Fluorine reacts with sulfur to form several different covalent compounds. Three of these compounds are \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). Draw the Lewis structures for these compounds, and predict the molecular structures (including bond angles). Would you expect \(\mathrm{OF}_{4}\) to be a stable compound?

To draw the Lewis structures for the given compounds, we need to know how many valence electrons each atom has. Fluorine has 7 valence electrons and sulfur has 6 valence electrons. For \(\mathrm{SF}_{2}\): • S: 6 electrons • F: 7 electrons × 2 = 14 electrons Total = 20 valence electrons For \(\mathrm{SF}_{4}\): • S: 6 electrons • F: 7 electrons × 4 = 28 electrons Total = 34 valence electrons For \(\mathrm{SF}_{6}\): • S: 6 electrons • F: 7 electrons × 6 = 42 electrons Total = 48 valence electrons

For each compound: \(\mathrm{SF}_{2}\): Place sulfur (S) in the center and add two fluorine (F) atoms, then distribute the valence electrons: S: 6 valence electrons - 2 bonds = 4 electrons. F: 7 valence electrons × 2 atoms = 14 electrons Total: 20 electrons Resulting structure: S || F-S-F \(\mathrm{SF}_{4}\): Place sulfur (S) in the center and add four fluorine (F) atoms, then distribute the valence electrons: S: 6 valence electrons - 4 bonds = 2 electrons. F: 7 valence electrons × 4 atoms = 28 electrons Total: 34 electrons Resulting structure: F | F - S - F | F \(\mathrm{SF}_{6}\): Place sulfur (S) in the center and add six fluorine (F) atoms, then distribute the valence electrons: S: 6 valence electrons - 6 bonds = 0 electrons. F: 7 valence electrons × 6 atoms = 42 electrons Total: 48 electrons Resulting structure: F | F - S - F | F - S - F | F

Using VSEPR theory, we can predict the molecular structures for these compounds: \(\mathrm{SF}_{2}\): Two electron domains (two bonding, no lone pairs) Molecular structure: Linear Bond angle: 180° \(\mathrm{SF}_{4}\): Four electron domains (four bonding, no lone pairs) Molecular structure: Tetrahedral Bond angle: 109.5° \(\mathrm{SF}_{6}\): Six electron domains (six bonding, no lone pairs) Molecular structure: Octahedral Bond angle: 90°

Oxygen has six valence electrons, and fluorine has seven valence electrons. To form \(\mathrm{OF}_{4}\), oxygen would need to form four single bonds, which means it would have a total of eight valence electrons, violating the octet rule. Therefore, \(\mathrm{OF}_{4}\) is not likely to be a stable compound.