

@Gordon I don't think the number of steps is relevant. In fact, as Karl has mentioned in their answer below, it is entirely irrelevant whether it is possible to even produce the compound in that single reaction.

@GaurangTandon Yes and no. If the actual synthesis of the product takes multiple steps then you can still calculate the heat of formation by adding up the details at each step. Yes the umber of steps doesn't matter, but real reactions of multiple steps can still get you to the answer.

Heat of formation is the enthalpy difference to the elements in standard state as defined by IUPAC, which is in most cases their most thermodynamically stable form at STP.
It is completely irrelevant if the respective chemical equation makes any sense, i.e. it would work out in a laboratory. However high the number of intermediate steps to reach the desired product is, energy is always conserved.

Always seems to be an exception. Black phosphorus is more thermodynamically stable  than white  phosphorus which was chosen as the standard state.

@MaxW Yes, i think white phosphorus is the only modification at STP which can be readily produced and has a well defined crystal structure and density. The polymeric forms (black, red, violett) are notorious for containing a huge amount of packing defects.

The heat of a reaction $\ce{A + B + ... -> X}$ is called the standard heat of formation of the product subject to a number of strict restrictions. These restrictions are:

Each of the reactants must be in their thermodynamically most stable state at the room temperature. For example, if you take bromine as a reactant, it must be in liquid state. If you take carbon, it must be graphite (s), etc. Also, polyatomic molecules must be written in complete molecules and not atoms (hence, $\ce{Br2}$ and not $\ce{Br}$)
Each of the reactants must themselves be elements and not compounds. For example, $\ce{C (s, graphite) + O2 (g) -> CO2 (g)}$ is correct, but $\ce{CO (g) + 1/2 O2 (g) -> CO2 (g)}$ is incorrect as $\ce{CO}$ is not an element.
There should be only one product and that too the one we're concerned about.
There should only be one mole of the product. Hence, $\ce{S (s) + 3/2 O2 (g) -> SO3 (g)}$ is correct, but $\ce{2S (s) + 3 O2 (g) -> 2 SO3 (g)}$ is incorrect (actually, the latter's heat of reaction is just double of the standard enthalpy of formation).

You can read more about the standard enthalpy of formation on the Wikipedia page.
Can you now tell by which rule your given reaction fails to be a standard formation reaction?

Also, here's a fun little exercise: can you tell which of the following reactions are standard formation reactions?

$\ce{PCl5}$

$\ce{P (white, s)  + 5/2 Cl2 (g) -> PCl5 (s)}$
$\ce{P (black, s)  + 5/2 Cl2 (g) -> PCl5 (s)}$
$\ce{P (white, s)  + 5 Cl (g) -> PCl5 (s)}$

$\ce{KCl}$

$\ce{K (s) + 1/2 Cl2 (g) -> KCl (s)}$
$\ce{K (g) + 1/2 Cl2 (g) -> KCl (g)}$ (a homogenous mixture!)
$\ce{2K (s) + Cl2 (g) -> 2KCl (s)}$

$\ce{O2}$

$\ce{KClO3 (s) ->[\Delta] KClO (s) + O2 (g)}$
$\ce{2Na2O (s) ->[\Delta] 4Na (s) + O2 (g)}$
$\ce{H2O (g) + CO2 (g) -> 1/6 C6H12O6 (l) + O2 (g)}$


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dangers of radiation question
At a recent flea market, someone was selling a surplus military device that was marked as having 900 millicuries and should be disposed of in a safe way.  I picked it up to read the words.  Whould 900 millicuries be a concerning amount of radiation?

Do you have a picture of the surplus military device, or can you describe it?

900 mCi is a lot of activity, the problem is that it is impossible to know how bad it will be without knowing which radionuclide is present.
If it is a tritium light with 900 mCi in it, then the external threat is close to zero. Just do not break open the glass vials in a tritium light.
If it is 900 mCi of radium-226 or cobalt-60 then I would suggest to stay a very long way from the object and call the fire brigade or police. They will not have the ability to resolve the problem. What they will do is to seal off the area and then send for expert advice and an emergency team from a nuclear site or some other place where people have experience of working with large amounts of radiaoctivity.
Even a pure beta emitter at 900 mCi is a lot of activity if you are dealing with a high energy beta emitter such as Sr-90 / Y-90.


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What is the effect of oxygen in welding?
Welders use a shielding gas to avoid oxygen interact with the electrode. I supposed is might be because of ferric oxide formation as the electrode is steel-made. But fusion temperature of ferric oxide is quite the same than iron so I'm not sure what is the reason for the shielding gas (usually argon). I have gone through some works but found nothing about it.
Any idea what is the effect of oxygen? Is there any reaction involved which plays a role against a good welding or just physical issues?

The notion is to have iron in the weld seam not iron oxide which is very much weaker mechanically.

Consider - welders add oxygen to turn a torch into a cutting tool.

According to https://www.bernardwelds.com/mig-welding-shielding-gas-basics-p152080 the purpose of shield gas in steel welding is:

...to prevent exposure of the molten weld pool to oxygen, nitrogen and hydrogen contained in the air atmosphere. The reaction of these elements with the weld pool can create a variety of problems, including porosity (holes within the weld bead) and excessive spatter.

There is also a picture of a weld that either didn't use a shield gas or didn't have enough. It's pretty obvious that such a weld isn't going to be very strong.

Edit: From this Wikipedia article explaining the purpose of inert gas in welding:

Their purpose is to protect the weld area from oxygen, and water vapour.

And in what appears to be a bit of a contradiction:

Oxygen is used in small amounts as an addition to other gases; typically as 2–5% addition to argon. It enhances arc stability and reduces the surface tension of the molten metal, increasing wetting of the solid metal. It is used for spray transfer welding of mild carbon steels, low alloy and stainless steels... Oxygen causes oxidation of the weld, so it is not suitable for welding aluminium, magnesium, copper, and some exotic metals.

When you consider that atmospheric oxygen is almost 21%, the two to five percent used in inert gas welding seems pretty small. However, though I can find no definitive reference, I'd have to say that the prevention of oxides of the metal(s) being joined is a consideration only when extremely reactive metals are being welded. Instead, it seem that the prevention of gaseous materials (probably mostly water vapor) within the weld seem to be of the most importance. Also, considering the temperatures involved in welding, the creation of oxides of nitrogen from atmospheric nitrogen and oxygen would have to be a concern.

As someone who has run out of gas while welding, I can confirm... it doesn't work for beans.


text:
How to acid-wash sand?
I need some acid-washed sand to smooth out boiling in diethyl ether synthesis. I aready have pre-washed sand for general filtration purposes, but it is not labeled as acid-washed.
Unfortunately, I have not found any reliable lab manual for acid washing of sand.
I found one reference instructing to wash 30 g of sand twice with 0.05 M sulfuric acid, then four times with distilled water.
However, other references instructing to use only hydrochloric acid and de-ionized water, but have not specified how much or the concentration.
Yet another reference says that sulfuric, hydrochloric or phosphoric acid can be used interchangeably.
I don't know how seriously should I take the acid washing step (for the ether synthesis from ethanol and sulfuric acid) and how to check whether it is properly washed (pH measurement?).
I know one can just purchase acid-washed sand, but I have excess of good quality sand and the professional acid-washed one is overly expensive.

Ask yourself "why would I want do acid-wash the sand for this", and you will see.

I need to remove stuff that can interfere with chemical reactions. But I don't know what acids can remove what stuff and how strong it should be, how many washes should be done etc. ...

@Karl yesterday Frankly, I don't know why the sand must be washed with acid to be used in ether synthesis. No tutorial said WHY, only that the sand must be acid washed. I can only imagine the reasons but I am not a professional chemist and even looking up that information is hard for me as I don't know where to look.

Your syntesis runs at a low pH, the idea is to wash out previously any impurities that might dissolve in an acidic reaction pot and disturb your reaction.

Ahem. Why do you want to make diethyl ether? You're aware of the fire hazards of it's high density vapour, and that it will knock you clean off your feet if you take too large a noseful? And gives terrible headache if you inhale smaller amounts over a few hours.

Thanks, Karl. Yes I am very well aware of the fire hazards. It's done inside a running fume hood and the apparatus has a wash bottle connected so there should be no ether leakage anyway. And the area is well ventilated. Plus all the standard lab safety precautions... Why? Because it's an extremely useful solvent with low boiling point. It is unfortunately not easily (or cheaply) available to amateur chemists and experimentators these days.

I have just started working with sand columns and was also surprised that I could not find any valid and trustworthy guidelines or process breakdown for acid-washing sand.
However, I found this paper, which may answer some of your questions. In short, acid washing (or leaching) is supposed to remove impurities (mostly iron impurities) from the sand.
From the introduction of the paper:

Iron is one of the most troublesome impurities in silica
  sand because it impairs the transparency of high-quality
  glass, e.g. for tableware, and optical glass; it impairs
  transmission in optical fibers; it adversely affects the production
  of pure silicon products, e.g. silicon metal for
  photovoltaic application, silicon carbide. One of the main tasks in the purification of silica sand
  is to further reduce iron compounds to a required level
  by a highly efficient method.

In table IV in the results section you can see a comparison between several acids with regard to their efficiency in removing iron and aluminum impurities. According to the data, sulfuric and phosphoric acids should be preferred. From what I understand, however, diluted HCl solution may be preferred by some, probably due to better price and availability, but that's just an assumption.
As for the procedure itself, I heard about adding a solution of 1% HCl to the sand, stirring and letting it stand for a few hours, and then washing with distilled water and decanting ~10 times, and then drying in an oven overnight.  I don't have any valid reference for it, though.

