Which of the following statements about water is FALSE?

We begin by carefully examining each option and recalling the basic facts about water, $$H_2O$$. In the condensed phases (liquid water and ice) the individual water molecules interact through hydrogen bonds. A hydrogen bond is an attractive interaction of the type $$X{-}H\cdots Y$$ where $$X$$ and $$Y$$ are highly electronegative atoms such as $$O$$, $$N$$ or $$F$$. Two distinct situations are possible:

1. Intermolecular hydrogen bonding  - the hydrogen bond forms between two different molecules.

2. Intramolecular hydrogen bonding  - the hydrogen bond forms within the same molecule, holding together two different parts of that molecule.

With these definitions in place, we now test every given statement.

Option A states: “There is extensive intramolecular hydrogen bonding in the condensed phase.” The water molecule has the simple bent structure $$\angle H{-}O{-}H \approx 104.5^\circ$$, and within a single molecule the two $$O{-}H$$ bonds are already covalent. Because the $$O{-}H$$ groups are attached to the same central oxygen, there is no geometrical possibility for another site inside the very same molecule to accept a hydrogen bond from one of those hydrogens. What actually happens is that a hydrogen of one molecule hydrogen-bonds to the oxygen of a different molecule:

$$\text{H}_2\text{O}\;{-}\;H\cdots O{-}\text{H}_2\text{O}$$

This is clearly intermolecular. Hence, Option A describes the situation incorrectly. There is no extensive intramolecular hydrogen bonding; the bonding present is intermolecular.

Option B reads: “Ice formed by heavy water sinks in normal water.” Heavy water is $$D_2O$$, where $$D$$ (deuterium) has atomic mass $$2$$ instead of $$1$$. The molar mass of $$D_2O$$ is therefore $$20 \text{ g mol}^{-1}$$ compared with $$18 \text{ g mol}^{-1}$$ for ordinary $$H_2O$$. Because of the greater mass, both liquid heavy water and the ice formed from it are denser than their normal counterparts. Thus a block of $$D_2O$$ ice has density $$\rho_{D_2O\; \text{ice}} \approx 1.14 \text{ g cm}^{-3}$$, which is larger than the density of liquid ordinary water $$\rho_{H_2O\; \text{(l)}} = 1.00 \text{ g cm}^{-3}$$. According to Archimedes’ principle, an object sinks when its density exceeds that of the surrounding fluid, so $$D_2O$$ ice indeed sinks in normal water. Therefore Option B is a true statement.

Option C asserts: “Water is oxidized to oxygen during photosynthesis.” In the light-dependent reactions of photosynthesis, water is split according to the overall process

$$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$

Here water loses electrons; loss of electrons is oxidation. Thus water is oxidized to molecular oxygen. So Option C is also true.

Option D declares: “Water can act both as an acid and as a base.” This is the well-known amphoteric nature of water. In the Brønsted-Lowry sense:

As an acid (proton donor) in the presence of a strong base: $$H_2O + NH_2^- \rightarrow NH_3 + OH^-$$

As a base (proton acceptor) in the presence of a strong acid: $$H_2O + HCl \rightarrow H_3O^+ + Cl^-$$

Therefore Option D is absolutely correct.

Summarizing, Options B, C and D are all true, whereas Option A is false because the hydrogen bonding in water is intermolecular, not intramolecular.

Hence, the correct answer is Option A.