To draw the Lewis structure for phosphorus triiodide (PI3), we start by determining the total number of valence electrons. Phosphorus (P) has 5 valence electrons and each iodine (I) atom has 7 valence electrons. Thus, the total is:
5 (from P) + 3 × 7 (from 3 I) = 5 + 21 = 26 valence electrons.
Next, we place phosphorus in the center as it is less electronegative than iodine. We then attach the three iodine atoms to phosphorus with single bonds. This consumes 6 electrons (2 for each bond), leaving us with:
26 – 6 = 20 valence electrons remaining.
Now, to complete the octets of the iodine atoms, we need to add 3 lone pairs (6 electrons) to each iodine, which utilizes another 18 electrons:
20 – 18 = 2 valence electrons remain.
These 2 electrons can be placed as a lone pair on the phosphorus atom. Therefore, the final Lewis structure of PI3 has phosphorus with one lone pair and each iodine having 3 lone pairs:
Regarding the necessity of an expanded valence shell, it is not needed for phosphorus in this case. Phosphorus can have an expanded octet when it forms bonds with more electronegative elements or exhibits the need for more bonds, but in PI3, it completes its valence without needing to expand. Therefore, phosphorus is stable with the lone pair and the three bonded iodine atoms.