2

$\mathrm{s}$ orbitals are spherically symmetric while $\mathrm{p}$ orbitals are not. For Boron, the electronic configuration is $\mathrm{1s^2 2s^2 2p^1}$.

In which p orbital does the most energetic electron lie? Is it $\mathrm{p}_x$, $\mathrm{p}_y$ or $\mathrm{p}_z$? Since all the three are degenerate, how does the electron decide which orbital to occupy? If it occupies any of the one orbitals, is a neutral atom not spherically symmetric? Or does the electron keep changing between $\mathrm{p}_x$, $\mathrm{p}$ and $\mathrm{p}_z$ rapidly (or in a superposition of occupying all the three orbitals at once with the probability of finding in any one of the orbitals equal to $1/3$)? The same question can be asked for $\mathrm{d}$ and $\mathrm{f}$ orbitals.

Apoorv Potnis
  • 1,278
  • 1
  • 16
  • 38
  • 2
    Since the three p orbitals are degenerate, the electron is on all three (a linear combination of the three orbitals). The wavefunction will be spherically symmetric if you do not introduce any external field or similar that would break the symmetry. Related: https://chemistry.stackexchange.com/questions/13766/why-are-atom-spherical-in-shape and https://pubs.acs.org/doi/abs/10.1021/ed042p397 – Greg Jan 13 '20 at 11:46

0 Answers0