Classify the bonding in each of the following molecules as ionic, polar covalent, or nonpolar covalent. a. \(\mathrm{H}_{2}\) b. \(K_{3} P\) c. Nal d. \(\mathrm{SO}_{2}\) e. HF f. \(\mathrm{CCl}_{4}\) g. \(\mathrm{CF}_{4}\) \(\mathbf{h} . \mathbf{K}_{2} \mathbf{S}\)

For each molecule, find the electronegativity difference using a table of electronegativity values. a. \(\mathrm{H}_{2}\): Electronegativity of H = 2.20 (both atoms) Difference = \(|2.20 - 2.20|\) = 0.00 b. \(K_{3}P\): Electronegativity of K = 0.82 Electronegativity of P = 2.19 Difference = \(|0.82 - 2.19|\) = 1.37 c. NaI: Electronegativity of Na = 0.93 Electronegativity of I = 2.66 Difference = \(|0.93 - 2.66|\) = 1.73 d. \(\mathrm{SO}_{2}\): Electronegativity of S = 2.58 Electronegativity of O = 3.44 Difference = \(|2.58 - 3.44|\) = 0.86 e. HF: Electronegativity of H = 2.20 Electronegativity of F = 3.98 Difference = \(|2.20 - 3.98|\) = 1.78 f. \(\mathrm{CCl}_{4}\): Electronegativity of C = 2.55 Electronegativity of Cl = 3.16 Difference = \(|2.55 - 3.16|\) = 0.61 g. \(\mathrm{CF}_{4}\): Electronegativity of C = 2.55 Electronegativity of F = 3.98 Difference = \(|2.55 - 3.98|\) = 1.43 h. \(K_{2}S\): Electronegativity of K = 0.82 Electronegativity of S = 2.58 Difference = \(|0.82 - 2.58|\) = 1.76

Using the electronegativity difference, classify each bond as ionic, polar covalent, or nonpolar covalent: a. \(\mathrm{H}_{2}\): Nonpolar covalent (difference = 0.00) b. \(K_{3}P\): Ionic (difference = 1.37, but the bond is between a metal and a non-metal) c. NaI: Ionic (difference = 1.73) d. \(\mathrm{SO}_{2}\): Polar covalent (difference = 0.86) e. HF: Ionic (difference = 1.78) f. \(\mathrm{CCl}_{4}\): Polar covalent (difference = 0.61) g. \(\mathrm{CF}_{4}\): Polar covalent (difference = 1.43) h. \(K_{2}S\): Ionic (difference = 1.76)