Show how a hydrogen 1\(s\) atomic orbital and a fluorine 2\(p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

Short Answer

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In the hydrogen fluoride (HF) molecule, the hydrogen 1\(s\) atomic orbital and fluorine 2\(p\) atomic orbital overlap in a head-on manner to form bonding and antibonding \(\sigma\) molecular orbitals. A bonding \(\sigma\) molecular orbital is created through constructive interference, and its lower energy stabilizes the HF molecule. In contrast, the antibonding \(\sigma^*\) molecular orbital arises through destructive interference and has higher energy. This antibonding orbital remains unoccupied in the ground state of the HF molecule.

Step by step solution

01

Understanding Molecular Orbitals

Molecular orbitals are formed by the interaction of atomic orbitals when two atoms combine to form a chemical bond. The atomic orbitals combine to form new molecular orbitals that belong to the system as a whole. These molecular orbitals can either be bonding (lower energy, more stable) or antibonding (higher energy, less stable). The bonding molecular orbital is formed by constructive interference between the wave functions of atomic orbitals, while the antibonding molecular orbital is formed by destructive interference between their wave functions.
02

Recognizing the Differences Between \(\sigma\) and \(\pi\) Molecular Orbitals

There are two types of molecular orbitals, namely \(\sigma\) and \(\pi\) orbitals: - \(\sigma\) orbitals: These molecular orbitals are cylindrically symmetrical around the bond axis (i.e., the line connecting the nuclei of the two atoms). They are formed by the head-on overlap (end-to-end) of the atomic orbitals. - \(\pi\) orbitals: These molecular orbitals are not cylindrically symmetrical around the bond axis. They are formed by the side-on overlap (side-by-side) of the atomic orbitals.
03

Evaluating the Hydrogen 1\(s\) and Fluorine 2\(p\) Orbitals in HF

In the hydrogen fluoride (HF) molecule, the hydrogen atom contributes with its 1\(s\) atomic orbital, while the fluorine atom participates with one of its three 2\(p\) atomic orbitals (either \(2p_x\), \(2p_y\), or \(2p_z\)). Since the 1\(s\) atomic orbital is spherically symmetrical, and the 2\(p\) atomic orbitals are dumbbell-shaped, the overlap between these orbitals will be head-on, leading to the formation of \(\sigma\) molecular orbitals.
04

Formation of Bonding and Antibonding Molecular Orbitals

When the hydrogen 1\(s\) and fluorine 2\(p\) atomic orbitals overlap in a constructive manner, a bonding \(\sigma\) molecular orbital is formed. This bonding molecular orbital has lower energy and is more stable than the parent atomic orbitals. The electron density in this orbital is concentrated between the two nuclei, leading to the stabilization of the HF molecule. When the hydrogen 1\(s\) and fluorine 2\(p\) atomic orbitals overlap in a destructive manner, an antibonding \(\sigma^*\) molecular orbital is formed. This antibonding molecular orbital has higher energy and is less stable than the parent atomic orbitals. The electron density in this orbital has a node between the two nuclei, leading to repulsion and destabilization of the HF molecule if electrons were to occupy it. However, in the ground state of the HF molecule, this antibonding orbital remains unoccupied. In summary, the hydrogen 1\(s\) and fluorine 2\(p\) atomic orbital overlap in the hydrogen fluoride (HF) molecule produces both bonding and antibonding \(\sigma\) molecular orbitals. The bonding \(\sigma\) molecular orbital is responsible for the chemical bond between hydrogen and fluorine, while the antibonding \(\sigma^*\) molecular orbital remains unoccupied in the ground state.

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Most popular questions from this chapter

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CO} \quad \text { b. } \mathrm{CO}^{+} \quad \text { c. } \mathrm{CO}^{2+} $$

An unusual category of acids known as superacids, which are defined as any acid stronger than 100\(\%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with SbF produces the superacid $\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-} :$ $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of $\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}$ can be prepared when 2.93 \(\mathrm{mL}\) anhydrous \(\mathrm{HF}\) (density $=0.975 \mathrm{g} / \mathrm{mL} )\( and 10.0 \)\mathrm{mL}\( SbFs (density \)=3.10 \mathrm{g} / \mathrm{mL}$ ) are allowed to react?

In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed? $$ \begin{array}{ll}{\text { a. } H_{2}} & {\text { c. } C_{2}^{2-}} \\ {\text { b. } B_{2}} & {\text { d. OF }}\end{array} $$

What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibonding orbitals which is lower in energy? Explain.

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. $\mathrm{H}_{2}^{+}, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}$ b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

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