text:
Why are naturally occuring liquids more rare than solids and gasses on Earth?
There are only a few naturally occurring liquids I can think of on earth, such as water, mercury and animal oils.
Why are naturally occuring liquids more rare than solids and gasses on Earth?
To be specific,

By Earth I mean the surface which is inhabited by humans, not the interior core or mantle.
I am referring to naturally occurring substances, i.e. naturally occurring liquids are far more rare than naturally occurring gasses or solids. Why is this?
I am referring to average conditions of temperature and pressure that prevail on our planet's surface, not man-made conditions or extremities.

Current Line of Thought
I looked into inter-molecular forces for an answer. Here's what I've come up with.
At Earth's temperature, substances held by dipole-dipole forces are solid and substances held together by weaker london forces are liquids or gasses, such as oils or noble gasses.
The majority of substances have dipole-dipole interactions, hence are solids. In comparison, london forces are weaker so require only a little energy to change phase, which is why non-polar and noble-gasses are gasses or liquids at the same temperature. However, the majority of substances or not noble gasses or non-polar, which is why solids are more common on Earth.
To sum, dipole-dipole are the most common inter-molecular forces. Substances with dipole-dipole forces are solids at earth's temperature, hence the majority of substances are solids.
I am aware that many exceptions arise to this general trend, such as water, however the fact remains that liquids are more rare than solids and gasses on Earth.
I still have a feeling I am wrong, so if anyone could provide a better answer or critique on my line of thought that would be great.
Updated the question with specifics to make it less broad

Comments moved to http://chat.stackexchange.com/rooms/39043/discussion-on-question-by-g-gip-why-are-naturally-occuring-liquids-more-rare-th.  Please do not have extended discussions in the comments.  Thanks!

Your argument misses two major points.
First, a great number of solids are ionic and metallic substances. Those show a really strong bonding mechanism between elementar particles.
And second, to get your theory to work, you have to take into account the size of molecules. To evaporate or melt anything, you have to break or at least disturb all the bonds each molecule has to its surroundings.
For something as small as H2O, not even hydrogen bonding is strong enough to make it solid. A large molecule like Polyethylene, on the other hand, is only held by weak London forces, but has melting points as high as 145°C, and cannot be evaporated.

I'm primarily after an answer to the main question of the article why are liquids more rare than solids or gasses on earth, so I can then understand better as to why I'm wrong.
Do you know the answer to the main question? If so, please edit to include it.

I'm not sure how your point relates to why natural liquids are more rare than solids and gasses. First, almost all ionic and metallic substances are solid. Yes, but aren't liquids still more rare than solids and gasses on Earth? Apologies If I misunderstood.

I just wanted to point out hat you've overlooked a large part of all solids in your argument. And are you talking about compounds in general or elements? For the latter: I'd say they are all solid, unless they have very good reason to be gaseous. Plus the two oddballs mercury and bromium.

Related (albeit remotely): http://chemistry.stackexchange.com/questions/41971/at-what-temperature-in-kelvin-are-most-of-the-elements-on-the-periodic-table-l

Hi Karl. Thanks for the response. I'm generally speaking about both naturally occurring elements and naturally occurring compounds, ones commonly seen in nature. But yes, that's precisely my point! Most are solid, with the exception of oddballs like mercury and bromium. My question is why is this the case? On colder planets like Jupiter you have methane seas. What is the molecular reason for most elements being  solid or gas, rather than liquid, in the conditions of temperature and pressure that prevail on our planet's surface?

The reason for your confusion is that you generalise. If you look at individual classes of substances (metals, ionic compounds, diatomic elements, ...), there are very clear trends with the molecular weight, polarity, etc of those substances, perfectly understandable for every high school kid. There is however no general theory for everything.

Is it then the case that there is lots of separate reasons for most elements being solid or gas, rather than liquid, in the conditions of temperature and pressure that prevail on our planet's surface, rather than a general single reason? For example, when you state "For the latter: I'd say they are all solid, unless they have very good reason to be gaseous", is it lots of separate reasons and not general reason for this behaviour?
I presumed there would be a general reason as this question was posed in an education chemistry video.

Indeed there is no general reason. Heavy particles tend to be solids. Nonpolar substances are gaseous, unless they're too heavy. Metals are solid. Polar substances are liquids, unless they're in fact not very polar (ammonium, carbon monooxide), or have a relatively low melting point, depending on molar weight. Ionic substances are solid. And so on. There are exceptions, but they only slightly bend the rules, not break them. If you look closer, everything starts to make sense.


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Na2HPO4 is amphoteric: write the two reactions
$\ce{Na_2HPO_4}$ is amphoteric, which means it can act as a base or as a acid depending on which substance they react with.
This is the reaction in which disodium phosphate appears to be an acid:
$$\ce{Na_2HPO_4 + H_2O \rightarrow H3O+ + Na_2PO_4-}$$
This is when it acts as a base:
$$\ce{Na_2HPO_4 + H_2O \rightarrow OH- + Na_2H_2PO_4+}$$
Did I write those two reactions currently? If not, what's wrong with my reasoning?

No it's not. I don't think there is such particle as $\ce{Na2PO4-}$ (let alone $\ce{Na2H2PO4+}$). Either write everything in ionic form, or write everything in molecular form.

Oh dear, consequences of not looking closely enough.. Shameful.

$\ce{Na2HPO4}$ is dissociated in solution. The amphoteric species is actually the hydrogen phosphate dianion:
Reaction as an acid:
$$\ce{HPO4^{2-} + H2O <=> PO4^{3-} + H3O+}$$
Reaction as a base:
$$\ce{HPO4^{2-} + H2O <=> H2PO4- + OH-}$$

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What gives variations in the color of cadmium sulfide paint pigment?
Oil paint manufacturers sometimes make several "shades" of a color, all of which list the exact same chemical ingredient.  For example, look at Gamblin's artist grade oils, in the yellow colors, linked here.
Two of those yellows, named Cadmium Yellow Medium, and Cadmium Yellow Deep, both list the pigment chemical as cadmium sulfide (CdS).

If they are the same chemical, then why are they different colors? I asked the paint company and they mentioned that it is the amount of sulfide, and the temperature and amount of time in the furnace.  I'm wondering, what does this translate to, in the chemistry?  Are there different crystal structures to CdS? Or is the furnace introducing some impurity like oxygen? Or is it something else?
I have similar questions about the cadmium red paints that are mixes of CdS and CdSe. But the yellow is the simpler case, so I ask about that first.

I guess there is something to do with quantum dots.

also see this:-https://www.youtube.com/watch?v=bNuoYm7Su4o

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Will this series of reactions work to produce potassium phosphate?
I want to end up with a solution or anhydrous sample of potassium phosphate with which I can grow larger crystals. I want a challenge and something to do so I don't want to buy potassium phosphate, I would rather make it myself.
I don't have access to a lot of chemicals but I do have a fair amount of monoammonium phosphate, (or ammonium dihydrogen phosphate), from a crystal growing kit. I plan to heat this until the ammonia is released to leave behind phosphoric acid to which I can then add water to have an acidic solution. I will then dissolve potassium carbonate from corroded alkali batteries into water and add this to the phosphoric acid. The products will then be carbon dioxide, water and potassium phosphate.
My question is will these reactions work and how easy/difficult will it be. I am sorry if this makes no sense at all because I have done my research for this on Wikipedia and it isn't exactly a 100% reliable source. Thanks.

Getting phosphoric acid by heating ammonium phosphate probably won't work since it requires temperatures above 200°C where phosphoric acid will start to vaporize and decompose.
A better approach might be the reaction of ammonium phosphate with potassium carbonate. The ammonium carbonate formed by this reaction decomposes at much lower temperatures (> 60°C). However, heating will produce corrosive, irritant and toxic products (ammonia, phosphoric acid, phosphorus oxide), so I do not recommend doing such experiments.

So monoammonium phosphate will react at room temperature with potassium carbonate to form ammonium carbonate and potassium phosphate but the products will decompose at temperatures above roughly 60 degrees Celsius. Also what will the phosphoric acid decompose into if I heat it beyond 60 degrees?

The idea is tho get rid of the ammonia by decomposition of ammonium carbonate which starts at 60°C (in dry state). But you have to test before if this actually will work. I would try to mix the ammonium phosphate with potassium carbonate and a small amount of water and then cautiously heat the mixture until it's dry.
Phosphoric acid begins to decompose at its boiling point of 213°C.  It loses water and forms species like pyrophosphoric acid and higher condensated products (not actually phosphorus oxide as I said before).


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Reaction of chlorine with hot/cold solution of hydroxide
When chlorine reacts with cold solution of say potasium hydroxide, the disproportionation goes to lower oxidation states:
$\ce{Cl2 + 2 KOH -> KCl + KClO + H2O} $
Whilst with hot solution the oxidation state of chlorine goes up to $+V$:
$\ce{3 Cl2 + 6 KOH -> KClO3 + 5 KCl + 3 H2O} $
Why is that? Does the heat make the oxidation to higher ox. states easier?

Heat makes pretty much everything easier.

The chlorate ($\ce{ClO3-}$) is formed by disproportionation of hypochlorite and hypochlorous acid:
$$\ce{ClO- + 2 HClO -> ClO3- + 2 HCl}$$
Higher temperature promotes the formation of hypochlorous acid through hydrolysis of hypochlorite and therefore speeds up the reaction.

I am afriad this answer is incorrect. There is no hypochlorous acid in a highly alkaline medium.

Additionally and generally, reactions in highly alkaline medium are not written in the way producing strong acid, furthermore not even ionized even if the equation takes the ionic form.

This is how the reaction is described in literature. The mechanism involves dissociation of HClO into hydroxyl and chlorine radicals. Note that hypochloric acid is a very weak acid and that this fact and increased temperature can very well lead to the formation of enough HClO to enable the reaction, even in very alkaline solutions. Of course you are free to balance out the reaction in a way that emphasizes more on educts and product of the reaction.


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How is food-grade calcium sulfate made?
I'm doing a chemistry project on ionic compounds found on food labels, and I can't find a clear article on how calcium sulfate is made for edible foods. I know it comes from calcium sulfate dihydrate. I have found articles on how they turn that into calcium sulfate for fertilizer, concrete, drywall, and other materials, but not for food products.

Use of calcium sulfate in food and pharmaceutical applications is widespread
and continues to expand. United States Gypsum Company offers two highly refined calcium sulfate products: USG Terra Alba and SNOW WHITE filler. Both fillers are food- and pharmaceutical-grade forms of calcium sulfate, a mineral that appears on the Food and Drug Administration’s GRAS (Generally Recognized as Safe) list of additives approved for nutritional and functional use in food products.

USG Terra Alba ($\ce{CaSO4 \cdot 2H2O}$), the dihydrate form of calcium sulfate, results from fine-grinding and air-separating a select, high-purity white gypsum that contains about $20$ percent water of crystallization.
SNOW WHITE filler ($\ce{CaSO4}$), the anhydrous form of calcium
sulfate, is produced by calcining and milling high-purity white
gypsum$^{[1]}$.

