(a) Predict the electron-domain geometry around the central \(\mathrm{S}\) atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). ( \(\mathbf{b}\) ) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

To find the electron domain geometries of \(\mathrm{SF}_2\), \(\mathrm{SF}_4\), and \(\mathrm{SF}_6\), we need to consider the number of bonding pairs and lone pairs around the central sulfur atom. Sulfur (S) is in group 16 and has six valence electrons. In each of the compounds, bond pairs will form as Sulfur bonds with Fluorine (F). - In \(\mathrm{SF}_2\), there are two bond pairs (S-F bonds) and two lone pairs (non-bonding electrons) on the Sulfur atom. - In \(\mathrm{SF}_4\), there are four bond pairs (S-F bonds) and one lone pair on the Sulfur atom. - In \(\mathrm{SF}_6\), there are six bond pairs (S-F bonds) and no lone pairs on the Sulfur atom. Now we can predict the electron-domain geometry for each compound using VSEPR theory: - In \(\mathrm{SF}_2\), the electron-domain geometry is tetrahedral, as there are four electron domains (two bond pairs and two lone pairs). - In \(\mathrm{SF}_4\), the electron-domain geometry is trigonal bipyramidal, as there are five electron domains (four bond pairs and one lone pair). - In \(\mathrm{SF}_6\), the electron-domain geometry is octahedral, as there are six electron domains (six bond pairs).

Given that \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure, we can infer that the central Iodine (I) atom forms double bonds with three Oxygen (O) atoms and a single bond with a negatively charged Oxygen atom. In \(\mathrm{IO}_6^{5-}\), we need to predict the molecular geometry of the molecule considering additional Oxygen atoms. First, let's determine the electron-domain geometry: Iodine (I) is in group 17 and has seven valence electrons. In \(\mathrm{IO}_6^{5-}\), it forms six I-O single bonds, and the molecule has a 5- charge. The central iodine atom needs to gain an additional five electrons to have a 5- charge, so it will have six bond pairs (I-O bonds) and one lone pair. The electron-domain geometry of \(\mathrm{IO}_6^{5-}\) will be pentagonal bipyramidal, as there are seven electron domains (six bond pairs and one lone pair). Now, we can predict the molecular geometry: Since there is one lone pair in the electron-domain geometry, the molecular geometry of \(\mathrm{IO}_6^{5-}\) will be slightly distorted from a perfect pentagonal bipyramidal structure. The lone pair will repel the I-O bond pairs, giving the molecule a "Pentagonal Pyramidal" shape.