Predict the molecular structure (including bond angles) for each of the following. (See Exercises 111 and \(112 .\) ) a. \(\mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{3}\) c. \(\mathrm{TeF}_{4}\) d. \(\mathrm{PCl}_{5}\)

For each molecule, we'll first find the total number of valence electrons and divide that by 2 to determine the electron domains around the central atom. Electron domains include bonding electrons and non-bonding (lone pair) electrons. We'll use this information to apply VSEPR theory in the following steps. a. XeCl2: Xe has 8 valence electrons, each Cl has 7 valence electrons. Total valence electrons = (8) + 2(7) = 22 Electron domains = 22 / 2 = 11 b. ICl3: I has 7 valence electrons, each Cl has 7 valence electrons. Total valence electrons = (7) + 3(7) = 28 Electron domains = 28 / 2 = 14 c. TeF4: Te has 6 valence electrons, each F has 7 valence electrons. Total valence electrons = (6) + 4(7) = 34 Electron domains = 34 / 2 = 17 d. PCl5: P has 5 valence electrons, each Cl has 7 valence electrons. Total valence electrons = (5) + 5(7) = 40 Electron domains = 40 / 2 = 20

Now that we have the electron domains around the central atom, we will use VSEPR theory to predict the molecular structure and bond angles. a. XeCl2: Electron domains = 11 (5 bonding, 1 lone pair) Molecular geometry: Trigonal bipyramidal with one lone pair. Molecular structure: Linear Bond angle: 180 degrees b. ICl3: Electron domains = 14 (6 bonding, 1 lone pair) Molecular geometry: Octahedral with one lone pair. Molecular structure: Square pyramidal Bond angle: Approximately 90 degrees c. TeF4: Electron domains = 17 (8 bonding, 1 lone pair) Molecular geometry: Trigonal bipyramidal with one lone pair. Molecular structure: See-saw (disphenoidal) Bond angle: Approximately 120 degrees and 90 degrees d. PCl5: Electron domains = 20 (10 bonding, 0 lone pair) Molecular geometry: Trigonal bipyramidal Molecular structure: Trigonal bipyramidal Bond angle: 120 degrees (equatorial plane) and 90 degrees (above and below equatorial plane)