Question

The major product formed in the following reaction \(K+O_2\rightarrow\) is

• A. K₂O
• B. K₂O₂
• C. KO₂
• D. K₂O₃

Answer 1

The Correct Option is C

$\text{1. Reaction with Oxygen}$

Alkali metals ($\text{Group 1}$) react readily with oxygen ($\text{O}_2$) upon heating or combustion. The nature of the major product formed depends on the size and polarizing power of the metal cation ($\text{M}^{+}$).

$$\text{Reaction: } \text{K} + \text{O}_2 \xrightarrow{\text{Excess Air}} \mathbf{\text{Major Product}}$$

$\text{2. Products Formed by Alkali Metals}$

Alkali metals form three main types of oxides:

Metal (M)	Product Type	Formula	Name	Oxygen Ion
$\text{Li}$	Normal Oxide	$\text{Li}_2\text{O}$	Lithium oxide	$\text{O}^{2-}$
$\text{Na}$	Peroxide	$\text{Na}_2\text{O}_2$	Sodium peroxide	$\text{O}_2^{2-}$
$\mathbf{\text{K}, \text{Rb}, \text{Cs}}$	Superoxide	$\mathbf{\text{MO}_2}$	Potassium superoxide	$\mathbf{\text{O}_2^-}$

 

$\text{3. Stabilization of the Superoxide Ion}$

The formation of the stable superoxide ($\text{O}_2^-$) ion requires a large cation ($\text{M}^{+}$) to stabilize the large superoxide anion ($\text{O}_2^-$) through efficient lattice energy.

Potassium ($\text{K}^{+}$) is a much larger ion than $\text{Li}^{+}$ or $\text{Na}^{+}$. Its large size allows for better packing and stabilization of the relatively large $\text{O}_2^-$ ion in the crystal lattice. * Due to this size match, when potassium metal is heated in excess air or oxygen, the thermodynamically most stable product is the superoxide, $\text{KO}_2$.

$\text{4. Balanced Equation}$

The balanced reaction for the formation of the major product is:

$$\text{K}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{KO}_2(\text{s})$$

$$\text{The major product formed is } \mathbf{\text{KO}_2}$$