How does electronegativity affect interactions between water molecules?

1 Answer
Apr 3, 2017

Charge separation results in a strongly dipolar molecule..........

Explanation:

Molecules which comprise two elements of vastly different electronegativity, give rise to strongly polar entities: and the best examples of this property is #HF#, #OH_2#, and #NH_3#. Clearly, the heteroatom (the halogen, oxygen, or nitrogen atom) is the more electronegative centre, and it STRONGLY polarizes electron towards itself to give systems that we could represent as #""^(delta+)H-F^(delta-)#, or #""^(delta+)H-O^(delta-)-H^(delta+)#, or #""^(delta+)H_3N^(delta-)#, which gives rise to a strong molecular dipole, and a polar, i.e. charge separated molecule.

In the condensed phase, the dipoles line up appropriately, with the result that the physical constants of these first row hydrides, as manifested by their melting/boiling point, are vastly different to their 2nd and 3rd congeners. The normal boiling point of water is #100# #""^@C# (ridiculously hi for a molecule of this size!), of ammonia, #-33.3# #""^@C#, hydrogen fluoride #19.5##""^@C#.

And let's compare these constants to those of #PH_3#, #H_2S#, and #HCl#, which are #-87.7# #""^@C#, #-60.0# #""^@C#, and #-85.0# #""^@C#. The difference in volatility, the reduced boiling point, may be attributed directly to the degree that the INTERMOLECULAR interaction is reduced, and here the dominant intermolecular interaction is hydrogen bonding.

Note that as always we define #"electronegativity"# as the ability of an atom in a molecule to polarize electron density towards itself. Chalcogen, and halogen atoms, Group 16, and Group 17 respectively, tend to be highly electronegative, and the volatility of their hydrides ultimately reflect this.