Gypsum is made suitable for commercial use by a process called calcination, which involves heating the mineral to remove approximately three-quarters of its water. Calcined gypsum can recrystallize into any shape with the simple addition of water.$^{[2]}$.
Presumably through repeated calcination, gypsum is made pure enough to meet FDA standards.

$^{[1]}$ USG, Calcium Sulfate Fillers
$^{[2]}$ Encyclopedia.com, USG Corporation


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Why do colored solutions become more transparent as volume is decreased?
For instance, I was drinking Gatorade, and as I drank more and more, the drink became lighter in color. Why?
I don't understand. I would think that the molarity stays constant and the color too.
I feel like it's a really basic question, but as finals approach, I seem to have forgotten everything I know.

That's the Beer - Lambert law in action. The color you see depends on absorbance, which is proportional to molarity and the thickness of your specimen.
Come to think of it, how could it be otherwise? Would you expect 1cm thick layer of liquid to look the same as 10cm? What about 1mm? Or 0.1mm?


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Are modern newspaper inks a health risk?
It's very common in India to see newspapers being used to absorb oils from oily snacks, or simply for temporary food packaging. Some articles say that covering food items in newspaper is a health risk. They also mention that using newspapers for these purposes are strictly disallowed. For example:

Eating food wrapped in newspapers can cause cancer!

Newspapers and cardboard boxes used for packaged foods are made of recycled paper that may be contaminated with harmful chemicals like diisobutyl phthalate and di-n-butyl phthalate which can cause digestive problems and also lead to severe toxicity.
Recycled paper also has printing ink residues trapped from previous prints. These trapped residues have found to contain hormone disruptors like benzophenones and mineral oils.

Dangers of eating newsprint

Canned food and drinks, newspapers used for packing food, cosmetics like eye makeup, hair dye, sindhoor, lead-based paints, ceramic glazes, industrial emissions, exposure at battery units, diety-making units, and tea are all potent media for lead poisoning.

Healthy eating : Watch out for poisonous packaging

Exposure to a class of organic chemicals called arylamines, such as benzidine, 2-Naphthylamine and 4-Aminobiphenyl, is associated with high risks of bladder and lung cancer. Apart from these, printing inks also contain colorants, pigments, binders, additives and photo-initiators (used for speeding up the drying process of the ink), which have harmful effects.

I am highly inclined to believe these articles and therefore I strongly advise against using newspapers as absorbent paper and packaging paper.
But I read some articles that say that the West has now implemented strict regulations and therefore modern newspaper inks are safe.
So I've come here to clear my confusion. Are they safe now?

What chemicals do modern newsprint and the ink contain exactly?
Are they still toxic?

Modern newspaper inks are probably less harmful than they used to be in the olden days, but they are still not quite "food-grade".

I think it should be okay if it has soy ink. But don't think small Indian newspaper companies uses that!

This appears to be asking for personal medical advice. I find it unlikely that your local food and drug administration would approve of commercial packaging of food products in random newspapers, so the regulatory answer would be no...

@JonCuster I'm only asking out of curiosity. I recently had an argument with my sister and I told her not to use the newspaper for placing oily snacks. She said "there's no visible ink spread so don't worry and modern paper ink is safe". So I could use a definitive answer here explaining to her why I was right/wrong. We do not sell food, nor package it with paper on a daily basis, I'm an engineer, and she's a medical student, FYI :)

@JonCuster Is asking about the toxicity of certain chemicals on-topic?

@NVZ - well, I would suggest you go find the safety data sheets for particular chemicals (they are very easy to Google).

@Jon different toxicity questions have been on-topic here before. If your reason for flagging for closure is that it's Googleable, then I think you should downvote. That's now what CV's are for.

@PhMgBr - well, I kind of agree, but in my mind there are two things I considered - (1) no specific material/compound was mentioned, so we are left guessing on ink compositions, and (2) it specifically mentioned wrapping up food. Now, if a specific compound or family was mentioned, OK, and if it is in the context of chemistry lab safety, OK. But neither seemed to apply. I am very uncomfortable suggesting somebody use Chemistry SE to happily wrap up their lunch in something...

@JonCuster I'm not new to how SE works and maybe I should've defined a particular chemical. But I didn't, because I couldn't reach a conclusion as to what exactly makes up modern newsprint and the ink it uses. Google gives me several choices, but not a definitive answer.

@JonCuster I would, in fact, find some answers if I keep digging, but in the end I'm a civil engineer, not chemical, so I believe it's best that I ask you people.

@PhMgBr Should I change my question to "what chemicals do modern newsprint and the newspaper ink contain and are they toxic?". Will that help keep it on topic?

@NVZ I think the main problem lies in the fact that we don't know what they exactly contain. We don't have one beside the table to run an analysis on, if we could. To save this thread, I think you should get a report from somewhere about which amount of which material is used here.

I was always under the impression that the health risks of using newspapers to wrap food came from biological, not chemical, components. The desire for lower toxicity of inks would more likely come as part of industry/manufacturing OSH, not because the paper might end up wrapped around two pound o' chup.

I think the question is fine here. This article - although not India - is related. http://articles.chicagotribune.com/2001-11-11/news/0111110385_1_flint-ink-color-ink-soybean-oil The best approach might well be to contact one of the newspaper companies and ask them about what ink they use.

I would wonder if recycling would introduce toxins into food packaging -- if for example toxic inks were printed on non-food packaging but that packaging got recycled and the recycled paper was used in napkins, etc. Unless specific measures are taken to avoid this, I would guess it happens.

Before anybody may wrap some food in a newspaper, it needs to be printed.
Given that workers in print shops are the first that are exposed to potentially dangerous compounds, this is a matter occupational safety their national regulations.
In Germany, the BG ETEM provides an extensive exclusion list for inks and solvents. Compounds covered by this list are not be be used!
In order to avoid an endless recital of compound names or CAS numbers, the exclusion list is based of GHS hazard statements and includes (among others):

materials with CMR properties:
H340, H341, H350, H350i, H351, H360, H360F, H360D, H360FD, H360Fd, H360Df
materials with specific organ toxicity, acute toxicity and other health risks:
H370, H372, H300, H310, H330, H301, H311, H331, EUH 029, EUH 031, EUH 032

Summary
Does this mean that German newspaper are safe for handle in production and daily use? Yes, unless printing houses violate existing regulations.
Is the same true for Indian newspapers? I have no idea, since I am not familiar with the legal situation in India.

Consider not only the carrier or "vehicle" as it is termed in the industry (soybean oil or petroleum distillate or derivative) but the pigments themselves. A bit of Web digging will provide the reader with ample information about the toxicity of some of these pigments.

Oily food would be expected to pick up
much more pigment from newsprint than
non-oily food such as green leafy
vegetables.
In the US and elsewhere, many produce
items such as tomatoes and apples are
coated with edible wax to extend shelf
life. These products would best be
placed in the oily food category.

One issue is that in much of the Third World newsprint is used by street vendors as a low cost or zero cost container for produce. If one vendor decided to use safe packaging and had to absorb the cost in order to remain competitive with other vendors using newspaper, such a vendor would struggle or even be out of business within a short time.
A broader view of the issue should include the use of recycled paper for food containers. The category includes bags, sheet wrap, uncoated trays, bowls and cups, even napkins. Although the areas of containers made from recycled paper in direct contact with food do not have inked surfaces, the paper itself contains residual amounts of inks. In addition, source material used for the manufacture of recycled paper earmarked for the food industry may include heavily inked magazine paper, inkjet and laser printed paper. I once inquired about the destruction process of the floppy diskette of yesteryear by a paper shredding service which offers such. Apparently, one or two individual diskettes on occasion are acceptable in the recycled paper stream. It is likely that other contaminants make their way into the paper.
The recycled paper industry depends greatly on the food container end use. I do not wish to damage this industry in any way; in fact, I would like to give it all of my encouragement. However, from a food safety standpoint I am raising issues that have been on my mind for quite some time. As a complete outsider to the industry, and one who has not looked into readily available information very much, I would have to admit I know little about the actual recycling process including source separation, purification or cleanup of the paper source for the food industry, as well as other process and product features.
Nevertheless I would like to see the availability of virgin paper, made in dedicated lines in one or more paper mills, for paper products for the food industry. Paper product manufacturers and vendors or retailers could offer virgin paper in their product lineup either as a more expensive alternative to recycled paper or as lower cost alternative, depending on the marketing situation. In the past, recycled paper was a buzzword and an expensive alternative. The situation may now be regarded differently, with a virtually pure, even USDA Organic, source of virgin paper for food packaging products, especially products such as bowls and trays in direct contact with prepared food. Instead of damage to the paper industry, the idea of pure virgin paper may represent a marketing opportunity in a new direction.
The use of newsprint for wrapping produce will not kill you immediately but would be expected to place undue stress on the liver and other organs and tissues in the body involved with the detoxification process. Nevertheless, many individuals live healthy lives to great ages despite lifelong exposure to newsprint. The entire picture is complex and enters the area of individual outcome, outside of the scope of this discussion. However, health problems in general should be incrementally lower over the long run, enough to counter the higher cost of virgin paper in terms of reduced medical costs, but government involvement may be needed in certain countries and in certain situations (see second paragraph below).
An associated issue is the use of BPA (Bisphenol A) as a colorant binder on thermal print paper. In the food retail industries (stores and restaurants) receipt paper is almost exclusively thermal. It is nearly impossible to avoid exposure to BPA, an endocrine disrupter and carcinogen, as it is absorbed directly into the skin within seconds upon handling receipts as well as released into the air as dust from the flexing and bending action of the receipts. Additional exposure is gained upon handling food after handling a receipt without washing hands.
Credit should be given to the supermarket chains that have switched to using BPA-free receipt paper; a question remains regarding the safety of the substitute for BPA. A switching back to the more labor intensive dot matrix printers used a generation ago would likely improve collective health enough to "pay" for itself over the long run. The issue of switching back, however, is the same as the issue described above of the use of newsprint to contain food. The vendor(s) offering the older, healthier alternative place themselves at a competitive disadvantage. This situation is precisely where government may need to step in with either regulation (or enforcement of existing regulation) or recommendation such as direct government-paid advertising. Another avenue to government involvement is a "seal" of approval; a widely used example is the USDA Organic symbol which may in fact be used as a marketing tool.
Custom printed perishable food package labels are typically made of thermal paper stock. Examples are self-stick labels showing "Sell Before" date, prices and other information on packages of produce, meat and take-out prepared food. Another major source of BPA is the lining material used in cans. A number of manufacturers and brands have made the switch to BPA-free cans.

Regardless of whether the newspaper uses the newer, safer inks or older, toxic ones, isn't wrapping food in newspaper something of a food safety risk? For one thing, newspaper isn't designed for food packaging and so it would not be subject to the same standards of 'cleanliness' or hygiene (can't think of a better word) as plastic takeaway containers, for example. During the manufacturing process and delivery, the newspaper could be exposed to many germs. Secondly, since newspaper is quite thin and porous, pathogens could easily pass through the newspaper wrapping and into the food. I would argue that the immediate risk of illness from a bacterial or viral infection from unsafe food handling methods is just as worrying as the potential cumulative effect of the dyes in the paper.

