Animated Solution for Chemistry - d and f-Block Elements: The nature of oxides ${\text{V}}_2{\text{O}}_3$ and $\text{CrO}$ is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is

Have you ever wondered why some metal oxides dissolve in acids, some in bases, and some stubbornly refuse to pick a side? The answer lies hidden in a simple yet powerful concept: the oxidation state.

In this problem, we are tasked with uncovering the true nature of two transition metal oxides: Vanadium(III) oxide (${\text{V}}_2{\text{O}}_3$) and Chromium(II) oxide ($\text{CrO}$). Are they acidic, basic, or amphoteric? Let's dive into the chemistry behind it.

Transition metals are fascinating because they can exhibit a wide variety of oxidation states. This chameleon-like behavior directly dictates the chemical properties of their compounds.

Here is the master key to unlocking this problem: Lower Oxidation States ($+2,+3$): The metal cation is relatively large and has a lower effective nuclear charge. It cannot polarize the oxygen electron cloud effectively. The bond remains highly ionic. When placed in water, the oxide ion (${\text{O}}^{2-}$) easily grabs a proton to form hydroxide ions (${\text{OH}}^{-}$). Hence, these oxides are basic. Intermediate Oxidation States ($+4$): The metal starts pulling harder on the oxygen's electrons. The bond becomes more covalent. These oxides sit on the fence and can act as both acids and bases, making them amphoteric. Higher Oxidation States ($+5,+6,+7$): The metal cation is tiny and highly charged. It violently distorts the oxygen's electron cloud (high polarizing power according to Fajans' rules). The bond becomes highly covalent. These oxides react with water to release protons (${\text{H}}^{+}$), making them strongly acidic*.

Let's apply our rule to ${\text{V}}_2{\text{O}}_3$. First, we need to find the oxidation state of Vanadium.

Let the oxidation state of $\text{V}$ be $x$. We know oxygen typically carries a $-2$ charge.

Vanadium is in a $+3$ oxidation state. According to our golden rule, $+3$ is considered a lower oxidation state. Therefore, ${\text{V}}_2{\text{O}}_3$ is a basic oxide. It will happily react with acids to form Vanadium(III) salts.

Now, let's look at $\text{CrO}$.

Let the oxidation state of $\text{Cr}$ be $y$.

Chromium is in a $+2$ oxidation state. This is the lowest common oxidation state for Chromium. Just like Vanadium in the previous step, this low oxidation state means the oxide is highly ionic and therefore basic.

Both ${\text{V}}_2{\text{O}}_3$ and $\text{CrO}$ feature metals in lower oxidation states ($+3$ and $+2$, respectively). Consequently, both of these oxides are basic in nature.

If we map this back to our question variables: $\text{X}$ (nature of ${\text{V}}_2{\text{O}}_3$) = basic $\text{Y}$ (nature of $\text{CrO}$) = basic

This perfectly matches option (d).

A Quick Thought Experiment: What if the examiner had asked about ${\text{CrO}}_3$ instead? In ${\text{CrO}}_3$, Chromium is at a whopping $+6$ oxidation state. The bonds are highly covalent, and ${\text{CrO}}_3$ is actually a strongly acidic oxide (it forms chromic acid in water). Always let the oxidation state be your guide!