The most common exceptions to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3},\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

For each compound or ion, we need to determine the total number of valence electrons for all the atoms involved. We can do this by looking at the group number in the periodic table. To account for negative ions, add the appropriate number of electrons. For \(\mathrm{PF}_5\): P has 5 valence electrons and F has 7. Therefore, the total number of valence electrons is \(5 + 5\times7 = 40\). For \(\mathrm{SF}_4\): S has 6 valence electrons and F has 7. Therefore, the total number of valence electrons is \(6 + 4\times7 = 34\). For \(\mathrm{ClF}_3\): Cl has 7 valence electrons and F has 7. Therefore, the total number of valence electrons is \(7 + 3\times7 = 28\). For \(\mathrm{Br}_3^{-}\): Br has 7 valence electrons. With the negative charge, we add another electron. Therefore, the total number of valence electrons is \(3\times7 + 1 = 22\).

Using the valence electrons, we will draw the Lewis structures for each compound. For central atoms that break the octet rule having more than 8 electrons, distribute the remaining electrons to satisfy the octet rule for the outer atoms. For \(\mathrm{PF}_5\): Place P in the center, and surround it with 5 F atoms. Place single bonds between each P-F pair, and complete the octet for all the F atoms. P will have 10 electrons around it. For \(\mathrm{SF}_4\): Place S in the center, and surround it with 4 F atoms. Place single bonds between each S-F pair, and complete the octet for all the F atoms. S will have 10 electrons around it. For \(\mathrm{ClF}_3\): Place Cl in the center, and surround it with 3 F atoms. Place single bonds between each Cl-F pair, and complete the octet for all F atoms. Cl will have 10 electrons around it. For \(\mathrm{Br}_3^{-}\): Place 3 Br atoms in the line and form 2 single bonds between the neighboring Br atoms. Distribute the remaining electrons to complete the octet rule for each Br atom, adding the extra electron to the central Br atom. The central Br atom will have 10 electrons around it. The Lewis structure of these ions and compounds all have a central atom with an expanded octet, that is, having more than eight electrons around them.

The elements that can have more than eight electrons are those from the third period and beyond in the Periodic Table. This includes P, S, Cl, and Br, as seen in the examples above. The phenomenon occurs due to d-orbitals available in the valence shell of these elements, allowing them to accommodate additional electrons beyond the usual octet. This expanded octet helps these elements form stable compounds with high coordination numbers and gives them various molecular geometries.