Bond Polarity, Shape and Boiling Point in a Series of Molecules
A student compares four substances: CH4, NH3, H2O and HF. All have small molecules with hydrogen bonded to a non-metal, yet their boiling points differ markedly: CH4 −162 °C, NH3 −33 °C, HF 20 °C, H2O 100 °C. The student argues that "since all four contain hydrogen, they should all hydrogen-bond, so boiling points should be similar." The student also claims that HF, having the most electronegative partner for hydrogen, must have the strongest hydrogen bonding and therefore the highest boiling point. Electronegativities: H 2.1, C 2.5, N 3.0, O 3.5, F 4.0.
1. Evaluate the student's claim that all four should hydrogen-bond. Use molecular shape and the presence of lone pairs in your reasoning.
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Hydrogen bonding requires H bonded to a highly electronegative atom (N, O or F) that carries lone pairs, and it requires the H to bear a substantial partial positive charge.
CH4. Carbon (2.5) is only slightly more electronegative than H (2.1), so the C–H bond is nearly non-polar and the H carries negligible δ+. Methane also has no lone pairs on carbon. Tetrahedral shape and no N–H/O–H/F–H bond mean no hydrogen bonding - only weak dispersion forces. Hence the very low boiling point.
NH3. N (3.0) is sufficiently electronegative; the trigonal pyramidal molecule has one lone pair on N, so N–H···N hydrogen bonds form. These are moderate, giving a boiling point above CH4 but below H2O and HF.
H2O. O (3.5) is more electronegative and the bent molecule has two lone pairs and two O–H bonds, so each molecule can donate and accept two hydrogen bonds - an extensive network. This explains the anomalously high boiling point.
HF. F is the most electronegative, so each H–F···F bond is individually strong. However, each HF molecule has only one H and three lone pairs, and the small molecule forms short chains rather than a three-dimensional network, so the total intermolecular attraction per molecule is less than in water.
The student's blanket claim is wrong: CH4 cannot hydrogen-bond, and "most electronegative" does not automatically give the highest boiling point because the number of hydrogen bonds per molecule matters.
2. Explain why H2O boils far above HF despite F being more electronegative than O, and predict the trend in surface tension across the series.
Read the worked answer
Why water out-boils HF. Boiling point reflects the total energy needed to separate molecules, not the strength of a single hydrogen bond. Water is bent with two O–H donors and two lone-pair acceptors, so it forms on average about four hydrogen bonds per molecule in a roughly tetrahedral network. HF has one H donor and three lone pairs, but steric and geometric constraints mean each molecule engages in fewer effective hydrogen bonds, forming zig-zag chains. The greater number of hydrogen bonds per water molecule outweighs the greater polarity of the H–F bond. This is why H2O (100 °C) exceeds HF (20 °C).
Surface tension trend. Surface tension rises with the strength and extent of intermolecular attraction: CH4 (dispersion only) is lowest, then NH3, then HF, then H2O highest. Water's extensive hydrogen-bond network gives it the largest surface tension of the four. A good answer links the trend explicitly to the number and strength of hydrogen bonds per molecule rather than to electronegativity alone.