Thanks. I think one problem is that ion contents depends on the sand in question so maybe we can wash the sand once with distilled water, check pH and from that estimate the amount of acid needed, use excess and multiple washes. HCl is also great as it evaporates so the traces are easily removed - but washing is still needed as we don't want to corrode the oven.


text:
What is the chemical reaction between coins (copper and nickel) and polyvinyl chloride?
I often carry coins in a cheap plastic sandwich bag. A green solid soon appears.  Is it copper oxide? How is the polyvinyl chloride (or polyethylene or polypropylene) greatly accelerating the reaction? Or are they?  Are there other chemicals being formed?
Coin collector websites all say vinyl-containing compounds, and maybe other plastics, damage coins. But they don't say how....
P.S.:  Coins are made of copper, nickel and zinc, right?

A cheap plastic bag shouldn't be made of PVC in the year 2018, i hope, and all other ordinary plastics surely don't react with copper. And I also doubt that PVC does! So this is most likely just a moisture problem.

Karl seems to be right, my bag is either polyethylene or (maybe) polypropylene...

Far, far too broad a question. I don't know what is going on in your situation.
I can, perhaps, clear up some misconceptions you may have. PVC is NOT commonly used in sandwich bags. PVDC is a close relative and used to be common as "saran wrap". But most sandwich bags are made out of polyethylene (PE). Chemically both PVC and PVDC degrade due to weathering and other things. Dehydrohalogenation is the reaction of organic chlorine compounds (both PVC and PVDC) which can be expressed in the partial formula C-HC-Cl → C=C + HCl (this is a crude simplification). Note that this reaction generates HCl, also known as hydrochloric acid (or muriatic acid). HCl is a powerful acid, attacking most metals. If you live by the ocean or in a desert chloride levels in the air (dust or mist) can be quite high and can accelerate corrosion dramatically. Coins are designed to be inert. Placed clean and dry into a sealed PE bag, they should not corrode. But. Whenever two different metals (or metal alloys) come into contact, a electrical potential is created. A random bag of coins will be sorta like a bag of little batteries. That is, you shouldn't be surprised if they react (discharge), especially in the presence of moisture. Neither copper nor nickel form green oxides, but both do form green chlorides and green carbonates. You need a source of chlorine for the chlorides, but the carbonates might form when simply exposed to air and moisture when a electrical potential exists. Sweat, sweaty hands, are potentially one source of chloride/chlorine, but as I stated up front I am not familiar with your situation, I carry (mixed) coins in my pocket for days at a time without any green forming.

You're right, Lio, my bag is polyethylene or polypropylene, not polyvinyl chloride.

So the green stuff, in conclusion, does not involve the bag, just the coins plus carbon dioxide/ monoxide plus moisture.      It is carbonate(s) of copper, nickel or zinc (I forgot about zinc).  And having multiple coins and different metals touch creates a tiny electrochemical voltage, accelerating the process?   Is that right?

"A random bag of coins will be sorta like a bag of little batteries." Very nice analogy. +1

Many polymers (including Polyvinyl Chloride) are water-resistant materials. The moisture that they trap causes metals such as copper to ionise on the surface and corrode (oxidise) into “rust” by readily reacting with oxygen.


text:
Considering the d-d transition how, does tetracyanidonickelate(II) ion exist as a colored complex?
As I refer on google, the resources indicated that,

the d block elements which have unpaired electrons as in their ions, when they are making complexes with ligands, they can absorb different frequencies in the spectrum due to degenerated d-d orbital transition.

But I am confused in understanding how there is a colour for $\ce{[Ni(CN)4]^2-}$ complex.

My question is: In the $\ce{Ni (II)}$ ion there are two unpaired electrons in the d orbitals. But while they are hybridizing there aren't any unpaired electrons, all are paired. In the solution state, the nickel ions do not exist as $\ce{Ni^2+}$ ions, but rather as $\ce{[Ni(CN)4]^2-}$ ions. But $\ce{[Ni(CN)4]^2-}$ has no unpaired electron. So it is unable to absorb energy from the spectrum and transit between d orbitals. (My logic is that there are no more spot in d orbitals to transit between after the complex occur)
But as the image below this complex is yellow in colour.

I can't find the mistake in my logic. I referred to many resources but I wasn't able to find any satisfactory answer.

Hi Osal, please cite the source (website, book, paper) of the quoted text. Google is never the source for any information (it's just a search engine). Thank you!

See the edit now...

Websites are like Wikipedia, ChemGuide, etc. Please mention one like that if you recall. If you don't recall, no problem, let it be like that.

https://chemistry.stackexchange.com/questions/4667/why-do-transition-elements-make-colored-compounds.        : but the quoeted
text had been not  mentioned directly, I quoted the summary that can get from it.

Hybridisation is a very flawed theory for metal complexes... Please, stop using it as soon as you can. Look up an MO diagram for a typical square planar complex.

@orthocresol♦:  Can you say the reason of why does hybridisation a flowed theory for metal complex. Instead of that, is there any mechanism to understand the electron arrangement of the metal complexes. Thanks.

https://chemistry.stackexchange.com/q/76726/16683

@orthocresol♦:  As you indicated in the link that; "All in all, hybridisation theory is simply inadequate - it does not explain the rich chemistry of the transition metals and their complexes. Even something simple, such as the d-d* transitions responsible for the colours of complexes, cannot be explained by it" So is there any accurated way for describe the colours of the complexes? Thanks

Molecular orbital theory or at the very least crystal field theory

I'd suggest you pick up a book on ligand field theory or MO theory; it'll be very helpful. But you're right though- a d8 square planar complex should be diamagnetic.

When the metal complex is formed the symmetry of the ion is no longer spherical but  octahedral ( four ligands in a plane and one north and one south) or square planar( four in a plane). The effect this change in symmetry has is to split the energy of the d orbitals into two groups of 3 and 2 (labelled $t_{2g},e_g$}. Transitions giving the colour occur between these two sets of d orbitals. You can see examples in the answers below.

I think the key piece that you are missing is that the ligand electrons don't just transfer into the metal orbitals. The metal and ligand orbitals combine to form bonding and antibonding orbitals (you should learn this even in valence bond theory). In transition metal complexes, the metal d electrons generally occupy non-bonding and anti-bonding orbitals, whereas the ligand electrons occupy  bonding orbitals. The non-bonding and anti-bonding orbitals closely resemble the original metal d orbitals and so we use the same names for them, which may be where you're getting confused.

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which are effectively isoelectronic anyways).
As you can see the frontier orbitals of this complex are all mostly on the metal complex , and I'd expect the excitation of electrons from lower-energy orbitals into these anti bonding orbitals to be the reason for the colors.

How we are sure that the color doesn't come from the transition between $p$ (filled) to $s$ (empty) orbital?

This can be explained by Crystal Field Theory (CFT)
According to CFT, when a strong field ligand brings about pairing on electrons in d orbital, splitting of d-orbitals occurs (loss of degeneracy, i.e. all 5 d orbitals are not of same energy level)

Now as the upper orbitals are not of the same energy as lower ones, transitions take place giving rise to colours.
CFT has the capability to explain the rise of colours in coordination compounds.
Read more here:
https://en.wikipedia.org/wiki/Crystal_field_theory

Your answer is regarding the mechanism of showing color of $\ce{[Ni(CN)4]^2-}$ complex. But the question is actually not that. You are describing standing on the Ni2+ ion. But my logic after existing the complex, the nickel will not exist as  Ni2+ ion, it is exist as the $\ce{[Ni(CN)4]^2-}$ complex at the compound. As I shown in the figure there are no more d orbitals in the $\ce{[Ni(CN)4]^2-}$ after having the complex. So, as I mentioned above how there is a transition between d orbitals while there is no any space in the d orbitals regarding the complex. That's the relevant.

Wrong energy scheme for a square planar complex. You’ve illustrated it for an octahedral geometry.


text:
Aren't "lack of specificity" and "nature of adsorbate" contradictory factors in physisorption?
It was mentioned in my course book under "Characteristics of physisorption"  that:

Lack of specificity: An adsorbent does not show any preference for a particular gas as the van der Waal's forces are universal.
Nature of adsorbate: In general an easily liquefiable gas is more readily adsorbed as it's critical temperature is higher as van der Waal's forces are stronger near critical temperature.

Aren't these two points contradictory? If not then please explain.

It does not show preference to any gas for chemical reasons.
Laws of physical chemistry apply, that gas which has higher critical temperature can be easily liquefied...
It is not specific to how reactive the gas is.


text:
Why do atoms have electron Affinity?
Atoms are stable so how can they pull electrons i.e how do they have any electron affinity at all?

Which is more stable in the gas phase, F or F-?  Another way to think about this is, why do we have anions and cations?

Actually, not all atoms are stable, at least not in the gas phase, which is where we measure the electron affinity. You should first read Why do atoms "want" to have a full outer shell? for detailed elaboration.
Now, electron affinity involves the gain of an electron to an isolated gaseous atom, but not all atoms actually want to gain an electron, i.e., they are already pretty stable. This is precisely the reason why elements like the noble gases or nitrogen have negative electron affinity i.e. you need to supply energy for them to accept an electron.
Also read: Are there any major exceptions when comparing electron affinity?

So, in short: "Atoms are stable so..." not always, in fact, such atoms are in the minority. "...how can they pull electrons..." read the first link. "how do they have any electron affinity at all?" we can always supply excess energy to make an atom accept an electron, forcibly. So, the only difference for stable vs non-stable atoms would be the sign of electron affinity.