Until perhaps the 1980s printed newspapers used lead type which was set in sheets -- some lead would get onto the paper this way while the ink itself while perhaps toxic did not contain lead. However, the colored comics in the USA used lead in the ink itself. I think lead has been eliminated from newspapers since that time in the USA; I don't know about India. But newspapers are probably still not the great of a thing to use in food handling.
Somewhat related is a woman came to a doctor complaining of various symptoms which suggested that she might have mercury poisoning and she indeed had high levels. She confessed to eating compulsively the pages of books and it turned out that logs or wood pulp which eventually became paper was treated with mercury to prevent growth of fungus. This not only contributed to mercury in rivers and oceans but also some mercury remained in paper and that is how she managed to poison herself.
I do not believe that mercury is still used as a fungicide but I have wondered if recycled paper, which may end up in paper napkins and plates, has various toxic metals that come from recycling color-printed packaging, for example into paper products that come into contact with food or human hands while humans are eating.

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How does the pressure change in the dimerization of nitrogen dioxide?
say we have a reaction N2O4 reversible 2NO2
for this reaction,if initial number of moles of N2O4 is 'a' can we write
t = 0 a -
at equilbrium a-x 2x
similarly can we do the same for pressure? i.e.
if initial pressure is p,then can we say?
t=0           p -
at equilbrium p-y 2y
if yes why? and is x=y,i.e is the number of moles of x dissociated equal to decrease in pressure of reactant?
and is this method valid for all reversible reaction irrespective of the stoichiometric coefficient of reactants and products

If you assume ideal gas in a constant volume this works, but x is not y. 'a' and 'p' are related via the ideal gas law(a = n): P = n * R*T/V .
Take in mind that the pressure actually increases as the reaction progesses as more particles are formed.


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Why is ionisation energy defined for gaseous species?
Most high school and college textbooks usually define the first ionisation energy as such:

The first ionisation energy is the energy required to remove 1 mole of electrons from one mole of gaseous atoms in the ground state to form 1 mole of gaseous singly-charged cations.

I've not been able to receive a satisfactory answer from my teachers, as they either admit they do not know, or treat it as a convention.
Is there a rationale for this definition? Thank you :)

Have a look at: http://chemistry.stackexchange.com/questions/36883/why-is-ionization-energy-definition-restricted-to-gaseous-atoms


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Reversible or irreversible process
If there is a process in which a gas is being expanded against a constant pressure (ie the pressure of the atmosphere is constant before the process, during the process and after the process), then is this kind of process always irreversible process or always reversible process?

It depends on the process.  If the gas is initially at a higher pressure than atmospheric, and you suddenly release the piston and allow the gas to expand, then the process is irreversible.  If the gas is initially at atmospheric pressure (with the piston unconstrained), and you gradually add heat to the gas (so that the average gas pressure is never significantly greater than atmospheric), then the process is reversible.


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Sulfurous acid tautomerisation
In the structure of $\ce{H2SO3}$

there is a lone pair on sulfur.
Why can't sulfur make a bond with one of the hydrogen atoms and a double bond with oxygen to avoid having a lone pair?
Such as given below:

Because here there will be sp3 hybridisation..... no lone pair, 4 sigma and 2 pi bonds.

What's wrong with having a lone pair though?

related http://chemistry.stackexchange.com/questions/38129/why-does-h-bond-to-s-in-hso3

there is nothing wrong but also whats wrong in sulfur having a direct bond with hydrogen and a double bond with oxygen.

I would very much doubt the existence of pi bonds in this molecule.

Most of what I would be saying now is stuff I already highlighted in another answer of mine to a related question.
$\ce{H2SO3}$ exists in two forms, one of which is equivalent to the structure you drew, the other is the structure you proposed. Taking the monoanions (because that was the focus of the other question and I am too lazy to redraw the picture) will give you these two structures:

(All horizontal bars are formal $-$ signs, never lone pairs.)
The crystal structure of $\ce{CsHSO3}$ was solved in 1980 by Johansson, Lindqvist and Vannerberg; they conclude that $\ce{CsHSO3}$ is not cesium hydrogen sulphite but rather cesium sulphonate.[1] See the other answer for details.
In cis-bisbipyridylbishydrogensulphito-κS-ruthenium(II) $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$, a crystal structure solved by Allen, Jeter, Cordes and Durham, the sulphur atom directly coordinates the ruthenium centre which is only possible if it has a lone pair; thus we must be dealing with hydrogensulphite.[2]
Therefore, depending on the conditions both forms are possible and observed.
In case of the free acid, it is likely more stable in one of the two forms. I do not have analytical or thermodynamic data at hand to tell you which tautomer is preferred.

References:
[1]: L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2]: L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.

Jan you are right but maybe you got my proposed structure wrong . i have just edited my question ,have a look at my proposed structure.

While not entirely rigorous, I guess a simple approach would be to just look at the bond strengths to gauge which tautomer is more stable. In the tautomerisation $\ce{SO(OH)2 -> HSO2(OH)}$, we lose one S-O and one O-H bond, but gain one S-H and one "S=O" bond. Now, we just need the data... which I am lazy to find. (The S-O bond is rather elusive.)

@orthocresol I’m looking forward to your competing answer ;D

@jasnoor No, except for the fact that I chose the conjugate base our structures are identical. Martin showed that the sulphite ion is better represented with three $\ce{S-O}$ single bonds though. I’m sure there are other questions/answers on the topic of $\ce{S=O}$ versus $\ce{{S+}-O^-}$ on here that I can’t find atm.

Honestly I don't really have anything to say, the idea just popped into my mind when reading your last sentence. I guess there might be some paper out there with calculations on the gas phase stabilities (the free acid doesn't exist in solution afaik), but again, lazy to find ><

@ortho but me wants sportsmanship ó.ò

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Propene react with ICl
When propene react with ICl , it gives an inoptically active compound ,why.
According to me there is chiral carbon so it should be optically active

The product is optically active compound. However reaction is not stereoselective and therefore both stereoisomers are formed without one being in excess to the other. Product is a racemic mixture of both isomers and therefore should not expose optical activity because effects of both isomers just cancel out.
Reagents or substrate must be optically active in first place in order to get optically active mixture of products.


text:
DFT Code for Atoms : Sources
Some time ago I implemented the restricted and unrestricted Hartree-Fock methods. I had a lot of fun and I learned a lot. It goes without saying that Szabo and Ostlund's book "Modern Quantum Chemistry" has been fundamental to achieve a working code.
I would like to repeat the experience implementing a DFT code. I would like to implement it for atoms with all-electron non-relativistic calculations (scalar relativistic calculation will maybe be an interesting extension as well as the generation of norm-conserving pseudopotentials). Unfortunately I did't came across a "DFT bible" such as Szabo and Ostlund's book.
Does anybody knows good references (at the same level of Szabo and Ostlund's for in-depth discussion of the practical implementation) for the practical implementation of DFT for atoms (radial SE, logarithmic grids, different functionals, ...)?
Note that I am not searching a general DFT introduction or an explanation of real PW+NCPP implementations.

This question is similar to this one but I don't think it is a duplicate. In the previous question I was asking about a book explaining the implementation details of DFT for periodic systems. DFT applied to periodic systems usually involves a PW basis set as well as pseudopotentials and FFTs. This corresponds to a simplified version of the PWSCF code of QuantumESPRESSO.
In this question I am asking for the references that will help me to write of an atomic code, i.e. no basis set expansion and no pseudopotentials (all-electron calculation, logarithmic grid, ...). This corresponds more and less to the atomic code of QuantumESPRESSO which is usually used to generate pseudopotentials.

you might find a few ideas here: http://www.etsf.eu/system/files/users/SottileF/file_101.pdf

Surprisingly, you may look into quantum espresso code again. It contains atomic code, used for pseudopotential generation, though I don't remember the exact location in the source tree.

@permeakra The source code is in espresso-5.x.x/atomic/scr ! ; ) I know everything has already been implemented, but ATOMIC contains around 100 files (relativistic calculations, PP generation, ...)... I would like to start to write a minimal code in order to understand the underlying theory. ATOMIC is already too complex for my needs. In addition the risk is that I will start to "copy" QE routines; I would like to write them from scratch based only on the theory of the references.

Maybe my answer there would have been more suitable here, but anyway, I think it answers partly both topics.

Marx, D.; Hutter, J. Ab initio molecular dynamics: basic theory and advanced methods; Cambridge University Press: Cambridge, 2012. This has some info on implementation of algorithm in CPMD code including flow chart of algorithm.
Thijssen, J. Computational Physics, 2nd Ed.; Cambridge University Press: Cambridge, UK ; New York, 2007. Chapter 6 comes with a short computer program downloadable from companion site written by author of the book.

Maybe these two help a bit.


text:
Acetaldoxime Synthesis: Hydroxylamine vs Hydroxylamine HCl
Wikipedia says acetaldoxime can be prepared by reacting Acetaldehyde & Hydroxylamine in the presence of NaOH. https://en.wikipedia.org/wiki/Acetaldoxime
But the reaction shown uses Hydroxylamine Hydrochloride instead of Hydroxylamine.
Does it not matter? Do both work? Would Hydroxylamine Sulfate work as well?

Hydroxylamine is a low-boiling liquid readily oxidized to explosive nitrogen-oxygen compounds.  It is stored as the hydroxylammonium hydrochloride salt for convenience and safety, since the salts do not oxidize as easily.  Additionally, hydroxylamine itself may explode when heated.
The sodium hydroxide is to render the freebase which then may participate in the reaction.  In principle, the hydroxylammonium bisulphate would work as well.
It is somewhat important that an exact stoichiometry be used for base to hydroxylamine to prevent aldol side-reactions if relevant (which for acetaldehyde, they are).  Additionally, this reaction is reversible in the presence of acid and water, so all the acid must be consumed, since water is a byproduct of the condensation.  Given the explosive nature of hydroxylamine, azeotropic distillation via Dean-Stark trap is not viable for dehydration.  Chemical dehydration may be effective.

Thanks! What you refer to as the bisulfate, is that this: https://en.wikipedia.org/wiki/Hydroxylammonium_sulfate

For commercial quantity use at scale I seem to get three options from BASF: The freebase, the HCl salt, the sulfate. http://www.intermediates.basf.com/chemicals/inorganic-specialities/hydroxylamine-free-base-and-salts I'm working out the economics of each option. Funnily they do ship the freebase in  spite of the risks, but just as a 50% solution in H2O

@curious_cat, that link points to the sulphate (not the bisulphate).  The sulphate may also work, but you will need two equivalents of base, and also gain two equivalents of hydroxylamine from the reagent.  The sulphate is (obviously) non-halogenated, so it will probably be cheaper to get rid of, even if the mass of waste is higher.

Thanks. I don't see the bisulphate made  / sold in tonnage quantities. So forced to choose from the freebase / chloride or sulfate.

