Which statement below matches the correct response with the proper reasoning when comparing the volatility of ch2cl2 with ch2br2? a. ch2cl2 is more volatile than ch2br2 because the dipole-dipole interactions in ch2cl2 are greater than the dipole-dipole interactions in ch2br2. b. ch2cl2 is more volatile than ch2br2 because of the large dispersion forces in ch2br2. c. ch2br2 is more volatile than ch2cl2because because the dipole-dipole interactions in ch2br2 are greater than the dipole-dipole interactions in ch2cl2d. ch2br2 is more volatile than ch2cl2 because of the large dispersion forces in ch2br2

b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂

Explanation:

CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE

Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.

There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.

I hope it helps!

they are all solids.