woelen

Your test covers all the compounds mentioned above, but in general I would say many, or most, not all. If a compound does not dissolve in xylene, nor in water, then the test is not conclusive. In that case, the compound can be either ionic, covalent or something in between. Covalent example: Polymeric compounds (e.g. plastics). Ionic example: CaF2.

You cannot easily say, which element forms less compounds. How would you count this? Most organics happen to be compounds, both of H and C. What I can state, however, is that the chemistry of C is even more rich than the chemistry of H. C has 4 electrons, available for bonding, while H only has one. In organic compounds, the C atom can be present in many different bondingschemes, while the H-atoms basically are always bound in one of few ways (most notably a single bond to other element).

AgF --> very soluble AgF2 --> decomposes water, formation of O2 AgCl --> white like snow AgBr --> very pale yellow AgI --> pale yellow, but somewhat darker than AgBr All these compounds change color, when exposed to light. AgCl turns light purple on exposure to light, AgBr and AgI turn greyish/yellow on exposure.

Why all these difficult things? If you have a mathematical description of your spiral (x, y, z) as function of running parameter t, then integrate along the path the length of the curve. This may require numerical methods of integration, but fortunately, integration is very easy and stable numerically. If you do not have a mathematical description, then determine many points (x, y) (or (x, y, z) is it is 3D) and estimate the length of the spiral, simply by taking the length of all small line segments. Better approximations are obtained by interpolating quadratically or cubically between the data points and integrating that.

All confusion is just a matter of language . Of course, hydrogen can form compounds with practically every element and in endless varieties. The only constant thing is that it always has a single bond, or is ionic as hydride. But, if you first only knew of organic compounds, H2S, H2O and NH3 and acids containing hydrogen, then of course one can find it striking to see that hydrogen has such a richer chemistry than you first knew of. If hydrogen bonds as H(+) ion, then it hardly exists as naked H(+) ion, it almost always is coordinated to something else. In the strong acids like HNO3, HCl and H2SO4 is is bonded covalently. When these acids ionize (e.g. in water, or liquid ammonia), then H(+) is coordinated to H2O or NH3: H3O(+) and NH4(+). The only compound, which I know of, and which has a naked proton, a real H(+) ion, is a recently discovered chloro-borane-based super acid. This is the strongest acid known, yet hardly corrosive to most materials.

Basically I agree with your net reaction, but the underlying mechanism is MUCH more complex. I do not think that if you add carefully stoichiometrically computed amounts of NaOH and CuCl2 to each other, that you end up with a precipitate of Cu(OH)2 and a solution of NaCl, but that you end up with a berthollide compound CuOH/Cl and still NaOH in solution. This experiment does not have anything to do with stability of NaCl or Na2SO4. These compounds simply do not exist in aqueous solution, you only have ions Na(+) and ions Cl(-). The Na(+) ions just are spectator ions, but the Cl(-) ions are not. They do very complicated things in the reaction. Only when there is quite some excess NaOH, the chloride-content of the precipitate will become very low and one can speak of Cu(OH)2.

No, the chloride (not chlorine) certainly plays an important role. It cannot be cancelled out with sodium ion. Chloride forms complexes with copper. Just try dissolving some CuSO4 in plain water and dissolve in a solution of NaCl (fairly concentrated NaCl). You'll definitely see a difference! Next, when you add NaOH, then a complicated precipitate is formed, which can be regarded as hydrated copper hydroxide/chloride. This is not the same as a mix of copper hydroxide and copper chloride, but it is a so-called berthollide compound. Berthollide --> indefinite stoichiometry, no precise formula can be given, only a formula like I give above with an uncertain parameter x. Daltonide --> definite stoichiometry (e.g. CuCl2, CuSO4, many other compounds). The formation of berthollide compounds is an annoying thing in chemistry, but unfortunately it is more common than purely daltonide compounds.

Cuprate ions are only formed at VERY high concentration of NaOH, so I do not think it is that. I'll try the experiment as well. It might be that a basic chloride is precipitated and that that gives rise to the special colors observed. If more NaOH is added, an ion exchange could occur (chloride in the precipitate, replaced by hydroxide ions). More to be said on this, when the experiment is done . EDIT: I did the experiment as follows: Take a solution of NaOH and add solution of CuCl2, while stirring/shaking after each few drops. At first, a nice bright blue precipitate is obtained, fairly dark. At a certain point, however, the color becomes lighter. The precipitate becomes lighter green/blue. Then, I added NaOH solution again, while shaking all the time. This results in darkening and shifting from green/blue to bright blue again, until a certain point is reached. From that point on, the color does not shift anymore. I explain this as follows: Initially, there is NaOH in excess, and addition of the CuCl2-solution causes formation of bright blue Cu(OH)2. At a certain point, when all NaOH is used up, then addition of more CuCl2 causes formation of basic copper chloride, something like CuCl(OH), or more general [math]CuCl_{2-x}(OH)_x[/math]. When lateron, more NaOH is added again, then the basic copper chloride looses its Cl(-) ions again and Cu(OH)2 is formed again. --------------------------------------------------------------- Now, to you observations. I think that your liquid was very uneven, with unused NaOH below and unused CuCl2 at the top. At the top, there is excess CuCl2 and then you see the basic copper chloride. But, when you shake, then the unused NaOH reacts with the excess and basic copper chloride, forming Cu(OH)2.