The energy necessary to detach an electron from an atom is called the ionisation energy. It has a large positive value for atoms with filled shells, such as the noble gasses, and is much smaller with atoms with one more electron, such as alkali metals in which the electron is easily released. The value is positive with the convention that energy is absorbed.
The electron affinity is the energy released when an electron is added to an atom $\ce{X(g) + e^- \to X^-(g)}$ and this energy released is, confusingly, by convention also a positive value. Thus for fluorine the electron affinity $\ce{F(g) + e^- \to F^-(g)} = 333$ kJ/mol) so this much energy is released, but is $-31$ for N atoms in which adding an electron is unfavourable (energy absorbing).
Thus the 'stability' of atoms as you call it is reflected in the energy released when an atom gains an electron ( or absorbed when an electron is removed and the ionisation energy is measured) and these reflect the energy of the different levels filled by the specific number of electrons that each type of atom contains.


text:
Do bases undergo dissociation or ionisation in aqueous solution?
What do bases undergo when dissolved in water: ionisation or dissociation or both? Some books say ionisation while others say dissociation. What is the correct answer?

How is ionisation different from dissociation in the first place?

Dissociation is the process of separation of charged particles that already existed in the compound. Ionization is the formation of new charged species which were not originally present in the compound.

It really depends on what base you are talking about. A Lewis base, like an amine, does not dissociate. Instead, it donates its electron pair to accept a hydrogen from a water molecule, forming its conjugate acid and a hydroxide ion.
An Arrhenius base, like sodium hydroxide, which is already composed of sodium ions and hydroxide ions will dissociate into free ions in water. To be clear—though in the solid form, the sodium atom has already been ionized, meaning it has donated its outermost electron. It is the electrostatic attraction between positive and negative ions that holds an ionic compound together.


text:
Solution to the rate law of first order reaction followed by second order reaction (consecutive)
I'm trying to derive the rate law of a first order followed by second order consecutive reaction:
$$ \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_1[\ce{A}]$$
$$ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_1[\ce{A}]-k_2{[\ce{B}]^2}$$
$$ [\ce{A}](0)=[\ce{A}]_0,\ce{[B]}(0)=0$$
The solution for A is trivial:
$$ \ce{[A]}=\ce{[A]}_0\exp(-k_1t)$$
The problem is B, whose rate law ODE is not linear:
$$ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_1[\ce{A}]_0\exp(-k_1t)-k_2{[\ce{B}]^2}$$
This is a nonlinear first order ODE, but can be rewritten to a linear second order ODE by substituting $[\ce{B}]$ with:
$$[\ce{B}]=\frac{\mathrm{d}y}{\mathrm{d}t}\frac{1}{k_2y}$$
Which gives:
$$ -\frac{\mathrm{d}^2y}{\mathrm{d}t^2}+k_1k_2[\ce{A}]_0\exp(-k_1t)y=0$$
The solution is quite complicate and contains Bessel functions of both the first and the second kind:
$$ [B] = \frac{C_1\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,J_1\left(\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-\ce{A}_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\,\sqrt{-A_{0}\,k_{1}\,k_{2}}-\ce{C}_2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,Y_1\left(-\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{2}\,\left(C_1\,J_0\left(\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)+C_2\,Y_0\left(-\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\right)}$$
There are two arbitrary constants but I only have one condition:
$$\ce{[B]}(0)=0$$
So my questions are:

Is my approach solving the rate law correct ?
If so, how do I get rid of both constants ?
Is there any approximated solution to this rate law ?

[update]:
The ODE is solved numerically with Runge-Kutta method. The symbolic solution seems correct.

Unless you really want an algebraic solution you can numerically solve $dB/dt$  quite easily and you know both initial conditions; at $t=0, [A]=[A_0],[B]=0$. Once you make a  2nd order eqn, you will have 2 initial conditions making 3 in all. I'm not sure what this could be, $[B]=0, t=\infty$ is probably not helpful, so this means knowing $dB/dt$ at some time. Your substitution is a cunning one :), its a pity it does not make a Bernoulli eqn.

@porphyrin I don't see how numerical solution is easy because of the two parameters k1 and k2. To solve that one needs to nest the numerical ode solver inside optimization iterations. The optimization of k1 and k2 will be a global fit of a 2D array.

Of course you have to give k1 and k2 values which you will also have to do for your algebraic solution when you want to plot it. The form of the chemical kinetics is that A will decay exponentially while B will rise and fall until no B is left but only the product  P. This  forms only from B. The product can be calculated by mass balance since $\ce{A0 + B0 =A +  B + P}$ at all times. If you have real data to fit then it is still routine  to do nonlinear optimisation with numerical solution of equations.

The  following code is in Python and uses numpy, scipy and matplotlib that are usually packaged with python.  The code will solve the equations $\mathrm{A} \underset{k_3}{\stackrel{k_1}\leftrightharpoons}  2\mathrm{B} \stackrel{k_2} \longrightarrow \mathrm{ P}$ but you can set $k_3$ to zero if you want. (its called km1 in the code). I've used log time spaced points so that a wide time range can be seen. The x axis is log in the plots.
I have used a built in solver, odeint(), buy you could write your own based on the Euler or Runge -Kutta methods.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def dX_dt(X, t, const):     # returns rates dEf/dt, only used in odeint()

    k1, km1, k2 = const     # unpack constants needed for calc'n
    A = X[0]
    B = X[1]
    dAdt=    -k1*A  +  km1*B*B                 #  rate equations here
    dBdt=     k1*A  - (km1+k2)*B*B

    f = np.array( [(dAdt), (dBdt)] )           # evaluate equations;
    return f

C0 = np.zeros(2,dtype=float)
C0[0] = 1.0e-3    # C0= [ A(0), B(0)   ]      # initial concentrations
numt = 200
t = np.logspace( -9,-4,  numt )    # time as initial power, final power, num points
k1  = 1e7  # 1st order rate const s^(-1)
km1 = 1e6  # 2nd dm^3mol^(-1)s^(-1)
k2  = 1e9  # 2nd  ....
const = [ k1, km1, k2 ]                            # rate constants for calc'n

soln = odeint(dX_dt, C0, t, args = (const,)  )     # solve equations
Aval, Bval = soln.T     # extract data

plt.plot(t,Aval,color='red',  label='A')
plt.plot(t,Bval,color='green',label='B')
plt.plot(t,C0[0]-Aval-Bval,color='blue',label='P')
plt.xscale('log')
plt.legend()

plt.tight_layout()
plt.show()

The figure shows the populations vs (log) time

I'm not sure what to do with your answer because I took your advice - numerical solution but I don't use python so I cannot comment on your code. I'm going to accept your answer because you suggested numerical solution.

You will find that it is quite easy to convert it into your preferred language. Most now do have the equivalent of the odeint integrator used above.  Alternatively download python (along with numpy, matplotlib & scipy ) as all are free to use.


text:
How can we measure the rate of reaction of EDTA chelation with zinc carbonate?
I am performing an experiment in order to find the activation energy of EDTA chelation. Consequently, I need to measure the rate of reaction of the process. I am high school student, and so I do not have access to sophisticated equipment, although they might be unnecessary. However, I have been attempting to measure the volume of gas produced, but I am not achieving any result. So, can you suggest any other method to measure the rate of reaction? I would greatly appreciate the help since there is not a sufficient amount of information on the internet to help me get a clear answer, I have been working on the reaction for more than a month and a half.

It's too fast for you.

Thank you but it's quite the opposite. I mean it would be sufficient if I get a few values however I am not getting any.

sigh ZnCO3 is almost insoluble, so you decompose it with acid and expect it would tell you anything about subsequent complexation? Or what? Your query doesn't seem to make much sense.

Procedure: (I am using powdered Zn)
ZnCO3 + C10H16N2O8 = C10H14N2O8Zn + CO2 + H2O

Pipette a 10 mL ( 0.05 mol dm-3) sample of the EDTA solution into a conical flask.
Add 5 mL of acidic buffer solution
Start the stopwatch when adding 4.49 g of ZnCO3. Given the Zn2+: EDTA ratio of 1:1
Stop the stopwatch when a certain volume of CO2 is reached      5.Repeat the experiment at 35 °C using a water bath.
Repeat steps 2,3,4,5 for the following temperatures:45 °C ,55 °C, 65 °C
Graph ln 1/t vs 1/T.
Find the activation energy of the reaction by using the Arrhenius equation.

Sorry, to break it to you but it's completely misdesigned. You won't get any Ea whatsoever in this way. Kinetics of this process isn't even driven by reaction, but by diffusion. You'd need to use soluble and not decomposing salt of Zn and have advanced equipment, as complexation would take less then second.

All of these comments should not be comments, but part of the original question or an answer.

I think that the reaction will be too fast for you to measure with the tools in a high school lab. I think that you are not measuring the zinc binding to EDTA but instead making a measurement of the dissolution rate of zinc carbonate in an aqueous solution of EDTA.
If you were to want to measure the rate of the binding rate of zinc by EDTA the best way might be to use a radioactive tracer. I would start with radioactive zinc in an anionic EDTA solution. I would bring this into contact with a non-radioactive solution of zinc. I would then keep on pulling out small samples from the mixture and filtering them through an ion exchange resin which is able to remove either the anionic EDTA complex or the cationic form of zinc.
I would use zinc nitrate or some other form of zinc which does not tend to form anionic complexes when in solution.
I know that this experiment has been done as a teaching lab experiment in a nuclear chemistry unit with radioactive europium, but it has a reputation as a difficult experiment to do.
I suggest you read Williams, K.R.; Choppin, G.R., 1974,  J. Inorg. Nucl. Chem., v. 36, no. 8, pp. 1849-1853. This has details of an experiment on the kinetics of metal exchange between EDTA complexes and "free unsolvated metal" for trivalent metals.
An alternative non radioactive method would be to use stopped flow experiments with a metal senstive indicactor present in the mixture. If we assume that the exchange of zinc between the solution and the indicator is very fast then this method could be of use.
Another method could be to take a solution of the zinc complex of EDTA and then treat it with C13 KCN, you could measure the C13 NMR spectrum again and again if it is a relatively slow process. But I think for you that you will be unable to deal with this problem with the chemcials and tools you have in a school lab. Sorry for having to pour some cold water around.

