Question

list these electron subshells in order of increasing energy.
5p, 5s, 6s, 4d
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 \(n + l\) values

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

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

When \(n + l\) values are equal, the sub - shell with the lower \(n\) value has lower energy.
For \(n + l = 6\):
Since \(4<5<6\) for \(4d\), \(5p\), \(6s\) (when comparing \(4d\) with \(5p\) and \(6s\)), and for \(n + l = 5\) (\(5s\))

Answer:

\(5s<4d<5p<6s\)