Animated Solution for Chemistry - s and p-Block Elements: What is the best description of the change that occurs when ${\text{Na}}_2\text{O}(s)$ is dissolved in water?

When solid sodium oxide, ${\text{Na}}_2\text{O}$, is dropped into water, a vigorous reaction takes place. The balanced chemical equation for this process is:

At first glance, it might be tempting to classify this as a complex redox reaction where electrons are being shuffled between different elements. However, to truly understand what is happening, we must look at the oxidation states of the elements involved.

Let's systematically assign oxidation numbers to every atom in the reactants and the products. According to standard rules, alkali metals like sodium ($\text{Na}$) always have an oxidation state of $+1$ in their compounds. Oxygen ($\text{O}$) is almost always $-2$ (except in peroxides or superoxides), and hydrogen ($\text{H}$) is $+1$ when bonded to non-metals.

In the reactants: - For ${\text{Na}}_2\text{O}$: $\text{Na}$ is $+1$, and $\text{O}$ is $-2$. - For ${\text{H}}_2\text{O}$: $\text{H}$ is $+1$, and $\text{O}$ is $-2$.

In the products: - For $\text{NaOH}$: $\text{Na}$ is $+1$, $\text{O}$ is $-2$, and $\text{H}$ is $+1$.

Notice a pattern? Absolutely nothing changed. The oxidation state of sodium remains $+1$, oxygen remains $-2$, and hydrogen remains $+1$ throughout the entire process. Because there is no change in oxidation numbers, this is not a redox reaction. This immediately eliminates the options suggesting that the oxidation number of sodium decreases or that of oxygen increases.

Lewis Acid-Base Interaction

If it's not a redox reaction, what exactly is driving this chemical change? To find out, we need to zoom in to the molecular level. Sodium oxide is an ionic compound. When it dissolves in water, it dissociates into sodium ions (${\text{Na}}^{+}$) and oxide ions (${\text{O}}^{2-}$).

The oxide ion, ${\text{O}}^{2-}$, is the star of the show here. It has a full octet of electrons, including multiple lone pairs, and carries a high negative charge density. This makes it an exceptionally strong Lewis base—a species that is highly eager to donate a pair of electrons.

Water, on the other hand, acts as a Brønsted-Lowry acid (a proton donor) in the presence of such a strong base. The oxide ion "attacks" one of the partially positive hydrogen atoms of the water molecule.

During this attack, the oxide ion donates a pair of its electrons to form a new covalent bond with the hydrogen atom. This forces the existing oxygen-hydrogen bond in the water molecule to break, pushing those shared electrons onto the water's oxygen atom.

The result of this electron pair donation is the formation of two hydroxide ions (${\text{OH}}^{-}$). Therefore, the most accurate description of the fundamental change occurring in this reaction is that the oxide ion donates a pair of electrons.