Question

which of the following orbitals cannot exist? answer: a 2p b 3d c 4d d 3f

Explanation:

• For a given principal quantum number \(n\), the angular - momentum quantum number \(l\) can take values from \(0\) to \(n - 1\).
• For \(n=2\), \(l = 0\) (s - orbital) and \(l = 1\) (p - orbital). So, \(2p\) exists.
• For \(n = 3\), \(l=0\) (s), \(l = 1\) (p), \(l=2\) (d). So, \(3d\) exists.
• For \(n = 4\), \(l = 0\) (s), \(l=1\) (p), \(l = 2\) (d), \(l=3\) (f). But for \(n = 3\), the maximum value of \(l\) is \(n-1=2\). So, \(3f\) cannot exist as \(l = 3\) is not allowed when \(n = 3\).
• For \(n=4\), \(l = 2\) (d - orbital) is allowed (\(l\) ranges from \(0\) to \(3\)), so \(4d\) exists.

Answer:

D. \(3f\)