Why is the flame test used for cations and not anions (1 Viewer)

[ekjchale#1]

^^^
Thanks in advance

 

[5uckerberg]

The idea of the flame test is that by putting the flame near the cation you will be able to see different colours of flame such as green for copper and then another colour for different metals. Plus, anions are typically gases so therefore, a flame test will not reveal anything as they are negatively charged.

 

[Run hard@thehsc]

I am not too sure on this, however through application I believe anions are obviously electronegative and hence as a result intend to hold onto their electrons. This would make it inefficient and time consuming to use a flame test on anions. In addition, anions are mostly polyatomic as well. Hence, their net negative charge arises from various constituent particles - As such using the flame test will not effectively allow us to differentiate between these....

 

[ekjchale#1]

I am not too sure on this, however through application I believe anions are obviously electronegative and hence as a result intend to hold onto their electrons. This would make it inefficient and time consuming to use a flame test on anions. In addition, anions are mostly polyatomic as well. Hence, their net negative charge arises from various constituent particles - As such using the flame test will not effectively allow us to differentiate between these....

fair this makes logical sense thx

 

[someth1ng]

I disagree with the comments here.

There is an assumption that the emission lines are in the visible spectrum. It happens to be true for many metals (e.g. Na, Ca). However, for anions (particularly the common ones like chloride, bromide, sulfate), the emission lines might not be or the excitation efficiency is low (i.e. difficult to excite). Remember, emissions come from exciting an electron up, then when it relaxes, a photon is released. With that said, there are cases where anions give have emissions in the visible spectrum like boric acid, H3BO3).

1 Angstrom = 0.1 nm, so divide the angstrom values by 10 to get wavelength in nm. NIST publish the values. A few notes. Make sure you look at the right cation/anion (you don't want the II values when using Cl or Na but they still may appear)
Sodium Lines (air: 5889.950 A, 5895.924 A): https://physics.nist.gov/PhysRefData/Handbook/Tables/sodiumtable2.htm
Chloride Lines (air: 8375.94 A, 8585.97 A): https://physics.nist.gov/PhysRefData/Handbook/Tables/chlorinetable2.htm

Chlorine does not emit much in the visible spectrum, it emits in the infrared.

 

[ekjchale#1]

I disagree with the comments here.

There is an assumption that the emission lines are in the visible spectrum. It happens to be true for many metals (e.g. Na, Ca). However, for anions (particularly the common ones like chloride, bromide, sulfate), the emission lines might not be or the excitation efficiency is low (i.e. difficult to excite). Remember, emissions come from exciting an electron up, then when it relaxes, a photon is released. With that said, there are cases where anions give have emissions in the visible spectrum like boric acid, H3BO3).

1 Angstrom = 0.1 nm, so divide the angstrom values by 10 to get wavelength in nm. NIST publish the values. A few notes. Make sure you look at the right cation/anion (you don't want the II values when using Cl or Na but they still may appear)
Sodium Lines (air: 5889.950 A, 5895.924 A): https://physics.nist.gov/PhysRefData/Handbook/Tables/sodiumtable2.htm
Chloride Lines (air: 8375.94 A, 8585.97 A): https://physics.nist.gov/PhysRefData/Handbook/Tables/chlorinetable2.htm

Chlorine does not emit much in the visible spectrum, it emits in the infrared.

thx for the clarification