CH301H - Principles of Chemistry I: Honors
Fall 2016, Unique 50015

Lecture Summary, 11 October 2016

The Hydrogen Atom Continued:  We will examine the radial and angular parts of the hydrogen atom wavefunction separately.  This is done for several reasons.  Solving the Schrodinger equation for the hydrogen atom exactly is most easily done through the separation of variables technique, which naturally results in a full wavefunction, psi, that is the product of the radial and angular components.  Second, the energy of the wavefunction is dependent only on n, so being able to evaluate r separately from theta and phi might be useful.  Third, it will be significantly easier to visualize the size and shape of the resulting atomic orbitals if we can look at the radial and angular parts separately.  The radial and angular parts of the wavefunction for the hydrogen atom are given in Table 5.2 in your book.  Your book has lots of figures of the resulting wavefunctions and probability distributions, and you should spend some time getting comfortable looking at these shapes and interpreting the information they reveal.  Here is also a good web resource:

Wikipedia page on atomic orbitals

s orbitals (l = 0): Table 5.2 shows that when l = 0, the angular part of the wavefunction is a constant - i.e. does not depend on angle.  This means the orbital is spherically symmetric about the nucleus.  The probability distribution function shows that as n increases, there are nodes in the radius where the electron cannot be located.  These figures also show that even when n > 1, there is a small probability that the electron will be found close to the nucleus.  

p orbitals (l = 1):  Table 5.2 shows that when l = 1, the angular part of the wavefunction does depend on theta and phi, so the amplitude of the wavefunction will vary not just by distance but also by angle.  The angular wavefunction describes an orbital with two lobes of different phase, meaning that there is one angular node between them.  When the wavefunction is squared (to find the probability distribution), we are left with the characteristic "dumbbell" shape of the p orbitals.  

d orbitals (l = 2):  These orbitals now have 2 angular nodes (spot a pattern?), resulting in 4 lobes with maximum amplitude separated by 90˚. By convention, four of the d orbitals have the same 4-leaf clover shape, just oriented differently in Cartesian coordinates.  The 5th orbital looks a bit strange, but you should convince yourself that it also has two angular nodes.  

The Schrodinger equation cannot be solved exactly for any atom with more than one electron.  It can, however, be approximated using several different techniques, and it has been shown to produce accurate predictions of experimental observations.  We will therefore use the atomic orbitals that we built up for our hydrogen atom for multi-electron systems.