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We will first learn the electron configuration for sulfur and fluorine atoms. Sulfur belongs to the 3rd period and 6th group of the periodic table, so its electron configuration is \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{4}\]. Fluorine belongs to the 2nd period and 7th group of the periodic table, so its electron configuration is \[1s^{2} 2s^{2} 2p^{5}\].

For each sulfur-fluorine compound, distribute the shared and unshared electron pairs between the sulfur and fluorine atoms. SF2: Sulfur has 6 valence electrons, and each fluorine atom has 7 valence electrons. In total, SF2 has 20 valence electrons. The Lewis structure is: F | S | F SF4: Sulfur has 6 valence electrons, and each fluorine atom has 7 valence electrons. In total, SF4 has 34 valence electrons. The Lewis structure is: F | S--F / F SF6: Sulfur has 6 valence electrons, and each fluorine atom has 7 valence electrons. In total, SF6 has 48 valence electrons. The Lewis structure is: F | F--S--F | F /

Now, we will use the VSEPR model to predict the molecular geometries and bond angles for each compound: SF2: The molecular geometry of SF2 is bent because there are two bonding electron pairs and one lone pair on the sulfur atom. The bond angle is approximately 104.5° since sulfur is in the same period as oxygen, which has a bent molecular geometry for water with a similar electron distribution. SF4: The molecular geometry of SF4 is seesaw because there are four bonding electron pairs and one lone pair on the sulfur atom. The bond angles are approximately 102° within the equatorial plane and 173° for the axial fluorine atoms. SF6: The molecular geometry of SF6 is octahedral because there are six bonding electron pairs and no lone pairs on the sulfur atom. All bond angles are 90°.

Oxygen, being in the same group as fluorine, has a similar electron configuration with 6 valence electrons. If we attempt to form OF4, oxygen would need to share eight electrons for bond formation. However, oxygen can share a maximum of four electrons (forming double bonds) because it has only four unpaired electrons available for bonding. Therefore, OF4 would not be a stable compound as oxygen requires more than four available bonding electrons to form a tetra-covalent compound with fluorine.