The amount of bromide and bromate are chosen, such that they cancel each other precisely: 5Br(-) + BrO3(-) + 6H(+) ---> 3Br2 + 3H2O This reaction is not instantaneously, it takes some time to react all of the reactants. The amount of acid added is in excess. Not all acid will be reacted. After the reaction, 66.7% of the acid remains. If the activation energy rises, then the reaction can be expected to run at a lower rate, but the total amount of energy produced (or consumed) in the reaction will not be affected. The role of the phenol is not clear to me. It might be that it is used as an indicator, which tells something about the concentration of the bromine. Phenol forms many highly colored species when it is oxidized, but someone with more knowledge about organical chemistry may give a more complete answer on this.

Yes, indeed, the highschool chemistry is oversimplified. That is not a problem (otherwise we would not be able to learn the subject), but what I think is a problem is that no one tells you that things are oversimplified. Many people think that the things they learn at school are an accurate description of reality. So, pupils should be made more aware of the implifications. ----------------------------------------------------------------- The very dark brown stuff is a mixed oxidation state complex of copper ions and chloride ions. It contains Cu(+), Cu(2+) and Cl(-) ions, in such a way, that the copper has an oxidation state somewhere between +1 and +2. Yes, that also is possible! Atoms can have non-integer oxidation states. Probably, the brown stuff can be described as Cl-Cu-Cl-Cu-Cl, with the copper ions having (average) oxidation state +1.5 and chlorine having oxidation state -1. If you add some oxidizer (more FeCl3, but dilute H2O2 also works), then all copper is oxidized to oxidation state +2 and the deep brown/black color disappears again.

Could you please be more specific in your description. If you want an adequate answer, then your description must be precise. 1) Give a precise description of the procedure. Give the order in which chemicals are added. Also give an indication of the amounts added. 2) State precisely which color is observed during which stage of the experiment. Also try to describe the colors precisely. There are many shades of green, like bluish green, bright green like grass, but also muddy green/brown colors. There are other threads on this subject on SFN, which are of very recent date. Just look a little further in the existing threads of the chemistry section. Especially this topic may be very helpful: http://www.scienceforums.net/forums/showthread.php?t=17423 But it is also a good idea to read this one: http://www.scienceforums.net/forums/showthread.php?t=15956 Next time: UTFSE

What you basically are talking about is hydrolysis and this effect is very common. Two examples: Na2S is a basic salt. When it is dissolved in water, the sulfide ions, S(2-), combine with H(+) from dissociated water to form HS(-) and the OH(-) remains in solution. A 1 M solution of Na2S is very alkaline, it feels slippery, almost as slippery as a 1M solution of NaOH. An other example is iron (III) chloride, FeCl3, which is very acidic, when dissolved in water. Fe(3+) ions tend to attract OH(-) ions, forming ions like [Fe(OH)](2+), leaving the H(+) ions in solution. A solution of FeCl3, hence, is quite acidic. [in reality things are quite more complex with hydrolysis of Fe(3+), but for understanding the concept, this answer should be sufficient].

No, I cannot agree with this explanation, because then with CuSO4 this effect would even be stronger. The sulfate ion is derived from the acid ion HSO4(-), bisulfate, which is much weaker than nitric acid. If Cu(HSO4)2 would result in copper-plating and CuSO4 would not, then I could agree, but CuSO4 does do the copper plating very well. I myself am thinking in the direction of oxidative passivation of iron. It is known that iron (and also many other metals like magnesium and aluminium) can be made more corrosion resistant, when treated with an oxidizing solution (e.g. potassium dichromate or dilute HNO3). This effect is used for making nails less susceptible to corrosion and it also is used in pyrotechnics, to make magnesium-containing compositions more stable on storage. For this purpose, the metal is treated with K2Cr2O7 for a while and then cleaned again and then used. It might be that the nitrate from the Cu(NO3)2 has something to do with this, but I only heard the bells ringing, I have no sound explanation for this.

