What is the molecular orbital bond order for OF^-?
Is the correct answer 1?
0.5(10-8) = 1
10 bonding electrons and 8 anti-bonding electrons.
To determine the molecular orbital bond order for OF^-, you need to follow these steps:
1. Write down the electron configuration for the atoms involved. Oxygen (O) has an atomic number of 8, so its electron configuration is 1s^22s^22p^4. Fluorine (F) has an atomic number of 9, so its electron configuration is 1s^22s^22p^5.
2. Draw the molecular orbital diagram by combining the atomic orbitals of the two atoms. In this case, you'll be combining the 2p orbitals of oxygen and fluorine.
3. Fill the molecular orbitals with electrons by referencing the electron configurations of the atoms. Refer to Hund's Rule, which states that electrons will half-fill degenerate orbitals before pairing up. Start filling from the lowest energy molecular orbital and move up.
4. Count the number of electrons in the bonding orbitals and the antibonding orbitals. Bonding orbitals bring electrons closer together and stabilize the molecule, while antibonding orbitals pull electrons apart and destabilize the molecule.
In the case of OF^-, the molecular orbital diagram would look like this:
σ*2p (antibonding)
σ2p (bonding)
π2p (bonding)
π*2p (antibonding)
Based on the electron configurations of oxygen and fluorine, their electrons would fill the two bonding orbitals (σ2p and π2p) and leave the antibonding orbitals (σ*2p and π*2p) empty.
Therefore, there would be a total of 10 bonding electrons and 0 anti-bonding electrons in OF^-. The bond order is calculated by taking the difference between the number of bonding and antibonding electrons and dividing it by 2.
Bond order = (Number of bonding electrons - Number of antibonding electrons)/2
In this case, the bond order would be (10 - 0)/2 = 5/2 = 2.5, not 1.
Hence, the correct bond order for OF^- is 2.5, not 1.