U3.1 · Intermolecular ForcesPractice · 分子间作用力

Boiling is a physical change: the $\ce{N2}$ molecules stay intact and just move apart. Breaking the N≡N bond takes 941 kJ/mol, far more than boiling liquid nitrogen needs. $\ce{N2}$ is nonpolar, so dispersion forces are the only forces between its molecules, and those are what boiling overcomes.

Boiling points (CRC): NH₃ −33.3 °C, PH₃ −87.8 °C, AsH₃ −62.5 °C. PH₃ and AsH₃ have only dispersion forces and weak dipole–dipole forces; AsH₃ has more electrons, so stronger dispersion forces and a higher boiling point. In NH₃, H is bonded to N, so the molecules hydrogen-bond: NH₃ has the fewest electrons but the highest boiling point.

$\ce{CH4}$ is a symmetric tetrahedron with a dipole moment of 0, so it has only dispersion forces. $\ce{CH3Cl}$ swaps one H for the more electronegative Cl; the bond dipoles no longer cancel (1.89 D), so dipole–dipole forces appear. Every molecule has dispersion forces; the H atoms in $\ce{CH3Cl}$ are on C, so there is no hydrogen bonding; and there are no ions, so no ion–dipole forces.

For molecules to hydrogen-bond with each other, H must be bonded directly to N, O, or F. Formic acid, HCOOH, has O–H; hydrazine, $\ce{H2N-NH2}$, has N–H; and hydrogen peroxide, HO–OH, has O–H, so all three hydrogen-bond. $\ce{CH3F}$ and acetaldehyde, $\ce{CH3CHO}$, contain F or O, but every H is bonded to C, so there is no hydrogen bonding between their molecules (acetaldehyde can still accept hydrogen bonds from water, which is a different situation).

From HCl to HI the molecules have more electrons and more polarizable electron clouds, so dispersion forces grow a lot. The dipole moment shrinks, but dipole–dipole forces are only a small part of the attraction here, so dispersion sets the boiling-point order. Boiling doesn’t break H–I bonds (H–I is actually the weakest of the three, 299 kJ/mol), and H on I can’t hydrogen-bond.