Question

following the aufbau principle, which orbital is filled after 3p? 3d 4p 4s

Explanation:

The Aufbau principle states that electrons occupy the lowest - energy orbitals first. The order of orbital filling is determined by the \(n + l\) rule (where \(n\) is the principal quantum number and \(l\) is the azimuthal quantum number). For the \(3p\) orbital, \(n = 3\) and \(l=1\), so \(n + l=4\). For the \(4s\) orbital, \(n = 4\) and \(l = 0\), so \(n + l=4\). When two orbitals have the same \(n + l\) value, the one with the lower \(n\) value is filled first. But in the case of \(3p\) (\(n = 3\)) and \(4s\) (\(n = 4\)), after \(3p\) (since \(3p\) is part of the \(n = 3\) shell and when \(3p\) is filled, the next orbital with the next - lowest energy is \(4s\)). The \(3d\) (\(n = 3,l = 2,n + l=5\)) and \(4p\) (\(n = 4,l = 1,n + l=5\)) have higher \(n + l\) values than \(4s\) for the post - \(3p\) filling.

Answer:

\(4s\)