Write Lewis structures and predict whether each of the following is polar or nonpolar a. \(\mathrm{HOCN}\) (exists as \(\mathrm{HO}-\mathrm{CN} )\) b. \(\mathrm{COS}\) c. \(\mathrm{XeF}_{2}\) d. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) e. \(\mathrm{SeF}_{6}\) f. \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{C} \text { is the central atom })\)

To draw the Lewis structures, we need to determine the number of valence electrons in each molecule, place the least electronegative atom in the center (except for hydrogen, which always goes on the outside), create bonds, and fill in the remaining valence electrons. a. \(\mathrm{HOCN}\): - Total valence electrons: \(1 + 6 + 5 + 4 = 16\) - Central atom: O - Lewis Structure: H-O-C≡N b. \(\mathrm{COS}\): - Total valence electrons: \(4 + 6 + 6 = 16\) - Central atom: C - Lewis Structure: O=C=S c. \(\mathrm{XeF}_{2}\): - Total valence electrons: \(8 + 2 \times 7 = 22\) - Central atom: Xe - Lewis Structure: F-Xe-F, Xe has 3 lone pairs around it. d. \(\mathrm{CF}_2\mathrm{Cl}_2\): - Total valence electrons: \(4 + 2 \times 7 + 2 \times 7 = 28\) - Central atom: C - Lewis Structure: F-C-Cl, 90-degree angles, with F and Cl opposite each other and two lone pairs on C. e. \(\mathrm{SeF}_{6}\): - Total valence electrons: \(6 + 6 \times 7 = 48\) - Central atom: Se - Lewis Structure: Se is surrounded by six F atoms which form a hexagonal configuration. No lone pairs on Se. f. \(\mathrm{H}_{2}\mathrm{CO}\): - Total valence electrons: \(2 \times 1 + 4 + 6 = 12\) - Central atom: C - Lewis Structure: H-C-O; C is bonded to each H and to O with a double bond.

a. \(\mathrm{HOCN}\): Linear (O in the center) b. \(\mathrm{COS}\): Linear c. \(\mathrm{XeF}_{2}\): Linear d. \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\): Tetrahedral e. \(\mathrm{SeF}_{6}\): Octahedral f. \(\mathrm{H}_{2}\mathrm{CO}\): Trigonal planar around carbon

a. \(\mathrm{HOCN}\): Polar due to the difference in electronegativity between H, O, C, and N, and a linear shape, which does not cancel out the dipole moment. b. \(\mathrm{COS}\): Polar due to the difference in electronegativity between O, C, and S, and a linear shape, which does not cancel out the dipole moment. c. \(\mathrm{XeF}_{2}\): Nonpolar due to the linear shape, which helps to cancel out dipole moments between F atoms. d. \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\): Nonpolar due to the symmetrical tetrahedral shape, which cancels out any polar bonds. e. \(\mathrm{SeF}_{6}\): Nonpolar due to the symmetrical octahedral shape, which cancels out any polar bonds. f. \(\mathrm{H}_{2}\mathrm{CO}\): Polar due to the difference in electronegativity between H, C, and O, and a trigonal planar shape, which does not cancel out the dipole moment.