atomic orbital hybridization
Example: N₂ and O₂

Example: O₂⁻, and O₂²⁺

Consider the following molecular ions:
For Li₂, Be₂, B₂, C₂, and N₂ use MO diagram 1. For O₂, F₂, and Ne₂ use MO diagram 2. Then answer the following questions.

Molecular Orbital Configurations
We can fill the MO diagrams for each ion by counting the total number of electrons in the valence shell. Ne₂⁺: 15 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\sigma_{2p}^{})^2 (\pi_{2p}^{})^4 (\pi_{2p}^{*})^4 (\sigma_{2p}^{*})^1 \] N₂⁺: 9 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\pi_{2p}^{})^4 (\sigma_{2p}^{})^1 \] O₂⁻: 13 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\sigma_{2p}^{})^2 (\pi_{2p}^{})^4 (\pi_{2p}^{*})^3 \] C₂²⁺: 6 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\pi_{2p}^{})^2 \] F₂⁺: 13 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\sigma_{2p}^{})^2 (\pi_{2p}^{})^4 (\pi_{2p}^{*})^3 \] O₂²⁻: 14 electrons in n = 2 \[ (\sigma_{2s}^{})^2 (\sigma_{2s}^{*})^2 (\sigma_{2p}^{})^2 (\pi_{2p}^{})^4 (\pi_{2p}^{*})^4 \] solution 1
Paramagnetic: has unpaired electrons
Diamagnetic: no unpaired electrons
Solution 2
Bond order (BO) can be computed using the completed MO diagram using the following formula:
\[ \text{BO} = \frac{1}{2} \left[ (\text{bonding e}^)  (\text{antibonding e}^) \right] \] The bonding orbitals are \(\sigma_{\text{2s}}^{}\), \(\pi_{\text{2p}}^{}\), and \(\sigma_{\text{2p}}^{}\). Adding electrons into bonding orbitals tends to stabilize or strengthen a bond (bond order increases). Removing electrons from bonding orbitals tends to destabilize or weaken a bond (bond order decreases). The antibonding orbitals are \(\sigma_{\text{2s}}^{*}\), \(\pi_{\text{2p}}^{*}\), and \(\sigma_{\text{2p}}^{*}\). Adding electrons into antibonding orbitals tends to destabilize or weaken a bond (bond order decreases). Removing electrons from antibonding orbitals tends to stabilize or strengthen a bond (bond order increases). So, we can compute the bond orders for both the neutral molecule and the molecular ions.
Only Ne₂⁺ and F₂⁺ have bond orders greater than their neutral counterparts (Ne₂ and F₂) because forming the cation requires removing an electron from an antibonding orbital. 

Consider carbon monoxide (CO).
hint
Start by determining the number of total electrons. Then fill the molecular orbitals (MOs) using the same rules for filling atomic orbitals (AOs). solution 1
The electron configuration for carbon and oxygen are:
There are a total of 10 electrons for carbon monoxide (CO) for which we can fill the MO diagram starting from the lowestenergy MO to the highestenergy MO. Click right arrow on the blank MO diagram to see the completed MO diagram for CO. Solution 2
Bond order (BO) can be computed using the completed MO diagram using the following formula: \[ \text{BO} = \frac{1}{2} \left[ (\text{bonding e}^)  (\text{antibonding e}^) \right] \] For carbon monoxide, there are 8 electrons in the bonding orbitals (\(\sigma_{\text{2s}}^{}\), \(\pi_{\text{2p}}^{}\), and \(\sigma_{\text{2p}}^{}\)) and there are 2 electrons in the antibonding orbitals (\(\sigma_{\text{2s}}^{*}\), \(\pi_{\text{2p}}^{*}\), and \(\sigma_{\text{2p}}^{*}\)). Therefore, the bond order is 3, which corresponds to a triple bond. \[ \text{BO} = \frac{1}{2} \left[ 8  2 \right] = 3 \] solution 3
Based on the completed MO diagram, there are no unpaired electrons, so CO is diamagnetic. solution 4
If neutral CO has 10 electrons, then CO⁴⁺ has 6 electrons. We can fill the electrons into the MO diagram for CO⁴⁺ the same way we do for AOs. Remember that we want to fill across with one spin electron before pairing (this is Hund's rule)! Click right arrow to see filledout MO diagram for CO and for CO⁴⁺. Based on the completed MO diagram for CO⁴⁺, there are two unpaired electrons, one unpaired electron in each \(\pi_{\text{2p}}^{}\) orbital, so CO⁴⁺ is paramagnetic. Note: You may notice that I don't fill the electrons on the AOs for carbon and oxygen for CO⁴⁺. The reason is that in that electrons in MO theory are shared completely between the atoms, so we don't need to think about where these electrons come from. solution 5
To form CO⁴⁺, 4 electrons are removed from the bonding orbitals of CO (\(\pi_{\text{2p}}^{}\) and \(\sigma_{\text{2p}}^{}\)), which will weaken the bond. The bond order would then decrease to \[ \text{BO} = \frac{1}{2} \left[ 4  2 \right] = 1 \] 
Click right arrow to see filledout MO diagram for CO (Solution 1) and for CO⁴⁺ (Solution 4). 