Question

which of the following cannot exist? 2d 3d 4f 4p

Explanation:

Step1: Recall orbital - subshell rules

The principal quantum number \(n\) can take values \(n = 1,2,3,\cdots\). The angular - momentum quantum number \(l\) can take values \(l=0,1,\cdots,n - 1\). The sub - shells are named as \(s\) (\(l = 0\)), \(p\) (\(l = 1\)), \(d\) (\(l = 2\)), \(f\) (\(l=3\)).

Step2: Analyze each option

For \(2d\): Here \(n = 2\), and for \(d\) sub - shell \(l = 2\). Since \(l\) must satisfy \(l\lt n\), when \(n = 2\), \(l\) can be \(0\) ( \(s\) sub - shell) or \(1\) ( \(p\) sub - shell), but not \(2\). So \(2d\) cannot exist.
For \(3d\): Here \(n = 3\) and \(l = 2\) for \(d\) sub - shell. Since \(l=2\lt n = 3\), \(3d\) can exist.
For \(4f\): Here \(n = 4\) and \(l = 3\) for \(f\) sub - shell. Since \(l = 3\lt n=4\), \(4f\) can exist.
For \(4p\): Here \(n = 4\) and \(l = 1\) for \(p\) sub - shell. Since \(l = 1\lt n = 4\), \(4p\) can exist.

Answer:

A. \(2d\)