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Chapter 11: Problem 87

Construct a molecular orbital diagram for \(\mathrm{HF}\), and label the molecular orbitals as bonding, antibonding, or nonbonding.

Short Answer

Expert verified
The molecular orbital diagram for HF consists of one bonding orbital formed from the overlap of Hydrogen's 1s and Fluorine's 2s atomic orbital. The 2p atomic orbitals of Fluorine create nonbonding molecular orbitals as they do not overlap with Hydrogen's atomic orbital.

Step by step solution

01

Determining Atomic Orbitals

The first step is to determine the atomic orbitals of the involved atoms. Hydrogen has 1 electron in the 1s atomic orbital. Fluorine has 2 electrons in the 2s atomic orbital and 5 electrons in the 2p atomic orbitals (2px, 2py, and 2pz). Therefore, Fluorine has a total of 7 valence electrons.
02

Constructing Molecular Orbital Diagram

Now, construct a molecular orbital diagram. The hydrogen 1s atomic orbital can only interact with the 2s atomic orbital of Fluorine, forming a molecular orbital. The 2p atomic orbitals of Fluorine do not overlap with the Hydrogen's 1s and therefore are considered as nonbonding orbitals.
03

Filling Electron in Orbital

Begin filling electrons into the molecular orbitals. Follow the aufbau principle (lower energy levels are filled first), Pauli’s exclusion principle (maximum of two electrons can occupy an atomic orbital and they must have opposite spins) and Hund’s rule (if two or more orbitals have the same energy (degenerate), one electron goes into each until all are half full, the remaining electrons can then be paired up). The one electron from Hydrogen and the one electron from Fluorine's 2s atomic orbital fill up the bonding molecular orbital, and the remaining 6 electrons from Fluorine will fill up the nonbonding orbitals.
04

Labelling the Molecular Orbitals

The final step is to label the molecular orbitals. The molecular orbital formed from the overlap of Hydrogen's 1s and Fluorine's 2s atomic orbital is a bonding orbital. The molecular orbitals formed from Fluorine's 2p atomic orbitals remain nonbonding as they do not overlap with anything.

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Most popular questions from this chapter

The best electrical conductor of the following materials is (a) \(\mathrm{Li}(\mathrm{s}) ;\) (b) \(\mathrm{Br}_{2}(\mathrm{l}) ;\) (c) \(\mathrm{Ge}(\mathrm{s}) ;\) (d) \(\mathrm{Si}(\mathrm{s})\).

Use the valence molecular orbital configuration to determine which of the following species is expected to have the lowest ionization energy: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{C}_{2}\) (c) \(\mathrm{C}_{2}^{-}\)

Delocalized molecular orbitals are found in (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{HS}^{-} ;\) (c) \(\mathrm{CH}_{4} ;\) (d) \(\mathrm{CO}_{3}^{2-}\) .

In which of the following ions would you expect to find delocalized molecular orbitals: (a) \(\mathrm{HCO}_{2}^{-} ;\) (b) \(\mathrm{CO}_{3}^{2-}\) (c) \(\mathrm{CH}_{3}^{+} ?\) Explain.

Write plausible molecular orbital diagrams for the following heteronuclear diatomic species: (a) \(\mathrm{NO} ;\) (b) \(\mathrm{NO}^{+}\) (c) \(\mathrm{CO} ;\) (d) \(\mathrm{CN} ;\) (e) \(\mathrm{CN}^{-} ;\) (f) \(\mathrm{CN}^{+} ;\) (g) BN.
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