Using the molecular dipoles/polarity of #H_2O#, #NH_3#, and #CH_4#, how do you explain why does #CH_4# does not mix with #H_2O#?

1 Answer
May 22, 2018

Answer:

In short: #"CH"_4# is nonpolar solute whereas water polar a solvent.

Explanation:

Consider the number of electron domains hence the molecular geometry for each of the three species.

Central atoms in the three molecules (#"O"#, #"N"#, and #"C"# respectively, all of which lies in the first three periods of the periodic table) form octets of eight valence electrons. That would be a combination of totally four covalent bonds and lone pairs- hence each of the central atoms has four electron domains. The particular molecular geometry would, therefore, depend on the number of bonding pairs per central atom.

  • Each oxygen atom forms two covalent bonds, one with each hydrogen. Therefore with two bonding pairs out of four electron domains, water molecules share a characteristic asymmetrical bent/V-shaped molecular geometry.

  • Each nitrogen atom has five valence electrons and forms three #"N"-"H"# bonds in an #"NH"_3# molecule. Therefore the nitrogen atom in an #"NH"_3# molecule has three bonding pairs out of its four electron domains and similarly, have a asymmetrical pyramidal geometry.

  • Each carbon atom in #"CH"_4# forms four #"C"-"H"# single bonds. With no lone pairs present, charges in the molecule spread out evenly, giving #"CH"_4# a symmetrical geometry.

With only one single type of covalent bond present, dipole strengths are supposed to be identical in each of the molecules. Dipoles add up vectorially in each particle; due to the symmetrical layout, they cancel out in #"CH"_4# methane molecules, meaning that these molecules carry no net dipole and are nonpolar.

The rule "like dissolves like" suggests that nonpolar solutes tend to have high solubility in nonpolar solvents but not polar solvents, and vice versa for polar solutes. As previously explained, the solute in question here, #"CH"_4#, is a nonpolar substance- whereas the solvent #"H"_2"O"# is highly polar. As a result, it is reasonable that #"CH"_4# barely dissolve in #"H"_2"O"#.