Periodic Table

Previous sections of these lectures have discussed the electronic structure of H, He and Li atoms. The general approach implemented in those sections is summarized as follows. First, we neglect the repulsive interaction between electrons and write the zeroth order ground state wave functions as antisymmetrized products of spin-orbitals (Slater determinants), e.g.,

$$\psi_{He}^{gr} = \frac{1}{\sqrt{2}} \left \vert \begin{matrix}1S(... ...S(2) \frac{1}{\sqrt{2}} \left [\alpha(1) \beta(2) - \beta(1) \alpha(2) \right],$$

$$\psi_{Li}^{gr} = \frac{1}{\sqrt{6}} \left \vert \begin{matrix}1S(... ... 1S(3)\alpha(3) & 1S(3)\beta(3) & 2S(3)\alpha(3) \\ \end{matrix} \right \vert,$$

with zeroth order energies,

$$E_{He}^{(0)} =2E(1S),$$   and$$\qquad E_{Li}^{(0)}=2E(1S) +E(2S),$$

represented by the following diagram:

It is important to note that these approximate wave functions are found by assuming that the electrons do not interact with each other. This is, of course, a very crude approximation. It is, nonetheless, very useful because it is the underlying approximation for the construction of the periodic table. Approximate zeroth order wave functions can be systematically constructed for all atoms in the periodic table by considering the energy order of hydrogenlike atomic orbitals in conjunction with Hund's Rules.

Hund's First Rule: Other things being equal, the state of highest multiplicity is the most stable.

Hund's Second Rule: Among levels of equal electronic configuration and spin multiplicity, the most stable level is the one with the largest angular momentum.

These rules establish a distinction between the zeroth order wave functions of ground and excited electronic state configurations. For example, according to Hund's rules the lithium ground state wave function is,

$$\psi^{gr} = \frac{1}{\sqrt{6}} \left \vert \begin{matrix}1S(1)\al... ...\ 1S(3)\alpha(3) & 1S(3)\beta(3) & 2S(3)\alpha(3) \\ \end{matrix} \right \vert,$$ (53)

and the first excited state wave function is,

$$\psi^{exc} = \frac{1}{\sqrt{6}} \left \vert \begin{matrix}1S(1)\a... ... 1S(3)\alpha(3) & 1S(3)\beta(3) & 2P(3)\alpha(3) \\ \end{matrix} \right \vert .$$

Note that the energy order of hydrogenlike atomic orbitals, $E_n=-\frac{z^2 e^2}{2 a n^2}$, is not sufficient to distinguish between the two electronic configurations. According to such expression, orbitals 2p and 2s have the same energy $E_2$. However, Hund's second rule distinguishes the ground electronic state as the one with higher angular momentum. This is verified by first order perturbation theory, since the perturbation energy of $\psi^{exc}$ is higher than the perturbation energy computed with $\psi^{gr}$. Exercise 46: Prove that according to first order perturbation theory, the energy difference $\Delta E$ between the two states is

$$\Delta E (\psi^{gr} \rightarrow \psi^{exc}) = 2(J_{1S, 2P} -J_{1S, 2S}) -(K_{1S, 2P} -K_{1S, 2S}),$$

where $J_{\phi_1, \phi_2} =<\phi_1^{(i)}\phi_2^{(j)} \vert\frac{e}{r_{ij}}\vert \phi_1^{(i)}\phi_2^{(j)}> \equiv$   Coulomb Intergral$,$

and $K_{\phi_1, \phi_2} =<\phi_1^{(i)}\phi_2^{(j)} \vert\frac{e}{r_{ij}}\vert \phi_2^{(i)}\phi_1^{(j)} > \equiv$   Exchange Integral. Exercise 47: Use Hund's Rules to predict that the ground states of nitrogen, oxygen and fluorine atoms are $^4S, \ ^3P \$   and$\ ^2P$, respectively.