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Chapter 10: Problem 59

Use molecular orbital theory to compare the relative stabilities of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{+}\)

Short Answer

Expert verified
By applying molecular orbital theory, it is observed that \(\mathrm{F}_{2}^{+}\) is more stable than \(\mathrm{F}_{2}\) because it contains less electrons in its anti-bonding orbitals.

Step by step solution

01

Determine Number of Electrons in Both Molecules

Each Fluorine in \(\mathrm{F}_{2}\) has 9 electrons, so \(\mathrm{F}_{2}\) in total has 18 electrons. On the other hand, since \(\mathrm{F}_{2}^{+}\) is a cation, has lost one electron, and so has a total of 17 electrons.
02

Distribution of Electrons in Molecular Orbitals

Distribute these electrons into the Molecular Orbitals (MOs) for both molecules according to the Aufbau principle, Hund's rule and Pauli exclusion principle. In \(\mathrm{F}_{2}\), 18 electrons occupy the orbitals in following order: \( \sigma(2s), \sigma^*(2s), \sigma(2pz), \pi(2px) = \pi(2py), \pi^*(2px) = \pi^*(2py) \) and \(\sigma^*(2pz) \). In \(\mathrm{F}_{2}^{+}\), 17 electrons occupy the same orbitals till \(\pi^*(2px) = \pi^*(2py)\), but only one electron is in \(\sigma^*(2pz)\) which is less than \(\mathrm{F}_{2}\)
03

Judging Stability

Assess the stability of the molecules. The less electrons in the anti-bonding orbitals, the more stable is the molecule or ion. Therefore, \(\mathrm{F}_{2}^{+}\) is more stable than \(\mathrm{F}_{2}\) because it has less electrons in its anti-bonding orbitals.

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