If s, p, d and f subshells have these kinds of boundary surface diagram, then why don't they include the probability of finding electrons in their preceding shells?

As in, if d subshells have boundary surface diagrams as stated by the great scientists, then it must include the nodes of the preceding subshells, like that of s and p shells.

  • 1
    $\begingroup$ May be add a picture of your question to clarify it. Make a diagram of your question in power point and share a picture. Orbitals are solutions of a differential equation proposed by Schrodinger. When this equation is solved for a single electron bound to a nucleus of positive charge. Under various mathematical restrictions, you get various "shapes" of orbitals. $\endgroup$ – M. Farooq Jun 22 '19 at 4:20
  • 3
    $\begingroup$ 10 is greater than 9; how come it doesn't have the figure 9 in it? Well, just like that. Same thing here. $\endgroup$ – Ivan Neretin Jun 22 '19 at 5:57
  • 5
    $\begingroup$ Is it me or the title is a totally different question than the actual question? $\endgroup$ – Greg Jun 22 '19 at 8:22

There are no requirements for orbitals as wave functions to respect geometrical details of other orbitals.

There are 3 meanings of the term orbital:

  1. A complex wave function $\Psi$ being a particular solution of the quantum wave equation
  2. A quantum state of an electron, following 1., described by particular quantum numbers $n$, $l$ and $m$, representing discrete values of electron energy, orbital angular momentum and one of component of angular momentum, respectively.
  3. 3D shape, following 1. and 2., representing statistical probability of the electron presence at given point by$|\Psi|^2$, determined by the conventional value by residual external probability and by the isoprobability 3D surface.

In one electron atomic system, like $\ce{H}$, $\ce{He+}$, orbitals are independent on each other.

In a multielectron system, an electron occurence probability depends on quantum states of other electrons, due kernel shielding and electron repulsion.

The typical example is the relative energy of ns and (n-1)d orbitals.

For quantum chemistry calculations, there are done simplifications like electron independence and orbital shielding factors ( s > p > d > f ).

Electrons in f orbitals do not shield much, what is the major contribution factor of lanthanoid contraction.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.