# How do I draw the MO diagrams for "O"_2^- and "CO"^(+) and find out which one is paramagnetic?

Aug 15, 2017

Well, the MO diagrams for ${\text{O}}_{2}$ and $\text{CO}$ (i.e. both neutral compounds) are below.

• For $\underline{{\text{O}}_{2}^{-}}$, just take the MO diagram of ${\text{O}}_{2}$ and add one electron into the ${\pi}_{2 p x}^{\text{*}}$ antibonding molecular orbital.
• For $\underline{{\text{CO}}^{+}}$, just take out one of the electrons from the $3 \sigma$ HOMO from the MO diagram of $\text{CO}$.

Both substances are paramagnetic, as they have at least one unpaired electron in a given orbital. ${\text{O}}_{2}^{-}$ has the unpaired electron in a ${\pi}^{\text{*}}$ antibonding orbital, and ${\text{CO}}^{+}$ has the unpaired electron in a $\sigma$ bonding orbital.

The unpaired electron orients itself in the direction of an applied magnetic field, and gives rise to a total magnetic moment ${\mu}_{S + L} = 2.00023 \sqrt{S \left(S + 1\right) + \frac{1}{4} L \left(L + 1\right)}$ in units of $\text{Bohr Magnetons}$. It is a function of

• the total electron spin $S = | {\sum}_{i} {m}_{s , i} |$
• the total orbital angular momentum $L = | {\sum}_{i} {m}_{l , i} |$

for the $i$th electron, where ${m}_{l}$ and ${m}_{s}$ are the angular momentum and spin quantum numbers, respectively.

The total magnetic moment is a value that can be obtained experimentally from magnetic susceptibility measurements.