Interestingly the sulfate vs HCl salt is pricing agnostic. i.e. Per equivalent of Hydroxylamine produced the cost is the same. The freebase on the other hand is 3x as expensive. Not sure why.

Explosion hazard.  Shipping hazard.  Liquid.

Another question: Just like the Hydrochloride Salt needs NaOH to render the freebase if one used the Sulfate salt would one use a co-reagent to render the freebase? What would be this co-reagent?

NaOH, two equivalents


text:
What is the structural difference between ketogenic amino acids and glucogenic amino acids?
We know some amino acids could go in ketogenosis path way ,some in glucogenic and some in both. I 'm wondering to know what the differences are between the amino acids that make this situation.

Please [edit] the question turning your attention to punctuation. 2) Out of the blue I do not know what ketogenic and glucogenic amino acids are. Please [edit] your question to tell us this difference. (Editing a question (substantially!) is a much better way to attract attention and receive answers than posting an NAA-answer which will be deleted.)

“Standard” amino acids are degraded to one of seven metabolic
  intermediates: pyruvate, α-ketoglutarate, succinyl-CoA, fumarate,
  oxaloacetate, acetyl-CoA, or acetoacetate.

The amino acids may therefore be divided into two groups based on their catabolic pathways:

Glucogenic amino acids;

whose carbon skeletons are degraded to pyruvate, α-ketoglutarate, succinyl-CoA, fumarate, or oxaloacetate and are therefore glucose precursors.

Ketogenic amino acids;

whose carbon skeletons are broken down to acetyl-CoA or acetoacetate and can thus be converted to ketone bodies or fatty acids.

The amino acids degraded to  acetyl-CoA and acetoacetyl-CoA are used in the citric acid cycle, but mammals cannot synthesize glucose from acetyl-CoA. This fact is the source of the distinction between glucogenic and ketogenic amino acids. Glucogenic amino acids can be converted to glucose, with oxaloacetate as an intermediate, but ketogenic amino acids cannot be converted to glucose.
Strictly speaking, whether an amino acid is regarded as being glucogenic, ketogenic, or both depends partly on the observer. This classification is not universally accepted, because different quantitative criteria are applied will identify the degradation pathways by the entry point into metabolism.
I would also attribute to things like availability of feasible pathways, enzymes etc (for instance alanine is glucogenic because its transamination product, pyruvate can be converted to glucose via gluconeogenesis)
Some amino acids have more than one pathway for catabolism, which explains why four of the amino acids are listed as both glucogenic and ketogenic examples are Isoleucine, phenylalanine, threonine, tryptophan, and tyrosine):

Below I will include snippets of degradation products of named amino acids:

Five amino acids, alanine, cysteine, glycine, serine, and threonine, are broken down to yield pyruvate:

Asparagine and Aspartate Are Degraded to Oxaloacetate:

Arginine, Glutamate, Glutamine, Histidine, and Proline Are Degraded to α-Ketoglutarate:

The reason I included these pathways, is to allow you to visualise structure, feasible metabolic pathways and enzymes so you have a clear picture on the differences between these amino acids.
And so on..
Hope this helps..
References
Voet and Voet biochemistry
J.Berg Biochemistry
Campbell and Farrel Biochemistry


text:
Is ciprofloxacin acidic or basic?
I would expect an aqueous solution of ciprofloxacin to be basic for the following reasons:

The carboxyl group has pKa = 6.09 and the secondary amino group has pKa = 8.74. Because 8.74 is further from neutral pH than 6.09, I would expect the pH of an aqueous solution to be above 7.
The tertiary amino groups are connected to an aromatic ring, so they should be very weakly basic and not affecting pH much.

However, the PubChem Compound Summary says the following about ciprofloxacin: "pH range for the 1.0% aqueous concentrates in vials is 3.3 to 3.9; the pH range for the 0.2% ready-for-use infusion solutions is 3.5 to 4.6" (see https://pubchem.ncbi.nlm.nih.gov/compound/ciprofloxacin#section=pH)
Is an aqueous solution of ciprofloxacin solution really acidic and if it is, what is the reason?
I also found an article citing several estimates of the pKa:s and the consensus seems to be that the amino group is stronger than the carboxyl group. (Babić et al. Determination of pKa values of active pharmaceutical ingredients. Trends in Analytical Chemistry, Vol. 26, No. 11, 2007.)
Here is a picture of ciprofloxacin from Wikipedia:

Here is a link to the article I mentioned. I cannot add the link to my question because I don't have enough reputation. https://bib.irb.hr/datoteka/305797.sandra.pdf

If you have to ask, it's probably both.

Your predictions are right; the molecule has more basic character.
The drug itself is practically insoluble in water, because the big organic molecule is near it's isoelectric point at the pure water's pH. However by protonating or deprotonating it and thus making it charged it's solubility is greatly enchanced.
This is also the case with those medical ready-for-use infusions:

...CIPRO IV solutions are available as sterile 0.2% ready-for-
  use infusion solutions in 5% Dextrose Injection. CIPRO IV contains lactic acid as a solubilizing agent and hydrochloric acid for pH adjustment. The pH range for the 0.2% ready-for-use infusion solutions is 3.5 to 4.6.

Reference

From the pKa values I calculate that even at neutral pH the carboxyl group should be charged 89,0% of the time and the amino group should be charged 98,2% of the time. That means that 99,8% of ciprofloxacin is charged at neutral pH. Why is this not enough for a good solubility?

You're right in that near the isoelectric point that the amines are mostly protonated and the carboxylate groups are mostly deprotonated. However, this means that most of the molecules are actually NOT charged because they balance each other off. This results in the solubility dropping because the molecule as a whole has no charge.

You are right. The important thing is that it has no net charge. Charged molecules reppel (Columbian interaction) each other and therefore they can not aggregate. On the other hand neutral moleculs can stick together and participate.

That makes sense. I did not think that the net charge would matter.

Btw, could it also be an important factor that the decreased pH makes the teartiary amino gropus more likely to be protonated?

Sorry, I don't exactly understand what do you mean.
Note that the tertiary amine groups are not the most basic group in the molecule because of the phenyl group as their substituent. In this case their predicted pKa values are negative (-0,2 and  -1,5), so it is probably safe to assume that they don't get protonated at all. Secondary  amine groups is the most basic in this molecule.
In case of more complex  molecules you may want to check  pKa values using pKa predictor web tool (like this one: https://epoch.uky.edu/ace/public/pKa.jsp )

Ok. That explains it. I did not think that the pKa:s would be that low. I read that aryl amines have a pKa of around 4 or 5 and I thought that would be the case here as well. If the pKa:s are -0,2 and -1,5, it is clear to me that no protonation takes place.

The commercial active pharmaceutical ingredient (API) is ciprofloxacin hydrochloride.  This may explain why you refer to the pH range as 3.3 to 3.9. It's not uncommon for non-literature references to mention API's without mentioning them as a salt.


text:
Why does Boron only need 6 valence electrons unlike the standard 8?
So I was a bit confused. Hydrogen and Boron seem to be the only outliers to the octet rule. Hydrogen makes sense because it has only one shell and 2 electrons complete its shell. Boron's configuration, however, is $1s^2 2s^2 2p^1$. Having 6 valence electrons would mean having a configuration of $1s^2 2s^2 2p^4$ which doesn't seem as stable as $1s^2 2s^2 2p^6$ or even as stable as $1s^2 2s^2 2p^3$ since one would think that there is repulsion going on between electrons in the first orbital of p in $1s^2 2s^2 2p^4$. Thanks in advance!

Boron wants the standard octet as badly as anyone else, and indeed reaches it, albeit sometimes by unusual means.

Also, lithium dimerizes in the gas phase.  Just like hydrogen, it fills its s shell, and adds no more.  Another outlier.

By this, I would assume that helium is also an outlier.

related http://chemistry.stackexchange.com/questions/43777/how-can-borane-exist http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane

Also http://chemistry.stackexchange.com/questions/32265/which-relatively-simple-molecules-violate-the-octet-rule http://chemistry.stackexchange.com/questions/34752/beryllium-hydride-violates-the-octet-rule

The principle of full octet is that most reactivity patterns are described well if there is a driving force / energetic gain for full octets in main-group atoms.
However, the only facet of an atom that dictates anything about the number of valence electrons is its nuclear charge.  Boron has a charge of 5.  This is balanced by 5 electrons.  Two of them are core electrons and the remaining 3 are valence electrons.  The valence electrons may participate in bonding through sharing with other atoms, to make three bonds.  Three bonds = six electrons.
That is the simple answer. The way this question is posed indicates you are very new to the study of chemistry, and this next bit goes well beyond an introductory chemistry course.
For a fourth bond to form, another atom would have to donate two electrons to the bond.  This happens in many cases, such as when an amine reacts.  In other cases, an atom already bonded may donate electrons, such as in boron trichloride.  Finally, borane gas itself can engage in 3-center two electron bonding which is really weird, but provides a full octet when two boron atoms share the bonding electrons in a boron-hydrogen bond.
Don't distress, boron has many ways to get is full octet.

No worries I understood. I've been taking AP Chem and we didn't really to deep into why the exceptions were the way they were which is why I was asking. Thanks a lot!


text:
Water conditioning with a descaler?
Products known as “descalers” are said by the manufacturers (such as this one) to use electrical impulses to transform calcium bicarbonate (which builds up in pipes and water-using appliances, causing premature failure, and also leaves calcium residue on glassware, clothes, and so on) into calcium carbonate (in a form known as aragonite), which does not adhere -- so the scale problem is solved.
Truth or snake oil?

So, you only need a descaler/softener if you have what is known as hard water, which is basically just water with a lot of minerals such as calcium and magnesium. You can do a test that shows how many GPGs (grains per gallon) your water has. According to http://home.howstuffworks.com/question99.htm,

If your water tests at 1 GPG (17.1 mg/L) or less, then you have soft water. Water around 1-3.5 GPG (17.1-60 mg/L) occupies a gray zone between soft and slightly hard water and 3.5-7 GPG (60-120 mg/L) is moderately hard. Hard water is around 7-10.5 GPG (120 - 180 mg/L), and very hard water is above that.

So if it does turn out that you have hard water, then you look into descalers and softeners. The reason for this is that the calcium and magnesium come out of the water as scale, which builds up on the insides of pipes, water heaters, etc., to the point where they can become completely clogged.
A descaler is used to remove this buildup. They do this because they are acids and release hydrogen ions when dissolved in water. These ions, in turn, eat away at the scale by pulling at them with the charge. A softener is used to clean the water as it comes, disposing of the minerals safely. Water softeners are mechanical appliances that are put into your home's water system. They exchange the minerals for something else, usually sodium, in what's called an ion exchange. A diagram of a water softener is shown below:

This website provides more information about the exact workings of a water softener.
Hope this helps!

Same thing here - you should tell where did you get the pic

https://waterpluspure.wordpress.com/2011/06/21/apa-yang-dimaksud-water-softener-dan-bagaiman/ has the image (I found this one searching Google Images as well)

You shouldn't post pics which aren't free. Maybe leave only links to them? I'm no expert in this stuff unfortunately.

