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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

Expert verified
In the hydrogen fluoride (HF) molecule, the hydrogen 1s atomic orbital and the fluorine half-filled 2pz atomic orbital overlap to form bonding and antibonding molecular orbitals, which are both \(\sigma\) (sigma) molecular orbitals. This overlap is symmetric around the bond axis, resulting in a strong covalent bond that stabilizes the hydrogen fluoride molecule.

Step by step solution

01

Understand orbital overlap

Orbital overlap is the process by which atomic orbitals of two atoms combine to form molecular orbitals in a new molecule. In the formation of a covalent bond, the atomic orbitals of the bonding atoms overlap, leading to the sharing of the electron pair between atoms. This results in a region of increased electron density between the atoms, creating a stable molecule.
02

Hydrogen 1s atomic orbital and Fluorine 2p atomic orbital

The hydrogen atom has only one electron, which occupies the lowest energy 1s atomic orbital. The fluorine atom has nine electrons, with two electrons occupying the 1s atomic orbital, two electrons occupying the 2s atomic orbital, and five electrons occupying the 2p atomic orbital. In the hydrogen fluoride molecule (HF), one electron from the hydrogen 1s orbital and one electron from the fluorine half-filled 2pz atomic orbital will combine to form a new molecular orbital.
03

Formation of bonding and antibonding molecular orbitals

When two atomic orbitals overlap, two new molecular orbitals are formed: a bonding molecular orbital and an antibonding molecular orbital. The bonding molecular orbital is formed when there's constructive interference between overlapping orbitals, resulting in a region of higher electron density and lower energy. Electrons in the bonding orbital stabilize the molecule. The antibonding molecular orbital is formed when there's destructive interference between overlapping orbitals, resulting in a region of lower electron density and higher energy. Electrons in the antibonding orbital destabilize the molecule.
04

Type of molecular orbitals formed in hydrogen fluoride

When hydrogen's 1s atomic orbital and fluorine's 2p atomic orbital overlap, a \(\sigma\) (sigma) type molecular orbital is formed. The reason for this is that a \(\sigma\) molecular orbital is symmetric around the bond axis, while a \(\pi\) (pi) molecular orbital is asymmetric with respect to the bond axis. In the hydrogen fluoride molecule, since the overlapping orbitals are symmetric around the bond axis, the resulting molecular orbitals will be \(\sigma\) molecular orbitals.
05

Conclusion

In the hydrogen fluoride (HF) molecule, the hydrogen 1s atomic orbital and the fluorine 2p atomic orbital overlap to form bonding and antibonding molecular orbitals, which are both \(\sigma\) (sigma) molecular orbitals. This overlap is symmetric around the bond axis, resulting in a strong covalent bond that stabilizes the hydrogen fluoride molecule.

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Most popular questions from this chapter

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed? a. \(\mathrm{H}_{2}\) c. \(\mathrm{C}_{2}^{2-}\) b. \(\mathrm{B}_{2}\) d. \(\mathrm{OF}\)

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

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 \(\mathrm{SbF}_{5}\) 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 HF (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

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