Problem 110
Question
(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure \(9.46,\) how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HF. Where are the nonbonding electrons?
Step-by-Step Solution
VerifiedKey Concepts
Valence Atomic Orbitals
- The 1s orbital of hydrogen contains one electron.
- Fluorine has its valence electrons distributed in the 2s () and 2p () orbitals.
Understanding the role of valence atomic orbitals is essential for predicting how different elements form bonds and how these bonds affect molecular structure and behavior.
Bonding and Antibonding Orbitals
On the other hand, antibonding orbitals result from destructive overlap, which creates a node between the nuclei where the probability of finding an electron is minimal, thus increasing the system's energy.
- In HF, hydrogen's 1s and fluorine's 2p orbitals primarily interact.
- This interaction produces one bonding (s-2p_z) orbital and one antibonding (s-2p_z)* orbital.
Energy-Level Diagram
- A lower energy bonding orbital s-2p_z).
- Nonbonding orbitals that maintain the energy of fluorine's 2s and two 2p orbitals.
- A higher-energy antibonding orbital, s-2p_z)*, which isn't occupied by electrons.
Lewis Structure
- These nonbonding pairs are represented as dots around fluorine.
- Lone pairs occupy space and affect molecular shape but don't contribute to bond order.
Bond Order
For HF, the bond order is:\[ Bond\ Order = \frac{Number\ of\ electrons\ in\ bonding\ orbitals\ - \ Number\ of\ electrons\ in\ antibonding\ orbitals}{2} \]With 2 electrons in the bonding orbital and none in the antibonding orbital for HF, the bond order is 1, indicating a single, stable bond existing between hydrogen and fluorine.
A higher bond order generally means a more stable and shorter bond. Therefore, as the bond order increases, the energy and strength of the bond also increase. This concept is crucial for predicting the characteristics and reactivity of molecules.