# What fraction of the orbitals in "1 mol" of "Mg" atoms in a metallic network are occupied at "0 K"?

## a) Determine the number of metallic bonding orbitals used by $\text{1 mol}$ of $\text{Mg}$ atoms. b) What is the Fermi level? Where does the most likely electronic transition occur? Determine the fraction of the orbitals in $\text{1 mol}$ of $\text{Mg}$ atoms in a metallic network are occupied at $\text{0 K}$?

Jan 26, 2016

a) This is asking you to apply the Linear Combination of Atomic Orbitals (LCAO).

LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO)

The idea is that $\setminus m a t h b f \left(n\right)$ number of atomic orbitals (AOs) yields $\setminus m a t h b f \left(n\right)$ number of molecular orbitals (MOs).

We can see this in context when we form the theoretical dimetal ${\text{Mg}}_{2}$ (just pretend it's in the gas phase if it bothers you to think about this; ${\text{Na}}_{2} \left(g\right)$ exists at very high temperatures):

1. Two $3 s$ AOs (one from each $\text{Mg}$) combine to form one ${\sigma}_{3 s}$ bonding MO and one ${\sigma}_{3 s}^{\text{*}}$ antibonding MO. That's a 2:2 conversion.
2. Six $3 p$ AOs (three from each $\text{Mg}$) combine to form one ${\pi}_{3 p x}$ bonding MO and one ${\pi}_{3 p x}^{\text{*}}$ antibonding MO, one ${\pi}_{3 p y}$ bonding MO and one ${\pi}_{3 p y}^{\text{*}}$ antibonding MO, and one ${\sigma}_{3 p z}$ bonding MO and one ${\sigma}_{3 p z}^{\text{*}}$ antibonding MO. Sum that up and you get a 6:6 conversion.
3. Since we used two $3 s$ and six $3 p$ AOs, that is $2 + 6 =$ eight AOs. We got out eight MOs---one each of the following: ${\sigma}_{3 s}$, ${\sigma}_{3 s}^{\text{*}}$, ${\pi}_{3 p x}$, ${\pi}_{3 p x}^{\text{*}}$, ${\pi}_{3 p y}$, ${\pi}_{3 p y}^{\text{*}}$, ${\sigma}_{3 p z}$, and ${\sigma}_{3 p z}^{\text{*}}$.

$\setminus m a t h b f \left(M {g}_{2}\right)$ MO DIAGRAM

Although the $3 p$ AOs are empty, as seen in the electron configuration for magnesium atom:

$\left[\text{Ne}\right] 3 {s}^{2}$

... pictorially this nevertheless looks like this:

With the molecular electronic configuration like so:

(sigma_(1s))^2(sigma_(1s)^"*")^2(sigma_(2s))^2(sigma_(2s)^"*")^2color(blue)((sigma_(3s))^2(sigma_(3s)^"*")^2)

where blue indicates the valence orbitals.

Naturally, this is for one ${\text{Mg}}_{2}$ diatomic molecule. We could have extended this to $n$ $\text{Mg}$ atoms, but I wanted to keep it as simple as possible.

For one $\text{Mg}$ atom, on the other hand (instead of two), divide the number of MOs by two to get four. Then, since we are talking about $\text{1 mol}$ of $\text{Mg}$ atoms, multiply the number of MOs/atom by $6.0221413 \times {10}^{23}$ to get:

$\setminus m a t h b f \left(\text{4 mol}\right)$s of $\text{MOs/Mg atom}$

b) Okay, so some terminology from Band Theory. It's not too bad.

THE HOMO-LUMO GAP

The Fermi level is where the highest-occupied molecular orbital (HOMO) currently lies at $\text{0 K}$.

That means the ${\sigma}_{3 s}^{\text{*}}$ antibonding orbital, the HOMO, is at the "Fermi level", the energy level that is closest to the band gap. The most likely electronic transition occurs from the HOMO to the lowest-unoccupied molecular orbital (LUMO), across this band gap.

This event constitutes conduction, and this energy gap is also called the HOMO-LUMO gap. The HOMO-LUMO gap is small for very conductive metals, as seen in the above diagram.

(When electron promotion occurs, it is said that each electron that moves into the empty orbitals above leaves a "hole" in the filled orbitals below the Fermi level, which is what is depicted on the right portion of the above diagram.)

FRACTION OF MOS FILLED

When we are at $\text{0 K}$, no electrons have been promoted yet (due to any thermal energy imparted due to an increase in temperature, for instance), so all the electrons in the MO diagram above are where they should be at $\text{0 K}$.

Having $2$ valence electrons per $\text{Mg}$ atom, we can fill $1$ MO. With $4$ MOs formed from the LCAO event, we would have needed $8$ valence electrons to fill $4$ MOs.

Hence, with $\text{1 mol}$ of $\text{Mg}$ atoms, we would fill $\text{1 mol}$ of MOs, out of the $\text{4 mol}$s we actually have.

Therefore, 2/8 = 25% of the MOs are occupied by electron pairs, i.e. filled.