Question

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

Explanation:

Step 1: Use the \(n + l\) rule

For \(3p\): \(n = 3\), \(l=1\), so \(n + l=3 + 1=4\)
For \(4s\): \(n = 4\), \(l = 0\), so \(n + l=4+0 = 4\)
For \(3d\): \(n = 3\), \(l = 2\), so \(n + l=3+2 = 5\)
For \(4p\): \(n = 4\), \(l=1\), so \(n + l=4 + 1=5\)
When \(n + l\) values are equal (e.g., \(3p\) and \(4s\), \(3d\) and \(4p\)), the sub - shell with lower \(n\) has lower energy.

Step 2: Compare sub - shells with same \(n + l\)

Since for \(n + l = 4\), \(3p\) (\(n = 3\)) has lower energy than \(4s\) (\(n = 4\)).
Since for \(n + l=5\), \(3d\) (\(n = 3\)) has lower energy than \(4p\) (\(n = 4\))

Answer:

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