How can you tell if an image on, say, Google Images, is free?

Usually they aren't, don't post them if you don't know.


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Volcano plots: How is the data fitted?
When studying about catalysts I normally come accross volcano plots like the one shown below:

What confuses me is how are the two lines fitted, how do you determine what points to include and what points not to include. For example I could have included the two points for Pt and Ir to obtain a slightly different shape of the volcano plot. Is there a guidline for this?

The most obvious approach for me would be to find a partition of the set of points into two subsets and two straight lines, one for each subset, such that the sum of the squared residuals over both partitions is minimized.

Correct me if I'm wrong but if it is not explicitly mentioned in the paper how the fitting is obtained, it is very likely 'drawn' as a visual guide instead of obtained through a rigorous statistical fitting. The use of 'descriptors' to create volcano plots is a somewhat inexact art, and the inherent errors in the computational method/approximations probably make a rigorous fitting irrelevant, besides a very general positioning of the peak.


text:
How to write the redox reaction of chlorate to chlorine and oxygen using half rections?
I know how to solve redox reactions, but this one is really confusing me since I can't really write half reactions for it.
$$\ce{ClO3- -> Cl2 + O2}$$
The answer is apparently:
$$\ce{2H+ + 2ClO3- -> 5/2 O2 + Cl2 + H2O}$$
This makes sense. I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine(V)'s so you have to use a water to make the charge balance work out. I just don't see how you'd come to this conclusion on your own.

This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing disproportionation with different elements being reduced/oxidised, if you could split the starting material into the two different elements, then you could circumvent the issue. Obviously the bonding in the chlorate ion is not ionic but this procedure allows you to construct "half-equations" and then an overall equation.
I would not suggest that you write this in an exam.
The two half-reactions are then:
$$\begin{align}
\ce{Cl^5+ + 5e-} &\longrightarrow \ce{1/2Cl2} \tag{1} \\
\ce{O^2-} &\longrightarrow \ce{1/2 O2 + 2e-} \tag{2}
\end{align}$$
$[(1) \times 2] + [(2) \times 5]$ gives
$$\ce{2 Cl^5+ + 5O^2- -> Cl2 + 5/2 O2}$$
In order to "re-form" your chlorate, you need three oxide ions per Cl ion, so you need to add one more oxide to both sides:
$$\ce{2 Cl^5+ + 6O^2- -> Cl2 + 5/2 O2 + O^2-}$$
Now you can get back your chlorate:
$$\ce{2 ClO3- -> Cl2 + 5/2 O2 + O^2-}$$
Now it's highly unlikely that there are going to be free oxide ions hanging around, so we just protonate them by adding $\ce{2 H+}$ on both sides. The oxide ion simply becomes water:
$$\ce{2 ClO3- + 2H+ -> Cl2 + 5/2 O2 + H2O}$$

I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine (V)'s so you have to use a water to make the charge balance work out.

Yeah, if all six oxygens became $\ce{O2}$ then they would lose 12 electrons. Therefore, one of the oxygens does not become $\ce{O2}$; it becomes $\ce{H2O}$ and remains as oxygen(-2). It is not about charge balance.

What I meant by "to make charge balance work out" was that if you didn't use a water to balance out the electrons then your net equation wouldn't have the protons on the left so the charge balance wouldn't work out.

Thanks a bunch orthocresol! I was trying to do exactly what you're doing but I didn't break the "half reactions" down to their fundamental level like you did (I was writing the numbers of each "ion" that come from ClO3 instead of just one ion to the half diatomic molecule). This seems to be the key here, so I will just remember to do that tomorrow :) This has been very helpful thank you!

@gannex I see what you mean by the charge balance, it's just a different way of looking at it I suppose. Anyway no problem. Again though I must reiterate that this is just a trick that allows you to get the half reactions - it has no physical interpretation nor does it reflect what's actually going on in the system. So be careful with it.

I guess you could do this:
(12 H+ + 2ClO3- + 10 e= Cl2 +6H2O)
+[2 Clo3 -  + 10e= Cl2 + 3O2 + 12 e] 5
= 12 clo3 +12H+ =6 cl2 + 15 o2+6 H2o

@gannex Yeah, I suppose you could do that too.

There is no real reason not to do it the traditional way. As a reminder: You do not have to oxidise oxides from chlorate, you can intermediately assume you are oxidising oxygen from water molecules. That allows you to formulate an oxidation half-equation without having to deal with cumbersome additional elements. But first the reduction:
$$\begin{align}\ce{2 \overset{\mathrm{+V}}{Cl}O3- + 10 e- \phantom{\ + 12 H+} &-> \overset{\mathrm{\pm0}}{Cl}_2}\tag{Red1}\\
\ce{2ClO3- + 10 e- + 12 H+ &-> Cl2}\tag{Red2}\\
\ce{2ClO3- + 10 e- + 12 H+ &-> Cl2 + 6 H2O}\tag{Red3}\end{align}$$
(In all these half-equations the first step is balancing oxidation states with electrons, the second is balancing charge with protons and the third is balancing mass with water.) Here goes for the oxidation:
$$\begin{align}\ce{2 H2O &-> O2 + 4e-}\tag{Ox1}\\
\ce{2 H2O &-> O2 + 4 e- + 4 H+}\tag{Ox2}\end{align}$$
(Here, we are done after two steps.)
And now, simply add up the two equations making sure the number of electrons is balanced; we need $2 \times (\text{Red3})$ and $5 \times (\text{Ox2})$ followed by cancelling out:
$$\begin{align}\ce{4 ClO3- + 24H+ + 10 H2O &->
2 Cl2 + 12 H2O + 5 O2 + 10 H+}\tag{Redox1}\\
\ce{4 ClO3- + 4 H+ &-> 2 Cl2 + 2 H2O + 5 O2}\tag{Redox2}\\
\ce{4 ClO3- + 4 H+ &-> 2 Cl2 + 5 O2 + 2 H2O}\tag{Redox3}\end{align}$$


text:
Charge, Size, and Movement in Aqeuous Ions as Electrolytes
Would a pair of aqueous $+2$ and $-2$ ions necessarily conduct electricity better than the same concentration of $+1$ and $-1$ ions? Furthermore, would more massive ions not conduct as well as less massive ions because they cannot travel as rapidly? It seems to me that ions move in the liquid to conduct electricity, because solid ionic crystals do not conduct electricity. If they move, then size and charge should both be factors affecting electrolyte strength. However, if ions are moving, why don't they ever achieve the most stable position and stop moving?

Would a pair of aqueous $+2$ and $−2$ ions necessarily conduct electricity better than the same concentration of $+1$ and $−1$ ions?

Not necessarily. Taking the concentration of 0.001 aqueous solution of $\ce{KOH}$ and $\ce{MgSO4}$ at $18$ degree Celsius, we find that the current conducted by $\ce{KOH}$ is $2.34 \space mA$ while the current conducted by $\ce{MgSO4}$ is $2.00 \space mA$. Why did this happen?
It happened because of the mass. And therefore here comes the answer to your second question.

Would more massive ions not conduct as well as less massive ions because they cannot travel as rapidly?

Yup, more massive ion means less current. It's because of simple equation (and I hope you are aware of these physics equations)$$qV=\dfrac{mv^2}{2}$$
Here $q$ is the charge on the ion. $V$ is potential difference between both the solutions. $m$ is the mass of the ion. And $v$ is the drift velocity. We know that $qV$ is constant. So we get $$v \propto \dfrac{1}{\sqrt{m}}$$ We can easily analyse that more the mass we have, less the speed we get.

If ions are moving, why don't they ever achieve the most stable position and stop moving?

They do achieve the most stable position and that's why batteries die.

Hydroxide and hydrogen ions are special in water solution, as they have anomalously high mobility. This is due to 'hopping'  mechanism https://en.wikipedia.org/wiki/Grotthuss_mechanism

But if you were sending a current through a solution, it could go on indefinitely.

@YunfeiMa You can take this scenario as recharging of battery.

@manshu If I ran prepared a solution of say, NaCl, and continued to run current through it, would it stop conducting after a certain amount of time?

@YunfeiMa Yeah, at the time when the potential between the two electrolytes becomes 0, the conduction will stop.


text:
The effect of reactant concentration on when equilibrium is established
Take the equation
$$\ce{A(g) <=> B(g)}$$
where the reaction is first order with respect to A. The rate law is thus
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k[\ce{A}]$$
Say you run the reaction, starting with $x$ moles of A. Equilibrium is established at time $y$. If you run the reaction again, this time with $2x$ moles of A, won’t equilibrium be stablished at time $y$ as well? You double the rate of reaction, but you have twice as many molecules to react, so wouldn’t time remain the same? I’ve been told that the equilibrium effect dominates over the kinetic effect; that is, it will take a time that is greater than $y$ for the reaction to reach equilibrium. I cannot fathom the mathematical basis of such a conclusion. Can someone elucidate?

You've misunderstood that statement about the "equillibrium effect" (which is imo a rather stupid term). It only means that after some time, if there is a significant backward reaction, you cannot predict the state of the reaction from the kinetics of the forward reaction any more.

I you have an equilibrium with significant amounts of $\ce{A}$ and $\ce{B}$ you need to consider the backwards reaction. If it is first order in $\ce{B}$, the rate law is
$$\frac{\mathrm d[\mathrm{A}]}{\mathrm dt}= -k_1 \cdot [\mathrm{A}] + k_2 \cdot [\mathrm{B}] \tag{1}$$
If the initial concentration of $\ce{A}$ is $\mathrm{[A]}_0$ and the initial concentration of $\ce{B}$ is $0$ the solution of the differential equation is
$$\frac{\mathrm{[A]}}{\mathrm{[A]}_0} = \frac{1}{k_1 + k_2} \cdot \left(k_2 + k_1 \cdot \mathrm e^{-\left(k_1 + k_2\right) t}\right) \tag{2}$$
The equilibrium concentration of $\ce{A}$ is obtained as the limit for $t \to \infty$:
$$\frac{\mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = \frac{k_2}{k_1 + k_2} \tag{3}$$
From $(2)$ and $(3)$ it follows that
$$\frac{\mathrm{[A]} - \mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = k_1 \cdot \mathrm e^{-(k_1 + k_2) t} \tag{4}$$
or, logarithmically,
$$t = -\frac{1}{k_1 + k_2} \left(\ln\left(\mathrm{[A]} - \mathrm{[A]_{eq}}\right) - \ln\left(\mathrm{[A]}_0\right) - \ln\left(k_1\right)\right) \tag{6}$$
The criterion for equilibrium $\left|\mathrm{[A]} - \mathrm{[A]_{eq}}\right| \ll 1$ lets us neglect the term $\ln\left(\mathrm{[A]}_0\right)$ in equation $(6)$.
This shows that the time it takes to reach the equilibrium is independent of the initial concentration of $\ce{A}$.


text:
Spontaneous and Favoured Chemical Reactions
I have some problems in understanding what is the difference between a reaction which is spontaneous and a reaction which is termodynamically favoured. I'm looking at the equation of Gibbs Free Energy:
$$\Delta G = \Delta G^0 + RT ln \prod a_i^{\nu_i}$$
I know that a reaction is spontaneous if $\Delta G <0$ and a reaction is Termodynamically favoured at a given T, by convenction, if $\Delta G^0<0$.
So the condition of $\Delta G^0<0$ is a necessary, but not sufficient condition to be spontaneous?
Thanks a lot!!

