Oxygen
a. What is the electron configuration for oxygen?
b. How many atomic orbitals does each oxygen atom have (hint: only think about the valence electron placements as core electrons are not involved in forming bonds)? Write out what are they (e.g. s, p, etc)
c. Total atomic orbitals for both oxygen atoms?
d. How many molecular orbitals would the two oxygen atoms create? Write out what are they (e.g. bonding or antibonding for each orbital combination)
e. How many core electrons and valence electrons does oxygen have?
f. If two oxygen atoms come together how many electrons total would be present?
g. Using your understanding of energy and atomic orbitals, predict where the electrons for oxygen would be located for the molecular orbital diagram on the next page. Then use it to answer questions
h and i. h. Calculate the bond order of O2.
i. Briefly describe what properties of molecular oxygen are apparent from the MO diagram.
The Correct Answer and Explanation is :
a. Electron Configuration for Oxygen:
The electron configuration for oxygen (O) is:
[
1s^2 2s^2 2p^4
]
Oxygen has 8 electrons in total, with 2 electrons in the first energy level (1s) and 6 electrons in the second energy level (2s^2 2p^4).
b. Atomic Orbitals for Oxygen:
When considering only the valence electrons (the ones involved in bonding), we look at the 2s and 2p orbitals, because core electrons in the 1s orbital do not participate in bonding. Therefore, each oxygen atom has the following atomic orbitals for bonding:
- 2s
- 2p_x, 2p_y, 2p_z (the three p orbitals in the second energy level)
c. Total Atomic Orbitals for Both Oxygen Atoms:
Each oxygen atom contributes four orbitals to the bonding process:
- 1 from the 2s orbital
- 3 from the 2p orbitals
Since there are two oxygen atoms, the total number of atomic orbitals involved in bonding will be:
[
2 \text{ (O atoms)} \times 4 \text{ (orbitals per atom)} = 8 \text{ atomic orbitals in total.}
]
d. Molecular Orbitals Created by Two Oxygen Atoms:
When two oxygen atoms bond, their atomic orbitals combine to form molecular orbitals. For simplicity, let’s assume we’re considering the formation of a (\text{O}_2) molecule and only focus on the valence electrons. The molecular orbitals that result from the combination of these atomic orbitals are:
- Bonding orbitals: Formed by the constructive interference of atomic orbitals.
- (\sigma_{2s}) (from the 2s orbitals)
- (\sigma^*_{2s}) (antibonding from the 2s orbitals)
- (\sigma_{2p_z}) (from the 2p_z orbitals)
- (\pi_{2p_x}) and (\pi_{2p_y}) (from the 2p_x and 2p_y orbitals)
- Antibonding orbitals: Formed by the destructive interference of atomic orbitals.
- (\pi^{2p_x}) and (\pi^{2p_y})
- (\sigma^*_{2p_z})
e. Core and Valence Electrons in Oxygen:
- Core electrons: The electrons in the 1s orbital (2 electrons) are considered core electrons.
- Valence electrons: The electrons in the 2s and 2p orbitals (6 electrons in total) are the valence electrons.
Therefore, oxygen has:
- Core electrons: 2
- Valence electrons: 6
f. Total Electrons When Two Oxygen Atoms Come Together:
Each oxygen atom has 8 electrons, so when two oxygen atoms come together, the total number of electrons is:
[
8 \text{ (electrons per O atom)} \times 2 = 16 \text{ electrons total.}
]
g. Electron Location in the Molecular Orbital Diagram:
For (\text{O}_2), the molecular orbitals would be filled starting with the lowest energy orbitals:
- First, the 2 electrons fill the (\sigma_{2s}) orbital.
- Next, 2 electrons fill the (\sigma^*_{2s}) antibonding orbital.
- Then, 2 electrons fill the (\sigma_{2p_z}) bonding orbital.
- The remaining 8 electrons will fill the (\pi_{2p_x}) and (\pi_{2p_y}) orbitals, with 4 electrons in these bonding orbitals, and the last 2 electrons will fill the (\pi^{2p_x}) and (\pi^{2p_y}) antibonding orbitals.
h. Calculating the Bond Order of O(_2):
The bond order is calculated using the formula:
[
\text{Bond Order} = \frac{1}{2} \left( \text{number of electrons in bonding orbitals} – \text{number of electrons in antibonding orbitals} \right)
]
For O(_2):
- Electrons in bonding orbitals: (2 (\sigma_{2s}) + 2 (\sigma_{2p_z}) + 4 (\pi_{2p_x}, \pi_{2p_y}) = 8)
- Electrons in antibonding orbitals: (2 (\sigma^{2s}) + 2 (\pi^{2p_x}, \pi^*_{2p_y}) = 4)
So the bond order is:
[
\text{Bond Order} = \frac{1}{2} \left( 8 – 4 \right) = \frac{1}{2} \times 4 = 2
]
The bond order for O(_2) is 2, indicating a double bond between the two oxygen atoms.
i. Properties of Molecular Oxygen from the MO Diagram:
From the molecular orbital diagram, we can see that:
- The bond order of 2 indicates a stable double bond between the two oxygen atoms.
- There are two unpaired electrons in the antibonding (\pi^*) orbitals, meaning O(_2) is paramagnetic (it has unpaired electrons and is attracted to a magnetic field).
- The presence of antibonding orbitals reduces the bond strength slightly compared to a pure bonding interaction, but the bond order still indicates a strong double bond between the atoms.
Thus, O(_2) is a stable, paramagnetic molecule with a double bond.