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Chapter 11: Problem 33

For the following pairs of molecular orbitals, indicate the one you expect to have the lower energy, and state the reason for your choice. (a) \(\sigma_{1 s}\) or \(\sigma_{1 s}^{*} ;\) (b) \(\sigma_{2 s}\) or \(\sigma_{2 p}\) (c) \(\sigma_{1 s}^{*}\) or \(\sigma_{2 s} ;\) (d) \(\sigma_{2 p}\) or \(\sigma_{2 p}^{*}\)

Short Answer

Expert verified
The orbitals with lower energy are \(\sigma_{1s}\), \(\sigma_{2s}\), \(\sigma_{2s}\), and \(\sigma_{2p}\). In each pair, the one without the '*' superscript has lower energy because anti-bonding orbitals (indicated by '*') have more energy. Among 's' and 'p' types, 's' usually has lower energy.

Step by step solution

01

Understand Molecular Orbits

Molecular orbitals are obtained by combining atomic orbitals, according to quantum physics. The type of atomic orbitals that create the molecular orbitals is indicated by the subscript (e.g., 's' or 'p'). The superscript '*' indicates that the orbital is an anti-bonding orbital with higher energy level. Therefore, for each pair of molecular orbitals, it can be judged which has lower energy based on these rules.
02

Compare The First Pair

Compare \(\sigma_{1s}\) and \(\sigma_{1s}^{*}\). The second one is an anti-bonding orbital, and the first one is not. Anti-bonding orbitals generally have higher energy than bonding orbitals. Therefore, \(\sigma_{1s}\) is assumed to have lower energy.
03

Compare The Second Pair

Compare \(\sigma_{2s}\) and \(\sigma_{2p}\). Both are molecular orbitals, but 's' orbitals usually have lower energy than 'p' orbitals. Therefore, \(\sigma_{2s}\) is assumed to have lower energy.
04

Compare The Third Pair

Compare \(\sigma_{1s}^{*}\) and \(\sigma_{2s}\). The first one is an anti-bonding orbital and the second one is not. Although the first one is of '1s' type and the second one is of '2s' type, the anti-bonding orbital will typically still have higher energy. Therefore, \(\sigma_{2s}\) is assumed to have lower energy.
05

Compare The Fourth Pair

Compare \(\sigma_{2p}\) and \(\sigma_{2p}^{*}\). The second one is an anti-bonding orbital, and the first one is not. Anti-bonding orbitals generally have higher energy than bonding orbitals. Therefore, \(\sigma_{2p}\) is assumed to have lower energy.

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