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

First, determine the number of valence electrons for each atom in the compound and add them together to find the total number of valence electrons. The number of valence electrons for an element corresponds to its group number in the periodic table. For \(\mathrm{SF}_{6}\): Sulfur (S): 6 valence electrons (Group 16) Fluorine (F): 7 valence electrons (Group 17) Total = 6 + 6(7) = 48 valence electrons For \(\mathrm{ClF}_{5}\): Chlorine (Cl): 7 valence electrons (Group 17) Fluorine (F): 7 valence electrons (Group 17) Total = 7 + 5(7) = 42 valence electrons For \(\mathrm{XeF}_{4}\): Xenon (Xe): 8 valence electrons (Group 18) Fluorine (F): 7 valence electrons (Group 17) Total = 8 + 4(7) = 36 valence electrons

Next, arrange the atoms in the compound. The central atom should be the one that does not follow the octet rule. For each compound, arrange the remaining atoms evenly around the central atom. For \(\mathrm{SF}_{6}\): S is the central atom with six F atoms surrounding it. For \(\mathrm{ClF}_{5}\): Cl is the central atom with five F atoms surrounding it. For \(\mathrm{XeF}_{4}\): Xe is the central atom with four F atoms surrounding it.

Connect the atoms using single covalent bonds, which consist of one pair of shared electrons. We will also subtract the electrons involved in these bonds from our total number of valence electrons for each compound. For \(\mathrm{SF}_{6}\): Connect each F atom to the S atom, using 2(6) = 12 electrons, leaving 48 - 12 = 36 electrons. For \(\mathrm{ClF}_{5}\): Connect each F atom to the Cl atom, using 2(5) = 10 electrons, leaving 42 - 10 = 32 electrons. For \(\mathrm{XeF}_{4}\): Connect each F atom to the Xe atom, using 2(4) = 8 electrons, leaving 36 - 8 = 28 electrons.

Finally, add lone pairs (pairs of electrons not involved in covalent bonding) to each atom to complete the Lewis structure, using the remaining valence electrons calculated in step 3. For \(\mathrm{SF}_{6}\): Each F atom receives 3 lone pairs (using 6(6) = 36 electrons). Lewis structure for \(\mathrm{SF}_{6}\): \[ \chemfig{F-[:30]S(-[:90]F)(-[:150]F)(-[:210]F)(-[:270]F)-[:330]F} \] For \(\mathrm{ClF}_{5}\): Each F atom receives 3 lone pairs (using 6(5) = 30 electrons). The Cl atom receives 1 lone pair (using 2 electrons). Lewis structure for \(\mathrm{ClF}_{5}\): \[ \chemfig{F-[:18]Cl(-[:90]F)(-[:162]F)(-[:234]F)(-[:306]F)<:18(F)} \] For \(\mathrm{XeF}_{4}\): Each F atom receives 3 lone pairs (using 6(4) = 24 electrons). The Xe atom receives 2 lone pairs (using 4 electrons). Lewis structure for \(\mathrm{XeF}_{4}\): \[ \chemfig{F-Xe(-[:90]F)(-[:180]F)(<:0|--[:30]F)(-[:270]F)} \]