Hydride Anomalies: Separating Electronegativity from Hydrogen Bonding
In a practical class, students are given a table of measured data for group 16 and group 15 hydrides and asked to account for the anomalies rather than simply quote them. The recorded values are: boiling points - H2O 373 K, H2S 213 K, H2Se 232 K, H2Te 269 K; NH3 240 K, PH3 185 K, AsH3 211 K. Bond angles - H2O 104.5°, H2S 92°, H2Se 91°, NH3 107.8°, PH3 93.6°. Dipole moments - H2O 1.85 D, H2S 0.97 D, NH3 1.47 D, PH3 0.58 D. One student argues that the whole pattern follows from electronegativity: because oxygen is more electronegative than sulfur, H2O must have the larger bond angle, the larger molecular dipole and therefore the higher boiling point, and H2S boils low simply because its molecular dipole is small. The teacher accepts part of this argument and rejects the rest, asking the class to say exactly which parts survive scrutiny.
1. Explain the trend in boiling point from H2S to H2Te, and account for the position of H2O and of NH3 in the same series.
Read the worked answer
From H2S (213 K) to H2Se (232 K) to H2Te (269 K) the boiling point rises steadily. All three are non-hydrogen-bonded, and molar mass and polarisability increase down the group, so London dispersion forces between molecules become stronger. H2O at 373 K breaks the trend completely: each water molecule has two O–H bonds and two lone pairs on oxygen, so it can act as both donor and acceptor and form up to four hydrogen bonds per molecule, building an extended network that needs far more energy to disrupt. H2S cannot do this - sulfur is larger and less electronegative, so the S–H···S interaction is too weak. NH3 also hydrogen-bonds, but each molecule has only one lone pair on a single small N atom, so the extent of hydrogen bonding per molecule is smaller; its boiling point (240 K) is therefore well above PH3 (185 K) but below H2O.
2. Both series show the bond angle falling towards about 90° in the third period (104.5°→92° for H2O→H2S; 107.8°→93.6° for NH3→PH3). Explain this drop without using electronegativity of the central atom as the cause.
Read the worked answer
The angle is set by the orbitals the central atom actually uses for bonding. In H2O and NH3 the compact 2s and 2p orbitals mix efficiently, so the central atom is close to sp3: about 109.5°, compressed in H2O by repulsion from two lone pairs (104.5°) and in NH3 by one lone pair (107.8°). In H2S and PH3 the valence orbitals are larger and more diffuse, the energetic advantage of mixing s with p is small, and the bonds are formed from almost pure p orbitals, which are mutually perpendicular; the s orbital stays largely as the lone pair. Hence the angles fall to about 92° and 93.6°, close to 90°. The drop is a hybridisation/VSEPR effect, not an electronegativity effect - note that the less electronegative central atoms S and P give the smaller angles, which is the opposite of what a simple electronegativity argument would predict.
3. Assess the student's electronegativity argument claim by claim, stating which parts are correct and which are not.
Read the worked answer
Claim 1, the O–H bond dipole is larger than S–H: correct, because oxygen is more electronegative and the bond is shorter. Claim 2, the H2O molecular dipole is larger: correct (1.85 D vs 0.97 D), but the reasoning needs care - the molecular dipole is a vector sum, and the wider H–O–H angle would by itself give a smaller resultant, so the larger bond dipole must dominate. Claim 3, the larger angle is caused by electronegativity: wrong; the angle follows from sp3 hybridisation and lone-pair repulsion, not from electronegativity. Claim 4, H2S boils low because its dipole is small: wrong reasoning that reaches a partly right conclusion. H2S has a higher molar mass (34 vs 18) and stronger dispersion forces than H2O, yet boils 160 K lower; the dominant factor is that H2O hydrogen-bonds and H2S cannot. Dipole magnitude alone would not predict the 373 K boiling point of water.