For each compound, draw the Lewis structure, determine the geometry using VSEPR theory, determine whether the molecule is polar, identify the hybridization of all interior atoms, and make a sketch of the molecule according to valence bond theory, showing orbital overlap. a. IF5 b. CH2CHCH3 c. CH3SH

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1. For each compound, start by drawing the Lewis structure: a. For IF5, count the valence electrons (Iodine has 7, each Fluorine has 7, total 42 electrons). Arrange them to satisfy the octet rule for each atom. b. For CH2CHCH3, identify the connectivity (CH2=CH-CH3) and count the valence electrons (Carbon has 4, Hydrogen has 1). c. For CH3SH, count the valence electrons (Carbon has 4, Hydrogen has 1, Sulfur has 6).
2. Determine the geometry using VSEPR theory: a. For IF5, with 5 bonding pairs and 1 lone pair on Iodine, the geometry is square pyramidal. b. For CH2CHCH3, the geometry around each carbon is determined by the number of bonds: sp2 hybridized carbons are trigonal planar, sp3 hybridized carbons are tetrahedral. c. For CH3SH, the geometry around the sulfur is bent due to the lone pairs.
3. Assess the polarity of each molecule: a. IF5 is polar due to the asymmetrical distribution of electrons and the presence of a lone pair. b. CH2CHCH3 is nonpolar overall, but the C=C bond is polar. c. CH3SH is polar due to the S-H bond and the bent shape.
4. Identify the hybridization of all interior atoms: a. In IF5, Iodine is sp3d2 hybridized. b. In CH2CHCH3, the first two carbons are sp2 hybridized, and the last carbon is sp3 hybridized. c. In CH3SH, the carbon is sp3 hybridized, and the sulfur is also sp3 hybridized.
5. Sketch the molecule according to valence bond theory, showing orbital overlap: a. For IF5, show the overlap of sp3d2 hybrid orbitals of Iodine with p orbitals of Fluorine. b. For CH2CHCH3, show the overlap of sp2 orbitals in the double bond and sp3 orbitals in the single bonds. c. For CH3SH, show the overlap of sp3 orbitals of Carbon and Sulfur with Hydrogen's s orbitals.