Animated Solution for Chemistry - s and p-Block Elements: On combustion of Li, Na and K in excess of air, the major oxides formed, respectively, are

Imagine taking a piece of an alkali metal and tossing it into a roaring fire with an endless supply of oxygen. You might expect them all to react the same way—after all, they belong to the same family in the periodic table. However, nature has a beautiful subtlety. When lithium, sodium, and potassium are combusted in excess air, they don't just form a generic oxide. They each form a distinctly different type of oxygen compound.

To understand why this happens, we need to dive into the microscopic world of ions and the architectural rules of crystal lattices.

In the realm of inorganic chemistry, there is a golden rule that governs the stability of ionic compounds: The Size Compatibility Principle.

Think of building a brick wall. If you use small, uniform bricks, the wall is incredibly sturdy. If you try to mix tiny pebbles with massive cinder blocks, the structure becomes unstable and prone to collapsing. The same logic applies to ions. A small positive ion (cation) forms a highly stable, tightly packed crystal lattice with a small negative ion (anion). Conversely, a large, bulky cation is required to stabilize a large, bulky anion.

When alkali metals react with oxygen, the oxygen can exist in three different anionic forms depending on how many electrons it grabs and how the atoms bond: 1. Oxide ion ($O^{2-}$): A single oxygen atom with two extra electrons. It is relatively small. 2. Peroxide ion ($O_2^{2-}$): Two oxygen atoms bonded together, sharing two extra electrons. It is medium-sized. 3. Superoxide ion ($O_2^{-}$): Two oxygen atoms bonded together, sharing only one extra electron. It is large and bulky.

Let's see how our metals pair up with these anions.

The Tiny Titan

Lithium ($Li$) is the first alkali metal and has the smallest atomic radius. When it loses an electron to become the $L{i}^{+}$ ion, it becomes exceptionally tiny.

Because of its small size, $L{i}^{+}$ has a very high positive charge density. According to our compatibility rule, this tiny cation perfectly matches the small oxide ion ($O^{2-}$). If lithium tried to bond with a larger peroxide or superoxide ion, its intense electric field would severely polarize (distort) the large anion's electron cloud, causing the unstable molecule to break apart into the simpler oxide form.

Therefore, burning lithium in excess air yields normal lithium oxide:

The Middle Ground

Moving one step down the periodic table, we find sodium ($Na$). The sodium ion ($N{a}^{+}$) is significantly larger than the lithium ion.

Because it is larger, its charge is spread out over a greater volume, meaning its polarizing power is weaker. This larger size allows it to comfortably accommodate and stabilize a larger anion in its crystal lattice. Thus, sodium pairs up perfectly with the medium-sized peroxide ion ($O_2^{2-}$).

Burning sodium in excess air yields sodium peroxide:

The Giants

Finally, we reach potassium ($K$). The potassium ion ($K^{+}$) is a giant compared to lithium and sodium.

To build a stable crystal lattice, this massive cation requires an equally massive anion. The small oxide ion would leave too much empty space, making the lattice unstable. Instead, potassium perfectly stabilizes the largest of the oxygen anions: the superoxide ion ($O_2^{-}$).

Burning potassium in excess air yields potassium superoxide:

By simply looking at the atomic radii of the elements, we can predict their chemical behavior with stunning accuracy.

- Lithium (small) forms the normal oxide: $L{i}_{2}O$ - Sodium (medium) forms the peroxide: $N{a}_2{O}_2$ - Potassium (large) forms the superoxide: $K{O}_2$

Matching these results with our given options, we find that the correct sequence is $L{i}_{2}O$, $N{a}_2{O}_2$, and $K{O}_2$.