Draw the Lewis Dot Structure for PI3 and Provide the Following Information: a) Number of Bond Pairs b) Number of Lone Pairs c) Molecular Geometry d) Hybridization of the Central Atom

To draw the Lewis dot structure for phosphorus triiodide (PI₃), we start by determining the total number of valence electrons. Phosphorus has 5 valence electrons, and each iodine atom has 7 valence electrons. Therefore, for PI₃:

Total valence electrons = 5 (P) + 3 × 7 (I) = 5 + 21 = 26

In the Lewis structure, phosphorus is the central atom bonded to three iodine atoms. We will place one bond between phosphorus and each iodine atom. Each bond consists of a pair of electrons. So, we use 6 electrons for the three P-I bonds:

PI₃ structure:

      I
      | 
  I - P - I

Now, we have 26 – 6 = 20 electrons remaining. Each iodine atom will hold 6 more electrons around it to complete their octets, which uses up all the remaining electrons:

Final Lewis structure:

      I:
      :I 
  I: - P - :I
      :

1. Number of Bond Pairs: There are 3 bond pairs, as phosphorus forms one single bond with each iodine atom.
2. Number of Lone Pairs: The central atom phosphorus has no lone pairs, while each iodine has 3 lone pairs.
3. Molecular Geometry: The molecular geometry of PI₃ is trigonal pyramidal due to the three bonds to iodine and the absence of lone pairs on phosphorus, resulting in a three-dimensional shape.
4. Hybridization of the Central Atom: The hybridization of phosphorus in PI₃ is sp³, as it forms four hybrid orbitals (three for bonding with iodine and one lone pair, which is not shown in the final structure).

In summary, the Lewis structure of PI₃ reveals critical information about its bonding characteristics and geometry.