{c}
{title2=1927}
{wiki}

Allow us to determine with good approximation in a multi-electron atom which <electron configuration> have more energy. It is a bit like the <Aufbau principle>, but at a finer resolution.

Note that this is not trivial since there is no explicit solution to the <Schrödinger equation> for multi-electron atoms <Schrödinger equation solution for the hydrogen atom>[like there is for hydrogen].

For example, consider carbon which has <electron configuration> 1s2 2s2 2p2.

If we were to populate the 3 <p-orbitals> with two electrons with spins either up or down, which has more energy? E.g. of the following two:
``
m_L -1  0  1
    u_ u_ __
    u_ __ u_
    __ ud __
``


Hund's rules

{title2=$m_l$}
{wiki}

Fixed <quantum angular momentum> in a given direction.

Can range between $\pm l$.

E.g. consider <gallium> which is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1:
* the electrons in <s-orbitals> such as 1s, 2d, and 3d are $l=0$, and so the only value for $m_l$ is 0
* the electrons in <p-orbitals> such as 2p, 3p and 4p are $l=1$, and so the possible values for $m_l$ are -1, 0 and 1
* the electrons in <d-orbitals> such as 2d are $l=2$, and so the possible values for $m_l$ are -2, -1, 0 and 1 and 2

The z component of the <quantum angular momentum> is simply:
$$
L_z = m_l \hbar
$$
so e.g. again for gallium:
* <s-orbitals>: necessarily have 0 z angular momentum
* <p-orbitals>: have either 0, $- \hbar$ or $+ \hbar$ z angular momentum

Note that this direction is arbitrary, since for a fixed <azimuthal quantum number> (and therefore fixed total angular momentum), we can only know one direction for sure. $z$ is normally used by convention.


Ciro Santilli @cirosantilli 34

 Incoming links: p-orbital