Animated Solution for Chemistry - d and f-Block Elements: The oxide that shows magnetic property is

Unpaired Electrons

When we look at a chemical compound and ask, "Is this magnetic?", what we are really asking on a quantum level is, "Does this compound have any unpaired electrons?" Electrons have a property called spin, which generates a tiny magnetic field. When electrons are paired up in an orbital, their opposite spins cancel each other's magnetic fields out, resulting in a diamagnetic (non-magnetic) substance. However, if there are unpaired electrons, their magnetic fields add up, making the substance paramagnetic or even ferromagnetic.

Let's analyze the given options one by one to hunt down these elusive unpaired electrons.

First, let's look at the oxides formed by s-block and p-block elements: ${\text{SiO}}_2$, ${\text{Na}}_2\text{O}$, and $\text{MgO}$.

In Silicon dioxide (${\text{SiO}}_2$), Silicon is in the $+4$ oxidation state. The ground state electronic configuration of Silicon ($Z=14$) is $[\text{Ne}]3s^{2}3p^2$. By losing all four of its valence electrons to form ${\text{Si}}^{4+}$, it achieves the highly stable, fully paired noble gas configuration of Neon ($[\text{Ne}]$). With zero unpaired electrons, ${\text{SiO}}_2$ is strictly diamagnetic.

Similarly, in Sodium oxide (${\text{Na}}_2\text{O}$) and Magnesium oxide ($\text{MgO}$), the metal ions are ${\text{Na}}^{+}$ and ${\text{Mg}}^{2+}$ respectively. Both Sodium ($[\text{Ne}]3s^1$) and Magnesium ($[\text{Ne}]3s^2$) lose their valence electrons to achieve the exact same Neon core configuration. Again, no unpaired electrons are left. Thus, both ${\text{Na}}_2\text{O}$ and $\text{MgO}$ are diamagnetic.

Transition Metal Mixed Oxides

Now we arrive at the interesting candidate: ${\text{Mn}}_3{\text{O}}_4$. Manganese is a d-block transition metal, and transition metals are famous for their variable oxidation states and partially filled d-orbitals.

${\text{Mn}}_3{\text{O}}_4$ is not a simple oxide; it is a mixed oxide. Structurally, it can be thought of as a combination of Manganese(II) oxide and Manganese(III) oxide, written as $\text{MnO}\cdot {\text{Mn}}_2{\text{O}}_3$. This means Manganese exists simultaneously in $+2$ and $+3$ oxidation states within the crystal lattice.

Let's write down their electronic configurations. The ground state of Manganese ($Z=25$) is $[\text{Ar}]3d^{5}4s^2$.

For the ${\text{Mn}}^{2+}$ ion, we remove the two outermost $4s$ electrons: According to Hund's rule, these 5 electrons will singly occupy all five d-orbitals, giving us 5 unpaired electrons.

For the ${\text{Mn}}^{3+}$ ion, we remove the two $4s$ electrons and one $3d$ electron: These 4 electrons will singly occupy four of the d-orbitals, giving us 4 unpaired electrons.

Because both ${\text{Mn}}^{2+}$ and ${\text{Mn}}^{3+}$ ions possess a significant number of unpaired electrons, their individual magnetic moments combine to give the entire ${\text{Mn}}_3{\text{O}}_4$ crystal strong magnetic properties (specifically, it exhibits ferrimagnetism at lower temperatures).

Therefore, among the given choices, ${\text{Mn}}_3{\text{O}}_4$ is the only oxide that shows magnetic property.