Hi.. thank you for the video :) Electronic configuration of copper is [Ar]3d10 4s1. Though 4s orbitals are filled before 3d orbitals, fully-filled 3d orbitals are more stable. Moreover, when the 3d and 4s orbitals have electrons, 3d orbitals are of lower energy than 4s orbitals, hence 3d should be written before 4s. Thanks :)
can zinc emit light in the non-visible spectrum? I'm just wondering where energy goes if the atom/ion is excited by some input of energy. Does it release thermal energy?
No, because no electron excitation is transpiring. With complex ions which have orbitals space for electrons to promote to, it is possible that the difference in energy level between the lower d sub-shell and higher d sub-shell is large enough that the corresponding wavelength of light required to be absorbed by an electron is outside of the visible light spectrum. Zinc is the Pluto of transition metals; it just doesn't belong :(
it depends on your source of energy input, I believe. For example, when we did flame tests in lab, certain elements released visible light when excited by thermal energy (the heat of the flame) and this was due to the exact method described in the video. Excited electrons jumped to higher energy orbital, absorbing light, and when they returned to ground state, they emitted photons in the visible spectrum.
Hi.. thank you for the video :) Electronic configuration of copper is [Ar]3d10 4s1. Though 4s orbitals are filled before 3d orbitals, fully-filled 3d orbitals are more stable. Moreover, when the 3d and 4s orbitals have electrons, 3d orbitals are of lower energy than 4s orbitals, hence 3d should be written before 4s. Thanks :)
Very simple and crisp explanation..Love you dude
Thanks brother. That really helped.
Question: Is the Cu2+ having an electron excited to a higher energy level a d-d transition? Or is it referred to as charge transfer?
can zinc emit light in the non-visible spectrum? I'm just wondering where energy goes if the atom/ion is excited by some input of energy. Does it release thermal energy?
No, because no electron excitation is transpiring. With complex ions which have orbitals space for electrons to promote to, it is possible that the difference in energy level between the lower d sub-shell and higher d sub-shell is large enough that the corresponding wavelength of light required to be absorbed by an electron is outside of the visible light spectrum. Zinc is the Pluto of transition metals; it just doesn't belong :(
The explanation is actually wrong: it is complementary color! Not released photon which results in color.
Steve Carter yes i was looking for this
it depends on your source of energy input, I believe. For example, when we did flame tests in lab, certain elements released visible light when excited by thermal energy (the heat of the flame) and this was due to the exact method described in the video. Excited electrons jumped to higher energy orbital, absorbing light, and when they returned to ground state, they emitted photons in the visible spectrum.
This video kicked ass thank you
this was really helpful!
You're great!! Thanks!!
Wait, why does it start with Ar?
jellyandme so he doesn't have to start writing the configuration from the beginning.
Instead he started from Argon's configuration .
jellyandme because argon is a noble gas and it comes before zinc....
@@alinajahmusic It is Argon's electron structure + the remaining e- shells of the element named.
Thank You..Very very helpful
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