Arrange the following molecules from most to least polar and explain your order: \(\mathrm{CH}_{4}, \mathrm{CF}_{2} \mathrm{Cl}_{2}, \mathrm{CF}_{2} \mathrm{H}_{2}, \mathrm{CCl}_{4}\), and \(\mathrm{CCl}_{2} \mathrm{H}_{2}\).

For each molecule, we will use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict their molecular geometry: \(\mathrm{CH}_{4}\): Tetrahedral \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\): Tetrahedral \(\mathrm{CF}_{2} \mathrm{H}_{2}\): Tetrahedral \(\mathrm{CCl}_{4}\): Tetrahedral \(\mathrm{CCl}_{2} \mathrm{H}_{2}\): Tetrahedral

Now, we will analyze the electronegativity values of the atoms bonded and look for any differences. 1. \(\mathrm{CH}_{4}\) - Carbon (C) to Hydrogen (H) difference: 2.55 - 2.20 = 0.35 2. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) - Carbon (C) to Fluorine (F) difference: 3.98 - 2.55 = 1.43 - Carbon (C) to Chlorine (Cl) difference: 3.16 - 2.55 = 0.61 3. \(\mathrm{CF}_{2} \mathrm{H}_{2}\) - Carbon (C) to Fluorine (F) difference: 3.98 - 2.55 = 1.43 - Carbon (C) to Hydrogen (H) difference: 2.55 - 2.20 = 0.35 4. \(\mathrm{CCl}_{4}\) - Carbon (C) to Chlorine (Cl) difference: 3.16 - 2.55 = 0.61 5. \(\mathrm{CCl}_{2} \mathrm{H}_{2}\) - Carbon (C) to Chlorine (Cl) difference: 3.16 - 2.55 = 0.61 - Carbon (C) to Hydrogen (H) difference: 2.55 - 2.20 = 0.35

Now, let's analyze each molecule individually considering both shape and electronegativity differences: 1. \(\mathrm{CH}_{4}\): Symmetrical sp³ hybrid orbitals; the slight electronegativity difference is canceled out due to the symmetric arrangement - Nonpolar. 2. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\): Symmetrical sp³ hybrid orbitals; the electronegativity differences are canceled out due to the symmetric arrangement - Nonpolar. 3. \(\mathrm{CF}_{2} \mathrm{H}_{2}\): Asymmetric sp³ hybrid orbitals; higher electronegativity difference between C and F than between C and H - Polar. 4. \(\mathrm{CCl}_{4}\): Symmetrical sp³ hybrid orbitals; the electronegativity differences are canceled out due to the symmetric arrangement - Nonpolar. 5. \(\mathrm{CCl}_{2} \mathrm{H}_{2}\): Asymmetric sp³ hybrid orbitals; higher electronegativity difference between C and Cl than between C and H - Polar.

Now that we know each molecule's polarity, we can arrange them from most to least polar: 1. \(\mathrm{CF}_{2} \mathrm{H}_{2}\) - Most polar (asymmetric and highest electronegativity difference) 2. \(\mathrm{CCl}_{2} \mathrm{H}_{2}\) - Next in polarity (asymmetric) 3. \(\mathrm{CH}_{4}\) - Nonpolar (symmetrical, small electronegativity difference) 4. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) - Nonpolar (symmetrical) 5. \(\mathrm{CCl}_{4}\) - Least polar (symmetrical) So, the order of polarity is: \(\mathrm{CF}_{2} \mathrm{H}_{2} > \mathrm{CCl}_{2} \mathrm{H}_{2} > \mathrm{CH}_{4} > \mathrm{CF}_{2} \mathrm{Cl}_{2} > \mathrm{CCl}_{4}\).