Animated Solution for Chemistry - d and f-Block Elements: The correct order of following $3d$-metal oxides, according to their oxidation numbers is (A) ${\text{CrO}}_3$ (B) ${\text{Fe}}_2{\text{O}}_3$ (C) ${\text{MnO}}_2$ (D) ${\text{V}}_2{\text{O}}_5$ (E) ${\text{Cu}}_2\text{O}$

Transition metals are famous for their ability to exhibit a wide variety of oxidation states. This chameleon-like behavior is due to the close energy levels of their $ns$ and $(n-1)d$ orbitals, allowing them to lose different numbers of electrons depending on the chemical environment. In this problem, we are tasked with finding the oxidation states of various $3d$-metals in their respective oxides and arranging them in decreasing order.

Before we dive into the calculations, we need a reliable anchor. In almost all standard metal oxides, oxygen is highly electronegative and pulls two electrons towards itself, giving it a stable oxidation state of $-2$.

Since all the molecules given in the question are neutral, the sum of the oxidation states of all the atoms in each molecule must equal exactly zero. Let's use this principle to unlock the oxidation state of the metal (let's call it $x$) in each compound.

1. Chromium in ${\text{CrO}}_3$ (A) Here, we have one chromium atom and three oxygen atoms. Setting up our equation: $x+3(-2)=0$ $x-6=0⟹x=+6$ Chromium is flexing its maximum oxidation state here, $+6$.

2. Iron in ${\text{Fe}}_2{\text{O}}_3$ (B) This molecule has two iron atoms and three oxygen atoms. The equation becomes: $2x+3(-2)=0$ $2x-6=0⟹x=+3$ Iron is in a stable $+3$ state.

3. Manganese in ${\text{MnO}}_2$ (C) With one manganese and two oxygens, the math is straightforward: $x+2(-2)=0$ $x-4=0⟹x=+4$ Manganese sits at $+4$.

4. Vanadium in ${\text{V}}_2{\text{O}}_5$ (D) Two vanadiums and five oxygens give us: $2x+5(-2)=0$ $2x-10=0⟹x=+5$ Vanadium is at a high $+5$ state.

5. Copper in ${\text{Cu}}_2\text{O}$ (E) Finally, two coppers and one oxygen: $2x+1(-2)=0$ $2x-2=0⟹x=+1$ Copper is in its lowest common oxidation state, $+1$.

Now that we have unmasked all the oxidation states, let's line them up from highest to lowest:

${\text{CrO}}_3$ (A): $+6$ ${\text{V}}_2{\text{O}}_5$ (D): $+5$ ${\text{MnO}}_2$ (C): $+4$ ${\text{Fe}}_2{\text{O}}_3$ (B): $+3$ * ${\text{Cu}}_2\text{O}$ (E): $+1$

The correct decreasing order is (A) > (D) > (C) > (B) > (E). This perfectly matches option (c). By systematically applying the rules of oxidation states, even the most complex-looking inorganic compounds reveal their secrets easily!