Thank you for answering me! Well, I guess I will need to change my research question. Thank you for listing some sources and methods I will try to read them, because I want to understand how this will work. However, my experiment counts as 20% of my grade so I will need to change it.


text:
Range of HMBC Coupling
I need to determine the constitution of an organic molecule by several spectra like 1H NMR, 13C NMR, COSY, HMBC ,and HSQC.
Is it possible to get a HMBC crosspeak between the marked proton and carbon atom?

Typically, both two- and three-bond 1H–13C couplings appear in HMBC spectra, with three-bond couplings tending to be more prominent. The presence of other atoms in the path, e.g. oxygen, doesn't affect this. So, the three-bond coupling you have indicated is very likely to show up in the HMBC spectrum, as a cross peak at (f1, f2) ≈ (180 ppm, 4 ppm).
If all you are looking for is structural confirmation + assignment of 1H data (which would be typical if this was a product you expected from an experiment), I'd suggest that you don't necessarily need 2D data. The multiplet structures should contain more than enough information.


text:
Is E1cB possible in the reaction of cyclopentadiene with acetone in presence of ethanolate?

When cyclopentadiene 1 and acetone 2 are reacted with sodium ethoxide in ethanol, what product is obtained? I was told that the correct product 3 is formed by nucleophilic addition of cyclopentadienyl anion to the carbonyl group.
However, when I thought about it more, I wondered why an E1cB reaction to form the dimethyl fulvene 4 isn't possible in this case. I believe it should be possible because the aromaticity of the cyclopentadienyl carbanion allows the rate-determining deprotonation of 3 to proceed under strongly basic conditions.
On the other hand, there would be more ring strain in the more unsaturated compound 4.
How do I decide if E1cB is possible or not here?

I believe you were right to conclude E1cB would occur. Both of your considerations are exactly on point. The amount of ring strain is minimal -- if we think that it's OK to deprotonate to form a sp2-like carbon in the first place, then we've already decided that the added ring-strain is not a huge issue. Of course, the added conjugation and the entropic component of eliminating a hydroxide ion also help to drive the reaction forward.
A literature source that confirms our thinking is shown below, for your specific example no less, and it looks like there are many derivatives for which the elimination is observed. [1]

I'll go further to note that Scifinder didn't turn up any reaction results for  for your alcohol product, or any similar derivatives, under equilibrium basic conditions. This suggests that elimination is actually quite likely, and that your answer key is wrong. It is however, possible to isolate the alcohol if acid-base equilibrium between the alkoxide intermediate and solvent/conjugate-acid is minimized.[2]

References

Ragauskas, A. J.; Stothers, J. B. 13C magnetic resonance studies. 119. Tricyclo[3.3.0.0] and [3.3.1.0]octanones from substituted norbornenones via cyclopropanation and homoketonization. Can. J. Chem. 1985, 63 (11), 2961–2968. DOI: 10.1139/v85-491.
Smits, G.; Audic, B.; Wodrich, M. D.; Corminboeuf, C.; Cramer, N. A β-Carbon elimination strategy for convenient in situ access to cyclopentadienyl metal complexes. Chem. Sci. 2017, 8 (10), 7174–7179. DOI: 10.1039/C7SC02986A.


text:
Why is there only a slight change in pH when soda loses its fizz?
I did an experiment for my chemistry class and did a control with sparkling water and the pH increased by a significant amount, but when done with sodas like cream soda, coke, or 7 Up, the pH change is so minimal that I can't say that there is a significant change in the pH level. Why do the pH level of these sodas to stay the same when they lose their fizz?

The sodas have other acids with pKa's lower than that of carbonic acid.

In addition to fizz (carbonic acid), Coca-Cola contains phosphoric acid (for an additional tangy taste). According to this leaflet put out by Coca-Cola, a 250 mL can of Coca-Cola contains 43 mg of phosphorus. With some quick calculation, you can find that this translates to $\ce{5.5538 \times 10^{-3} M}$ $\ce{H3PO4}$, which would have a pH of approximately 2.44. Pretty acidic! The other sodas are sure to contain other acids as well.


text:
Absence of anti-Stokes line in Raman spectrum

If we have a molecule 1H-25X which shows a band at $3000~\mathrm{cm}^{-1}$ in the IR spectrum and you want to record a Stokes and anti-Stokes Raman spectrum with a laser of 1000 nm, why is the anti-Stokes band not observed?

My guess is that if you would calculate the wavelength of the anti-Stokes line, then you would get 769 nm. In this region there might be fluorescence, which could interfere with the spectrum. Or am I wrong?

Hi Fechem, welcome to Chemistry! Take the tour to get familiar with this site. Regular text can be formatted with Markdown, and mathematical expressions and chemical equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

Consider the size of the vibrational quantum vs thermal energy of $\approx 210 \mathrm{cm}^{-1}$ at room temperature. Not many molecules are going to be excited in $v=1$; you can work out the fraction $\approx \exp(-3000/210)$.  Excitation at 1 micron will probably not excite an electronic state as this energy is so low, so no fluorescence, but of course this will depend on exactly what the molecule is.


text:
Is it possible to calculate the Kc for an irreversible reaction?
Is it possible to calculate the Kc for an irreversible reaction? For instance, can we Calculate the $K_c$ for:
$$\ce{NaCl + AgNO3 -> NaNO3 + AgCl}$$

This question is a short, specific, conceptual query, and would also be a good dupe target for closing any future questions about it. Voting to reopen.

TL;DR: Yes.
ALL reactions have an equilibrium constant. So do "irreversible" ones. However, The ratio of products and reactants can vary wildly.
$$
K_c=\frac{[\text{Products}]}{[\text{Reactants}]}
$$
For small values of $K$ ($<10^{-3}$):
There is a lot of reactant, while barely any product has formed. Hence, The reaction has barely proceeded and it is a negligible or impossible reaction.
$$\ce{Cu + FeSO4 -> No reaction}$$
For moderate values of $K$ ($>10^{-3}, <10^3$):
There is a roughly comparable amount of product in respect to the reactants. Hence, this is a reversible reaction.
$$\ce{NH3 ->N2 + 3H2}$$
For large values of $K$ ($>10^{3}$):
There is a lot of product, while barely any reactants are left. Hence, The reaction has nearly completed and it is an irreversible reaction.
$$\ce{NaOH + HCl -> NaCl + H2O}$$
As you can see, there is an equilibrium constant for your reaction.
It is just extremely large.


text:
Why are two three-membered rings preferred to a single five-membered ring in the reaction of a dihalide with alcoholic KOH?

The given answer is 1,1-dicyclopropylmethanone:

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.
Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?

Entropy requirements for the transition states maybe? 5-Chloro-2-pentanone also forms the cyclopropane under these conditions: http://www.orgsyn.org/demo.aspx?prep=cv4p0597

Carbanions are very unstable species I think the product will be kinetically controlled.

@Mithoron the answer is not wrong, the dicyclopropylketone can be isolated in ~55% yields: http://www.orgsyn.org/demo.aspx?prep=CV4P0278

@DennisCao Is it possible that the carbonyl carbon here conjugates with the cyclopropyl rings and that conjugation leads to stability which makes it the thermodynamic product?

@Abcd hmm I doubt that would be significant enough, and we should be thinking about the transition state stability not the product stability

@DennisCao But in such reactions thermodynamic product takes the lead right?

@DennisCao See this answer: https://chemistry.stackexchange.com/a/32827/51553

"Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate."

@abcd but this isn't an acid-catalyzed reaction...

Some weight for kinetic control: https://en.m.wikipedia.org/wiki/Vinylcyclopropane_rearrangement

I think the product may be formed by the following pathway.
First, the carbanion will form when the strong base abstracts the proton from the α-position. The lone pair generated would obviously have two points of attack, either at the nearer carbon to form the cyclopropyl ring or at the farther carbon to form a five membered rkng where the whole controversy lies.
While apparently, the position five membered ring appears to be more stable than a small highly stained three membered ring, a very interesting phenomenon occurs if such a three membered, strained ring forms.
This is similar the basic resonace observed in cyclopropane.
The carbonyl functional group also triggers the reaction foreward towards formation of the product.
Let us begin by observing the resonance structures of the carbonyl functional group.
Clearly, out of the two contributing structures, positive charge may be seen residing on the carbonyl carbon while negative charge may be seen on the carbonyl oxygen.

A combination of these two resonance hybrids leads to generation of partial positive charge on carbon and partial negative charge on the net resonance hybrid.
Next, we have to observe the structure of the 1-methylcycloropyl carbocation.
Due to ring strain the bonds are bent and a special kind of resonance involving the σ C-C single bond is observed.

In fact the bonds are actually sp5 hybridized i.e. having large p character to ease the ring strain.
The enormous stablity can be alluded by the following resonce diagram.
Moreover, the immense stablity of tris(cyclopropyl)methyl carbocation, is actually due to this σ-resonace. It is actually the one of the most stable carbocations known.

As the number of cyclopropyl groups increase, the stablity increases.

