The structure of \(\mathrm{TeF}_{5}^{-}\) is Draw a complete Lewis structure for \(\mathrm{TeF}_{5}^{-}\), and explain the distortion from the ideal square pyramidal structure. (See Exercise 106.)

First, we need to determine the electron configuration and number of valence electrons for Tellurium (Te) and Fluorine (F). For Te: The electron configuration of Te is [Kr] 4d10 5s2 5p4. As Te belongs to group 16, it has 6 valence electrons. For F: The electron configuration of F is [He] 2s2 2p5. As F belongs to group 17, each Fluorine atom has 7 valence electrons.

There are 5 Fluorine atoms and 1 Tellurium atom in the structure. Together, they bring in a total of: (6 valence electrons for Te) + (5 × 7 valence electrons for F) = 41 valence electrons. However, since the overall charge of the molecule is -1, we will have one extra electron. Therefore, the total number of valence electrons in the \(\mathrm{TeF}_{5}^{-}\) ion will be 42.

To draw the Lewis structure, we start by placing the Te atom in the center and attaching the five F atoms. We will then share electrons between Te and F atoms to make single bonds. Finally, we will distribute the remaining electrons as lone pairs. - Te is in the center, so place it in the middle. - Connect each F atom to Te with a single bond. It will use up 10 valence electrons (5 bonds × 2 electrons per bond). Now, we have 32 (42 - 10) valence electrons left to distribute as lone pairs. - Each F atom has 3 remaining lone pairs of electrons, so that will use up 30 valence electrons (5 F atoms × 3 pairs × 2 electrons per pair). - The remaining 2 valence electrons will form a lone pair on Te. The Lewis structure for the \(\mathrm{TeF}_{5}^{-}\) ion looks like this: F | F-Te-F-LP | F Where "LP" represents the lone pair on Te. Each F atom has 3 lone pairs (not shown).

In an ideal square pyramidal structure, all five F atoms surrounding the Te atom would have equal bond angles of 90 degrees between them. However, in \(\mathrm{TeF}_{5}^{-}\), the lone pair of electrons on the Te atom exerts a repulsive force on the bonding electrons between the Te and F atoms. This repulsion causes a distortion in the structure, resulting in bond angles that are either larger or smaller than 90 degrees. Additionally, the presence of the lone pair on the central Te atom creates a steric hindrance, which can create further distortion in the molecular structure of \(\mathrm{TeF}_{5}^{-}\). In summary, the presence of the lone pair on the central Te atom in the \(\mathrm{TeF}_{5}^{-}\) ion causes distortion from the ideal square pyramidal structure due to the repulsive forces exerted on bonding electrons and steric hindrance.