You are missing a ln before the product.

Your equation shows in what direction an equillibrium  reaction will move to achieve td. equillibrium. A "spontaneous" reaction must of course be thermodynamically possible, but it also requires a viable kinetic pathway, which cannot be predicted from such simple thermodynamics.

Thanks for Your reply!  Therefore a reaction which is thermodynamically unfavoured in the considered direction will have a small amount of products with respect to the reactants but it will occur anyway. The condition of unfavoured reaction can be overcome with kinetics considerations (i.e. catalyzers)?


text:
What is the graphical formula when nitrobenzene is added to HCL and tin?

On the image i want to work out the structure of compound Q. I know that it is a reduction reaction. However why is it that the answer to this question is not C6H5NH2, but is actually (C6H5NH3)+Cl-

You are on the right track! The reduction product of nitrobenzene is indeed aniline. But there's $\ce{HCl}$ on top of the arrow and aniline is a base. Protonation gives $\ce{PhNH3+}$ and $\ce{Cl-}$ is around too.

@KlausWarzecha Okay. So am I right in saying that if I have the salt that you specified and I add NaOH, will I get C6H5NH2 ? Also, what is the easiest way to recognize a base such as this?

Yes, exactly. Addition of $\ce{NaOH}$ will release the aniline. One has to keep in mind that the reducing agent is actually $\ce{Sn^{2+}}$, which is formed by the reaction of the cheap tin metal with hydrochloric acid. Workup with   $\ce{NaOH}$ is a standard step for this kind of reduction. As far as the base is concerned: proton goes where lone pair waits. Looking for amines helps :)


text:
Apparent solubility of Ag2C2O4 in a buffer solution

We take the following quantities and mix them together:

$\pu{20 mL}$ solution of $\ce{AgNO3}$ $\pu{0.001 mol//dm3}$
$\pu{20 mL}$ solution of $\ce{H2C2O4}$ $\pu{0.200  mol//dm3}$
$\pu{40 mL}$ buffer solution $\mathrm{pH} = 5.00$

What is the apparent solubility of the insoluble salt formed, knowing: $K_\mathrm{a1}=6.50\cdot 10^{-2}$, $K_\mathrm{a2}=6.10\cdot 10^{-5}$ (for $\ce{H2C2O4}$) and $K_\mathrm{s}=1.10\cdot 10^{-12}$ (for $\ce{Ag2C2O4}$)?

The initial concentrations for $\ce{Ag+}$ and $\ce{H2C2O4}$ are:
\begin{align}
[\ce{Ag+}]&=\pu{2.50E-4 mol//dm3}\\
[\ce{H2C2O4}]&=\pu{5.00E-2 $$mol//dm3}$$ \text{(excess)}
\end{align}
We're dealing with the following equilibria:
\begin{align}
\ce{Ag2C2O4 &<=> 2Ag+ + C2O4^{2-}}\\
\ce{H2C2O4 &<=> HC2O4- + H+}\\
\ce{HC2O4- &<=> C2O4^{2-} + H+}
\end{align}
The pH of the solution is above $\mathrm{p}K_\mathrm{a2}$, therefore $\ce{H2C2O4}$ is found mostly as $\ce{C2O4^{2-}}$. By using the Henderson–Hasselbalch equation, we find that $[\ce{C2O4^{2-}}]=\pu{4.30E-2 mol//dm3}$.
Since we started with an excess of acid, we can consider that the variation of concentration for that particular species (after precipitation completes) is negligible. By using $K_\mathrm{s}$, we find that:
$$2S=[\ce{Ag+}]=\pu{5.06E-6 mol//dm3}$$
Is my approach correct? I'm confused because of their approach, where they invoke the following equation:
$$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$
Which doesn't really make sense since we started with an excess of $\ce{H2C2O4}$.
In my opinion, this equation would make sense if we made a solution only with the buffer and an arbitrary quantity of insoluble salt.

Your work looks good to me. (I just followed your logic and didn't check the math.) I also don't understand the equation $$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$

@MaxW Thanks for checking out! If we consider a certain case, the equation can be used to work out the apparent solubility: in a flask containing a buffer solution of arbitrary $pH$, we add an arbitrary quantity of $\ce{Ag2C2O4}$.
We get the equilibria as seen in the post and that relation represents a simple conservation law.
The equation becomes:
$$\frac{1}{2}[\ce{Ag+}]=\frac{1}{2}\sqrt{\frac{K_{s}}{[\ce{C2O4^{2-}}]}}=\frac{[\ce{C2O4^{2-}}][\ce{H+}]^2}{K_{a1}K_{a2}}+\frac{[\ce{C2O4^{2-}}][\ce{H+}]}{K_{a2}}+[\ce{C2O4^{2-}}]$$
Which is solvable and gives us the apparent solubility. :D

Ah.. yes since there are two silver ions, the molar solubility of the salt would be half the silver ion concentration. So $$S(\ce{Ag2C2O4})=2.53\cdot 10^{-6} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$

Oups, my solution forgot to take that into account. Corrected.
And as a matter of fact, considering their approach which applies to the case I presented, we get: $$S=6.84\cdot10^{-5} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$
Close, but not quite.


text:
Why are there 3 columns in Group VIII of the Periodic Table?
I'm currently reading in-depth about the layout of the Periodic Table, and I wondered why the table has 3 columns in its Group VIII:

As I understand, this is an old notation of the table, now deprecated. But why was Group VIII a triple-sized group?
The answer might be trivial, but it's interesting.

Note that this is an old naming system; the current IUPAC names for the groups are simply 1 through 18.

The reason for having them all in one group was, that they are very similar in each of the three periods, much more than to their neighbours on the left and right. Co is even heavier than Ni, so the order was somewhat unclear in the beginning.

Does anybody know what the bottom number means?

@gsurfer04 The number written below each element is the mass number of that element.

The bottom number, below the electron structure.

@gsurfer04 You’re generally more likely to get an answer if you post that kind of query as its own question. Nevertheless (a year and a half late): those numbers appear to be ionisation potentials, expressed in eV.

A bit late, but @gsurfer04 the bottom number would appear to be ionisation energy (from https://www.nist.gov/system/files/documents/2019/07/10/nist_periodictable_june_2019-crop.pdf )

In the beginning of the $20$th century, the periodic table had $8$ columns, and not $18$ as today. Electrons and of course $spdf$ electrons were unknown. The numerous transition elements were included in this table by doubling the line between Argon and Krypton, then between Krypton and Xenon. It was going this way.
The $4$th period was made of two lines. So each square (each box) contained two elements written on two lines. The first seven atoms $\ce{_{19}K - _{25}Mn}$ occupied the columns $\ce{I - VII}$, and their oxidation state may always have been equal to the number of the column. Then the three atoms $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ did not belong to any column. We will come back to them later on. Then the seven next atoms $\ce{_{29}Cu - _{35}Br}$ had to be put in columns $1$ to $\ce{VII}$, under the corresponding elements : $\ce{_{29}Cu}$ in the same box and under $\ce{_{19}K}$, $\ce{_{30}Zn}$ in the same box and under $\ce{_{20}Ca}$, etc. up to $\ce{_{25}Mn}$ above $\ce{_{35}Br}$. With this system all atoms had at least one oxidation state equal to the number of their column. The same procedure was adopted for $5$th period, with $\ce{_{37}Rb}$ and $\ce{_{47}Ag}$ in the first column,$\ce{_{38}Sr}$ above $\ce{_{48}Cd}$, etc.
This table was useful, except nobody knew what to do with the three atoms  $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ having all valence $2$, and of course of the six corresponding atoms of the platinum group, immediately under them : $\ce{_{44}Ru}$ to $\ce{_{46}Pd}$ and $\ce{_{76}Os}$ to $\ce{_{78}Pt}$. There was plenty of discussion at that time about what to do with them. Usually these elements were "rejected" in the 8th column, with the noble gases. or they were put in a separate column made specially for them. This problem was finally solved with the discovery of the $spdf$ electrons.


text:
Intermolecular Forces in Teflon vs Polyethylene
From Wikipedia, polyethylene has a melting point of around $400K$, while Teflon (polytetrafluoroethylene) has a melting point of $600 K$, which is much higher. Besides the increased London Dispersion Forces caused by the larger fluorine atoms, could the stronger bonding in Telfon be explained by Teflon molecules being polar?
The $\ce{C-F}$ bond is strongly polar, so there is definitely a dipole there. However, whether a molecule is polar or not depends on whether or not the individual dipole vectors cancel. The flexibility of $\ce{C-C}$ single bonds allow for the polymer to behave like any other saturated alkane: free rotation to produce a molecule that more resembles a noodle than a pencil.
When we imagine an individual molecule of Telfon without Van Der Waal forces, we see that it is statistically the case that the molecule is twisted and is thus polar, as the vectors don't cancel. However, when we take into account these dipole-dipole attractions (and the London Dispersion Forces) between molecules, the molecules line up and straighten (for it is more energetically favorable). However, once they straighten up, line up, and pack, the dipole-forces/polarity are gone. Thus, do dipole-dipole forces exist and play a role in the stronger bonding in (polytetrafluoroethylene) as compared to polyethylene?

PTFE chains form a helix in the crystalline state, afaik. So i guess there might be some polar interaction inside each strand. But I'd chalk up most of the melting point difference to teflon being just much heavier. Btw.: The perfluorated C-C bond is rather stiff, compared to a regular polyethylene backbone.

Teflon has LOWER intermolecular forces. Teflon has a super-low self friction and gekos hate it, perfluorohexane boils at a lower temperature than hexane, etc.

There are many, many things that influence melting points and glass transition points other than van der Waals forces or bond polarity. Do you even know the typical molecular size of the polymer chains? That might matter.


text:
Error in formula for the solution to a chemical problem
The problem in my textbook goes like this:

Burning $0.68\ \mathrm{g}$ of an unknown substance, we obtained $1.28\ \mathrm{g}$ of sulfur oxide (IV) and $0.36\ \mathrm{g}$ of water. Find the chemical formula of the burned substance, providing that it was a complete combustion.