Now to observe a combination of both.
In the final product the M- effect of the carbonyl group will favour the extended delocalisation of electrons of the σ-bond over the oxygen.
Thus, the enourmous stablity concept of cyclopropane may be attributed to the compound under question by the following diagram:

To draw the resonance structures in a simplified way we first show complete charge seperation between the carbon and oxygen of the carbonyl group. Then the structure becomes similar to a cyclopropylmethyl carbocation showing the required stablising resoance of σ-bond.
Due to this unique resonance effect, also known sometimes as dancing resonance, the product becomes exceptionally stable, more stable than the 5 membered ring obtained in the other case.
Hope this helps.
References:
What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

Honestly, I think you are overthinking it. The $\ce{C+-O-}$ resonance structure cannot have any significant contribution to the carbonyl group. You could argue that there is donation from C–C σ of cyclopropane into C=O π*, and I'd agree with that, but I am highly skeptical that a dicyclopropane product is more thermodynamically stable than a five-membered ring, especially when the alternative can undergo elimination and isomerisation to give an α,β-unsaturated ketone.

By far the simplest explanation, and the one I am most willing to believe, is that the three-membered ring is formed faster than the five-membered ring, and that the cyclisation is irreversible.

@orthocresol I wanted to show 'donation from C–C σ of cyclopropane into C=O π* ', as stated by you. I initially showed the carbonyl resonance, so that is is more comparable and identifiable to the cyclopropyl system taken as standard. Wanted to depict extended resonance.

@orthocresol Yes, maybe the kinetic parameter dominates.

Three membred ring is formed faster due to proximity effect.

@orthocresol Could you please expand that into an answer? That would be helpful. Thanks!

Enolization is promoted greatly by the basic environment. α-Carbon due to the evident resonance with O- gets a negative charge, becoming able to attack the closest terminal carbon to form the 3-membered ring. Then the enolization step is repeated for another neighbour of the carbonyl group.
Further, the 3-membered ring is preferred kinetically. If we believe in the most stable E-configuration of the carbon chain around the enolic C=C bond, then formation of the 5-membered one becomes impossible due to steric reason (it cannot contain double bond in E-configuration). Thus, the 5-membered ring is deprecated thermodynamically by the intermediate product.

Number the carbon atoms from 1 to 7 in order from left to right.  When a carbanion function forms at carbon atom #3, it can then attack either carbon atom #1 or carbon atom #7.  To get carbon #7 into position for the latter attack, however, you need a bond rotation around the bond between carbon atoms 4 and 5.  That could become restricted if a carbanion also forms at #5 (it would be conjugated with the carbonyl group, like the carbanion at #3), which is quite possible in a strongly basic medium.  Attack by the carbanion at #3 on carbon #1, despite forming a strained ring, could then be kinetically favored.

Building up from your answer, I guess we can also show the fault in the OP's proposed pathway. It was suggested that after the formation of a 5-membered ring, an elimination could occur in the next step to form 2-ethenylcyclopentan-1-one. However, elimination is favored at higher temperatures, and since no heating has been mentioned, both the moles of alc. KOH will simply function as base, abstracting the most acidic hydrogens(which are #3 and #5). Even if the abstraction were to occur step-wise(like at a low base concentration),the product won't be favourable.Suppose the H at C5 is removed...

...and the carbanion thus generated would perform a nucleophilic attack on C1, expelling the Cl and forming a substituted 5 membered ketone.Then,the next abstraction can occur at either of the positions alpha to the ketone group.Out of the two positions,one of them will result in the formation of a secondary carbanion, while the latter to a tertiary one. Since the former is more stable, this should be the result of the 2nd abstraction, and then if it finally does a substitution to remove the second Cl, it will lead to the formation of some sort of fused bicyclic compound,which will be unstable

I do mention that a second deprotonation at C-5 disfavors forming the five-membered ring because conjugation restricts the needed bond rotation.

Yeah, but you assumed both the deprotonations took simultaneously, which might occur at a higher base concentration. I have tried to show that even if we had a low base concentration, so in the sense that both the deprotonations were stepwise, then also the OP's proposed pathway isn't favorable. Just adding another case to your answer

@YusufHasan I think it is due to sigma resonance.

It’s an interesting thought, but I don’t think a double deprotonation is feasible; the first deprotonation already generates an enolate, so the delocalisation of the second anion is minimal at best (it should have a pKa close to that of propene, which is probably ~40, if not even higher).


text:
Calculate pH of a weak acid and weak base neutralization reaction
I'm an industrial pharmacist needing some help with some chemistry basics.
I'm trying to calculate the theoretical pH of an effervescent preparation containing 2 mmol citric acid ($\ce{H3C6H5O7}$), 4 mmol sodium bicarbonate ($\ce{NaHCO3}$), and 0.4 mmol sodium carbonate ($\ce{Na2CO3}$). These powders are dissolved in 100 ml of pH 6.0 water.
Citric acid: $\mathrm{p}K_\mathrm{a1}$ = 3.13, $\mathrm{p}K_\mathrm{a2}$ = 4.76, $\mathrm{p}K_\mathrm{a3}$ = 6.39
Sodium bicarbonate: $\mathrm{p}K_\mathrm{a}$ = 10.329, 6.351 (carbonic acid)
Sodium carbonate: $\mathrm{p}K_\mathrm{b}$ = 3.67
I don't know how to calculate pH in a case like this, where both acid and base are weak and polyprotic. If someone can do an example calculation, I'd really appreciate it.

Please try to have a go at solving a problem, post your attempt to solve it as part of the question. Stack Exchange is like life, you get out of it what you put into it. Keep in mind that it is not a homework answer machine, if you can show us what you have tried to do so far you will be likely to get more help here.

How accurate do you really need to be?

I don't need it to be very accurate. I just want to understand the procedure to determine the pH (and possibly the concentration of other ions, the equilibrium of the reaction) when given the amount and pKa/pKb of reactants.

You have a system of an acid, an ampholyte and a base. As you are interested in the pH value, your main question is what does "produce" or "consume" them. The proton concentration for your specific system is (citric Acid, sodium Bicarbonate, sodium Carbonate):
$$\ce{[H+] = [OH-] + \underbrace{[H2A-] + 2[HA^2-] + 3[A^3-]}_{citric acid's bases} \underbrace{+ [B^2-] - [H2B]}_{bicarbonate's deprotonated\\ and protonated form} \underbrace{- [CH+] - 2 [CH2^2+]}_{carbonate's sinlgy and\\ doubly protonated form}}$$
The multipliers arise from the fact, that a n-times deprotonated/protonated ion yields/consumes n protons.
Now the question is, where you get these concentrations from. It is by solving a linear equation system for each subsystem.

Citric acid, to be deprotonated three times:
\begin{align}
\ce{[H3A] + [H2O] &<=> [H2A-] + [H3O+]\\
[H2A-] + [H2O] &<=> [HA^2-] + [H3O+]\\
[HA^2-] + [H2O] &<=> [A^3-] + [H3O+]}
\end{align}
This gives you the following linear equation system
\begin{align}
k_{a1} &= \ce{\frac{[H2A-][H3O+]}{[H3A]}} \\
k_{a1} \ce{[H3A]} &- \ce{[H2A-][H3O+]} = 0\\
k_{a2} &= \ce{\frac{[HA^2-][H3O+]}{[H2A-]}} \\
k_{a2} \ce{[H2A-]} &- \ce{[HA^2-][H3O+]} = 0\\
k_{a3} &= \ce{\frac{[A^3-][H3O+]}{[HA-]}} \\
k_{a3} \ce{[HA-]} &- \ce{[A^3-][H3O+]} = 0\\
\ce{[H3A] + [H2A-] &+ [HA2-] + [A3-] = [H3A]0} = c_1\\
\end{align}
Which solves to:
\begin{align}
\ce{[H3A]} &= \frac{c_1 x^3}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[H2A-]} &= \frac{c_1 k_{a1} x^2}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[HA^2-]} &= \frac{c_1 k_{a1} k_{a2} x}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[A^3-]} &= \frac{c_1 k_{a1} k_{a2} k_{a3}}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3}
\end{align}

Sodium bicarbonate, to be protonated or deprotonated:
$$
\ce{[HB^-] + [H2O] <=> [B^2-] + [H3O+]\\
[HB^-] + [H2O] <=> [H2B] + [OH-]}
$$
\begin{align}
k_{a4} &= \ce{\frac{[B^2-][H3O+]}{[HB-]}} \\
k_{a4} \ce{[HB-]} &- \ce{[B^2-][H3O+]} = 0\\
k_{b1} &= \ce{\frac{[H2B][OH-]}{[HB-]}} \\
k_{b1} \ce{[HB-]} &- \ce{[H2B][OH-]} = 0\\
\ce{[H2B] &+ \ce{[HB-] + [B^2-] = [B]0}} = c_2\\
\end{align}
Which solves to:
\begin{align}
\ce{[H2B]} &= \frac{c_2 x k_{b1}}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 x k_{b1}}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[HB-]} &= \frac{c_2 x k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 k_W}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[B^2-]} &= \frac{c_2 k_{a2} k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
        = \frac{c_2 k_{a2} k_W/x}{x k_{b1} + k_W + k_{a4} k_W/x}\\
\end{align}

Sodium carbonate, to be doubly protonated:
$$
\ce{[C] + [H2O] <=> [CH+] + [OH-]\\
[CH+] + [H2O] <=> [CH2^2+] + [OH-]}
$$
\begin{align}
k_{b2} &= \ce{\frac{[CH+][OH-]}{[C]}} \\
k_{b2} \ce{[C]} &- \ce{[CH+][OH-]} = 0\\
k_{b3} &= \ce{\frac{[CH2^2+][OH-]}{[CH+]}} \\
k_{b3} \ce{[CH+]} &- \ce{[CH2^2+][OH-]} = 0\\
\ce{[C] &+ [CH+] + [CH2^2+] = [C]0} = c_3\\
\end{align}
Which solves to:
\begin{align}
\ce{[B]} &= \frac{c_3 (k_W/x)^2}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH+]} &= \frac{c_3 k_{b2} k_W/x}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH^2+]} &= \frac{c_3 k_{b2} k_{b3}}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2}\\
\end{align}

Putting it all together (it looks a bit different, though, as it is copied from a simplified form of Mathematica):
$$x=\frac{k_W}{x} +
c_1 \frac{3 k_{a1} k_{a2} k_{a3}+2 x k_{a1} k_{a2}+x^2 k_{a1}}{k_{a1} k_{a2} k_{a3}+x k_{a1} k_{a2}+x^2 k_{a1}+x^3} +
c_2 \left(\frac{k_{a4} k_W}{k_{a4} k_W+x^2 k_{b1}+x k_W}-\frac{x^2 k_{b1}}{k_{a4} k_W+x^2 k_{b1}+x k_W}\right) -
c_3 \frac{2 x^2 k_{b2} k_{b3}+k_W x k_{b2}}{x^2 k_{b2} k_{b3}+k_W x k_{b2}+k_w^2}$$
Solving this little monster for $x$ and calculating its negative decadic logarithm gives you the wanted pH of your system, which is 5.65.

