Chemical Bonding: Question 4

Worked solution

Part (a): Why water boils at a much higher temperature than hydrogen sulfide

Both molecules have a bent shape with two lone pairs on the central atom, so both have permanent dipole-dipole forces and van der Waals’ (London dispersion) forces between molecules. The key difference is hydrogen bonding.

• In $\text{H}_2\text{O}$, oxygen is very electronegative (much more so than hydrogen) and each O atom carries $2$ lone pairs. Because H is bonded directly to this small, highly electronegative atom, the O–H bond is extremely polar, giving H a large $\delta^+$ charge. This allows hydrogen bonds to form between the $\delta^+$ H of one molecule and a lone pair on the O of a neighbouring molecule.
• In $\text{H}_2\text{S}$, sulfur is far less electronegative than oxygen, so the S–H bond is much less polar, and sulfur (despite also having lone pairs) is not electronegative enough for hydrogen bonding to occur. $\text{H}_2\text{S}$ molecules are held together only by the weaker permanent dipole-dipole forces and van der Waals’ forces.

Hydrogen bonds are considerably stronger than permanent dipole-dipole or van der Waals’ forces (though still much weaker than a covalent bond), so much more energy is needed to separate water molecules from one another than to separate $\text{H}_2\text{S}$ molecules. This extra energy requirement is why water’s boiling point ($100\,^\circ\text{C}$) is so much higher than that of $\text{H}_2\text{S}$ ($-60\,^\circ\text{C}$).

Part (b): How a hydrogen bond forms between water molecules

Three conditions are needed for hydrogen bonding to occur:

1. A hydrogen atom must be covalently bonded directly to a small, highly electronegative atom, nitrogen, oxygen or fluorine.
2. That electronegative atom must carry at least one lone pair of electrons.
3. A neighbouring molecule must have an electronegative atom (N, O or F) with a lone pair positioned close to the $\delta^+$ hydrogen.

In water, the very polar O–H bond gives each hydrogen atom a significant $\delta^+$ charge, while each oxygen atom carries $2$ lone pairs. A hydrogen bond forms when the $\delta^+$ hydrogen of one $\text{H}_2\text{O}$ molecule is attracted to a lone pair on the oxygen of a different, neighbouring $\text{H}_2\text{O}$ molecule:

$\text{O–H} \cdots \text{O}$

(where the solid line is the covalent O–H bond within one molecule, and the dotted line is the hydrogen bond to the next molecule). This O–H⋯O arrangement is approximately linear, close to $180^\circ$, since this maximises the overlap between the lone pair and the $\delta^+$ hydrogen.

Part (c): Why ice is less dense than liquid water

In solid ice, each water molecule forms hydrogen bonds to $4$ neighbouring molecules (via its $2$ $\delta^+$ hydrogens and its $2$ lone pairs). Because hydrogen bonds are directional and relatively long compared with a covalent bond, this locks the molecules into a rigid, open hexagonal lattice in which molecules are held further apart, on average, than in the liquid.

In liquid water, some hydrogen bonds are constantly breaking and re-forming as molecules move, which allows the molecules to pack more closely together on average than in the fixed open lattice of ice.

Since the same mass of water occupies a larger volume as ice than as liquid, ice has a lower density than liquid water, which is why ice floats on liquid water.

Final answers

• (a) Water has additional hydrogen bonding (not present in $\text{H}_2\text{S}$) on top of van der Waals’ and permanent dipole-dipole forces, requiring more energy to separate its molecules, giving a much higher boiling point.
• (b) Conditions: H bonded directly to N/O/F; that atom has a lone pair; a neighbouring molecule’s N/O/F lone pair is nearby. The O–H⋯O hydrogen bond is approximately linear ($\approx 180^\circ$).
• (c) Ice’s rigid, open hydrogen-bonded lattice holds molecules further apart than in the liquid, so ice has a lower density and floats.