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Author: Subject: Igniting metals vs. flame test
FluoroPunch
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[*] posted on 11-9-2019 at 19:18
Igniting metals vs. flame test


There's a little chemical curiosity that's been perplexing me and I believe I've figured it out.

We know that lithium's flame test is crimson red, and that lithium salts will impart the same red color to the flame. This is usually explained as being due to the lithium ions present. What was curious to me was that when burning lithium, it glows white. My friend believed the lithium was impure but even in Nurdrage's video the lithium burns white. Why is this so?

I believe when lithium is burning freely in air, it is oxidizing, only oxygen atoms are sticking to the lithium, and taking an electron from it. None of the lithium atoms are getting very excited (or vaporized). However, if a lithium salt, e.g. lithium chloride is added to methanol and burned there will be a red color. This is proportional to the concentration. The lithium ions are vaporizing with the methanol and in the flame plasma their electrons become excited and give off the characteristic red wavelength. Similarly, when throwing it in water, a flame is produced which can excite the lithium atoms and give off the red color.

Does this make sense? Am I missing anything? Any thoughts would be great.
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[*] posted on 11-9-2019 at 20:02


My guess would be that the burning metal is hot enough that its blackbody radiation overwhelms anything else.



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[*] posted on 12-9-2019 at 03:36


From wikipedia: "When placed over a flame, lithium compounds give off a striking crimson color, but when it burns strongly the flame becomes a brilliant silver."

What Mayko said is correct.




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[*] posted on 16-9-2019 at 12:34


Wow, that makes a lot of sense. Blackbody radation didn't even come to mind. If the temperature is high enough it makes sense that the white light would be given off, since the red is fairly faint and barely visible in daylight.
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[*] posted on 6-11-2019 at 20:56


I want to point out that the crimson color in the lithium flame test comes from neutral lithium atoms, not lithium ions. The crimson color in lithium comes from the 2p->2s transition of the valence electron, moving outside the helium core; the lithium ion does not have this electron, which has been stripped off to make the ion.

Similarly the sodium yellow line is a 3p->3s transition of the valence electron, moving about a neon core.




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[*] posted on 15-11-2019 at 20:01


Quote: Originally posted by annaandherdad  
I want to point out that the crimson color in the lithium flame test comes from neutral lithium atoms, not lithium ions. The crimson color in lithium comes from the 2p->2s transition of the valence electron, moving outside the helium core; the lithium ion does not have this electron, which has been stripped off to make the ion.

Similarly the sodium yellow line is a 3p->3s transition of the valence electron, moving about a neon core.


I'm not convinced. If that were true you couldn't get the various colors you see when you dissolve metal salts in methanol. By your reasoning, lithium chloride in methanol wouldn't result in a red color because the lithium is ionized. Yet it happens so...
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[*] posted on 15-11-2019 at 21:45


Well to be precise the flame color is caused by a variety of spectral lines with a variety of intensities, that depend on circumstances such as temperature. But the main crimson color in lithium is the transition
2p->2s which is at 671 nanometers. The spectral line itself and its crimson color and its wavelength can be seen in a spectrometer; as for the fact that it's the 2p->2s transtion, I get that from the NIST data base of atomic energy levels, which gives the 671 nanometer wavelength. But that transition involves the single valence electron that lithium has; the 2p and 2s refer to the states of that electron. In the lithium ion, that electron has been stripped off. It's a transition in lithium I, not lithium II (the Li+ ion). There's a similar story with the yellow sodium line which is 3p->3s.





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[*] posted on 16-11-2019 at 03:01


https://en.wikipedia.org/wiki/Spectral_line#Pressure_broaden...
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[*] posted on 16-11-2019 at 08:30


Quote: Originally posted by annaandherdad  
Well to be precise the flame color is caused by a variety of spectral lines with a variety of intensities, that depend on circumstances such as temperature. But the main crimson color in lithium is the transition
2p->2s which is at 671 nanometers. The spectral line itself and its crimson color and its wavelength can be seen in a spectrometer; as for the fact that it's the 2p->2s transtion, I get that from the NIST data base of atomic energy levels, which gives the 671 nanometer wavelength. But that transition involves the single valence electron that lithium has; the 2p and 2s refer to the states of that electron. In the lithium ion, that electron has been stripped off. It's a transition in lithium I, not lithium II (the Li+ ion). There's a similar story with the yellow sodium line which is 3p->3s.



Ok fair enough but how does that explain the ions color then? How do lithium ions in plasma have an emission if that electron is gone?
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[*] posted on 16-11-2019 at 14:10


That's my point, the ions don't produce the 671 nanometer crimson line, neutral atoms do.

The ions are helium-like lithium, that is, with a closed 1s shell. I don't think they produce any spectral lines in an ordinary flame. They would if an electron could be ripped off of them into a sufficiently highly excited state, but the very first excited state is something like 70ev above the ground state so it doesn't seem likely. The helium-like closed shell is very stable.




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[*] posted on 16-11-2019 at 15:28


So you're saying that if you put lithium or sodium ions into a flame, they get reduced to the atoms before they give those characteristic glows?



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[*] posted on 16-11-2019 at 15:48


The characteristic colors are dominated by one spectral line, for example, the yellow sodium D line at 589-590 nanometers (it's actually a close doublet). This is the 3p->3s transition of the valence electron, which isn't present in the sodium Na+ ion. It's a transition in the sodium neutral atom, not the Na+ ion. Similar story with lithium.

Potassium is different since the analogous transition, 4p->4s, is in the infrared so you can't see it. The lilac color you see comes from other transitions, but even there I'll bet they are transitions in the neutral potassium atom, not the ion.

I'm not saying how the ions in a methanol solution (for example in the standard way of doing the flame test) get turned into neutral atoms in the flame, or if you vaporize a small amount of salt on the end of a platinum wire. What I'm saying is that the main colors are caused by transitions in the neutral atom.




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[*] posted on 16-11-2019 at 17:02


Quote: Originally posted by annaandherdad  
The characteristic colors are dominated by one spectral line, for example, the yellow sodium D line at 589-590 nanometers (it's actually a close doublet). This is the 3p->3s transition of the valence electron, which isn't present in the sodium Na+ ion. It's a transition in the sodium neutral atom, not the Na+ ion. Similar story with lithium.

Potassium is different since the analogous transition, 4p->4s, is in the infrared so you can't see it. The lilac color you see comes from other transitions, but even there I'll bet they are transitions in the neutral potassium atom, not the ion.

I'm not saying how the ions in a methanol solution (for example in the standard way of doing the flame test) get turned into neutral atoms in the flame, or if you vaporize a small amount of salt on the end of a platinum wire. What I'm saying is that the main colors are caused by transitions in the neutral atom.


And yet when there's no neutral atom present there's still color. A methanol flame isn't that hot anyways. So like DraconicAcid said, for your explanation to hold that would mean the atoms are somehow being reduced? Fire plasma is kinda mysterious so maybe the electrons are doing strange things ...idk
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[*] posted on 16-11-2019 at 18:22


I have a suggestion for someone with a spectrometer. Just make a strong sodium flame and look at it. If there's a nice yellow line at 589-590 nanometers then that's coming from neutral sodium atoms.

See https://physics.nist.gov/cgi-bin/ASD/energy1.pl?encodedlist=...

It's the excitation from the ground state of neutral Na to the first excited state. The energy difference (see website) is 2.102-2.104 eV. Convert to wavelength and you get 589-590nm. The transition is 3p->3s.

Then we can speculate or research the topic about how the neutral sodium atoms get into the flame from the Na+ ions in the methanol.




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