As you mention, further thinking about the problem makes it possible to simplify it a litle bit.
\begin{align}
\ce{[H+]  &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [B^2-] - [H2B] - [CH+] - 2 [CH2^2+]} \\
\ce{[H+] &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [CO3^2-]_B - [H2CO3]_B - [HCO3-]_C - 2 [H2CO3]_C} \\
x &= \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 - c_3 x k_{a4} + 2 c_2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
\end{align}
This does not change the resulting pH value and thus it is not more accurate in my eyes, it only simplifies (admittedly quite a bit) the equation to be solved.
If we assume your initial pH value to be exactly 6 and that is caused by CO2, then we can add it to the equation.
$$x = \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 + (c_4 - c_3) x k_{a4} + 2 (c_2 + c_4) k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} $$
Where $c_4$ can be calculated by rearranging the proton concentration equation for solely carbonic acid:
\begin{align}
x &= \frac{k_W}{x} + c_4 \frac{x k_{a4} + 2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
c_4 &= \left(x - \frac{k_W}{x}\right) \left(\frac{x^2 + x k_{a4} + k_{a4} k_{a5}}{x k_{a4} + 2 k_{a4} k_{a5}}\right) \\
c_4 &= 3.21119 \cdot 10^{-6}
\end{align}
Without initial pH 6: 5.6467
With initial pH 6: 5.6465

Thanks for the answer.  There's a few points I'd like to get some clarification.
1/ How do you get that [H+] is equal to that sum?
2/ Where does the formula to get ions concentration come from?

3/ Also, isn't it sodium carbonate that could dissociate twice (instead of sodium hydrogen carbonate)? So B(1)H+ is H2CO3, B(2)H+ is HCO3-, B(2)H2+ is H2CO3?

@Gnad yes, I mixed this up. For your other questions, I'll try to add details.

@Gnad I needed to change everything and added a little more information about the process.

1/ Regarding the [H+] equation, I understand that citric acid creates [H+] and the bases consume [H+], but why also add [OH-] to the amount of [H+]? 2/ The sodium bicarbonate and sodium carbonate 3 equations sets are essentially the same. Take the sodium bicarbonate equations, and multiply both numerator and denominator by Kb2/x (keep in mind that Kb1 and Kb3 are the same, and Ka4Kb2=Kw). 3/ I think the accurate way to determine the amount of [H+] consume by the bases is 2(c3-[CO32-])+c2-[HCO3-] (basically initial concentration minus final concentration, with corresponding multiplier).

Also, how do you calculate for x with such a monster equation (which software do you use)?

I admire your fortitude in calculating this. However you overlooked one point. The OP stated that the water from which the solution was made had a pH of 6. I'd guess from CO2 from air, but there is no way to be sure. Thus the buffer capacity at pH 6 is unknown.

@pH13-YetanotherPhilipp Anyway, thank you very much for taking your time answering this for a stranger on the internet. I didn't expect to get such a detailed answer. For the other questions I asked, I've read your edit and I think I've got it sorted out.

@MaxW you are totally right, but this is only solvable theoretically, if it is told where this pH comes from.

@Gnad Mathematica. 1/ the OH- comes from autoprotolysis of water. 2/ you are right, it is possible to combine them, but for clarity I let it as it is. 3/ is not fully clear to me right now. If you want, we can elaborate this in chat.

@pH13-YetanotherPhilipp I follow the chat in your profile and posted my thought. Please do check it out=.

Not really an answer but a long comment.
User ph13 answered the question with the generalized formula, then solved the equations computationally. A true tour de force.
However the particular problem could be simplified a bit using some chemical intuition. Using ph13's notation:
(1) The pKa1 of citric acid is so much greater than pKa1 of bicarbonate that essentially no fully protonated citric acid will be left. Likewise it is clear that the solution will be acidic enough that no unprotonated citric acid will be present.
$$\ce{[H3A]} \approx 0 \text{ and }\ce{[A^{3-}]} \approx 0$$
so
$$ \ce{[H2A^-] + [HA^{2-}]} = c_1$$
(2) There is enough citric acid so that all the carbonate will be protonated at least once. So carbonate can be neglected in the figuring total carbonate species.
$$\ce{[B^{2-}]} \approx 0$$
so
$$\ce{[HB^-] + [H2B]} = c_2$$
Likewise
$$\ce{[C]} \approx 0$$
so
$$\ce{[CH^+] + [CH^{2+}]} = c_3$$
(3) Lastly separating sodium bicarbonate species concentrations ($\ce{[B^{2-}], [HB^-], [H2B]}$)  and the sodium carbonate species concentrations ([C] + [CH^+] + [CH2^+]) makes no sense. In solution a carbonate ion, the bicarbonate ion and the carbonic acid can't tell from what solid they came. The gist is that the total of all carbonate species must equal 0.44 millimoles.
(4) As I pointed out in a comment on ph13's answer the 100 ml of pH 6 water is a mystery in this. Some acid is present in the original solution and it isn't clear which one, nor how much. Essentially the buffer capacity of the 100 ml of water is unknown.
(5) A pH of 5.65 is somewhat fortuitous. It is the pH at which carbon dioxide is in equilibrium between its atmospheric phase and being dissolved in water. Any more basic and the solution would draw CO2 from atmosphere. Any more acidic and the solution would effervesce CO2. Carbonic acid has an odd equilibrium in water.
(6) Method of preparation could also have an effect. If you first add citric acid to water then sodium bicarbonate the solution will effervesce $\ce{CO2}$. If you then add sodium carbonate the resulting solution will be more basic than 5.65.

Still the equations are messy and would be easiest solved numerically (trial and error or by computer).

1/ Actually , according to my calculation (done from the equations provided by @pH13), [A3−] is around 2.75 mM and [H2A−] is around 1.98 mM. Now when we notice that the amount of sodium bicarbonate + carbonate is more than twice and less than 3 times of citric acid, I think we would expect the reaction to have an equilibrium between [HA2-] and [A3−].
3/ Agreed. That's what I was trying to say to @pH13.
4+5/ So let's assume that pH 6 of initial water is CO2 absorbed from atmosphere. How does that extra acidic affect the calculation?

Thanks for the correction about $\ce{[A^{3-}]}$. Nice plot here https://en.wikipedia.org/wiki/File:Citric_acid_speciation.png // ph13's calculation is as good as can be done with the information given.


text:
What is the molecularity of the RDS of a zero order complex reaction?
If we have a zero order complex reaction (although uncommon in chemical practices), say the decomposition of ammonia:
$$\ce{2NH3 -> N2 + 3H2}$$
This is a zero order complex reaction. In my textbook, it says:

For complex reaction, the order is given by the slowest step and molecularity of the slowest step is the same as the order of the overall reaction. (sic)

At the same time, molecularity of an elementary reaction cannot be zero.
How is this possible and what am I missing?

Terminology
There are two term which you must understand and use carefully.

Molecularity - The [integer] number of reactant molecular entities that are involved in the 'microscopic chemical event' constituting an elementary reaction. (From IUPAC Gold Book)
Order of reaction - If the macroscopic (observed, empirical or phenomenological) rate of reaction (v) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $\ce{k = [A]^\alpha[B]^\beta\text{ ...}}$ (expressing in full the dependence of the rate of reaction on the concentrations A, B, ...) where $\alpha$, $\beta$ are constant exponents (independent of concentration and time) and k is independent of A and B etc. (rate constant, rate coefficient), then the reaction is said to be of order $\alpha$ with respect to A, of order $\beta$ with respect to B, ... , and of (total or overall) order $n = \alpha +\beta + ...$ The exponents $\alpha$, $\beta$, ... can be positive or negative integral or rational nonintegral numbers.