It's quite easy to see that $0.2\ \mathrm{mol}$ of both $\ce{SO2}$ and $\ce{H2O}$ have been produced, there are $0.04\ \mathrm{mol}$ of hydrogen in the water and $0.02\ \mathrm{mol}$ of sulfur in the oxide. Calculating the masses and summing them up, we get $0.68\ \mathrm{g}$, meaning the original substance contained only sulfur and hydrogen, and it's $\ce{H2S}$.
However, the solution volume offers this equation as part of the solution process:
$$\ce{S_{$x$}H_{$y$}O_{$z$} + \frac{x + y}{4} O2 -> $x$ SO2 + \frac{$y$}{$z$} H2O}$$
The volume is full of typos (apparently a plus sign missing here), but still there must be some explanation for the coefficients which I don't get. Why do the coefficients for $\ce{O2}$ and $\ce{H2O}$ equal $(x+y)/4$ and $y/z$? If there's zero oxygen in the first term, won't we get division by zero?
How do we come up with coefficients of this sort? Maybe these particular coefficients are erroneous? (the solutions volume is quite low in quality)

It is supposed to be $y/2$ not $y/z$. That arises from balancing hydrogen. Another typo - perhaps it was a scan and something went wrong during the character recognition.

Please use \ce{...} in the maths environment for chemical formulae. Upside: It automatically does things such as subscript numbers in formulae: $\ce{H2O}$ automatically gives $\ce{H2O}$

The equation is a little dubious, as I mentioned in the comments.

How do we come up with coefficients of this sort?

Let's go from scratch. We don't know the coefficients at this point so we'll just call them $a$, $b$, and $c$. Note that the coefficient of $\ce{S$_x$H$_y$O$_z$}$ is taken to be 1. It doesn't have to be 1; however, for the sake of simplicity you might as well take it to be 1. If you take it to be 2, then you just have to double all the other coefficients.
$$\ce{S$_x$H$_y$O$_z$ + $a\,$O2 -> $b\,$SO2 + $c\,$H2O}$$
In the reactants, we have $x$ sulfur, $y$ hydrogen, and $z + 2a$ oxygen. In the products, we have $b$ sulfur, $2c$ hydrogen, and $2b + c$ oxygen.
Our aim is to express $a$, $b$, and $c$ in terms of $x$, $y$, and $z$. Equating all three, we get
$$\begin{align}
x &= b \\
y &= 2c \\
z + 2a &= 2b + c
\end{align}$$
which immediately tells you that $b$ and $c$ are equal to $x$ and $y/2$ respectively. Substituting these into the last equation,
$$\begin{align}
z + 2a &= 2x + \frac{y}{2} \\
a &= \frac{4x + y - 2z}{4}
\end{align}$$
So the full equation is:
$$\ce{S$_x$H$_y$O$_z$ + $\frac{4x + y - 2z}{4}$O2 -> $x\,$SO2 + $\frac{y}{2}\,$H2O}$$

...and I'd far prefer to write the last equation as $\ce{S$_x$H$_y$O$_z$ + $(x + \frac14 y - \frac12 z)$O2 -> $x,$SO2 + $\frac12 y,$H2O}$, to get rid of the confusing cancellation in the $\ce{O2}$ coefficient.

They are very erroneous. No oxygen in the reactant would give you a division by zero as you state, so the last fraction is definitely wrong. Here’s how I would proceed:

We know that there is only one source of sulphur which fully reacts to $\ce{SO2}$; so the amount of sulphur atoms must be equal to $\ce{SO2}$ hence the $x$ before $\ce{SO2}$ is correct.
For every two hydrogens a water molecule is formed, but the amount of oxygen present in the original compound is irrelevant. Hence that fraction should be $\frac{y}{2}$, not $\frac{y}{z}$.
That leaves us the oxygen. Well, it is easy to see that we need one oxygen atom for two hydrogens and two oxygen atoms for a sulphur atom. What about oxygens originally present in the sample? They need to be taken away (subtracted) from the oxygen that we add in, leaving:
$$2x + \frac{y}{2} - z\ \text{oxygen atoms}\\
\frac{2x + \frac{y}{2} - z}{2}\ \text{oxygen molecules } (\ce{O2})\\
\frac{2x + \frac{y}{2} - z}{2}= \frac{2(2x + \frac{y}{2} - z)}{4} = \frac{4x + y - 2z}{4}$$
Thankfully, this is in full agreement with Orthocresol’s result.

Thank you, Jan! It took me awhile to understand the "They need to be removed from the oxygen that we add in" bit, but once I got it, it all clicked together.


text:
What would the phosphorus-NMR of trifluorodiphenylphosphorane look like?
I'm fairly new to NMR especially $\ce{^{19}F}$ and $\ce{^{31}P}$ NMR, we've not been formally taught NMR yet but we're expected to know $\ce{^{19}F}/\ce{^{31}P}$ NMR for our exam for basic molecules.
I know $\ce{PPh2F3}$ will have a trigonal bipyramidal geometry, I know in $\ce{PPh2F3}$ the phenyl groups will be equatorial due to the torsional angle and sterics. This leaves two fluorines axial and one fluorine equatorial, but  what I don't understand is why this gives a doublet of triplets signal (for $\ce{^{31}P}$). Any explanation to why you get this pattern would be helpful, taking into account my fundamental understanding of $\ce{^{31}P}$ NMR is not so great.
also what would the $\ce{^{19}F}$ NMR be for the same molecule?

Coupling to 1 equatorial fluorine gives 1+1 = 2 peaks 2) Coupling to 2 axial fluorines means that both of those two peaks is split into 2+1 = 3 peaks. Hence two triplets.

Technically it should be called a "doublet of triplets", not two triplets, because two triplets would imply that there are two different phosphorus environments (obviously not true since there's only one phosphorus). As for the 19F NMR, it's a bit more fiddly, but you just need to work through it slowly. There is F-P coupling across one bond as well as F-F coupling across two bonds, but the two axial fluorines, which are equivalent, do not couple to each other.

This earlier answer for an analogous compound may be helpful.

The thing is that the ax. and eq. fluorines are not equivalent, so each of them will have their own peak, and both of these peaks will be split accordingly.

Well, these are the fundamentals of NMR that are important:

Signals will be split if they couple to neighbouring atoms with a spin $\ne 0$ (though coupling to spin ½ nuclei is most common especially in organic NMR).
Every different environment of a coupling nucleus will produce a distinct coupling.
Depending on the number of nuclei in a given environment, the number of signals (coupling pattern) is given by the equation $2nI+ 1$, where $n$ is the number of nuclei and $I$ is that nucleus’ spin.

In your example you have correctly identified two different fluorine environments with one and two fluorines in them respectively.
The single equatorial fluorine will give a doublet because $2nI +1 = 2\ (n = 1, I = \frac{1}{2})$.
For the same reason, the axial fluorines will give a triplet: $2nI + 1 = 3\ (n =2, I = \frac{1}{2})$.
If one coupling gives a triplet and the other a doublet, the final result is either a doublet of triplets or a triplet of doublets. (The larger coupling constant i.e. the greater splitting is given first.)

The fluorine NMR would be more complex: They can couple to both phosphorus and the other fluorine. It is important to know that the axial fluorines are homotopic, i.e. the molecule can be rotated exactly exchanging their places — this means that the two give an identical signal. The equatorial one cannot be brought to overlap with the axial fluorines by any mean of symmetry; the two environments are heterotopic, i.e. they will produce different signals (and typically couple).

The axial fluorines will couple to the equatorial fluorine and phosphorus.
The equatorial fluorine will couple to the axial ones and phosphorus.

I’ll leave you to work out the coupling patterns. (Hint: The less bonds between two coupling nuclei, the larger the coupling constant usually is.)

A final note: It is due to electronic reasons that the phenyl groups occupy the equatorial positions rather than going axial. A four-electron three-centre bond is more stable for electronegative elements than electropositive ones.

This is very helpful. Okay so if the axial fluorines couple to the equatorial fluorines you will get a triplet (2+1) and then couple to the phosphorus will split this into two triplets? The equatorial fluorine will couple to the axial fluorines and give a doublet (1+1)  and which will then subsequently couple with phosphorus and split into two doublets. Overall giving a doublet of triplets and doublet of doublets?

@Petrichorr Yes, a doublet of doublets (the axial fluorines) and a doublet of triplets (the equatorial one). If the ppm scale works like it does for carbon and hydrogen, then the doublet of doublets should be shifted downfield when compared to the doublet of triplets.

What do you mean by shifted downfield? That the peaks will be of greater intensity?

No, have a lower ppm value @Petri ;)


text:
Why are silyl chlorides more readily hydrolysed than alkyl chlorides?
I know that the $\ce{Si-Cl}$ bond has a slightly higher energy, but silyl chlorides are much more readily hydrolyzed compared to alkyl chlorides. I do not fully understand why that is. My thoughts so far are:

There is a better match in energy between the orbitals of the $\ce{Si-Cl}$ bond and water.
Reactions proceed when there is a favorable interaction between HOMO and LUMO.
In this case, the HOMO is the lone pair on the water molecule. The LUMO of a $\ce{Si-Cl}$ bond is closer to it in energy compared to a $\ce{C-Cl}$ bond.

Is this the right argument to take?

Using $\ce{CCl4}$ and $\ce{SiCl4}$ as the basis for comparison, there are several reasons.

The greater electronegativity difference between Si and Cl leads to a larger $\delta^+$ on silicon, making it more electrophilic and susceptible to attack by water.
Larger size of Si means that there is less steric hindrance towards the approach of water.
It is possible that the hydrolysis of $\ce{SiCl4}$ proceeds via an alternative, faster pathway that involves a five-coordinate intermediate (as opposed to $\ce{CCl4}$, in which five-coordinate carbon does not occur).


text:
On the oxyacids of +III pnictogens and +IV chalcogens
Phosphorus in its $\mathrm{+III}$ oxidation state is known to exist as the phosphonic acid tautomer rather than phosphorous acid one. All salts isolated containing $\ce{H2PO3-}$ are (to the best of my knowledge) hydrogen phosphonates rather than dihydrogen phosphites. Similarly, all $\ce{HPO3^2-}$ salts should be phophonates rather than hydrogen phosphate. Only as organic esters (which cannot tautomerise easily as carbocations are less stable) are both forms observed: trimethyl phosphite and dimethyl methylphosphonate both exist.
A similar picture is true for sulfur in the $\mathrm{+IV}$ state. Sulfurous acid and sulphonic acid are both known as far as I know, and both structures have been isolated in salts. Similarly, methyl methylsulfonate and dimethyl sulfite are both known to the best of my knowledge.
The corresponding elements of the second period do not show this tautomerism. For oxygen, this is due to its high negativity preventing the formation of ‘oxyacids’ altogether. For nitrogen, this is because an $\ce{N=O}$ double bond is more stable than two $\ce{N-O}$ single bonds so that nitrite ($\ce{NO2-}$) is already devoid of protons and cannot tautomerise. However, remnants of this behaviour still exists in organic compounds where both the nitro group $\ce{-NO2}$ and organic nitrites $\ce{-ONO}$ are known.
What is the case for the heavier homologues of both phosphorus and sulfur? Do the elements arsenic/antimony/bismuth and selenium/tellurium/polonium exhibit both an -onic acid and an -ous acid or is one form preferred over the other? Is there a general trend going down the groups or are secondary effects (e.g. lanthanide contraction) more important?
Answers backed with references to crystal structures of anions (or the free acids) and/or corresponding quantum-mechanical calculations (both in references or self-performed) are preferred.
Note that the fully oxidised pnictogen(V) or chalcogen(VI) compounds are not part of this question,

Disclaimer: This isn't really the answer Jan is looking for.

