Question

according to the aufbau principle, which orbital is filled after 4s?

Explanation:

Step1: Recall Aufbau principle

Electrons fill orbitals in order of increasing energy. The energy order is determined by \(n + l\) rule. For \(4s\), \(n = 4\), \(l=0\), so \(n + l=4\).

Step2: Calculate \(n + l\) for each orbital

• For \(4p\): \(n = 4\), \(l = 1\), \(n + l=4 + 1=5\)
• For \(4d\): \(n = 4\), \(l = 2\), \(n + l=4+ 2=6\)
• For \(4f\): \(n = 4\), \(l = 3\), \(n + l=4 + 3=7\)
• For \(5s\): \(n = 5\), \(l = 0\), \(n + l=5+0 = 5\). But when \(n + l\) values are equal (e.g., \(4p\) and \(5s\)), the orbital with lower \(n\) is filled first. After \(4s\) (\(n + l=4\)), the next is \(3d\) (\(n = 3\), \(l = 2\), \(n + l=5\)) but since it's not in the options, among the given options, after comparing \(n + l\) values: \(4p\) (\(n + l = 5\)) has higher energy than \(4s\) (\(n + l=4\))

Answer:

\(4p\)