Equilibrium reaction
For a reaction like the decomposition of ammonia you could write a bimolecular reaction like:
$$\ce{2NH3 -> N2 + 3H2}\tag{1}$$
Often chemists consider such a reaction reaching equilibrium such as:
$$\ce{2NH3 <=> N2 + 3H2}\tag{2}$$
where there is a forward reaction which (naively) could be consider as the forward reaction
$$\ce{2NH3 ->[$k_\mathrm{f}$] N2 + 3H2}\tag{2.a}$$
with the forward rate given by $r_\mathrm{f} = k_\mathrm{f}[\ce{NH3}]^2$ and the reverse reaction
$$\ce{N2 + 3H2 ->[$k_\mathrm{r}$] 2NH3}\tag{2.b}$$
with the reverse rate given by $r_\mathrm{r} = k_\mathrm{r}\ce{[N2][H2]}^3$.
However, if you consider the reverse reaction, you can see that there are four molecules involved. It is highly unlikely that four molecules would ever collide in just the right way to cause a reaction. So the reverse reaction occurs in multiple steps with intermediate species involved. The exponents $x$ and $y$ would be determined experimentally and would almost certainly not be integer values. Thus more realistically:
$$r_\mathrm{r} = k_\mathrm{r}[\ce{N2}]^x[\ce{H2}]^y\tag{3}$$
Catalytic reaction
Now, back to the more salient point of the question. The decomposition of ammonia is typically not done just by heating ammonia to some high temperature to form an equilibrium, but rather it is done selectively with a catalyst and a lower temperature.
$$\ce{2NH3 ->[catalyst] N2 + 3H2}\tag{4}$$
The catalytic surface absorbs ammonia preferentially (and very quickly compared to the reaction time of the ammonia decomposition reaction) and can only absorb a small amount of ammonia. Hence the rate of decomposition doesn't depend on the concentration of ammonia so the rate of decomposition is essentially constant until almost all of the ammonia is consumed. So with a catalyst there is a zero order kinetic reaction with the equation:
$$r_\mathrm{f} = k_\mathrm{f}\tag{5}$$

So, to recap: since the decomposition of Ammonia is conducted with a catalyst that is inherently preferential to the amount of Ammonia that decomposes on its surface, and which is why the concentration of Ammonia is irrelevant, we cannot talk of its molecularity. Am I correct?

Not quite. The catalyst reaction has a molecularity of zero.

How is that possible? Molecularity cannot be zero or a non-integer value.

There is molecularity and there is kinetics. So if A + B -> C, the reaction is bimolecular. But rate for c, $r_c = k_f[A]^x[B]^y$ where x and y are not necessarily integers.

@MaxW Process, has zeroth order kinetics, because it's not reaction, but diffusion controlled. Actual reaction is multistep and definitely doesn't have molecularity of zero.

@Mithoron  - Thanks for pointing out that in my comments above I conflated molecularity and reaction order. They are of course two different concepts. I think I have corrected the answer accordingly now.

@MaxW Molecularity has to be a small positive number, not zero, which is in accord with the definition in your answer. A catalytic reaction is by definition at least bimolecular as a reacting molecule and substrate are always needed.


text:
Why is 3-chloro-5-(1-methylethenyl)cyclopenta-1,3-dien-1-ol named so?
I'm afraid I can't find an image for this organic compound, but when applying lowest locant rule in the numbering (1-methylethenyl) gets 2 and Cl gets 4. (2,4) is less than (3,5). Why then is lowest locant rule not applied here? Also here (1-methylethenyl) is an alkene, and isn't priority order of alkenes before halogens which have no priority?

The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The principal characteristic group of the compound that is given in the question is the $\ce{-OH}$ group. It is expressed as the suffix “-ol”. According to Rule (c), a low locant is assigned first to this group, i.e. “-1-ol”.
In the next step, low locants are assigned to the double bonds according to Rule (e), i.e. “cyclopenta-1,3-dien-1-ol”.
That leaves only one possible locant set for the remaining substituents, i.e. “3-chloro” and “5-(prop-1-en-2-yl)”. The rules (f) and (g) are not needed to make a decision.
Therefore, the complete name for the compound that is given in the question is 3-chloro-5-(prop-1-en-2-yl)cyclopenta-1,3-dien-1-ol.

Note that the preferred prefix for the “isopropenyl” or “1-methylethenyl” (completely: “1-methyleth-1-en-1-yl”) is now “prop-1-en-2-yl”.


text:
IUPAC-Naming of Dioxin
In the molecule 2,3,7,8-tetrachlorodibenzo-p-dioxin, the "p" means para--because each chlorine is para to an oxygen?
In IUPAC nomenclature shouldn't "para" be replaced by a number?

The IUPAC recommended name is given in the IUPAC Gold Book: 2,3,7,8-tetrachlorodibenzo[b,e][1,4]-dioxin

In the molecule 2,3,7,8-tetrachlorodibenzo-p-dioxin, the "p" means para because each chlorine is para to an oxygen?

No. The p (for “para”) indicates the dioxine isomer.

In IUPAC nomenclature, shouldn’t “para” be replaced by a number?

Yes. The preferred IUPAC name for this isomer is 1,4-dioxine.
Therefore, the corresponding name of the parent structure of the compound that is given in the question is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine. The complete name is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine.
The preferred IUPAC name, however, is 2,3,7,8-tetrachlorooxanthrene.


text:
Is an acid strong when its conjugate base is strong or weak?
It is often said that stronger acids have weak conjugate bases and vice versa. Yet also in organic chemistry if an acid is proven to have a stable conjugate base it is said to dissociate more, which means it is stronger. Aren't these contradictory statements? What exactly should a strong acid have- a strong or a weak conjugate base?

if an acid is proven to have a stable conjugate base it is said to dissociate more, which means it is stronger This is wrong.

@AvnishKabaj but many sources say so.. like in this website under Acidity it says so http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch01/ch1-8.html

@AvnishKabaj What don't you like about that statement?

Why are these two statements contradictory? If you have a stable conjugate base, then the base is weaker.

@Zhe I agree that acids with stable conjugate bases are stronger but over here I think that the op is interpreting that statement as "stronger acids have stronger conjugate bases". That's why I said that the statement is wrong.

@Hema Your statements don't contradict. A stable conjugate base implies a weak conjugate base. If it is stable, it means it is not as prone to grabbing a proton as an unstable base would be.

@AvnishKabaj So apparently the issue with this quotation is "it." What is "it?" If "it" is the acid, the statement is correct. If "it" is base, the statement is not.

@Zhe I think so

Since $\mathrm{pK_a+pK_b=14}$ and strong acids are defined as dissociating completely in water, then a strong acid must be stronger than hydronium, meaning its $\mathrm{pK_a<-1.74}$. For its conjugate base then, this means $\mathrm{pK_b>15.74}$. So yes, a strong acid and a weak conjugate base are two sides of the same coin.


text:
What is the relationship between the constant of equilibrium (Kc) and the position of equilibrium?
I understand that the constant of equilibrium is a ratio of the concentration of the products to the reactants and that this value allows one to estimate whether the position of equilibrium lies to the right (in which there will be more product at equilibrium) or lies to the left (where there will be more reactant at equilibrium). I'm also aware that Kc is affected only by a change in temperature and not a change in concentration, pressure etc.
I know the position of equilibrium is affected by changes in concentration, pressure and temperature (unlike Kc which is only affected by change in temperature). I understand Le Chatelier's Principle and how the position of equilibrium shifts in order to minimise the change, and that this maintains the equilibrium constant.
My problem arises when trying to relate the two. The position of equilibrium shifts so that the equilibrium constant will be maintained however I don't understand how this works or why!
When we say that the position of equilibriums shifts (to the right for example), what does this actually mean? I know that the amount of product will increase but what is actually happening. Is the rate of the forwards reaction increasing? In other words, what actually is the position of equilibrium? If Kc is the ratio of product concentration to reactant concentration then what actually is the position of equilibrium?
Thank you in advance!

What is actually happening is that some molecules react to form some other molecules. As for the position of equilibrium, this is not a thing at all.

What you are probably referring to is the degree of dissociation $\alpha$ when you ask about the position of equilibrium. At a given temperature the equilibrium constant is just that, constant, but as the pressure changes, say for a gaseous reaction then $\alpha$ changes.An example shows this.
In the gas phase reaction $\ce{A2 <=> 2A}$ with partial pressures $p_{A_2}$ and $p_{A}$ respectively, the  equilibrium constant is $\displaystyle K_p=\frac{p_A^2}{P_{A_2}}$ and is defined at standard conditions of 1 atm.  If $P$ is the total pressure then the two partial pressures are $\displaystyle p_A=\frac{2\alpha}{1+\alpha}P$ and $\displaystyle p_{A_2}=\frac{1-\alpha}{1+\alpha}P$ where and  so after some simplification $\displaystyle K_p=\frac{4\alpha^2}{1-\alpha^2}P$.
Thus, as the equilibrium constant is constant at any total pressure $P$, the degree of dissociation must change as the pressure changes to compensate, thus the 'position of equilibrium' changes.
[Notes: The amount of A$_2$ reacted is $1-\alpha$ and the amount of A produced is $2\alpha$, thus the mole fractions of each are $(1-\alpha)/(1+\alpha)$ and  $(2\alpha)/(1+\alpha)$ and so the partial pressure are as given above. The total amount needed for the mole fraction is $1-\alpha+2\alpha=1+\alpha$. You can derive related expressions in terms of concentrations.]

As long as the "equilibrium" shifts it is - by definition - no equilibrium. The equilibrium is reached as soon as nothing changes anymore. In theory this only occurs after an infinite amount of time.
As soon as the equilibrium is reached, the position of the equilibrium position is described by $K_c$.
And the system does not shift in order for $K_c$ to stay constant. The system shifts in order to reach the point of minimal energy. $K_c$ is only a way of thermodynamically describing that point.


text:
dilution of Calcium Hydroxide
I have a 30% solution of Calcium Hydroxide.  I want to make a 1% solution.  How would I go about doing this?  I tried using C1 x V1 = C2 x V2

How about you show your calculation as your approach seems to be right

You can't possibly have a 30% solution of calcium hydroxide. Other than that, your approach is right.

right formula, wrong reality. I agree with Ivan.

@user65526 Welcome to chemistry SE! If you haven't yet, take the [tour] and visit the [help]. I would agree with Avnish, try to write up your attempt using Mathjax and then other users can fix any formatting issues to show you how it is done.

