Predict the molecular structure (including bond angles) for each of the following. (See Exercises 105 and 106.) a. \(\mathrm{ICl}_{5}\) b. \(\mathrm{XeCl}_{4}\) c. \(\mathrm{SeCl}_{6}\)

For each molecule, identify the central atom. This is usually the atom with the lowest electronegativity. In these molecules, Iodine (I), Xenon (Xe), and Selenium (Se) will be the central atoms.

Count the total number of valence electrons for each molecule. Use the periodic table to find the number of valence electrons for each atom. For \(\mathrm{ICl}_{5}\): Iodine (I) has 7 valence electrons, and Chlorine (Cl) has 7 valence electrons. Therefore, the total number of valence electrons for \(\mathrm{ICl}_{5}\) is: \(1(7) + 5(7) = 42\) For \(\mathrm{XeCl}_{4}\): Xenon (Xe) has 8 valence electrons, and Chlorine (Cl) again has 7 valence electrons. Thus, the total number of valence electrons for \(\mathrm{XeCl}_{4}\) is: \(1(8) + 4(7) = 36\) For \(\mathrm{SeCl}_{6}\): Selenium (Se) has 6 valence electrons, and Chlorine (Cl) has 7 valence electrons. Therefore, the total number of valence electrons for \(\mathrm{SeCl}_{6}\) is: \(1(6) + 6(7) = 48\)

Use the VSEPR theory to determine the electron-pair geometry for each molecule by counting the number of bonding pairs and lone pairs on the central atom. For \(\mathrm{ICl}_{5}\): The central atom (Iodine) has 5 bonding pairs with Chlorine atoms and 1 lone pair. It forms an octahedral electron-pair geometry with 6 electron groups (5 bonding pairs and 1 lone pair) and has a molecular geometry of square pyramid. For \(\mathrm{XeCl}_{4}\): The central atom (Xenon) has 4 bonding pairs with Chlorine atoms and 2 lone pairs. It forms an octahedral electron-pair geometry with 6 electron groups (4 bonding pairs and 2 lone pairs) and has a molecular geometry of square planar. For \(\mathrm{SeCl}_{6}\): The central atom (Selenium) has 6 bonding pairs with Chlorine atoms and no lone pairs. It forms an octahedral electron-pair geometry with 6 electron groups (6 bonding pairs) and has a molecular geometry of octahedral.

We can predict the bond angles in each molecule based on the molecular geometry. For \(\mathrm{ICl}_{5}\): In a square pyramidal geometry, the bond angles between the equatorial atoms are 90°, and the angle between the axial and equatorial atoms is 90°. For \(\mathrm{XeCl}_{4}\): In a square planar geometry, the bond angles between the Cl atoms are 90°. For \(\mathrm{SeCl}_{6}\): In an octahedral geometry, the bond angles between the Cl atoms are 90°. In conclusion, the molecular structures and bond angles are as follows: a. \(\mathrm{ICl}_{5}\) has a square pyramidal molecular structure with bond angles of 90°. b. \(\mathrm{XeCl}_{4}\) has a square planar molecular structure with bond angles of 90°. c. \(\mathrm{SeCl}_{6}\) has an octahedral molecular structure with bond angles of 90°.