Complete the exercises below. The SF₅⁻ ion is formed when SF₄ (g) reacts with fluoride salts containing large cations, such as CsF(s). Draw the Lewis structures for SF₄ and SF₅⁻, and predict the molecular structure of each.

Verified step by step guidance

1. Step 1: Identify the total number of valence electrons for each molecule. For SF₄, sulfur (S) has 6 valence electrons and each fluorine (F) has 7 valence electrons. Therefore, SF₄ has a total of 6 + 4(7) = 34 valence electrons. For SF₅⁻, add one more electron for the negative charge, resulting in 34 + 1 = 35 valence electrons.
2. Step 2: Draw the Lewis structure for SF₄. Place sulfur in the center and arrange the four fluorine atoms around it. Connect each fluorine to sulfur with a single bond. Distribute the remaining electrons to satisfy the octet rule for each fluorine atom. Sulfur can have an expanded octet, so any remaining electrons can be placed on sulfur.
3. Step 3: Draw the Lewis structure for SF₅⁻. Place sulfur in the center and arrange the five fluorine atoms around it. Connect each fluorine to sulfur with a single bond. Distribute the remaining electrons to satisfy the octet rule for each fluorine atom. The extra electron from the negative charge should be placed on sulfur, allowing it to have an expanded octet.
4. Step 4: Predict the molecular geometry of SF₄ using VSEPR theory. SF₄ has 4 bonding pairs and 1 lone pair on the central sulfur atom. This results in a seesaw shape due to the lone pair occupying an equatorial position in a trigonal bipyramidal arrangement.
5. Step 5: Predict the molecular geometry of SF₅⁻ using VSEPR theory. SF₅⁻ has 5 bonding pairs and no lone pairs on the central sulfur atom. This results in a trigonal bipyramidal shape, as all positions are occupied by bonding pairs.