A solution of FeCl3 dissolves copper metal fairly quickly. It is used as PCB etchant in electronics and the hydrated form, FeCl3.6H2O can be purchased at many electronics parts and hobby shops. FeCl3 can act as an etchant, because of two reasons: Fe(3+) is a fairly strong oxidizer. Cl(-) is a strong complexing agent for copper (II) ions. The reaction in solution often is written as 2Fe(3+) + Cu --> 2Fe(2+) + Cu(2+) This, however, is a too strong simplification. If you put copper in a solution of iron (III) sulfate then it does not dissolve. The chloride plays an essential role: 2Fe(3+) + Cu + 4Cl(-) --> 2Fe(2+) + CuCl4(2-) In reality even this is simplified, because iron also is present as complex with chloride. So, a slightly better description of the reaction may be 2FeCl4(-) + Cu --> 2Fe(2+) + CuCl4(2-) + 4Cl(-) When all iron (III) is converted to iron ()II), then the CuCl4(2-) ion in turn also oxidizes copper metal quite well, where copper (I) species are formed: CuCl4(2-) + Cu --> 2CuCl2(-) These copper (I) species form very dark-colored mixed oxidation-state species with copper (II) species, when all iron (III) is used up. This causes the PCB-etchant liquid to turn very dark, when it is near exhausted. You cannot simply say that copper is more noble than iron and so the reaction cannot proceed. This only is true for Fe(2+) and Cu. Fe(3+) is a more oxidizing ionic form of iron. With chloride ions present, the nobleness of copper is reduced, due to complex formation, as described above, and that effect makes Fe(3+) a sufficiently strong oxidizer to dissolve copper. -------------------------------------------------------------------------- Try the following to make your FeCl3 work: Use a very concentrated solution, or add a lot of NaCl also. The chloride concentration needs to be really high. The complex formation is one of the main driving forces behind the reaction and that complex only is formed at very high concentration of Cl(-). If your solution is very turbid, then you probably have impure FeCl3 with a lot of basic Fe(3+) compounds, e.g. FeCl2(OH), FeCl(OH)2 and Fe(OH)3 and all kinds of hydrated forms of that. If this is the case, then add a few drops of HCl, until the solution is clear.

Of course you have my permission . I think it is good to show to the pupils that seemingly simple things in chemistry are not always that simple. When I myself was in high school (more than 20 years ago), I though that the information in schoolbooks was an accurate description of reality. Lateron I learnt that this only is a strongly simplified approximation.

YT, is your Cu(NO3)2.3H2O a commercial sample or did you make it yourself? Home-made copper nitrate is very hard to make free of nitric acid. I once made nickel nitrate and even after I obtained it as a solid, thin fumes could be observed above the solid. Another thing is that copper nitrate is very deliquescent. I have heard of more people that their copper nitrate had liquefied after a year of storage. I purchased 100 gram of this stuff two years ago in a drugstore and it still is nice dry and crystalline. I put it in a container with a screw cap. This container is put in a plastic bag, which is tightly closed. This plastic bag with the container inside is put in a larger outside container with screw cap again. The plastic bag inside is essential. It prevents fresh (and humid) air to cycle in and out of the container on varying air pressure and temperature.

Sorry for the double post, I cannot edit my previous post anymore. I also did an experiment with slightly acidified solution of Cu(NO3)2. The solution was acidified with a few drops of dilute (2 mol/l) HNO3. These little drops of acid result in almost immediate copper plating of the iron particles and the liquid becomes dark green. Apparently, slight acidification of the nitrate solution makes the copper-plating much faster. The green color also can be explained. HNO3 is reduced by iron and/or copper metal, which is converted to NO and/or NO2. This in turn results in formation of the deep brown [Fe(NO)(H2O)5](+) complex. What remains puzzling, however, is that neutral copper nitrate does not result in copper plating.

The nails I use probably contain quite some zinc. This often is used to make them less susceptible to rusting, but a zinc nail (or one, containing quite some zinc) will be copper plated in any solution, containing copper (II) ions, even in copper nitrate solution. The metals zinc, cadmium, magnesium and aluminium most likely will cause the formation of copper, regardless of the anion. I also did another experiment with another nail. This gives a dark green/black solution when Cu(NO3)2 is used and it becomes lightly copper-plated in a solution of CuSO4. So, using nails is indeed very inconsistent. The problem is that we do not know the impurities. This apparently also differs from country to country. --------------------------------------------------------------- To me it is a real surprise to see, that the anion has such a strong influence on the displacement reaction between iron and copper. The effect of chloride I already got aquainted with, I've seen that before many times, but the effect of the nitrate really puzzles me. As you see, the highschool text books oversimplify things. Even with seemingly simple things like displacement reactions there is a lot more to tell about this than what the school books suggest. I'll certainly dive into this more deeply.

If that is what you want, then simply dissolve as much boric acid as possible in methanol and keep that in a bottle. You can dissolve a few grams of boric acid in 100 ml of methanol. No gas needed at all. When you want the green flame, then you can put the liquid on a (small) fire. Do not put the entire vial in a fire, as that may rupture due to pressure buildup and may explode. As my website shows, using ethanol is less succesfull. The flame is not pure green, but orange with green rims.