Holleman, Wiberg "Lehrbuch der Anorganischen Chemie", de Gruyter, Vol. 101, notes that:
Selenium
Selenium forms a weaker, but more stable acid based on oxidation state $ +4$. No explicit mention of the structure, which suggests to me that $\ce{SeO(OH)2}$ is the structure.
Tellurium
Tellurous acid's structure is described as unknown.
Arsenic
For arsenous acid (the $+3$ oxidation state based acid) the tautomeric equilibrium is entirely on the side of $\ce{As(OH)3}$ (as opposed to $\ce{HAsO(OH)2}$). The reason given is that $\ce{As}$ does not form double bonds with $\ce{O}$, which is outdated: Current thinking is that only the first row forms double bonds, i.e. no double bonds in $\ce{H2SO4}$ or $\ce{HPO(OH)2}$. Organic esters exist that can be described as $\ce{RAsO(OH)2}$.
Antimony
The $+3$ oxidation state based acid is described as forming $\ce{H+ + [Sb(OH)4]-}$ in water. No mention of tautomerism.
The book gives sources only sparingly and on a per-section basis. Relevant items appear to be:

R. Paetzold: "Neuere Untersuchungen an Selen-Sauerstoff-Verbindungen", Fortschr. Chem. Forsch. 5 (1966) 590-630.
W.A. Dutton, W.Ch. Cooper: "The Oxides and Oxyacids of Tellurium", Chem. Rev. 66 (1966) 657-675.
M.A. Ansari, J.M. McConnachie: "Tellurometalates", Acc. Chem. Res. 26 (1993) 574-578.
J.D. Smith: "Arsenic, Antimony and Bismuth", Comprehensive Inorg. Chem. 2 (1973) 547-683.
C.A. McAuliffe: "Arsenic, Antimony and Bismuth", Comprehensive Coord. Chem. 3 (1987) 237-298.
GMELIN: "Arsenic", Syst.-Nr. 17, up to now 1 book, ULLMANN, Vol. 5: "Arsenic and Arsenic Compounds", A3 (1985) 113-141.
GMELIN: "Antimony", Syst.-Nr. 18, up to now 6 books, ULLMANN, Vol. 5: "Antimony and Antimony Compounds", A3 (1985) 55-76.

My thoughts on $\ce{As, Sb}$ are that in the MO picture, $\ce{H}$ has insufficient overlap to form a bond that would compete with the alternatives.

I have performed DFT calculations (PW6B95-D3/def2-QZVP//PBE-D3/def2-TZVP) on gas-phase $\ce{Pn(OH)3}$ and $\ce{HPnO(OH)2}$. Of course, these are gas-phase  energies on single conformers, not free enthalpies. For $\ce{Pn} = \ce{N}$, a 10 kcal/mol favorable energy difference towards $\ce{HPO(OH)2}$ was found. For $\ce{As}$, about 35 kcal/mol towards $\ce{As(OH)3}$ was found. The Mayer bond order analysis was not conclusive.


text:
Why is MnSO4 colourless?
I get it that transition metal salts have colour due to d-d transition from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$ orbitals in octahedral and the opposite in tetrahedral but why would $\ce{Mn^2+}$ not have any characteristic colour? Is it so because the orbitals are degenerate in $\ce{MnSO4}$? or because d-d transition only takes place when there is completely empty d orbital?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by using MathJax for chemical formulae. For more information on how to do so yourself, check out the [help], this meta-post or this one. And there are even more hints available.

I'm already a member at this site. Thank you @Jan

@Jan Because this email is unregistered. I am Quark.

It does have a colour, but it is very faint. It has to do with the selection rules governing d-d transitions in a metal complex: http://wwwchem.uwimona.edu.jm/courses/selrules.html in particular the $\Delta S = 0$ rule.

@orthocresol Could you elaborate the selection rules here in a simple language for me to be able to understand?

@orthocresol Oh I got the reason. So there must be no electron with same spin in that orbital, to which the electron is being transited to? Right? Because if it were present it should have opposite spin which will violate Hund's rule?

Manganese(II) sulphate is not colourless. It is a faint pink. Compare this image taken from Wikipedia, where a full list of authors is available:

However, it is true that the colour is rather faint, and you need a white background to properly see it. That’s because manganese(II) is a $\mathrm{d^5}$ system and a high-spin one of those meaning that all five d-orbitals are occupied by a single electron:

Exciting an electron directly is forbidden due to the spin-retaining rule: The excited electron must occupy an already populated orbital, which is only possible if it flips its spin, which is forbidden by selection rules. Special mechanisms are needed to account for the (obviously possible but faint) transition here; the page Orthocresol linked in the comments names spin-orbit coupling.

So basically, the orbitals should be empty for transition to take place,right?

@user29557 Not necessarily. For example if you added one more spin down electron to the lower level, then that electron could be promoted to the upper level, as long as it retains its spin state in the upper level (no change in spin).

@orthocresol Oh. So, if we had 8 electrons, there would be a characteristic colour, right?

@user29557 Indeed - so an example of an 8-electron ion would be $\ce{Ni^2+}$, and $\ce{NiSO4}$ has what I personally think is a very pretty colour (there's a picture on Wikipedia).

@user29557 $\ce{Ni^2+}$ is a typical $\mathrm{d^8}$ system, so yeah.

What is spin orbit coupling?

@user29557 Frankly, I have no clue. ^^' Could be the basis of a good question …

@user29557 Quantum mechanical coupling of two angular momenta (orbital angular momentum $L$ as well as spin $S$) to form one larger angular momentum. Don't worry about it.


text:
How to know which kind of distortion takes place due to Jahn-Teller effect
I have gone through a few videos and a few articles (including Wikipedia's) about Jahn-Teller distortion effect and I'm not able to follow if there is a rule which governs which kind of distortion will take place in which compound. If I want to remember which compound has which distortion, should I remember a rule or memorise the whole list?

The Jahn-Teller theorem does not predict the exact form of the distortion that will take place - it only states that such systems are unstable with respect to distortion. Honestly I would say the only one worth memorising is $\ce{Cu^2+}$, which is of course an axial elongation.

I recently came across a question which asked the electronic configuration for a chromium compound and asked where the electron would be in dz^2 or dx^2-y^2 orbital. And I wasn't able to answer. Isn't this related to  Jahn Teller?

If it is Cr(II) then probably yes. I don't fully understand that question though :/

If you read and watched videos, you probably already know that the Jahn-Teller effect causes stabilisation in the case of unevenly populated orbitals. Reducing the symmetry of the system renders the orbitals no longer symmetrically equivalent thereby creating an energy difference used to the compound’s advantage.
Let’s separate two cases: 1) octahedral complexes with $\mathrm{d^4}$ and $\mathrm{d^9}$ configuration, in which the $\mathrm{e_g}$ orbitals are unevenly populated; and 2) all the other cases.
Uneven $\mathbf{e_g}$ population
In an octahedral complex with four or nine d-electrons, there will be either one or three within the $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). It is unfavourable for orbitals of the same energy (as symmetrically equivalent orbitals have) to have different populations so the system distorts itself. The distortion will always be along a single axis and per defintion this axis is always the $z$ axis. Thereby, ligands on the $z$-axis (negative charges in the most simple crystal field model) move away from the metal centre stabilising any orbital with a contribution in $z$ direction and most significantly the $\mathrm{d}_{z^2}$ one.
The effect is greatest in this case because the $\mathrm{e_g}$ orbitals actually point towards the ligands thus rendering them most susceptible to distance effects.
All other cases
In all other cases, the effect is much weaker. This is mainly because no other orbitals are known to point directly at the ligands; even for tetrahedral complexes the $\mathrm{t_2}$ orbitals point in-between them. Therefore, moving a ligand away slightly has much less effect on the electronics of the system.
What exactly happens really depends on the system itself and how it can be stabilised. If you are speaking of an octahedral complex, it will usually distort itself to $D_{\mathrm{4h}}$, i.e. equivalently to the former case because that is the distortion that keeps the highest possible symmetry. In that case again, one would say that the ligands in $z$ direction are moved away. Otherwise you need to figure it out by drawing, thinking and guessing or by studying crystal structures.

Tl;dr
Instead of memorising a list remember the first case and $z$-distortion. Ignore all the other cases for being weak.

Very lucid answer @Jan but for questions like this- jahn teller effect is not observed in high spin complexes of : a) d^4 b)d^9 C)d^7 D) d^8 what strategy should be used? As you've written here options a and b can be ruled out but how to decide between C and D.is there any other way to find whether jahn teller distortion happens when provided with specific orbitals like these along with their spins?

@Harini Draw out the hypothetical octahedral state. Only in the case of $\mathrm{d^8}$ can you achieve fully evenly populated orbitals. All other cases require degenerate orbitals of uneven population: $\mathrm{d^4}$ and $\mathrm{d^9}$ have $\mathrm{e_g}^1$ and $\mathrm{e_g}^3$, respectively; $\mathrm{d^7}$ has $\mathrm{t_{2g}}^5$. This last case is contained in all other cases of my answer.


text:
Calculating ppm of an ion in a solution
Calculate the ppm of $\ce{Fe^2+(aq)}$ if $\pu{0.0055 g}$ of $\ce{Fe(NO3)2}$ is dissolved in $\pu{2.0 kg}$ of water.
I have tried converting g of iron nitrate into moles, using mole ratio between iron nitrate and iron(II) ions and then converting to grams of iron(II) ions. I then took these grams and divided by grams of solution ($0.0005 + 20000$).

You are wrong. Assuming that you are doing ppm in terms of mass: There are $0.0055$ grams of $\ce{Fe(NO3)2}$. Its molar mass is $179.8548$ grams/mol. Thus, you have $3.058022\!\times\! 10^{-5}$ moles. Multiplying this by the molar mass of iron gives you $0.00170775$ grams of $\ce{Fe}$. Thus, you have $0.0037922$ grams of $\ce{NO3}$. The concentration of iron is
$$\mathrm c = \ce{\frac{Fe}{Water+Fe(NO3)2}} = \frac{0.00170775}{0.0055+2000}= 8.538726518 \times10^{-7}$$
Thus, $\ce{Fe}$ has a $\pu{ppm}$ of $0.85$

It is not well received here to answer questions that should be closed for being homework. Although admittedly somehow this one wasn’t closed …

@Jan It showed effort. That is why I answered it. And I suppose that is the reason why it wasn't closed.

You can use $\ce{} for chemical formulae, it automatically puts the symbols upright for you. For example: $\ce{NO3-}$ --> $\ce{NO3-}$

not really …$%MathJax comments hurray! =D$

You have way too many significant figures and the ppm of iron should be $\frac{Fe}{Water+\ce{Fe(NO3)2}}$ but the mass of the iron nitrate is really insignificant.

@MaxW You are right, I didn't consider sig figs and I should have included Iron as well.