To calculate the ground state magnetic moment of an atom or ion, you need to determine the number of unpaired electrons present in its electron configuration. The electron configuration of fluorine (F) is 1s² 2s² 2p⁵. Let's break down the process step by step:
1. Determine the number of valence electrons: For fluorine (F), the valence electrons are in the 2s and 2p orbitals, which gives us a total of 7 valence electrons.
2. Determine the electron configuration: Starting with the lowest energy level (1s), fill the orbitals with electrons following the Aufbau principle, Pauli exclusion principle, and Hund's rule. For fluorine, the electron configuration is: 1s² 2s² 2p⁵.
3. Determine the number of unpaired electrons: In the 2p sublevel of fluorine, there are three unpaired electrons, as each p orbital can hold a maximum of two electrons, and one of the three orbitals remains unpaired.
4. Calculate the magnetic moment: The magnetic moment (μ) is calculated using the formula: μ = √[n(n+2)], where n represents the number of unpaired electrons. In this case, n = 3, so the magnetic moment of fluorine is:
μ = √[3(3+2)] = √[15] ≈ 3.87 Bohr magnetons.
Therefore, the ground state magnetic moment of fluorine (F) is approximately 3.87 Bohr magnetons.
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