Let’s imagine you have 10 mL of this 30% $\ce{Ca(OH)2}$ solution. You are right to plug into the equation that you specified.
$$C_1*V_1 = C_2*V_2$$
$$30\%*10 mL = 1\%*V_2$$
$$\begin{split}
V_2 & = \frac{30\%*10 mL}{1\%}\\
& = 300 mL
\end{split}$$
This means that the final volume will be 300 mL. However, since there are already 10 mL of solution, you only need to add 300 mL - 10 mL = 290 mL of solution.

I'm glad you started off Let’s imagine.... Wikipedia list solubility as 1.73 g/L (20 °C). That's a long way from a 30% solution. // You're also wrong that you could add 290 ml water and 10 ml 30% Ca(OH)2 and get 300 ml of solution. You'd need to take 10 ml of base and dilute to 300 ml.

@MaxW :) (characters needed)

@MaxW Your comment is confusing me: What's the difference between 10 ml 30% Ca(OH)2 and 10 ml base?

None. You'd of course need "about" 290 ml of water. But as an analytical chemist I was trying to point out that volumes aren't really additive. I greatly prefer to think of diluting the 10 ml to 300 ml total rather than adding 10 ml and 290 ml to get 300 ml.

Ivan and Max W it is a 30% solution, which of course is mixed constantly.

@user65526 Good to hear! Is there anything else I can help with? If not, would you mind accepting my answer by clicking the check mark in the upper-left of my answer? I would really appreciate it!

If the Ca(OH)2 isn't totally dissolved then there isn't a "solution" but a "mixture."

true.  Sorry for putting solution


text:
Does this cycle of reactions appear in nature?
Does anyone know a real-world example of a cycle exactly like this:

or in other words, this:
$$\begin{array}{ccc}
\ce{A + C1 -> C2}\\
\ce{X + C2 -> C3}\\
\ce{C3 -> B + C4}\\
\ce{C4 -> Y + C1}\\
\end{array} $$
where the reaction $\ce{A -> B}$ is exergonic (i.e., involves a decrease in free energy) while $\ce{X -> Y}$ is endergonic (i.e., involves a free energy increase)?
The idea is that the above cycle, presumably catalyzed so that all the reactions go fairly fast under normal conditions, 'couples' the exergonic reaction to the endergonic reaction, thus driving the endergonic one.
I would love an example from biochemistry.

Isn't this just the Krebs cycle?

@Zhe - no, the Krebs cycle is much more complicated!

You didn't set any stipulations on complexity. Maybe state that up front?

I want the reaction cycle to be just like the one I showed!  Four species cycling around, two inputs, two outputs!  I'll say "exactly like this".

@JohnBaez I think in a sense you are just describing an organometallic catalytic cycle such as the Suzuki Coupling: Wikipedia page


text:
Aldol Condensation with Cyclic Molecule

How does one solve this mechanism? Why doesn't the negatively charged oxygen on the non-cyclic molecule remove an alpha hydrogen from the cyclic one?

It certainly could undergo the acid-base equilibrium you described, if there's enough energy in the system. Your problem is omitting an important piece of information, which is the temperature at which the reaction is taking place. Lithium enolates (the right compound) are usually formed at low temperatures, down to -78 C. The hope/assumption is that the aldol reaction happens faster than the acid-base reaction between the two reactants. This is certainly hard to control in this particular example, and thus the yield of the reaction is not great, according to this example mentioned in a review article on aldols:

Furthermore, it's likely that in doing this reaction, the experimentalist would form the lithium enolate separately before mixing it with the cyclohexanone. You can read about the importance of order of addition here: https://en.wikipedia.org/wiki/Aldol_reaction#Order_of_addition
Here's the mechanism, note that the problem didn't provide you with a viable source of H+.

I would cool a solution of LDA in THF to -78 oC. Then slowly add the ethyl methyl ketone. I would then allow it to warm up slowly to 20 oC. Then I would cool it down again to -78 oC. Then I would slowly add the 2-methylcyclohexanone to the mixture and allow it to warm up slowly.


text:
Does chlorine trifluoride react with candlewax?
I've run across references to chlorine trifluoride quite a lot. It pops up fairly regularly on the Worldbuilding SE, and whatever idea people have with it is usually shot down very quickly - For obvious reasons.
One of the things that comes up in almost any web search about chlorine trifluoride is that it doesn't react with "candle wax" pretty much at all. (Example here)
Is this true? If it is, why? What is so special about candle wax that makes it immune to this substance that reacts extremely well with almost anything on, in, or around Earth?
(Worth noting: I'm not a chemist, so a basic overview would be appreciated in addition to any extensive scientific answer. Said extensive scientific answer is also appreciated because I'm sure there's others that would like the answer to this as well.)

I'm utterly amazed to find out ClF3 doesn't react with candle wax - can anybody give a better reference? A quick google shows up "Paraffin is by far the most frequently used candle wax on a worldwide basis today" at http://candles.org/elements-of-a-candle/wax/ and the paper "The explosive reaction of chlorine trifluoride with paraffin hydrocarbons" at https://www.sciencedirect.com/science/article/pii/S0082078406806053

https://pubs.acs.org/doi/abs/10.1021/jo00381a017 also says it reacts with saturated hydrocarbons. Only thing hampering the reaction^ might solid state and maybe low solubility of product, but it definitely reacts.

If it is true (and I doubt that) it is probably because there is a passivating reaction on the surface of the wax creating a layer of perfluorinated hydrocarbon (it is known not to react with PTFE).

These are exactly the reasons why I asked. It didn't sound right, but it's so often cited that I want to know if it's true, and why

I know that a polyethene container can be treated with dilute F2 in nitrogen to give a flourinated surface. The idea is to get the best of both HDPE and PTFE by having a surface coating of PTFE on HDPE. I worry with candle wax that if it is heated then the PTFE like coating will be lost from the surface and the reaction will start again.

I've read through the article quoted and can see no reference to candle wax not reacting

It says "One of the few substances known to be completely unreactive with chlorine trifluoride is ordinary candle wax" in the last paragraph.

Having reread the ClF3 section of Ignition, there may be some confusion about the exact conditions where reactions don't happen. Clarke reports that ClF3 and teflon don't react under static conditions, when flowing teflon is eroded rapidly. I suspect any observation about the reactivity of candle wax would be similar as ClF3 probably creates a teflon layer on the surface of the wax when static but both would react vigorously if there were any flow.

I don't think paraffin would last long, since PTFE doesn't seem to last in the presence of CTF.  From IGNITION! (John D. Clark), which you really, really need to read if you are interested in chlorine trifluoride or rocket propellents in general:

The results were excellent, but the difficulties were infuriating.
  Ignition was beautiful —so smooth that it was like turning on a hose.
  Performance was high —very close to theoretical. And the reaction was
  so fast that you could burn it in a surprisingly small chamber. But.
  If your hardware was dirty, and there was a smear of oil or grease
  somewhere inside a feed line, said feed line would ignite and cleverly
  reduce itself to ashes. Gaskets and O-rings generally had to be of
  metal; no organic material could be restrained from ignition. Teflon
  would stand up under static conditions, but if the CTF flowed over it
  with any speed at all, it would erode away like so much sugar in hot
  water, even if it didn't ignite. So joints had to be welded whenever
  possible, and the welds had to be good. An enclosure of slag in the
  weld could react and touch off a fire without even trying. So the
  welds had to be made, and inspected and polished smooth and
  reinspected, and then all the plumbing had to be cleaned out and
  passivated before you dared put the CTF into the system. First there
  was a water flush, and the lines were blown dry with nitrogen. Then
  came one with ethylene trichloride to catch any traces of oil or
  grease, followed by another nitrogen blow-down. Then gaseous CTF was
  introduced into the system, and left there for some hours to catch
  anything the flushing might have missed, and then the liquid chlorine
  trifluoride could be let into the propellant lines.


text:
What does "intensity of light" mean?
In this lecture from MIT, the professor defines the intensity of a wave as the square of the amplitude of the wave. But, at the same time, the professor defines the intensity of light as the number of photons per second. Are these definitions equivalent?
I know that light has both properties of a particle and properties of a wave, but the two different definitions of intensity have different units, which seems confusing to me.
Additionally, when I searched up intensity on Wikipedia, I was unable to find a definition of intensity that matched the definition provided in the lecture. Some of the definitions of intensity that I did find on Wikipedia were "power per unit area," "power per unit solid angle," and "luminous flux per unit solid angle."
Thus, my question is: what do the terms "intensity," "intensity of light," and "intensity of a wave," mean? What are their definitions?

This would be much better suited to Physics.

@orthocresol Perhaps, but the MIT lecture was from a chemistry course. Do you think I should post this question in Physics Stack Exchange? If so, is there a way to move the question to Physics? Or do I need to create a separate, but identical question in Physics?

Why are lasers dangerous to look into? How do they work?

As each  photon has a certain energy, ($E=h\nu$ ) then the power /unit area is the number of photons striking that area per second, so intensity. (btw 'square law detectors' i.e. your eyes , photodiodes etc. detect intensity not amplitude. )

All those things are the same. You're mixing the wave and particle descriptions of light, so naturally, there are different definitions of intensity depending on the perspective.

I'm voting to close this question as off-topic because it is about physics but is now too old to migrate.

@jonsca Really? 3-5 months is not "too old to migrate" I feel.

@Abcd It's a system imposed requirement.  We just plain cannot do it after a period of 60 days.

@jonsca I would rather blame it on the mods. After seeing so many comments that its about physics they should have moved it there. Poor OP can now not get an answer!