Question

Which of the following liberates ${\mathrm{O}}_2$ upon hydrolysis?

• A. ${\mathrm{Pb}}_3{\mathrm{O}}_4$
• B. ${\mathrm{KO}}_2$
• C. ${\mathrm{Na}}_2{\mathrm{O}}_2$
• D. ${\mathrm{Li}}_2{\mathrm{O}}_2$

Answer 1

The correct option is B

${\mathrm{KO}}_2$

The explanation for the correct option:

(b) ${\mathrm{KO}}_2$

1. It is superoxide.
2. Superoxide is diatomic oxygen, an inorganic radical anion, oxygen radical and a member of reactive oxygen species
3. The oxidation state of oxygen is known to be $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.
4. Superoxide liberates oxygen when it reacts with water.

$$\begin{gathered} 2{\mathrm{KO}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{KOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ \mathrm{Potassium}\mathrm{Potassium}\mathrm{Hydrogen}\mathrm{Oxygen} \\ \mathrm{superoxide}\mathrm{hyroxide}\mathrm{peroxide} \end{gathered}$$

The explanation for the incorrect option:

(a) ${\mathrm{Pb}}_3{\mathrm{O}}_4$

1. It is insoluble in water, so no reaction happens

$$\begin{gathered} {\mathrm{Pb}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{No}\mathrm{reaction} \\ \mathrm{Lead} \\ \mathrm{tetroxide} \end{gathered}$$

(c) ${\mathrm{Na}}_2{\mathrm{O}}_2$

1. It is a peroxide
2. The oxidation state of oxygen in peroxide is known to be $-1$
3. It liberates hydrogen peroxide (${\mathrm{H}}_2{\mathrm{O}}_2$) when reacts with water.

$$\begin{gathered} {\mathrm{Na}}_2{\mathrm{O}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right) \\ \mathrm{Sodium}\mathrm{Sodium}\mathrm{Hydrogen} \\ \mathrm{peroxide}\mathrm{hydroxide}\mathrm{peroxide} \end{gathered}$$

(d) ${\mathrm{Li}}_2{\mathrm{O}}_2$

1. It is also a peroxide and liberates hydrogen peroxide (${\mathrm{H}}_2{\mathrm{O}}_2$) when reacts with water.

$$\begin{gathered} {\mathrm{Li}}_2{\mathrm{O}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{LiOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right) \\ \mathrm{Lithium}\mathrm{Lithium}\mathrm{Hydrogen} \\ \mathrm{peroxide}\mathrm{hydroxide}\mathrm{peroxide} \end{gathered}$$

Therefore, option (b) is correct, ${\mathrm{KO}}_2$ will liberate ${\mathrm{O}}_2$ upon hydrolysis.