# What is the MO diagram of "O"_2 in its ground state, and how do I calculate the bond order?

##### 1 Answer
Jul 4, 2017

The ground state of oxygen molecule is the triplet state, which has two unpaired valence electrons in its ${\pi}^{\text{*}}$ antibonding molecular orbitals:

From this, you should be able to write the electronic configuration by inspection:

overbrace(color(blue)((sigma_(1s))^2 (sigma_(1s)^"*")^2))^"deep core orbitals " overbrace(color(blue)((sigma_(2s))^2 (sigma_(2s)^"*")^2))^"outer core orbitals" underbrace(color(blue)((sigma_(2p_z))^2 (pi_(2p_x))^2(pi_(2p_y))^2 (pi_(2p_x)^"*")^1 (pi_(2p_y)^"*")^1))_"valence orbitals"

It's exactly how you would approach the atomic configuration, with changed notation. Instead of notating atomic orbitals, we notate molecular orbitals.

As you (should) have been taught, $\sigma$ orbitals are made from head-on overlap, and $\pi$ orbitals are made from side-long overlap.

By conservation of orbitals, there must be the same number of molecular orbitals as atomic orbitals.

Its bond order is intuitive. It has a double bond, so its bond order is anticipated to be $\boldsymbol{2}$ (why?). And this can be verified:

$\textcolor{b l u e}{{\text{Bond Order") = 1/2 ("Bonding e"^(-) - "Antibonding e}}^{-}}$

= 1/2 [underbrace(overbrace(2)^(sigma_(1s))+overbrace(2)^(sigma_(2s))+overbrace(2)^(sigma_(2p_z))+overbrace((2 xx 2))^(pi_(2p_(x//y))))_("Bonding e"^(-)'s) - (underbrace(overbrace(2)^(sigma_(1s)^"*")+overbrace(2)^(sigma_(2s)^"*")+overbrace((2 xx 1))^(pi_(2p_(x//y)^"*")))_("Antibonding e"^(-)'s))]

$= \textcolor{b l u e}{2}$