Question

question 1. arrange the following subshells in order of increasing energy. 2p, 3p, 3s, 3d and 4s

Explanation:

Step1: Recall the energy order of orbitals

The energy of orbitals is determined by the \(n + l\) rule. For \(2p\), \(n = 2\), \(l=1\), so \(n + l=3\). For \(3p\), \(n = 3\), \(l = 1\), \(n + l=4\). For \(3s\), \(n = 3\), \(l = 0\), \(n + l=3\). For \(3d\), \(n = 3\), \(l = 2\), \(n + l=5\). For \(4s\), \(n = 4\), \(l = 0\), \(n + l=4\).
When \(n + l\) values are equal (e.g., \(2p\) (\(n + l = 3\)) and \(3s\) (\(n + l=3\))), the orbital with lower \(n\) has lower energy. So \(2p<3s\). When \(n + l\) values are \(4\) (e.g., \(3p\) (\(n + l = 4\)) and \(4s\) (\(n + l=4\))), \(3p<4s\) (because for same \(n + l\), lower \(n\) has lower energy).

Step2: Arrange the orbitals

Based on the \(n + l\) rule and the tie - breaker (lower \(n\) for same \(n + l\)):
The order of increasing energy is \(2p<3s<3p<4s<3d\)

Answer:

\(2p\), \(3s\), \(3p\), \(4s\), \(3d\)