Write orbital diagrams (boxes with arrows in them) to represent the electron configurations—without hybridization—for all the atoms in PH₃. Circle the electrons involved in bonding. Draw a three-dimensional sketch of the molecule and show orbital overlap. What bond angle do you expect from the unhybridized orbitals? How well does valence bond theory agree with the experimentally measured bond angle of 93.3°?

Verified step by step guidance

1. Identify the electron configurations for phosphorus (P) and hydrogen (H). Phosphorus has an atomic number of 15, so its electron configuration is 1s² 2s² 2p⁶ 3s² 3p³. Hydrogen has an atomic number of 1, so its electron configuration is 1s¹.
2. Draw the orbital diagrams for phosphorus and hydrogen. For phosphorus, represent the 3s and 3p orbitals with boxes and fill them with arrows to indicate electrons: 3s (↑↓), 3p (↑) (↑) (↑). For hydrogen, draw a single box for the 1s orbital with one arrow: 1s (↑).
3. Circle the electrons involved in bonding. In PH₃, each hydrogen atom forms a bond with one of the unpaired electrons in the 3p orbitals of phosphorus. Circle one electron in each of the three 3p orbitals of phosphorus and the single electron in the 1s orbital of each hydrogen.
4. Draw a three-dimensional sketch of the PH₃ molecule. Show the phosphorus atom at the center with three hydrogen atoms bonded to it. Indicate the overlap between the 3p orbitals of phosphorus and the 1s orbitals of hydrogen, forming three P-H sigma bonds.
5. Discuss the expected bond angle from unhybridized orbitals. In PH₃, the unhybridized 3p orbitals of phosphorus are oriented at 90° to each other. However, due to the presence of a lone pair on phosphorus, the bond angle is slightly less than 90°. Compare this with the experimentally measured bond angle of 93.3° and discuss how valence bond theory accounts for this deviation.