Lewis structures for expanded-octet (hypervalent) molecules
In a simple Lewis structure, atoms usually obey the octet rule: the central atom has
\(8\) valence electrons (bonding + lone pairs). Some atoms from the
third period and below (like \(\ce{P}\), \(\ce{S}\), \(\ce{Cl}\), \(\ce{Br}\), \(\ce{Xe}\))
can be surrounded by more than \(8\) electrons. These are
expanded-octet or hypervalent molecules, such as
\(\ce{PF5}\), \(\ce{SF6}\), \(\ce{ClF3}\), \(\ce{BrF5}\), \(\ce{XeF2}\), and \(\ce{XeF4}\).
1. Basic Lewis-structure rules
-
Choose the central atom.
In these examples, the central atom is the less electronegative atom:
\(\ce{P}\), \(\ce{S}\), \(\ce{Cl}\), \(\ce{Br}\), or \(\ce{Xe}\).
-
Count total valence electrons.
Add the valence electrons of all atoms (and adjust for any overall charge):
\[
N_{\text{valence}} = \sum N_{\text{valence, atoms}} \;(\pm\; \text{charge}).
\]
-
Draw a skeleton with single bonds.
Connect each surrounding atom to the central atom with one single bond.
Each single bond uses \(2\) electrons:
\[
N_{\text{bond}} = 2 \times (\text{number of single bonds}), \qquad
N_{\text{left}} = N_{\text{valence}} - N_{\text{bond}}.
\]
-
Complete the octets of the terminal atoms.
Use lone pairs on the surrounding \(\ce{F}\) (or other terminal atoms) until each of them
has an octet (8 electrons). Subtract these electrons from \(N_{\text{left}}\).
-
Place remaining electrons on the central atom.
Any electrons still left after terminal octets are satisfied become lone pairs on
the central atom. If the central atom ends up with more than \(8\) electrons
(for example, 10 or 12), the molecule has an expanded octet.
-
Check formal charges.
If needed, you can estimate whether the structure is reasonable by checking
formal charges:
\[
\text{FC} = N_{\text{valence}} - \bigl(N_{\text{nonbonding}} + \tfrac{1}{2}N_{\text{bonding}}\bigr).
\]
For the molecules used in this tool, the simple structures with only single bonds
and complete octets on the terminal \(\ce{F}\) atoms already give acceptable
formal charges (usually \(0\) on the central atom and each \(\ce{F}\)).
2. Short example: \(\ce{PF5}\)
For \(\ce{PF5}\) (phosphorus pentafluoride):
-
Valence electrons:
\[
N_{\text{valence}} = 5_{\ce{(P)}} + 5 \times 7_{\ce{(F)}} = 5 + 35 = 40\ e^-.
\]
-
Five \(\ce{P–F}\) single bonds use
\[
N_{\text{bond}} = 5 \times 2 = 10\ e^-,
\qquad
N_{\text{left}} = 40 - 10 = 30\ e^-.
\]
-
Each \(\ce{F}\) needs three lone pairs (\(6\ e^-\)) to complete its octet:
\[
5 \times 6 = 30\ e^-,
\]
so all remaining electrons go on the \(\ce{F}\) atoms and
\(\ce{P}\) keeps only the bonding pairs. Phosphorus is surrounded by
\(5\) bonds \(= 10\) electrons \(\Rightarrow\) expanded octet.
The calculator follows exactly this electron-counting procedure: it sums valence
electrons, assigns bonding pairs and lone pairs, and then reports how many
electrons surround the central atom in the final Lewis structure.
