The molecular shape of the PF5 compound can be predicted from the orbital hybridization of the P atom. Determine the orbitals involved in the hybridization of the P atom. complete with pictures

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To determine the hybridization of the phosphorus (P) atom in phosphorus pentafluoride (PF5), we start by analyzing the electron configuration and geometry of the molecule.

  1. Electron Configuration of Phosphorus:

    • The electron configuration of phosphorus (atomic number 15) is: \[ 1s^2 2s^2 2p^6 3s^2 3p^3 \]
    • The 3s and 3p orbitals are involved in bonding.
  2. Bonding in PF5:

    • In PF5, phosphorus forms five equivalent bonds with five fluorine atoms. To accommodate these five bonds, phosphorus undergoes hybridization.
  3. Hybridization:

    • The hybridization can be explained as follows:
      • The 3s orbital and three 3p orbitals mix to form five equivalent sp³d hybrid orbitals.
      • The electron configuration changes as phosphorus promotes one of its 3s electrons to the 3d orbital, allowing the P atom to form five bonds.
  4. Geometry of PF5:

    • The molecular shape of PF5 is trigonal bipyramidal:
      • Three of the sp³d hybrid orbitals lie in a plane (equatorial), and two of them are positioned above and below this plane (axial).
  5. Visualization:

    • While I cannot provide pictures directly, I can describe how to sketch the arrangement:
      • Draw a central phosphorus atom.
      • From the phosphorus atom, draw five lines radiating outward to represent the bonds with the five fluorine atoms.
      • Position three bonds in a horizontal plane (equatorial positions) like a triangle.
      • Position the remaining two bonds above and below this plane (axial positions).

In summary, the orbitals involved in the hybridization of the phosphorus atom in PF5 are one 3s orbital and three 3p orbitals, leading to the formation of five sp³d hybrid orbitals, resulting in a trigonal bipyramidal molecular geometry.

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