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Order of decreasing energy: 1d, 3f, 3p,1s?
Accepted Answer
The order of decreasing energy for the given orbitals is:
1. 3F
2. 3P
3. 1D
4. 1S
This is based on the following principles:
Principal quantum number (n): Higher values of n indicate higher energy levels. Therefore, 3 is higher in energy than 1.
Angular momentum quantum number (l): Higher values of l indicate higher energy levels for orbitals with the same n. Therefore, F (l=3) is higher in energy than P (l=1), D (l=2), and S (l=0).
Degeneracy: Orbitals within the same subshell (same n and l) are degenerate, meaning they have the same energy. However, this degeneracy is lifted in multi-electron atoms due to electron-electron interactions. Therefore, 1D is higher in energy than 1S.
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