Probably a company like Alfa Aesar does not sell to the general public. Chemical supply houses usually only sell to licensed companies, universities and government labs. A home chemist definitely has no license. If you want chemicals for home experimentation, then you have to look at raw chemical photography shops (they have a lot of chemicals), ceramics and pottery suppliers, but also in local hardware stores quite some interesting things can be found. Ebay indeed also can be nice source of chemicals. There are also special element supplyers, who sell very pure elements of all kinds. This is a nice source for very pure metals. Forget about the official chemical supply houses. These struggling suppliers almost are useless for the hobby chemist. Besides that, their prices are 10 to 100 times as high as with other suppliers.

This is plain stupid, ridiculous and mad . Coffee has to be hot and drinking coffee is a highly risky undertaking and you know in advance. The only thing I can do if I read this is laugh .

What you say here only is true for molecules without free electron pairs. Free electron pairs (which are not involved in bonds), also count. That is why water has a bent structure as well. Your visualization is not wrong, but why would you expect it to go further in higher dimensions? It is for precisely the same reason that you cannot go through higher dimensions. You, me and all of us are 3D people and all matter around us exists in 3D, so the structuring of molecules also is confined to 3D.

Well Jdurg, you have all single-letter elements, you also have most two-letter elements, now it becomes time to get some elements with three letters in their symbol .

I've done the three experiments with copper sulfate, copper (II) chloride and copper nitrate. The effect for each of the three salts is very different. First I did experiments with different types of nails, but these results are very inconsistent. Some nails are copper plated quickly, also in nitrate solution, other nails are not copper-plated. Apparently these nails have different composition, so Ferdinand's observation's cannot reliably reproduced with the nails, I have in my barn. =============================================== So, I took coarse powdered 99.9% iron (lab grade powder, not the cheap filings or powder used for magnetism experiments). With this iron powder I obtained very interesting results. Copper sulfate solution: The iron grains are copper plated fairly quickly. Within a few tens of seconds, the color of the grains shifts from grey to red/salmon. The liquid remains blue. After 4 hours, the liquid has become a little turbid and brown/yellow. This looks like hydrolysed iron (III). The observations can nicely be explained: metal displacement: Fe + Cu(2+) --> Fe(2+) + Cu The Fe(2+) lateron is slowly oxidized by oxygen from the air, forming basic Fe(3+) compounds. Copper nitrate solution: Even after 4 hours, the liquid still is nice bright blue and clear. The iron still is grey. It has become slightly darker grey, but there definitely is no copper plating. So, here I perfectly reproduce Ferdinand's observation, now with lab grade chemicals. This observation puzzles me indeed. I'll look more into this and come back on this later with variations (e.g. pH and concentration). Copper (II) chloride: This is most remarkable. Within 10 seconds, all iron is displaced by copper. A coarse grainy red/brown precipitate is formed and the liquid becomes very light green (color of iron (II) chloride in solution). After shaking for a minute, the liquid turns brown/yellow and turbid. This indicates oxidation of the iron (II), and formation of basic iron (III) compounds, which indeed are brown/yellow. See my webpage on iron (III) species and their hydrolysis. Next, I checked what the brown/red precipitate is. I rinsed it two times with clean water and then added 10% HCl. The precipitate does not dissolve. This means that it is copper metal. If it were Cu2O, then it would dissolve quickly in the dilute HCl. The observations with copper (II) chloride can also be understood very well. The complex CuCl4(2-), present in this solution is a fairly strong oxidizer, which is capable of oxidizing Fe to Fe(2+), itself being reduced to copper metal and free Cl(-) ions: Fe + CuCl4(2-) --> Fe(2+) + Cu + 4Cl(-) The Fe(2+) in turn is oxidized by oxygen from the air. This experiment also shows that CuCl2 does not oxidize the iron to Fe(3+). As YT already mentioned, that would not be the expected thing, because FeCl3 is used as PCB etchant, which dissolves copper. So, with Fe(3+), copper metal is oxidized to copper ions, and the Fe(3+) is converted to Fe(2+). With Fe and copper ions, however, the copper ions are converted to metal and the Fe is converted to Fe(2+).

When you immerse a PCB in a solution of FeCl3, then you'll get some complex mix of copper (I) and copper (II) ions. Indeed, with an iron immersed in copper (II) chloride solution, I do not expect formation of a copper layer on the iron, but I still expect it to react. I expect formation of iron (II) or even iron (III), and formation of copper (I) species. Probably the liquid will becomes very dark brown, due to formation of mixed-valency complexes. I think Ferdinand's experiment is interesting to do and see the effect of the anion on the total mix. I'll perform the experiment tonight or tomorrow and I'm really curious what will happen. I expect differences with all three different salts. I'll try to give an explanation, as soon as I have experimental results.