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Place these in decreasing order of energy: 1d, 3f, 3p, 1s
Accepted Answer
The order of decreasing energy is: 3F > 3P > 1S > 1D. This is based on the following factors:
Principal quantum number (n): Higher n values correspond to higher energy levels. In this case, 3 > 1, so 3F, 3P are higher in energy than 1S, 1D.
Angular momentum quantum number (l): For a given n, higher l values correspond to higher energy levels. So, within the same principal level (n = 3), 3F (l = 3) is higher in energy than 3P (l = 1).
Hund's Rule: For multi-electron atoms, the lowest energy state is the one with the maximum number of unpaired electrons with parallel spins. This principle can affect the relative energies of orbitals within a subshell (e.g., 1S and 1D).
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