What is the magnetism of F2+, and how
many unpaired electrons does it possess?
Would F2+ act the same way as F+? In that case it would be paramagnetic and have 2 electrons.
To determine the magnetism and number of unpaired electrons in F2+ (fluorine cation), we need to look at its electronic configuration.
The electronic configuration of a neutral fluorine atom (F) is 1s^2 2s^2 2p^5, with 7 valence electrons. When F loses one electron to form F+, its electronic configuration becomes 1s^2 2s^2 2p^5, with 6 valence electrons.
F2+ indicates that two fluorine atoms have lost one electron each. Therefore, the electron configuration of F2+ can be represented as the combination of two F+ ions: (1s^2 2s^2 2p^5) + (1s^2 2s^2 2p^5).
Combining the two configurations, we obtain (1s^2 2s^2 2p^5) (1s^2 2s^2 2p^5).
In order to determine the magnetism and number of unpaired electrons, we need to first construct the molecular orbital diagram for F2+.
In the molecular orbital diagram, we fill the molecular orbitals with the available electrons, starting from the lowest energy level. Each orbital has a maximum capacity of two electrons, each with opposite spins.
The molecular orbital diagram for F2+ would look like this:
σ2s* ↑↓
σ2s ↑↓
π2p ↑↓
π*2p ↑↓
σ2p ↑
The arrows in the molecular orbital diagram represent the electrons. Upward arrows indicate one electron with a positive spin, and downward arrows indicate one electron with a negative spin.
By counting the number of unpaired electrons, we see that F2+ has two unpaired electrons. Therefore, F2+ is indeed paramagnetic, and it possesses two unpaired electrons.
So, in summary, F2+ does not act the same way as F+. It has two unpaired electrons and exhibits paramagnetism.