Question

list these electron subshells in order of increasing energy.
4d, 5s, 4p, 3d
note for advanced students: you may assume these subshells are all in an atom with many electrons, not a hydrogen atom.

Explanation:

Step1: Determine the \(n + l\) value for each subshell

• For \(4d\): \(n = 4\), \(l=2\), so \(n + l=4 + 2=6\)
• For \(5s\): \(n = 5\), \(l = 0\), so \(n + l=5+0 = 5\)
• For \(4p\): \(n = 4\), \(l = 1\), so \(n + l=4 + 1=5\)
• For \(3d\): \(n = 3\), \(l=2\), so \(n + l=3 + 2=5\)

Step2: Compare sub - shells with the same \(n + l\) value

When \(n + l\) values are equal, the subshell with the lower \(n\) value has lower energy.

• For \(n + l = 5\):
• \(3d\) (\(n = 3\)) has lower energy than \(4p\) (\(n = 4\)) which has lower energy than \(5s\) (\(n = 5\)) (since for same \(n + l\), lower \(n\) has lower energy).
• \(4d\) has \(n + l=6\) which is higher than \(n + l = 5\) for the other three subshells.

Answer:

\(3d<4p<5s<4d\)