Animated Solution for Chemistry - Periodicity in Properties: Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

The periodic table is not just a tabular display of chemical elements; it is a profound map of chemical behavior. When we look at the elements, we aren't just seeing letters and atomic numbers; we are looking at a predictive engine that can tell us how these elements will interact with the world. One of the most beautiful demonstrations of this predictive power is understanding the acid-base nature of oxides.

In this problem, we are tasked with arranging four oxides—Aluminum oxide ($A{l}_2{O}_3$), Magnesium oxide ($MgO$), Sodium oxide ($N{a}_{2}O$), and Potassium oxide ($K_{2}O$)—in the increasing order of their basic nature. To solve this, we don't need to memorize a random list. We just need to understand the fundamental principles of periodicity.

Metals, Non-metals, and Oxides

Before we dive into the specific elements, let's establish a ground rule. Metallic oxides are generally basic, while non-metallic oxides are generally acidic.

Why is this the case? It all comes down to electronegativity and how these oxides interact with water. Metals are electropositive; they love to give away their electrons. When a metallic oxide like $N{a}_{2}O$ dissolves in water, the bond between the metal and the oxygen is highly ionic. The water molecules easily pull the ions apart, and the oxide ion ($O^{2-}$) reacts with water to form hydroxide ions ($O{H}^{-}$). The release of $O{H}^{-}$ is the very definition of a base!

On the other hand, non-metals hold onto their electrons tightly. Their bonds with oxygen are covalent. When a non-metallic oxide reacts with water, it tends to form an acid by releasing $H^{+}$ ions.

Therefore, the basic nature of an oxide is directly proportional to the metallic character (or electropositivity) of its central atom. The more metallic the element, the more basic its oxide.

Across Period 3

Let's locate our first three elements: Sodium ($Na$), Magnesium ($Mg$), and Aluminum ($Al$). If you visualize the periodic table, you will find them sitting side-by-side in Period 3. Sodium is in Group 1, Magnesium in Group 2, and Aluminum in Group 13.

As we move from left to right across a period, what happens to the atoms? The nuclear charge increases because we are adding protons to the nucleus, but the electrons are being added to the same principal energy level. This increased nuclear pull draws the electron cloud closer, making the atom smaller and holding the valence electrons more tightly.

Because the electrons are held more tightly, it becomes harder for the atom to lose them. In other words, metallic character decreases as we move from left to right across a period.

Applying this to our oxides: Sodium is a highly reactive alkali metal. Its oxide, $N{a}_{2}O$, is strongly basic. Magnesium is an alkaline earth metal. It is less metallic than Sodium, so $MgO$ is basic, but not as strongly basic as $N{a}_{2}O$. Aluminum is further to the right. Its metallic character is diminished to the point where its oxide, $A{l}_2{O}_3$, sits on the fence. It is amphoteric, meaning it can act as both an acid and a base depending on what it reacts with.

So, just by looking at Period 3, we can establish our first sequence: $A{l}_2{O}_3<MgO<N{a}_{2}O$

Down Group 1

Now we have one element left to place: Potassium ($K$). Potassium is located right below Sodium in Group 1. It belongs to Period 4.

What happens as we move down a group? We are adding entirely new electron shells. The valence electrons are getting further and further away from the pull of the nucleus. Even though the nuclear charge is increasing, the shielding effect of the inner electron shells completely overpowers it.

Because the valence electrons are so far away and well-shielded, they are incredibly easy to lose. Therefore, metallic character increases as we move from top to bottom down a group.

Comparing Sodium and Potassium, Potassium is larger and more electropositive. It is a stronger metal than Sodium. Consequently, its oxide will be more basic. When $K_{2}O$ dissolves in water, it forms Potassium Hydroxide ($KOH$), which is an exceptionally strong base.

This gives us our second crucial piece of information: $N{a}_{2}O<K_{2}O$

We have analyzed the horizontal trend and the vertical trend. Now, it is time to synthesize this information into a single, cohesive sequence.

From our horizontal analysis across Period 3, we know that basic nature decreases from left to right: $A{l}_2{O}_3$ (Amphoteric) $<MgO$ (Basic) $<N{a}_{2}O$ (Strongly Basic)

From our vertical analysis down Group 1, we know that basic nature increases from top to bottom: $N{a}_{2}O$ (Strongly Basic) $<K_{2}O$ (Very Strongly Basic)

By chaining these two inequalities together, we arrive at the final, undeniable sequence: $A{l}_2{O}_3<MgO<N{a}_{2}O<K_{2}O$

This perfectly matches option (a).

This problem is a classic example of why understanding trends is infinitely more powerful than rote memorization. By mastering the concepts of effective nuclear charge, atomic radius, and electropositivity, you can predict the chemical behavior of almost any compound on the periodic table.

Next time you encounter a question about acidic or basic strength, don't panic. Just draw a mental map of the periodic table, locate the elements, and let the fundamental laws of chemistry guide you to the answer.