\(\mathrm{SF}_{6}, \mathrm{ClF}_{5}\), and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

First, we need to determine the total number of valence electrons for each compound. \(\mathrm{SF}_{6}:\) Sulfur has 6 valence electrons, and each fluorine has 7 valence electrons. Since there are 6 fluorine atoms, the total number of valence electrons for this compound is \(6 + (6 \times 7) = 48\). \(\mathrm{ClF}_{5}:\) Chlorine has 7 valence electrons, and each fluorine has 7 valence electrons. Since there are 5 fluorine atoms, the total number of valence electrons for this compound is \(7 + (5 \times 7) = 42\). \(\mathrm{XeF}_{4}:\) Xenon has 8 valence electrons, and each fluorine has 7 valence electrons. Since there are 4 fluorine atoms, the total number of valence electrons for this compound is \(8 + (4 \times 7) = 36\).

In each compound, the central atom will be the one with the least electronegative element. In all three cases, the central atom is given: \(\mathrm{SF}_{6}:\) Sulfur is the central atom. \(\mathrm{ClF}_{5}:\) Chlorine is the central atom. \(\mathrm{XeF}_{4}:\) Xenon is the central atom.

Now we will add single bonds between the central atom and the surrounding fluorine atoms. Then, we will complete the octets of the fluorine atoms with lone pairs. \(\mathrm{SF}_{6}:\) Add 6 single bonds between sulfur and the 6 fluorine atoms and complete the fluorine octets. This uses up 48 electrons, so there are no remaining electrons. \(\mathrm{ClF}_{5}:\) Add 5 single bonds between chlorine and the 5 fluorine atoms and complete the fluorine octets. This uses up 40 electrons, so there are 2 remaining electrons. \(\mathrm{XeF}_{4}:\) Add 4 single bonds between xenon and the 4 fluorine atoms and complete the fluorine octets. This uses up 32 electrons, so there are 4 remaining electrons.

Finally, we will place any remaining electrons on the central atom. \(\mathrm{SF}_{6}:\) No remaining electrons. \(\mathrm{ClF}_{5}:\) Place the 2 remaining electrons on chlorine as a lone pair. \(\mathrm{XeF}_{4}:\) Place the 4 remaining electrons on xenon as two lone pairs. The Lewis structures for these compounds are: \(\mathrm{SF}_{6}:\) ``` F F \ / --S------ / \ F F ``` \(\mathrm{ClF}_{5}:\) ``` F F \ / --Cl----- / \ F F \ F ``` \(\mathrm{XeF}_{4}:\) ``` F F \ / --Xe----- / \ F F ```