atomic orbital hybridization
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Example: N₂ and O₂
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Example: O₂⁻, and O₂²⁺
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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.
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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. |
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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 lowest-energy MO to the highest-energy 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 filled-out 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 filled-out MO diagram for CO (Solution 1) and for CO⁴⁺ (Solution 4). |