In the excess of air, the major oxides formed by Li, Na and K are `Li_(2)O, Na_(2)O_(2)` and `KO_(2)` respectively. Arrange all the given oxides in the decreasing order of their stability.

Options

A. `Li_(2)O gt Na_(2)O_(2)gt KO_(2)`
B. `KO_(2)gt Na_(2)O_(2)gt Li_(2)O`
C. `Na_(2)O_(2)gt KO_(2)gt Li_(2)O`
D. `Li_(2)O gt KO_(2)gt Na_(2)O_(2)`

Explanation

To determine the stability of the oxides formed by lithium (Li), sodium (Na), and potassium (K) in the presence of excess air, we need to analyze the oxides: \( Li_2O \), \( Na_2O_2 \), and \( KO_2 \).

### Step-by-Step Solution:

1. **Identify the Oxides**:
- Lithium forms \( Li_2O \) (lithium oxide).
- Sodium forms \( Na_2O_2 \) (sodium peroxide).
- Potassium forms \( KO_2 \) (potassium superoxide).

2. **Understand the Types of Oxides**:
- \( Li_2O \) is a simple oxide.
- \( Na_2O_2 \) is a peroxide, which contains the peroxide ion \( O_2^{2-} \).
- \( KO_2 \) is a superoxide, which contains the superoxide ion \( O_2^{-} \).

3. **Stability of Oxides**:
- The stability of these oxides generally decreases as we move from simple oxides to peroxides and then to superoxides.
- This is because larger cations can stabilize larger anions better. Thus, the stability order is influenced by the size of the metal ions and the type of anion.

4. **Comparison of Stability**:
- \( Li_2O \) is the most stable because lithium is the smallest cation and can effectively stabilize the oxide ion.
- \( Na_2O_2 \) is less stable than \( Li_2O \) because the peroxide ion is larger and less stable than the oxide ion.
- \( KO_2 \) is the least stable due to the presence of the superoxide ion, which is even larger and less stable compared to the peroxide ion.

5. **Final Arrangement**:
- Based on the analysis above, the decreasing order of stability is:
\[ Li_2O > Na_2O_2 > KO_2 \]

### Final Answer:
The decreasing order of stability of the oxides is:
\[ Li_2O > Na_2O_2 > KO_2 \]