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May 24

Why is Aluminum so difficult to separate?

While the article on Aluminum talks about the fact that Aluminum wasn't really separated into pure or near-pure form until the 19th century, there doesn't seem to be an obvious reason given? Is it *somehow* tied to the other odd characteristic of Alumnimum, that when it oxidizes it goes the opposite way from Iron Oxide and flaking.Naraht (talk) 03:35, 24 May 2024 (UTC)[reply]

This has to do with the fact that aluminum forms such a strong bond with oxygen that it can only be reduced to the metal by means of electrolysis or thermite process (plus the fact that, as also with zinc and a few other metals, the melting point of the oxide is higher than the boiling point of the metal, so some way of reducing the former must be used to prevent the metal from simply boiling away). 2601:646:8082:BA0:A400:D9D:C5FD:AB24 (talk) 03:55, 24 May 2024 (UTC)[reply]

Is it that odd to have a non-spalling oxide? In ultradry pure oxygen (partial pressure of H₂O ≤2×10⁻⁵ mmHg), even Li, Na, and K passivate (cite: Russell and Lee's book on nonferrous metals). :) Double sharp (talk) 12:51, 24 May 2024 (UTC)[reply]

See: Hall–Héroult process -- Before then (1886), aluminum had been more expensive than gold.^{[1] [2] [3]} --136.54.106.120 (talk) 22:01, 24 May 2024 (UTC)[reply]

Neither the Hall–Héroult process article nor the links given immediately above seem to explain it as clearly as the comparison between Melting point of Aluminum oxide vs. boiling point of alumninum metal mentioned by the first poster. I'm sure it is somewhat more complex than that, but that would seem to be an incredibly difficult. I think the *first* statement is equivalent to basically that 6 Na + 2 AlO₃ doesn't lead to 3 Na₂O + 2 Al. (or any of the other Alkali Metals)Naraht (talk) 23:13, 30 May 2024 (UTC)[reply]

The Hall–Héroult process utilizes electrolysis; prior to the advent of electricity, extraction of aluminum was more difficult and expensive. 136.54.106.120 (talk) 04:49, 31 May 2024 (UTC)[reply]

Hexavalent chromium and heptavalent manganese

For the compounds with hexavalent chromium and heptavalent manganese (or the high oxidation states of other transition metals, such as hexavalent molybdenum, heptavalent technetium, hexavalent tungsten, heptavalent rhenium, etc.) and a single chemical element, there are CrO3 and Mn2O7, but CrF6 and MnF7 seem to not exist, but are there CrS3, Mn2S7, CrCl6, MnCl7, etc.? Is O the only element which has compound with only two elements and with high oxidation states of transition metals? +6 is one of the main oxidation state of chromium and +7 is one of the main oxidation state of manganese. 61.224.168.169 (talk) 10:29, 24 May 2024 (UTC)[reply]

In general, really high oxidation states require O or F, essentially due to high electronegativity. Cr(VI) and Mn(VII) are high enough to be considered under this rubric. Which one is favoured depends on other factors, e.g. F has higher electronegativity but its monovalency instead of O's divalency leads to steric hindrance being more important. It can go either way: for Os, OsO₄ is well-known and OsF₈ nonexistent, but for Pt, it's PtF₆ that's well-known and PtO₃ that's not well-characterised.

I don't think W(VI), for example, is really "high". Of course, it is the highest possible (just count valence electrons), but no one would call Na(I) high either. The early 4d and 5d metals tend to be happier in high oxidation states than their 3d counterparts. MoS₃, MoCl₆, WS₃, WBr₆, Tc₂S₇, and Re₂S₇ all exist. And WF₆ is not even a great oxidising agent, whereas MnF₄ is already ferocious (and the highest fluoride, too). Late d-metals are indeed much more oxidising in their highest oxidation states: once you reach Ir and Pt, I would definitely agree that VI is high (PtF₆ famously oxidises dioxygen and xenon). But even then you see the pattern, since Co and Ni cannot even get as far as VI. Double sharp (talk) 12:49, 24 May 2024 (UTC)[reply]

I am not sure what "happy" or "ferocious" mean here. MoS₃ and Re2S7 are of course not Mo(VI) nor Re(VII), as inorganic chemists well know. High oxidation states tend to require pi-donor ligands, and oxo (O^2-) is generally considered the strongest pi-donor.--Smokefoot (talk) 14:38, 30 May 2024 (UTC)[reply]

An anthropomorphic way to indicate their strength as oxidising agents.

My mistake, yes: these sulfides are not really in the group oxidation state. Apologies for the brain fart. (Though it seems that the actual oxidation state in Re₂S₇ is not so clear.) At least we still have the halides for Mo(VI) and W(VI). I think what I was really trying to get at is that to me, whether an oxidation state is high is more about its oxidising power than about the absolute value, e.g. Ag(III) is high but Cr(III) isn't. Double sharp (talk) 04:04, 31 May 2024 (UTC)[reply]

How do background gases influence combustion

If say, you ignite wood in a gas with the same partial pressure oxygen as air, but ten times the partial pressure of nitrogen. JoJo Eumerus mobile (main talk) 19:04, 24 May 2024 (UTC)[reply]

Some gases like dibromotetrafluoroethane vapour can suppress burning. Graeme Bartlett (talk) 22:41, 24 May 2024 (UTC)[reply]

The oxygen molecules move around randomly; to react with the fuel they need to bump into it. My guess is that the bump rate only depends on the partial pressure. --Lambiam 05:44, 25 May 2024 (UTC)[reply]

An inert gas would still absorb some of the heat from the combustion, lowering the flame temperature. That should suppress the fire somewhat. Actual flame retardants like abovementioned haloalkane aren't inert, but actively suppress the fire with the chemical reactions they undergo when heated or exposed to a flame. PiusImpavidus (talk) 10:21, 25 May 2024 (UTC)[reply]